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304621 | Proofs with real numbers and discrete mathematics!
Can anyone assist me in figuring out the way to tackle the question Let [imath]x[/imath] be a real number such that [imath]x > 0[/imath]. Prove that [imath]x + (9/x) ≥ 6[/imath]. I understand that its true because, for example, let [imath]x = 1[/imath] is greater than [imath]6[/imath]... but how would i prove this? | 304293 | Working with proofs help?
I'm trying to study for my midterm and doing some random practise questions to work with proofs. However I'm stuck on, as the only way I know how to prove it is through plugging in numbers, however as I understand that is not solid proof. Here it is: If [imath]x[/imath] is a real number and [imath]x > 0[/imath]. Prove that [imath]x+\frac{9}{x} \geq 6[/imath]. Any help would be greatly appreciated. |
87305 | Proving Integral Inequality
I am working on proving the below inequality, but I am stuck. Let [imath]g[/imath] be a differentiable function such that [imath]g(0)=0[/imath] and [imath]0<g'(x)\leq 1[/imath] for all [imath]x[/imath]. For all [imath]x\geq 0[/imath], prove that [imath]\int_{0}^{x}(g(t))^{3}dt\leq \left (\int_{0}^{x}g(t)dt \right )^{2}[/imath] | 1083831 | How to show [imath]\int_0^1 f^3 dx < ( \int_0^1 f dx )^2 [/imath]
Assume [imath]f[/imath] is [imath]C^1([0,1])[/imath] and [imath]f(0)=0[/imath] and [imath] 0 < f' \le 1 [/imath] then I want to show that [imath]\int_{0}^{1} f^3 dx < \left( \int_{0}^{1} f dx \right)^2 [/imath] my tries : I want to use a similar way like the proof of the poincare inequality [imath]f(x)^3=\int_0^1 3 f'(x) f^2 (x) dx [/imath] now I use the bound of [imath]f'[/imath] and [imath]f[/imath] and Holder inequality and integrating from both side of the inequalities but i reach inequalities different of the asked one. Can some one help me. thanks. |
14970 | No maximum(minimum) of rationals whose square is lesser(greater) than [imath]2[/imath].
Suppose [imath]A[/imath] is the set of all rational numbers [imath]p[/imath] such that [imath]p^2 <2[/imath] and [imath]B[/imath] is the set of all rational numbers [imath]p[/imath] such that [imath]p^2 > 2[/imath]. We want to show that [imath]A[/imath] contains no largest element and [imath]B[/imath] contains no smallest element. In Rudin's Principles of Mathematical Analysis, he associates [imath]q = p- \frac{p^2-2}{p+2} = \frac{2p+2}{p+2}[/imath]. Did he just come up with this using trial and error? | 918831 | In the following proof that square root of 2 is non-terminating, where does the expression for q come from?
I understand the following proof but how did the author come up with the following expression? [imath]q = p - \frac{p^2 -2}{p + 2}[/imath] [imath]q[/imath] being defined that way is crucial for the proof but what's so special about that choice of [imath]q%[/imath]? Why choose that? The expression for [imath]q^2 - 2[/imath] follows from the first part but I don't understand how to get [imath]q[/imath] itself. The best I could do was realize that, for this particular case, [imath]q = \sqrt{2} + 1[/imath] but I can't see how that's significant and ow that could have led to the expression I'm curious about. Also what if I need to do the same thing for an arbitrary root [imath]\sqrt{n}[/imath]. How do I come up with such a [imath]q[/imath]? P.S.: I'm talking about the second part of the proof- not where he proves that [imath]\sqrt{2}[/imath] is irrational but the part "... A contains no largest number and ...". |
121407 | How to compute [imath]\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}[/imath]?
I got stuck at the computation of the sum [imath] \sum\limits_{k=0}^n (-1)^k{2n-k\choose k}. [/imath] I think there is no purely combinatorial proof here since the sum can achieve negative values. Could you give me solution, it seems to involve generating functions. | 1495776 | Find the closed expression for binomial sum
I need help in finding closed expression for the following sum: [imath]a_n = \sum\limits_{k=0}^n (-1)^k \binom{2n - k}{k}[/imath]. By inspecting the first several elements of the sequence I came up with the hypothesis for expression for [imath]a_n[/imath]: [imath]a_{3j} = 1,\space a_{3j+1} = 0,\space a_{3j+2} = -1,[/imath] where [imath]j=0,1,2,\dots.[/imath] I was not able to prove this by induction. Is it an easy exercise? What other approaches can be taken to solve this problem? I am mostly looking for as basic solutions as possible since I am new to combinatorics. |
88719 | A group [imath]G[/imath] with a subgroup [imath]H[/imath] of index [imath]n[/imath] has a normal subgroup [imath]K\subset H[/imath] whose index in [imath]G[/imath] divides [imath]n![/imath]
I would be very thankful if someone could give me a hint with proving this. It's a very common exercise in abstract algebra textbooks. If [imath]G[/imath] is a group with a subgroup [imath]H[/imath] of finite index [imath]n[/imath], then [imath]G[/imath] has a normal subgroup [imath]K[/imath] contained in [imath]H[/imath] whose index in [imath]G[/imath] is finite and divides [imath]n![/imath]. I found the proof on this Wikipedia page at some point (although the proof appears to be no longer there), but I got lost in one of the details. | 427607 | Show there is a normal subgroup [imath]K[/imath] of [imath]G[/imath] with [imath]K\subset H[/imath] and such that the order of [imath]K[/imath] divides [imath]n![/imath]
Let [imath]G[/imath] be a finite group and suppose [imath]H[/imath] is a subgroup of [imath]G[/imath] having index [imath]n[/imath]. Show there is a normal subgroup [imath]K[/imath] of [imath]G[/imath] with [imath]K\subset H[/imath] and such that the order of [imath]K[/imath] divides [imath]n![/imath] . any suggestion? Thanks. |
33863 | How to prove that the number [imath]1!+2!+3!+...+n![/imath] is never square?
How to prove that the number [imath]1!+2!+3!+...+n! \ \forall n \geq 4[/imath] is never square? I was told to count permutations but I cannot figure out what we are permuting.... Thanks! | 1711402 | Is the sum of first [imath]n[/imath] factorials a square?
Find the values of [imath]n\geq 1[/imath] for which [imath]1!+2!+\cdots+n![/imath] is a square. I wanted to go through some small [imath]n[/imath].. [imath]1!=1[/imath] [imath]1!+2!=3[/imath] not a square [imath]1!+2!+3!=9[/imath] a square [imath]1!+2!+3!+4!=33[/imath] not a square [imath]1!+2!+3!+4!+5!=153[/imath] not a square [imath]1!+2!+3!+4!+5!+6!=873[/imath] not a square Here comes a pattern... All [imath]n![/imath] for [imath]n\geq 5[/imath] ends with [imath]0[/imath] because [imath]5\cdot 2[/imath] is factor of [imath]5![/imath]. So, [imath]1!+2!+\cdots+n![/imath] for [imath]n\geq 5[/imath] ends with [imath]3[/imath] so can not be a square.. See that unit digit of a square can only be one of [imath]0,1,4,5,6,9[/imath].. So, only possible [imath]n[/imath] are [imath]1,3[/imath] I just want to know if there are any gaps. |
26037 | Intuitive Understanding of the constant "[imath]e[/imath]"
Potentially related-questions, shown before posting, didn't have anything like this, so I apologize in advance if this is a duplicate. I know there are many ways of calculating (or should I say "ending up at") the constant e. How would you explain e concisely? It's a rather beautiful number, but when friends have asked me "what is e?" I'm usually at a loss for words -- I always figured the math explains it, but I would really like to know how others conceptualize it, especially in common-language (say, "English"). related but not the same: Could you explain why [imath]\frac{d}{dx} e^x = e^x[/imath] "intuitively"? | 391572 | Can someone please explain [imath]e[/imath] in layman's term?
I never really understood what [imath]e[/imath] means and I'm always terrified when I see it in equations. What is it? Can somebody dumb it down for me? I know it's a constant. Is it as simple as that? |
266998 | How to compute the determinant of a tridiagonal matrix with constant diagonals?
How to show that the determinant of the following [imath](n\times n)[/imath] matrix [imath]\begin{pmatrix} 5 & 2 & 0 & 0 & 0 & \cdots & 0 \\ 2 & 5 & 2 & 0 & 0 & \cdots & 0 \\ 0 & 2 & 5 & 2 & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\ 0 & \cdots & \cdots & \cdots & \cdots & 2 & 5 \end{pmatrix}[/imath] is equal to [imath]\frac13(4^{n+1}-1)[/imath]? More generally: How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)? [imath]M_n(a,b,c) = \begin{pmatrix} a & b & 0 & 0 & 0 & \cdots & 0 \\ c & a & b & 0 & 0 & \cdots & 0 \\ 0 & c & a & b & 0 & \cdots & 0 \\ \vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\ 0 & \cdots & \cdots & 0 & c & a & b \\ 0 & \cdots & \cdots & \cdots & \cdots & c & a \end{pmatrix}[/imath] Here [imath]a,b,c[/imath] can be taken to be real numbers, or complex numbers. In other words, the matrix [imath]M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}[/imath] is such that [imath]m_{ij} = \begin{cases} a & i = j, \\ b & i = j - 1, \\ c & i = j + 1, \\ 0 & \text{otherwise.} \end{cases}[/imath] There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type [imath]M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)[/imath] (where we could use [imath]\det(M) = \det(A) \det(B)[/imath] for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly. Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small [imath]n[/imath]: [imath]\begin{align} \det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\ \det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\ \det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc \end{align}[/imath] But there is no readily apparent pattern and the computation becomes very difficult when [imath]n[/imath] gets large. | 571664 | Determinant of a n x n Matrix - Main Diagonal = 2, Sub- & Super-Diagonal = 1
I'm stuck with this one - Any tips? The Problem: Let [imath]n \in \mathbb{N}.[/imath] The following [imath]n \times n[/imath] matrix: [imath]A = \left( \begin{array}{ccc} 2 & 1 & & & & ...\\ 1 & 2 & 1 & & & ...\\ & 1 & 2 & 1 & & ...\\ & & 1 & 2 & 1 & ...\\ & & & 1 & ... & 1\\ ... & ... & ... & ... & 1 &2 \end{array} \right) [/imath] e.g. for the main diagonal = 2, the sub- and superdiagonal = 1 . Show with Induction that [imath]\det(A) = n + 1[/imath]. My solution approach: Laplace Expansion starting with the 2 in the bottom right corner [imath](a_{n+1,n+1})[/imath]. But how can I tell wether its positive or negative? After that I'm stuck with the 1 [imath](a_{n,n+1})[/imath](the sub matrix matrix becomes ugly and I get a recursively solution). How can I formalize this in a proper manner? |
53877 | Is there anything like GF(6)?
Are there any galois fields which consist of product of two primes, as in [imath]$\mathrm{GF}(2\cdot 3) = \mathrm{GF}(6)$[/imath]? | 2425194 | Every field with 6 elements is isomorphic to [imath]\Bbb Z_2\times \Bbb Z_3[/imath]
Every field with 6 elements is isomorphic to [imath]\Bbb Z_2\times \Bbb Z_3[/imath] I think it's false because: [imath]\Bbb Z_6[x] / (3x +2)[/imath] is such that [imath]3x+2[/imath] is irreducible in [imath]\Bbb Z_6[/imath] so [imath]\Bbb Z_6 [x]/ (3x +2)[/imath] is a field with 6 elements. This is isomorphic to [imath]\Bbb Z_2\times \Bbb Z_3[/imath] [imath]\approx \Bbb Z_6[/imath], as the elements are actually the same. We also know that if [imath]f:A\rightarrow B[/imath] is an isomorphism, if A is field then B is field and as [imath]\Bbb Z_6 [x]/ (3x +2)[/imath] [imath]\approx \Bbb Z_6[/imath], with [imath]\Bbb Z_6 [x]/ (3x +2)[/imath] field, then [imath]\Bbb Z_6[/imath] is a field (contradiction as 6 isn't prime). Then not every field with 6 elements is isomorphic to [imath]\Bbb Z_2\times \Bbb Z_3[/imath], because [imath]\Bbb Z_6[x] / (3x +2)[/imath] is not isomorphic to [imath]\Bbb Z_2\times \Bbb Z_3[/imath]. |
305826 | [imath]\sup (A\cdot B)=\sup A \cdot \sup B[/imath]
I would like to prove that [imath]\sup (A\cdot B)=\sup A \cdot \sup B[/imath], where [imath]A,B[/imath] are sets of positive real numbers. I've already proved [imath]\sup (A+B)=\sup A +\sup B[/imath] and [imath]\sup (c\cdot A)=c\cdot \sup A[/imath][imath](c\gt 0)[/imath] easily without any problems. However, I'm stuck in the multiplication [imath]A\cdot B[/imath] case. I need help. Thanks a lot | 46738 | Product of sets and supremum
Let [imath]A[/imath] and [imath]B[/imath] be nonempty sets of positive real numbers that are bounded above. Also let [imath]AB = \{ab: a \in A, b \in B \}[/imath]. Prove that [imath]AB[/imath] is bounded above and [imath]\sup(AB) = (\sup A) (\sup B)[/imath]. So [imath]\sup A[/imath] and [imath]\sup B[/imath] exist by completeness. An upper bound for [imath]AB[/imath] is [imath](\sup A)(\sup B)[/imath]. Let [imath]\alpha = \sup A[/imath] and [imath]\beta = \sup B[/imath]. We want to show that if [imath]c[/imath] is an upper bound for [imath]AB[/imath] then [imath]\alpha \beta \leq c[/imath]. For [imath]a \in A[/imath], [imath]ab \leq c[/imath] for all [imath]b \in B[/imath]. So [imath]c/b[/imath] is an upper bound for [imath]A[/imath]. Thus [imath]\alpha \leq c/b[/imath]. It follows that [imath]\alpha \beta \leq c[/imath]. Is this correct? |
146723 | Question about Fredholm operator
[imath]X,Y[/imath] are Banach spaces and [imath]A\in B(X,Y)[/imath] is a Fredholm operator (that is, the dimensions of ker([imath]A[/imath]) and coker([imath]A[/imath]) are both finite), then are closed linear subspaces ker([imath]A[/imath]) and Im([imath]A[/imath]) complemented? (A closed linear subspace [imath]H[/imath] in a Banach space [imath]Z[/imath] is called complemented iff there is a closed linear subspace [imath]G[/imath] such that [imath]H+G=Z[/imath] and [imath]H \cap G=0[/imath]) | 818030 | Fredholm Index: Finite Corank [imath]\Rightarrow[/imath]Closed Range
Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic. Why do we thus require the range to be closed? Or is a subspace with finite codimension necessarily closed? Moreover, what could happen if the range wouldn't be closed but would be of finite codimension though? I'm thinking of some example like: [imath]l^p_0\subsetneq l^p,1\leq p\leq\infty[/imath] |
365 | Tiling a [imath]3 \times 2n[/imath] rectangle with dominoes
I'm looking to find out if there's any easy way to calculate the number of ways to tile a [imath]3 \times 2n[/imath] rectangle with dominoes. I was able to do it with the two codependent recurrences f(0) = g(0) = 1 f(n) = f(n-1) + 2g(n-1) g(n) = f(n) + g(n-1) where [imath]f(n)[/imath] is the actual answer and [imath]g(n)[/imath] is a helper function that represents the number of ways to tile a [imath]3 \times 2n[/imath] rectangle with two extra squares on the end (the same as a [imath]3 \times 2n+1[/imath] rectangle missing one square). By combining these and doing some algebra, I was able to reduce this to f(n) = 4f(n-1) - f(n-2) which shows up as sequence A001835, confirming that this is the correct recurrence. The number of ways to tile a [imath]2 \times n[/imath] rectangle is the Fibonacci numbers because every rectangle ends with either a verticle domino or two horizontal ones, which gives the exact recurrence that Fibonacci numbers do. My question is, is there a similar simple explanation for this recurrence for tiling a [imath]3 \times 2n[/imath] rectangle? | 675781 | Finding the recurrence relation?
If I let [imath]n \geq 1[/imath] be an integer and use a [imath]2 \times n[/imath] board [imath]D_n[/imath] containing [imath]2n[/imath] cells, each side has a length of 1. T The brick can be vetical or horizontal containg [imath]2[/imath] cells(explained in the bottom part of the above picture). The tiling of board ([imath]D_n[/imath]) is placed with bricks so that : 1.item the bricks exactly cover [imath]D_n[/imath] and 2.item no two bricks overlap. So for [imath]n \geq 1[/imath], we let [imath]x_n[/imath] become number of different tilings of the board [imath]D_n[/imath]. Now I have to determine the value of [imath]a_n[/imath] I know there are only two ways to cover the left-most two squares: either a single vertical brick, or two horizontal bricks and that if you cover those two left ones with a single vertical brick, after that you have to choose any covering of the other [imath]n-1[/imath] columns and the only way to do this is precisely [imath]x_{n-1}[/imath]. So, there are [imath]x_{n-1}[/imath] arrangements in which you cover the left two by a vertical brick and if you need two horizontal bricks, Then you've knocked out two columns, and need to pick a covering of the remaining [imath]n-2[/imath] columns. So now what would [imath]x_n[/imath] equal I think [imath]a_n = a_(n-1) + a_(n-2)[/imath] I am I right? |
154505 | Prove that [imath]\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)[/imath] when [imath]A,B,C[/imath] are angles of a triangle
Prove that [imath]\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)[/imath] when [imath]A,B,C[/imath] are angles of a triangle This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing [imath]4\sin(A)\sin(B)\sin(C)[/imath] to [imath]4\sin(B+C)\sin(A+C)\sin(A+B)[/imath] by making substitutions by reorganizing [imath]A+B+C=\pi[/imath]. I then did the same thing to the other side to get [imath]-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))[/imath] and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well) | 364011 | If [imath]A+B+C=\pi[/imath] does it imply that [imath]\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C[/imath]
If [imath]A+B+C=\pi[/imath] does it imply that [imath]\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C[/imath] If yes, how? |
176892 | Prove trigonometry identity for [imath]\cos A+\cos B+\cos C[/imath]
I humbly ask for help in the following problem. If \begin{equation} A+B+C=180 \end{equation} Then prove \begin{equation} \cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2) \end{equation} How would I begin the problem I mean I think [imath]\cos C [/imath] can be [imath]\cos(180-A+B)[/imath]. But I am unsure what to do next. | 2763117 | If [imath]\alpha=\beta+\omega[/imath], then [imath]1 + \cos\alpha+\cos\beta+\cos\omega = 4 \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\omega}{2}[/imath]
How can I prove the following? The RHS is not about SIN If [imath]\alpha = \beta + \omega[/imath], then [imath]1 + \cos\alpha+\cos\beta+\cos\omega = 4 \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\omega}{2}[/imath] I've already tried that one: [imath]\cos (\alpha+\omega) + \cos\omega +\cos\alpha + \cos 2(\alpha+\omega)[/imath] What more can I do? |
73421 | A non-noetherian ring with all localizations noetherian
If for a ring [imath]A[/imath] every localization [imath]A_\mathfrak{p}[/imath] by a prime [imath]\mathfrak{p}\subseteq A[/imath] is noetherian, is it true that [imath]A[/imath] is noetherian? I believe not but I can't find a good counterexample. | 2108362 | If the localization of a ring at each prime ideal is Noetherian, does this imply that ring is Noetherian?
If the localization [imath]R_p[/imath] of a ring [imath]R[/imath] at each prime ideal [imath]p[/imath] in A is Noetherian, does this imply that [imath]A[/imath] is Noetherian? What we call such rings which is not Noetherian but localization at each prime ideal is Noetherian ? Can somebody provide me any counterexample of (1) and also a good reference? |
156878 | Integrate and measure problem.
If [imath]f \in L^{p_0}(X,M,\,u)[/imath] for some [imath]0<p_0 \le\infty[/imath], then [imath]1. \lim_{p\to0}\int_X |f|^p \, d\mu=\mu(\{x \in X \mid f(x) \ne0\}).[/imath] And if additional assume [imath]\mu(X)=1[/imath], then I wanna prove that [imath]f \in L^p(X,M,\,u)[/imath] for some [imath]0<p \le p_0[/imath], and the equation below. [imath]2. \lim_{p\to0}\|f\|_p=e^{\int_X\log|f|\,d\mu}[/imath] I want to know how can I conclude those results. First, I'm trying to use integrate over the set on which [imath]0<|f(x)|\le1[/imath] and use MCT. and the set on which [imath]|f(x)|>1[/imath] use LDCT to prove that. But I can't conclude to measure [imath]\mu[/imath] and how can I approach second fact? | 363673 | Convergence of [imath]L^p[/imath] norm as [imath]p \downarrow 0[/imath]
Consider a measurable space [imath](\Omega, \mathscr{F}, P)[/imath] with [imath]P(\Omega) = 1[/imath]. Define for measurable functions [imath]X[/imath] the following [imath]\| X \|_p := \left(\int |X|^p dP\right)^{1/p}[/imath]. We know that for [imath]p \in [1, \infty)[/imath] that this is a norm, the [imath]L^p[/imath] norm. Let [imath]S = \{p : 0 < p < \infty \text{ and } \|X \|_p <\infty \}[/imath] and assume that [imath]S \neq \emptyset[/imath]. Prove [imath] \lim_{p \downarrow 0} \| X \|_p = \exp \{ \int \log |X| \; dP \} [/imath] defining [imath]\exp\{ - \infty \} := 0[/imath]. Disclaimer: This was a homework problem in a graduate measure theory. The teacher never distributed solutions, and I have struggled to prove it since I first saw it. |
1467 | Cardinality of all cardinalities
Let [imath]C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}[/imath]. What is [imath]\left|C\right|[/imath]? Or is it even well-defined? | 2153078 | A question about infinity orders in set theory
I would like to ask this question and i hope it is not obvious or rediculus. We know about cardinal numbers in set theory and that this ''numbers'' are equivalence classes. In other words two sets are equivalent(or equipotent) if there exists a bijection from one set to another.This relation is an equivalence relation. Thus the equivalence classes are the cardinal numbers. We also know that there are infinitely many cardinal numbers. For instance [imath]\aleph_0<c<2^c<2^{2^c}<....< [/imath] where [imath]c=2^{\aleph_0}[/imath] My question is that can we form a set that has as its elements all cardinal numbers? If such set exists then what is its cardinality?(maybe it does not exist because in some way we may derive the existence of the set of all sets..im not sure) Thank you in advance! My question is not an exact duplicate of the mentioned question.In the question Cardinality of the set of all cardinals the user assumes that there exist a set of all cardinals and i ask if there exists such set. |
56582 | Analogue of spherical coordinates in [imath]n[/imath]-dimensions
What's the analogue to spherical coordinates in [imath]n[/imath]-dimensions? For example, for [imath]n=2[/imath] the analogue are polar coordinates [imath]r,\theta[/imath], which are related to the Cartesian coordinates [imath]x_1,x_2[/imath] by [imath]x_1=r \cos \theta[/imath] [imath]x_2=r \sin \theta[/imath] For [imath]n=3[/imath], the analogue would be the ordinary spherical coordinates [imath]r,\theta ,\varphi[/imath], related to the Cartesian coordinates [imath]x_1,x_2,x_3[/imath] by [imath]x_1=r \sin \theta \cos \varphi[/imath] [imath]x_2=r \sin \theta \sin \varphi[/imath] [imath]x_3=r \cos \theta[/imath] So these are my questions: Is there an analogue, or several, to spherical coordinates in [imath]n[/imath]-dimensions for [imath]n>3[/imath]? If there are such analogues, what are they and how are they related to the Cartesian coordinates? Thanks. | 432193 | Unfolding the [imath]n[/imath]-dimensional sphere
Is there an extension to [imath]n[/imath] dimensions of the usual spherical coordinates mapping a three-dimensional sphere to a two-dimensional rectangle? [Duplicate]: Analogue of spherical coordinates in [imath]n[/imath]-dimensions |
58931 | Does [imath]\sum\limits_{k=1}^n 1 / k ^ 2[/imath] converge when [imath]n\rightarrow\infty[/imath]?
I can prove this sum has a constant upper bound like this: [imath]\sum_{k=1}^n \frac1{k ^ 2} \lt 1 + \sum_{k=2}^n \frac 1 {k (k - 1)} = 2 - \frac 1 n \lt 2[/imath] And computer calculation shows that sum seems to converge to 1.6449. But I still want to know: Dose this sum converge? Is there a name of this sum (or the series [imath]1 / k ^2 [/imath])? | 1039699 | Proving the sum of the reciprocals squared converges
I'm investigating the Basel Problem, and the sum to consider is: [imath]\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{n^2}[/imath] How can I show this converges? Using graphs/computer software is also fine, but how would I do it? Is there a way using calculus? I've looked on Google for a while and I found a few proofs, but I didn't understand any of them. |
244460 | Ring of trigonometric functions with real coefficients
Let [imath]R[/imath] be the ring of functions that are polynomials in [imath]\cos t[/imath] and [imath]\sin t[/imath] with real coefficients. Prove that [imath]R[/imath] is isomorphic to [imath]\mathbb R[x,y]/(x^2+y^2-1)[/imath]. Prove that [imath]R[/imath] is not a unique factorization domain. Prove that [imath]S=\mathbb C[x,y]/(x^2+y^2-1)[/imath] is a principal ideal domain and hence a unique factorization domain. Determine the units of the rings [imath]S[/imath] and [imath]R[/imath]. (Hint: Show that [imath]S[/imath] is isomorphic to the Laurent polynomial ring [imath]\mathbb C[u,u^{-1}][/imath].) | 1311675 | Show that [imath]\mathbb{C}[x,y]/(x^2+y^2-1)[/imath] is a UFD.
I am trying to prove that the ring [imath]\mathbb{C}[x,y]/(x^2+y^2-1)[/imath] is a UFD. I have an hint, that suggests to find an isomorphism between [imath]\mathbb{C}[x,y]/(x^2+y^2-1)[/imath] and [imath]\mathbb{C}[e^{it},e^{-it}][/imath], but to be honest I don't see it. Please, could you give me a hand? I would to solve my problem completely, any kind of suggestion is fully appreciated. Update: user Daniel McLaury helped me in showing the isomorphism - so now it remains only to see that this ring is actually an UFD. Thanks in advance. Cheers |
104810 | How to prove convex+concave=affine?
Suppose [imath]f:\mathbb{R}^n\to \mathbb{R}[/imath] is both convex and concave, how to prove that [imath]f[/imath] is linear? or exactly speaking, [imath]f[/imath] is affine? I thought for the whole day, but I cannot figure it out. When I was working on this problem, I met another problem, are all the convex function continuous? If not, is there any counter example? Actually, I can prove for one dimensional case, in which [imath]f:\mathbb{R}\to \mathbb{R}[/imath]. However, I cannot generalize it into n dimensional cases. By the way, I use definition for convex(concave) like this: [imath]f(t\vec{x}+(1-t)\vec{y})\leq(or \geq) tf(\vec{x})+(1-t)f(\vec{y}), \forall t\in[0,1].[/imath] Thank you so much! | 1236356 | [imath]f[/imath] convex and concave, then [imath]f=ax+b[/imath]
Let [imath]f[/imath] be a real function defined on some interval [imath]I[/imath]. Assuming that [imath]f[/imath] both convex and concave on [imath]I[/imath], i.e, for any [imath]x,y\in I[/imath] one has [imath]f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y),\, \, \lambda\in (0,1) .[/imath] I would like to show that [imath]f[/imath] is of the form [imath]f=ax+b[/imath] for some [imath]a,b[/imath]. I was able to prove it when [imath]f[/imath] is differentiable, using the relation [imath]f'(x)=f'(y).[/imath] Anyway, I was not able to provide a general proof (without assuming that [imath]f[/imath] is differentiable, and without assuming that [imath]0\in I[/imath]). Any answer will be will be appreciated. Edit: It is little bit different from tte other question How to prove convex+concave=affine?. Here [imath]f[/imath] is defined on some interval, so [imath]o[/imath] not necessary in the domain. Please remove the duplicate message if this possible |
54397 | The identity cannot be a commutator in a Banach algebra?
The Wikipedia article on Banach algebras claims, without a proof or reference, that there does not exist a (unital) Banach algebra [imath]B[/imath] and elements [imath]x, y \in B[/imath] such that [imath]xy - yx = 1[/imath]. This is surprising to me, but maybe the proof is straightforward; anyone have a proof and/or a reference? More generally, I would have naively thought that I could embed any ring into a Banach algebra. I guess there are actually serious restrictions to doing this; are these issues discussed anywhere? | 387916 | [imath]AB - BA = I[/imath] in Hilbert Space
Let H be a Hilbert space and [imath]A[/imath] and [imath]B[/imath] be bounded operators in [imath]H[/imath]. How can I prove that [imath]AB - BA = I[/imath] is not possible ? Probably this is as easy as in the matrices case, but I couldn't prove it. Can you help me please ? Thank you :) |
115800 | How to prove that an operator is compact?
Consider [imath]T\colon\ell^2\to\ell^2[/imath] an operator such that [imath]Te_k=\lambda_k e_k[/imath] with [imath]\lambda_k\to 0[/imath] as [imath]k \to \infty[/imath] how to prove that it is compact? | 415292 | How to show that a sequence in [imath]l^{2}[/imath] has a convergent subseqence after the action of an operator.
Problem: Let [imath]l^{2}=\{(x_{1},x_{2},...)| x_{n} \in \mathbb{R}, \forall n,(x_{1}^{2}+x_{2}^{2}+...)<\infty\}[/imath], [imath]\|x\|=(x_{1}^{2}+x_{2}^{2}+...)^{\frac{1}{2}}[/imath]. Given [imath]\lambda_{n}\in\mathbb{R}[/imath]([imath]n=1,2,3...[/imath]), [imath]\lim\limits_{n \rightarrow \infty} \lambda_{n}=0[/imath]. Define [imath]A:l^{2}\rightarrow l^{2}[/imath] as follows: For [imath]x=(x_{1},x_{2},...)\in l^{2}[/imath], [imath]Ax=(\lambda_{1}x_{1},\lambda_{2}x_{2},...)[/imath]. Show that for a bounded sequence:[imath]\{x^{(j)}\}_{j=1}^{\infty}[/imath] in [imath]l^{2}[/imath], [imath]\{Ax^{(j)}\}_{j=1}^{\infty}[/imath] has a subsequence which converges in [imath]l^{2}[/imath]. I am new in fuctional analysis and I have no idea how to handle this problem. I am grateful to any help! Thanks a lot in advance! |
65239 | Right identity and Right inverse implies a group
Let [imath](G, *)[/imath] be a semi-group. Suppose [imath] \exists e \in G[/imath] such that [imath]\forall a \in G,\ ae = a[/imath]; [imath]\forall a \in G, \exists a^{-1} \in G[/imath] such that [imath]aa^{-1} = e[/imath]. How can we prove that [imath](G,*)[/imath] is a group? | 365348 | Homework - Prove that a given set is a group
I have a homework question and I don't know how to approach this exercise. The exercise is the following: Let's suppose [imath]G[/imath] is a set with binary function * defined for its members, which is: closings; associative; there's [imath]e\in G[/imath], so that [imath]a\star e=a[/imath] where [imath]a\in G[/imath]; for each [imath]a\in G[/imath], there's a [imath]b\in G[/imath] so that [imath]a\star b=e[/imath]. Prove that [imath]G[/imath] is a group. I have no Idea how approach this exercise. pay attention that 3,4 are Noncommutative. |
261157 | Show [imath]S = f^{-1}(f(S))[/imath] for all subsets [imath]S[/imath] iff [imath]f[/imath] is injective
Let [imath]f: A \rightarrow B[/imath] be a function. How can we show that for all subsets [imath]S[/imath] of [imath]A[/imath], [imath]S \subseteq f^{-1}(f(S))[/imath]? I think this is a pretty simple problem but I'm new to this so I'm confused. Also, how can we show that [imath]S = f^{-1}(f(S))[/imath] for all subsets [imath]S[/imath] iff [imath]f[/imath] is injective? | 863268 | Suppose [imath]f: X \rightarrow Y[/imath], and is one-to-one, and let [imath]A \subseteq X[/imath], prove that [imath]f^{-1}[f[A]] = A[/imath].
Suppose [imath]f: X \rightarrow Y[/imath], and is one-to-one, and let [imath]A \subseteq X[/imath], prove that [imath]f^{-1}[f[A]] = A[/imath]. EDIT: Actually, this identity should hold even if [imath]f[/imath] is not one-to-one (injective), right? This is completely intuitive and logical, but I can't think of a proof strategy to use to prove it? I'm familiar with induction and element chasing, but they don't seem to work here. Here is the beginning of my element chasing proof to prove that [imath]f^{-1}[f[A]] \subseteq A[/imath]. Start with arbitrary element: [imath]a \in f^{-1}[f[A]][/imath] \ Apply definition of preimage: [imath]a \in \{x \in X | f(x) \in f[A] \}[/imath] \ Simplify and convert to boolean expression: [imath]a \in X \land f(a) \in f[A][/imath] \ Apply definition of image: [imath]a \in X \land f(a) \in \{y \in Y | y = f(x), x \in A \}[/imath] \ I'm stuck here... |
105107 | Prove: [imath]\int_0^\infty \sin (x^2) \, dx[/imath] converges.
[imath]\sin x^2[/imath] does not converge as [imath]x \to \infty[/imath], yet its integral from [imath]0[/imath] to [imath]\infty[/imath] does. I'm trying to understand why and would like some help in working towards a formal proof. | 794476 | Does [imath]\int _1^{\infty }\left(\sin \left(x^2\right)\right)dx[/imath] converge or diverge?
I'm in need of some assistance regarding a question in my Calculus textboox: Find if the following converges or diverges without calculating the integral: [imath]\int _1^{\infty }\left(\sin \left(x^2\right)\right)dx[/imath] I tried using several methods, including the convergence test but with no luck. Any help is appreciated, Thx! |
30811 | Prove that [imath]\lfloor \sqrt{p} \rfloor + \lfloor \sqrt{2p} \rfloor +...+ \lfloor \sqrt{\frac{p-1}{4}p} \rfloor = \frac{p^2 - 1}{12}[/imath]
Problem Prove that [imath]\lfloor \sqrt{p} \rfloor + \lfloor \sqrt{2p} \rfloor +...+ \lfloor \sqrt{\frac{p-1}{4}p} \rfloor = \dfrac{p^2 - 1}{12}[/imath] where [imath]p[/imath] prime such that [imath]p \equiv 1 \pmod{4}[/imath]. I really have no idea how to start :(! The square root part really messed me up. Can anyone give me a hint? Thank you | 364451 | [imath]\sum_{i=1}^{k}[\sqrt{ip}]=\frac{p^2-1}{12}[/imath], [imath]p[/imath] is a prime of the form [imath]4k+1[/imath]
[imath]\sum_{i=1}^{k}[\sqrt{ip}]=\frac{p^2-1}{12}[/imath], [imath]p[/imath] is a prime of the form [imath]4k+1[/imath] How to prove this? |
27425 | What am I doing when I separate the variables of a differential equation?
I see an equation like this: [imath]y\frac{\textrm{d}y}{\textrm{d}x} = e^x[/imath] and solve it by "separating variables" like this: [imath]y\textrm{d}y = e^x\textrm{d}x[/imath] [imath]\int y\textrm{d}y = \int e^x\textrm{d}x[/imath] [imath]y^2/2 = e^x + c[/imath] What am I doing when I solve an equation this way? Because [imath]\textrm{d}y/\textrm{d}x[/imath] actually means [imath]\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}[/imath] they are not really separate entities I can multiply around algebraically. I can check the solution when I'm done this procedure, and I've never run into problems with it. Nonetheless, what is the justification behind it? What I thought of to do in this particular case is write [imath]\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x[/imath] [imath]\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c[/imath] then by the fundamental theorem of calculus [imath]y^2/2 = e^x + c[/imath] Is this correct? Will such a procedure work every time I can find a way to separate variables? | 1356981 | Why does the "separation of variables" method for DEs work?
Heyho, I am using the separation-of-variables method for quite a while now, but what was always bothering me a bit, is why is it possible to do those operations. I'll give a concrete example (source Wikipedia): [imath]\frac{dy}{dx}=xy^2 + x \Rightarrow \frac{dy}{1+y^2} = x \:dx \Rightarrow \int{\frac{dy}{1+y^2}} = \int{x \:dx} \Rightarrow \cdots[/imath] and so on. My problem lies in step 2. Why can I just treat the differential operator like a variable? |
11028 | Convergence of integrals in [imath]L^p[/imath]
Stuck with this problem from Zgymund's book. Suppose that [imath]f_{n} \rightarrow f[/imath] almost everywhere and that [imath]f_{n}, f \in L^{p}[/imath] where [imath]1<p<\infty[/imath]. Assume that [imath]\|f_{n}\|_{p} \leq M < \infty[/imath]. Prove that: [imath]\int f_{n}g \rightarrow \int fg[/imath] as [imath]n \rightarrow \infty[/imath] for all [imath]g \in L^{q}[/imath] such that [imath]\dfrac{1}{p} + \dfrac{1}{q} = 1[/imath]. Right, so I estimate the difference of the integrals and using Hölder end up with: [imath]\left|\int f_{n} g - \int fg\right| \leq \|g\|_{q} \|f_{n} - f\|_{p}[/imath] From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed? | 1987200 | Convergence of the integral of product of [imath]L^p[/imath] and [imath]L^q[/imath] functions
Let [imath]p[/imath] and [imath]q[/imath] be finite Holder conjugates, and let [imath]f_n[/imath] be a sequence of [imath]p[/imath]-integrable functions converging pointwise to [imath]f[/imath]. Suppose there is some uniform constant [imath]C[/imath] such that [imath]\|f_n\|_p < C[/imath] for all [imath]n[/imath]. Prove that if [imath]g \in L^q[/imath], then [imath] \int f_ng \to \int fg. [/imath] The "obvious" estimate where one uses Holder's inequalty does not work here since we do not have [imath]f \in L^p[/imath] or that [imath]\|f_n - f\||_p \to 0[/imath]. I can't find a dominating function, so I do not know how to use the convergence a.e. criterion. |
61482 | Proving the identity [imath]\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2[/imath] without induction
I recently proved that [imath]\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2[/imath] using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs. | 366829 | Visual/intuitive proof of why [imath]\sum k^3 = (\sum k)^2[/imath], where [imath]k[/imath] goes from 1 to [imath]n[/imath]?
I understand that one could prove this by first proving the analytic expressions of the sigma terms through induction, and then square the [imath]\displaystyle\sum_{k=1}^n k[/imath] term to show LHS = RHS. Are there any other easier to understand (preferably visual) proofs one could show? Thanks. |
143468 | Entire function bounded by a polynomial is a polynomial
Suppose that an entire function [imath]f(z)[/imath] satisfies [imath]\left|f(z)\right|\leq k\left|z\right|^n[/imath] for sufficiently large [imath]\left|z\right|[/imath], where [imath]n\in\mathbb{Z^+}[/imath] and [imath]k>0[/imath] is constant. Show that [imath]f[/imath] is a polynomial of degree at most [imath]n[/imath]. | 518335 | Analytic function bounded by polynomial
Prove that a function which is analytic in the whole plane and satisfies the inequality [imath]|f(z)|<|z|^n[/imath] for some [imath]n[/imath] and all sufficiently large [imath]|z|[/imath] reduces to a polynomial. The function is analytic, so [imath]f^{n}(z)[/imath] exists for all [imath]n[/imath], all [imath]z[/imath]. We have the Cauchy's integral formula for higher derivatives [imath]f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz.[/imath] So [imath]\int_C\frac{f(z)}{(z-a)^{n+1}}dz<|z|^n[/imath] for all large [imath]|z|[/imath]. How does that help? |
22399 | If p is an odd prime, prove that [imath]1^2 \times 3^2 \times 5^2 \cdots \times (p-2)^2 \equiv (-1)^{(p+1)/2}\pmod{p}[/imath]
I also have to prove this for [imath]2^2 \times 4^2 \times 6^2 \cdots \times (p-1)^2 \equiv (-1)^{(p+1)/2} \pmod{p}[/imath] I made some progress so far and got stuck. I said that since p is odd, [imath](p+1)/2[/imath] is even. Then we can say that [imath](-1)^{\mathrm{even}}= 1[/imath], so [imath](-1)^{(p+1)/2}\pmod{p}[/imath] can be written as [imath]1\pmod{p}[/imath]. I know that the product of odds is odd and the product of evens is even, but I cant prove that the left side of this equation in either case is congruent to [imath]1 \pmod{p}[/imath]. Any help here would be greatly appreciated. | 2837357 | Prove that [imath]1^2\cdot3^2\cdot5^2\cdots(p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p[/imath]
If [imath]p[/imath] is an odd prime, prove that [imath]1^2\cdot3^2\cdot5^2\cdots(p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p[/imath] Then [imath]1^2\cdot3^2\cdot5^2\cdots(p-2)^2={1/16} (p-2)^2(3p-1)^2\tag{1}[/imath] Now since [imath]p[/imath] is odd then [imath](p+1)/2[/imath] is even and [imath](-1)^{(p+1)/2}=1[/imath]. Now I tried substract [imath]1[/imath] from [imath](1)[/imath] to find out if the result is a multiple of [imath]p[/imath]. But I can't get the answer any suggestions. |
64705 | If [imath]E[/imath] has measure zero, then does [imath]E^2[/imath] have measure zero?
Currently I'm just working through measure theory just to see if its something I would like to take. Unfortunately I am stuck on this problem from Carothers. If [imath]m^*(E)=0[/imath], then [imath]m^*(E^2)=0[/imath]. Where [imath]m^*[/imath] denotes outer measure and [imath]E^2=\{x^2:x\in E\}.[/imath] I toyed with the idea that [imath]I_k < 1 \Rightarrow I^2_k < I_k[/imath]. However I am at a loss as to how to set up a chain of inequalities (which is what I am assuming I need). | 348741 | Outer Lebesgue measure
Let [imath]E\in\mathbb{R}[/imath] and let [imath]\lambda^*(E)[/imath] denotes the outer Lebesgue measure of [imath]E[/imath]. Let [imath]F:=\{x^2:x\in E\}[/imath]. If [imath]\lambda^*(E)=0[/imath], how can we show that [imath]\lambda^*(F)=0[/imath]? |
39895 | The direct sum [imath]\oplus[/imath] versus the cartesian product [imath]\times[/imath]
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines [imath]A^n := A\times A\times\ldots\times A[/imath]; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form [imath] \mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t} [/imath] where [imath]\mathbb{Z}^n[/imath] now means [imath]\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}[/imath]. From an intuition perspective, and in the sense of sets, is this more or less the same as [imath]\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}[/imath]? (Bear in mind I am normally using these ideas in relation to homology groups.) | 1994437 | Why [imath]\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z[/imath] is used instead of [imath]\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z[/imath]?
Since both notations are the same, why [imath]\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z[/imath] is used instead of [imath]\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z[/imath] ? |
148850 | Every proper subspace of a normed vector space has empty interior
There is a conjecture: "The only subspace of a normed vector space [imath]V[/imath] that has a non-empty interior, is [imath]V[/imath] itself." (here, the topology is the obvious set of all open sets generated by the metric [imath]||\cdot||[/imath]). I have a proof for the case [imath]V[/imath] is finite dimensional. Because, let [imath]V[/imath] have dimension [imath]n[/imath] and a subspace [imath]S[/imath] of [imath]V[/imath] have dimension [imath]m < n[/imath]. Let [imath]\{v_1,v_2,\ldots,v_m\}[/imath] be a basis for [imath]S[/imath], extended to the basis [imath]\{v_1,\ldots,v_m,\ldots,v_n\}[/imath] of [imath]V[/imath]. Now, suppose that [imath]p = b_1v_1+\cdots+b_mv_m[/imath] is an interior point of [imath]S[/imath]. Now, consider the norm [imath]N(a_1v_1 +\cdots+a_nv_n)=\max(|a_1|,\ldots,|a_n|)[/imath]. Then, there is an [imath]r>0[/imath], such that [imath]||x-p||<r[/imath] and [imath]x[/imath] is in [imath]V[/imath] [imath]\implies[/imath] [imath]x[/imath] is in [imath]S[/imath], since on a finite dim. space, all norms are equivalent. Let [imath]m<k\leq n[/imath],and chose [imath]v[/imath] in [imath]V[/imath] as: [imath]v=\left(b_1+\frac{r}{2}\right)v_1+\left(b_2+\frac{r}{2}\right)v_2+\cdots+\left(b_m+\frac{r}{m}\right)v_m+\frac{r}{2}v_k[/imath] Then, [imath]N(v-p)\leq\frac{r}{2}<r[/imath], so [imath]v[/imath] is in [imath]S[/imath] and by the subspace property of [imath]S[/imath], [imath]v_k[/imath] is in [imath]S[/imath] too, a contradiction to [imath]m<n[/imath]. I have primarily 2 questions: (1) Is there a simpler method to proof the conjecture for the finite dimensional case? (2) Is the conjecture true for the infinite dimensional case? Sorry, if the question admits a very trivial answer. The motivation behind my question , is the fact that an open interval in [imath]\mathbb{R}[/imath] is not open in [imath]\mathbb{R}\times\mathbb{R}[/imath], etc. | 2614056 | Vector subspace of a normed space
Let [imath]X[/imath] be a normed space and let [imath]Y \neq X[/imath] be its vector subspace. Prove that Y does not contain any nonempty open set in X. Could someone help please with the proof? Thanks in advance! :) |
7266 | Why does this process, when iterated, tend towards a certain number? (the golden ratio?)
Take any number [imath]x[/imath] (edit: x should be positive, heh) Add 1 to it [imath]x+1[/imath] Find its reciprocal [imath]1/(x+1)[/imath] Repeat from 2 So, taking [imath]x = 1[/imath] to start: 1 2 (the + 1) 0.5 (the reciprocal) 1.5 (the + 1) 0.666... (the reciprocal) 1.666... (the + 1) 0.6 (the reciprocal) 1.6 (the + 1) 0.625 1.625 0.61584... 1.61584... 0.619047... 1.619047... 0.617647058823.. etc. If we look at just the "step 3"'s (the reciprocals), we get: 1 0.5 0.666... 0.6 0.625 0.61584... 0.619047... 0.617647058823.. This appears to always converge to 0.61803399... no matter where you start from. I looked up this number and it is often called "The golden ratio" - 1, or [imath]\frac{1+\sqrt{5}}{2}-1[/imath]. Is there any "mathematical" way to represent the above procedure (or the terms of the second series, of "only reciprocals") as a limit or series? Why does this converge to what it does for every starting point [imath]x[/imath]? edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1). However, I think I could easily make it the golden ratio by taking the +1 "steps" of the original series, instead of the reciprocation steps of the original series: 2 1.5 1.666... 1.6 1.625 1.61584... 1.619047... 1.617647058823.. which does converge to [imath]\frac{1+\sqrt{5}}{2}-1[/imath] Explaining either of these series is adequate as I believe that explaining one also explains the other. | 383456 | limit, low or high bound, convergence for recursive sequence
given is the following sequence: [imath]a_1 > 0[/imath] [imath]a_n = \frac{1}{1+a_n}[/imath] I succeeded in finding a (possible?) limit by guessing that the sequence is limited by a; then the sequence [imath]a_n[/imath] converges to a; however, the sequence [imath]a_n = \frac{1}{1+a_n}[/imath] also converges to a. This means: [imath]a = \frac{1}{1+a}[/imath]. Solving this equation gives the 'possible' limit. However, is this the end? Shouldn't I first show THAT the sequence actually IS convergent? If so: HOW should I do this? I can show: [imath]0 < a_n < 1[/imath] I cannot show: [imath]a_n[/imath] is monotonous. |
55638 | No continuous function switches [imath]\mathbb{Q}[/imath] and the irrationals
Is there a way to prove the following result using connectedness? Result: Let [imath]J=\mathbb{R} \setminus \mathbb{Q}[/imath] denote the set of irrational numbers. There is no continuous map [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] such that [imath]f(\mathbb{Q}) \subseteq J[/imath] and [imath]f(J) \subseteq \mathbb{Q}[/imath]. http://planetmath.org/encyclopedia/ThereIsNoContinuousFunctionThatSwitchesTheRationalNumbersWithTheIrrationalNumbers.html | 367856 | Continuous function that take irrationals to rationals and vice-versa.
Can someone help me? How can I prove that there isn't an everywhere continuous function [imath]f:\mathbb R \rightarrow \mathbb R[/imath] that transforms every rational into an irrational and vice-versa? |
732 | Proof that the irrational numbers are uncountable
Can someone point me to a proof that the set of irrational numbers is uncountable? I know how to show that the set [imath]\mathbb{Q}[/imath] of rational numbers is countable, but how would you show that the irrationals are uncountable? | 767622 | Is [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] countable?
I need to prove that [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] is countable or uncountable. I believe it is uncountable. I am not sure how to prove it. [imath]\mathbb{R}[/imath] is known to be uncountable and [imath]\mathbb{Q}[/imath] is countable. By reason when I take the difference of the two it would be uncountable. How do I prove this? |
296240 | Residue of [imath]\frac{-i}{y-\beta i}e^{A + By + Cy^{-1}}[/imath] in [imath]y=0[/imath].
I have been trying to find the residue of [imath]\frac{-i}{y-\beta i}e^{A + By + Cy^{-1}}[/imath] in [imath]y=0[/imath]. But I'm stuck. I would be really grateful for some help. | 295401 | Finding the residue of function with Laurent series [imath]\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{y^n(A+By+Cy^{-1})^k}{\beta (\beta i)^n \ k!}[/imath]
I have been trying to find the residue of [imath]f(\omega) = \frac{e^{i \omega a} e^{\frac{-b \omega}{\omega + ib}}}{i \omega}[/imath] at the essential singularity [imath]\omega = -ib[/imath] for a while, but it is giving me a headache and I would really appreciate some help. For inspiration I have been looking at Residue of [imath]z^2 e^{1/\sin z}[/imath] at [imath]z=\pi[/imath] and How to compute the residue of a complex function with essential singularity This is my aproach so far: I moved the singularity to 0 by substituting [imath]y = \omega + \beta i[/imath], then I rewrote and simplified the exponent to [imath]A + By +Cy^{-1}[/imath] which gives us the expression: [imath]\frac{-i}{y-\beta i}e^{A + By + Cy^{-1}}[/imath] where we want the residue at [imath]y=0[/imath] Then I used that [imath]\frac{-i}{y-\beta i} = \sum_{n=0}^{\infty}\frac{y^n}{(\beta i)^n\beta }[/imath] for [imath]|y|<|\beta|[/imath] and wrote the exponential as an infinite sum to get: [imath]\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{y^n(A+By+Cy{^-1})^k}{\beta (\beta i)^n \ k!}[/imath] Now unless I'm mistaken, this should be the Laurent series of [imath]f[/imath], and we should be able to find the residue from this. I just can't wrap my head around the double sum and the triple binomial formula. I would be very grateful for hints on how to proceed, or on a better approach if there is one. |
17320 | Derivation of the Partial Derangement (Rencontres numbers) formula
I'm looking for the method by which the partial derangement formula [imath]D_{n,k}[/imath] was derived. I can determine the values for small values of N empirically, but how the general case formula arose still alludes me. Any links/books or an explanation will be appreciated. The formula is: [imath]D_{n,k} = {n \choose k}!(n-k)[/imath] Links: Mathworld | 877785 | The probability that exactly / at-least [imath]k[/imath] numbers are in the correct position
Given a sequence of [imath][1,\dots,n][/imath] in random order: Let [imath]P_k[/imath] be the probability that exactly [imath]k[/imath] numbers are in the correct position Let [imath]Q_k[/imath] be the probability that at least [imath]k[/imath] numbers are in the correct position How can we calculate [imath]P_k[/imath] and [imath]Q_k[/imath]? These are the facts that I've managed to establish so far: [imath]P_n=\frac{1}{n!}[/imath] [imath]P_{n-1}=0[/imath] [imath]P_k=Q_k-Q_{k+1}[/imath] [imath]Q_n=P_n[/imath] [imath]Q_{n-1}=Q_n[/imath] [imath]Q_k=\sum\limits_{i=k}^{n}P_i[/imath] Partial answers (for example: [imath]P_0,P_1,Q_1[/imath]) or any other insights will also be appreciated... Thanks |
116526 | The Supremum and Bounded Functions
I'm trying to show that this is true: Let [imath]X[/imath] be a set and suppose [imath]f[/imath] and [imath]g[/imath] are bounded (real-valued) functions defined on [imath]X[/imath]. Then, [imath] \sup_{x \in X}|f(x)g(x)| \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)| [/imath] I think I'm pretty close but I'm not sure about the last step. First, since [imath]f[/imath] and [imath]g[/imath] are bounded, all involved suprema exist and are finite. If [imath]a = \sup|f(x)|[/imath] and [imath]b = \sup|f(x)|[/imath] then it is true that [imath] a \geq |f(x)| \;\;\;\;\;\; b \geq |g(x)| [/imath] for every [imath]x \in X[/imath]. Since none of the quantities involved are negative, this implies [imath] a b \geq |f(x)|\cdot |g(x)| \implies \sup|f(x)|\sup|g(x)| \geq |f(x)|\cdot |g(x)| [/imath] Can I say now that since this last inequality holds for all [imath]x[/imath] that [imath] \sup_{x \in X}|f(x)g(x)| = \sup_{x \in X} \left(|f(x)|\cdot |g(x)|\right) \leq \sup_{x \in X}|f(x)|\sup_{x \in X}|g(x)|? [/imath] Thanks. | 462364 | How can I show that [imath]\sup(AB)\geq\sup A\sup B[/imath] for [imath]A,B\subset\mathbb{R}[/imath] where [imath]A\cup B[/imath] is positive and bounded?
The question is based on the following exercise in real analysis: Assume that [imath]A,B\subset{\Bbb R}[/imath] are both bounded and [imath]x>0[/imath] for all [imath]x\in A\cup B[/imath]. Show that [imath] \sup(AB)=\sup A\sup B [/imath] where [imath] AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}. [/imath] Since [imath]0<a\leq\sup A[/imath] and [imath]0<b\leq\sup B[/imath] for all [imath]a\in A[/imath] and [imath]b\in B[/imath], we have [imath] ab\leq\sup A\sup B [/imath] for all [imath]ab\in AB[/imath] which implies that [imath]\sup AB\leq\sup A\sup B[/imath]. I have trouble with another direction: [imath] \sup AB\geq\sup A\sup B [/imath] I was trying to show that for every [imath]\epsilon >0[/imath], [imath]\sup AB-\epsilon \geq \sup A\sup B[/imath]. If one uses the definition of supremum, one has the estimates that for every [imath]\epsilon>0[/imath], [imath] \sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b [/imath] for some [imath]a\in A,\ b\in B[/imath]. It follows that [imath] \sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2 [/imath] which seems quite close to what I want. How can I go on? |
207335 | Prove [imath]\sup(f+g) \le \sup f + \sup g[/imath]
Suppose [imath]$D$[/imath] is a nonempty bounded subset of reals. Let [imath]$f:D \to \mathbb R$[/imath] and [imath]$g:D \to \mathbb R$[/imath]. Define [imath](f+g)(x)=f(x)+g(x)[/imath]. Prove [imath]\sup(f+g)(D) \le \sup f(D) + \sup g(D)[/imath] (also prove that [imath]\sup (f+g)[/imath] exists). I understand why this is the case, just not how to prove it. Left side is pretty much [imath]\sup (f(x)+g(x))[/imath] and right side is [imath]\sup (f(x) + g(y))[/imath] for [imath]x,\,y \in D[/imath]. Basically [imath]f+g[/imath] has to use the same variable and [imath]f(D)+g(D)[/imath] use different ones. But I don't know how to go about proving this. The second part of the question is to find a specific example where strict inequality holds. Let [imath]D=\{a,b\}[/imath] and [imath]f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1[/imath]. [imath]\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.[/imath] [imath]\sup (f+g)(D) = 1[/imath] (if we choose a, [imath]f+g = 1+0,\, b,\, f+g=0+1[/imath]). | 497760 | Suprema Properties and Real Value Functinos
Suppose [imath]f : A \to \mathbb{R}[/imath] and [imath]g: A \to \mathbb{R}[/imath] are real valued functions. Define [imath](f+g)[A] = \{f(x) + g(x): x \in A\}[/imath] and [imath]f[A] + g[A] = \{f(x) + g(y): x, y \in A\}[/imath]. What is the relationship between [imath]\sup(f[A] + g[A])[/imath] and [imath]\sup((f+g)[A])[/imath]? Repeat exercise for [imath]\inf(f[A] + g[A])[/imath]. Essentially one side is [imath]f(x) + g(x)[/imath] and the other is [imath]f(x) + g(y)[/imath]. I am thinking they are equal, but I'm not sure. I need a proof too. |
74265 | The sum of the first [imath]n[/imath] squares is a square: a system of two Pell-type-equations
This question comes from trying to see why 24 is the only non-trivial value of [imath]n[/imath] for which [imath]1^2+2^2+3^2+\cdots+n^2[/imath] is a perfect square. To this end, let [imath]m,n \in \mathbb N[/imath] be such that [imath]1^2+2^2+3^2+\cdots+n^2 = m^2[/imath], or [imath]n(n+1)(2n+1) = 6m^2.[/imath] When [imath]n=24[/imath] the left hand side is [imath]24\times 25 \times 49[/imath] and there are two things that make it work as a solution: [imath]7^2+1=2\times5^2[/imath] [imath]7^2-1=12\times2^2.[/imath] We can write these algebraically as [imath]x^2=2y^2-1[/imath] [imath]x^2=12z^2+1[/imath] and solve them simultaneously (with [imath]x,y,z\in \mathbb N[/imath]). These are instances of Pell's equation and each individually has an infinite number of solutions. How do we show there is only one (non-trivial) value of [imath]x[/imath] that is common to the solutions of both equations? | 303778 | Techniques to prove that there is only one square in a given sequence
What techniques/methods can be used to prove that the sequence produced by [imath]n\cdot (n+1)\cdot (2\cdot n+1)/6[/imath] contains only one square ([imath]4900[/imath]) greater than 1? While this particular sequence is an interesting example, I'm interested in techniques that can be generalized to any sequence with a polynomial generating function. In general, this is equivalent to asking for the solution to the Diophantine equation: [imath] a^2 = n\cdot (n+1)\cdot (2\cdot n+1)/6. [/imath] |
2815 | Find taxicab numbers in [imath]O(n)[/imath] time
This is a final exam question in my algorithms class: [imath]k[/imath] is a taxicab number if [imath]k = a^3+b^3=c^3+d^3[/imath], and [imath]a,b,c,d[/imath] are distinct positive integers. Find all taxicab numbers [imath]k[/imath] such that [imath]a,b,c,d < n[/imath] in [imath]O(n)[/imath] time. I don't know if the problem had a typo or not, because [imath]O(n^3)[/imath] seems more reasonable. The best I can come up with is [imath]O(n^2 \log n)[/imath], and that's the best anyone I know can come up with. The [imath]O(n^2 \log n)[/imath] algorithm: Try all possible [imath]a^3+b^3=k[/imath] pairs, for each [imath]k[/imath], store [imath](k,1)[/imath] into a binary tree(indexed by [imath]k[/imath]) if [imath](k,i)[/imath] doesn't exist, if [imath](k,i)[/imath] exists, replace [imath](k,i)[/imath] with [imath](k,i+1)[/imath] Transverse the binary tree, output all [imath](k,i)[/imath] where [imath]i\geq 2[/imath] Are there any faster methods? This should be the best possible method without using any number theoretical result because the program might output [imath]O(n^2)[/imath] taxicab numbers. Is [imath]O(n)[/imath] even possible? One have to prove there are only [imath]O(n)[/imath] taxicab numbers lesser than [imath]2n^3[/imath] in order to prove there exist a [imath]O(n)[/imath] algorithm. Edit: The professor admit it was a typo, it should have been [imath]O(n^3)[/imath]. I'm happy he made the typo, since the answer Tomer Vromen suggested is amazing. | 290635 | Algorithm to find the numbers expressible as the sum of two positive cubes in two different ways
I have known this from beginning that [imath]1729[/imath] is the smallest number expressible as the sum of two cubes in two different ways: [imath] 12^3 + 1^3 [/imath] and [imath] 10^3+9^3 [/imath] I am a Software Developer and if someone can tell me the logic to write a program for printing such types of number will be greatly helpful. |
56335 | Proving [imath]\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}[/imath] with induction
I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing. [imath]\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}[/imath] Perhaps someone can help me out I don't understand how to move forward from here: [imath]\sum\limits_{k=1}^{n+1}{\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n+1}}\ge \sqrt{n+1}}[/imath] proof and explanation would greatly be appreciated :) Thanks :) EDIT sorry meant GE not = fixed :) | 533961 | Prove by induction the following inequality for all n∈N
[imath]\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}[/imath] I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis. Please help :) I've solved equality's before using induction, but this is the first time I've done an inequality, and although I'm sure the process is fairly similar, I think the square root is messing me up. |
264889 | Double limit of [imath]\cos^{2n}(m! \pi x)[/imath] at rationals and irrationals
I stumbled upon this "relation" (is the name correct?): [imath] \lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x) = \begin{cases} 1,&x\text{ is rational}\\ 0,&x\text{ is irrational}\end{cases} [/imath] How is it called and why is it so? I'm really not asking for a proof since I fear it would be too complicated for me to understand, but rather for an "intuition". | 684344 | How to solve double limit [imath]\lim_{m\to\infty}[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}][/imath]
Please provide some hints as to how to solve questions with double limits such as this: [imath]\lim_{m\to\infty}\left[\lim_{n\to\infty}(\cos(m!\cdot \pi\cdot x))^{2n}\right][/imath] One of the things I did was convert the original function to: [imath]e^{n\ln(\cos(m!\cdot\pi\cdot x)^2)}[/imath] and then change cosine into sine and take [imath]t=1/m[/imath], and try to use [imath]\lim \frac{\sin(m!\cdot\pi\cdot x)}{m!\cdot\pi\cdot x}[/imath] but that only messed it up even further. I obviously can't use L'Hopital as not both the numerator and denominator go to zero. Another thing was to try to use the power series expansion, but that seemed even more complicated as there is still the power of 2n to deal with. Please help! Thanks. |
30317 | [imath]\arcsin[/imath] written as [imath]\sin^{-1}(x)[/imath]
I know that different people follow different conventions, but whenever I see [imath]\arcsin(x)[/imath] written as [imath]\sin^{-1}(x)[/imath], I find myself thinking it wrong, since [imath]\sin^{-1}(x)[/imath] should be [imath]\csc(x)[/imath], and not possibly confused with another function. Does anyone say it's bad practice to write [imath]\sin^{-1}(x)[/imath] for [imath]\arcsin(x)[/imath]? | 417330 | What 's the differece between [imath]\cot(x)[/imath] and [imath]\arctan(x)[/imath]?
I know that [imath]\displaystyle \cot(x)=\frac{1}{\tan(x)}[/imath] and [imath]\space \displaystyle \arctan(x)=\tan(x)^{-1}=\frac{1}{\tan(x)}[/imath] What is the difference between these two function? Is [imath]\cot(x)[/imath] the reciprocal function of [imath]\space \tan(x) \space[/imath] and [imath]\arctan(x)[/imath] is the inverse function of [imath]\tan(x)[/imath]? And, so the assumption that [imath]\space \displaystyle \arctan(x)=\tan(x)^{-1}=\frac{1}{\tan(x)}[/imath], is incorrect? |
29777 | Closed form for the sequence defined by [imath]a_0=1[/imath] and [imath]a_{n+1} = a_n + a_n^{-1}[/imath]
Today, we had a math class, where we had to show, that [imath]a_{100} > 14[/imath] for [imath]a_0 = 1;\qquad a_{n+1} = a_n + a_n^{-1}[/imath] Apart from this task, I asked myself: Is there a closed form for this sequence? Since I didn't find an answer by myself, can somebody tell me, whether such a closed form exists, and if yes what it is? | 634694 | Concerning the sequence [imath] x_1=1 ,\space x_{n+1}=x_n+\dfrac{1}{x_n} [/imath]
Let [imath](x_n)[/imath] be a real sequence satisfying [imath] x_1=1 ,\space x_{n+1}=x_n+\dfrac{1}{x_n} ,\forall n \in \mathbb N[/imath] , then can we find an expression for [imath][ x_n[/imath]] ( box-function) in terms of [imath]n[/imath] [imath],\forall n \in \mathbb N[/imath] ? ( I have only been able to find that [imath]x_n^2>2n , \forall n>2[/imath] and [imath]lim \dfrac {x_n^2}{n}=2[/imath]) |
39170 | How come such different methods result in the same number, [imath]e[/imath]?
I guess the proof of the identity [imath] \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x [/imath] explains the connection between such different calculations. How is it done? | 475990 | Why is[imath] (1+\frac{1}{n})^n=e[/imath] when n goes to infinity?
Why is [imath]\lim\limits_{n\to\infty}(1+\frac1n)^n=e[/imath]? I think it involves [imath]\sum\limits_{n=0}^\infty\frac1{k!}=e[/imath] but not sure how to get from one to the other. |
3668 | What is the value of [imath]1^i[/imath]?
What is the value of [imath]1^i[/imath]? [imath]\,[/imath] | 546080 | What does [imath]i^i [/imath] equal and why?
I've been reading up on why the value of 0^0 is controversial (see Zero to the zero power - is [imath]0^0=1[/imath]?) and I wondered: is it possible for [imath]i^i[/imath] to have a value? I plugged it into a TI-83 calculator and it returned 0.2078795764 (!) How is this possible and why is the result a real number not a complex number? Update I realize how to use Euler's Identity of [imath]e^{i\theta} = \cos(\theta) + i\sin(\theta)[/imath] and I understand the oldest answer's value of [imath]\theta = \pi/2[/imath] to solve for [imath]i^i[/imath], but doesn't this mean I can pick any odd multiple of [imath]\pi/2[/imath] (such as [imath]3\pi/2, 5\pi/2[/imath], etc.) as a value of [imath]\theta[/imath]? Does that mean that [imath]i^i[/imath] is somehow "periodic" like the cosine curve is? |
81583 | How do I prove that [imath]x^p-x+a[/imath] is irreducible in a field with [imath]p[/imath] elements when [imath]a\neq 0[/imath]?
Let [imath]p[/imath] be a prime. How do I prove that [imath]x^p-x+a[/imath] is irreducible in a field with [imath]p[/imath] elements when [imath]a\neq 0[/imath]? Right now I'm able to prove that it has no roots and that it is separable, but I have not a clue as to how to prove it is irreducible. Any ideas? | 458650 | [imath]x^p-x-a[/imath] is irreducible over [imath]F_p[/imath] for p-prime, [imath]a\in F_p, a\neq 0[/imath]
Actual Question is to prove that, for a prime p and [imath]a\in F_p[/imath], [imath]a\neq 0[/imath], [imath]f(x)=x^p-x+a[/imath] is irreducible. This is an exercise Question in Dummit Foote 13.5.5. Hint : Prove that if [imath]\alpha[/imath] is a root of [imath]f(x)[/imath] then so is [imath]\alpha+1[/imath]. Proof: Suppose [imath]f(\alpha)=0[/imath] i.e., [imath]\alpha^{p}-\alpha+a=0[/imath] Consider [imath](\alpha+1)^{p}-(\alpha+1)+a=\alpha^p+1-\alpha-1+a=\alpha^{p}-\alpha+a=0[/imath] . So, I proved that if [imath]\alpha[/imath] is a root of [imath]f(x)[/imath] then so is [imath]\alpha+1[/imath]. As further approach I have seen that no element of [imath]F_p[/imath] is a root because if [imath]\alpha\in F_p[/imath] is a root then so is [imath]\alpha+1[/imath] and so is [imath]\alpha+2[/imath] and so would be [imath]0\in F_p[/imath] which is a contradiction as [imath]f(0)=0-0+a=a\neq 0[/imath]. So, No element of [imath]F_p[/imath] is a root. Just by this he wants to say that [imath]f(x)[/imath] is irreducible i.e., if [imath]f(x)[/imath] has no root in [imath]F_p[/imath] then it is irreducible. I am unable to proceed further in this line. Any help would be appreciated. Thank You. P.S : I have seen just now that this question was already answered. I thought of deleting this Question. But Mr.AymanHourieh gave another nice answer. So, i Request Not to close this as this would just give another answer. Thank You |
7172 | Injective functions with intermediate-value property are continuous. Better proof?
A function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] is said to have the intermediate-value property if for any [imath]a[/imath], [imath]b[/imath] and [imath]\lambda \in [f(a),f(b)][/imath] there is [imath]x \in [a,b][/imath] such that [imath]f(x)=\lambda[/imath]. A function [imath]f[/imath] is injective if [imath]f(x)=f(y) \Rightarrow x=y[/imath]. Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps: An injective function with the intermediate-value property must be monotonic. A monotonic function possesses left- and right-handed limits at each point. For a function with the intermediate-value property the left- and right-handed limits at [imath]x[/imath], if they exist, equal [imath]f(x)[/imath]. I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice. Can there be a shorter or more elegant proof? | 1959311 | If [imath]f[/imath] is increasing on [imath][a,b] [/imath]and [imath]f[/imath] has the intermediate value property then [imath]f[/imath] is continuous at [imath]x[/imath] for every [imath]x \in [a,b][/imath]
I plan to use the fact that if [imath]f[/imath] is monotone and non constant [imath][a,b][/imath], then it is continuous on [imath][a,b][/imath] if its range [imath]R_f=\{f(x)|x \in [a,b]\}[/imath] is the closed interval with endpoints [imath]f(a)[/imath] and [imath]f(b)[/imath]. I dont know how to go from here. |
7881 | Preimage of generated [imath]\sigma[/imath]-algebra
For some collection of sets [imath]A[/imath], let [imath]\sigma(A)[/imath] denote the [imath]\sigma[/imath]-algebra generated by [imath]A[/imath]. Let [imath]C[/imath] be some collection of subsets of a set [imath]Y[/imath], and let [imath]f[/imath] be a function from some set [imath]X[/imath] to [imath]Y[/imath]. I want to prove: [imath]f^{-1}(\sigma(C))=\sigma(f^{-1}(C))[/imath] I could prove that [imath]\sigma(f^{-1}(C)) \subset f^{-1}(\sigma(C))[/imath] since complements and unions are 'preserved' by function inverse. But how do I go the other way? EDIT: One way to go the other way would be to argue that any set in [imath]\sigma(C)[/imath] must be built by repeatedly applying the complement, union and intersection operations to elements of [imath]C[/imath] and all these operations are preserved when taking the inverse. The problem I am facing with the approach is formalizing the word "repeatedly". [not-homework] | 222810 | Borel [imath]\sigma[/imath] algebra on a topological subspace.
Let [imath]T[/imath] be a topological space, with Borel [imath]\sigma[/imath]-algebra [imath]B(T)[/imath] (generated by the open sets of [imath]T[/imath]). If [imath]S\in B(T)[/imath], then the set [imath]C:=\{A\subset S:A\in B(T)\}[/imath] is a [imath]\sigma[/imath]-algebra of [imath]S[/imath]. My question is, if I also generated the Borel [imath]\sigma[/imath]-algebra [imath]B(S)[/imath] treating [imath]S[/imath] as a topological subspace, with the inherited topology from [imath]T[/imath], is it true that [imath]B(S)=C[/imath]? |
86209 | Steinhaus theorem (sums version)
This is a question from Stromberg related to Steinhaus' Theorem: If [imath]A[/imath] is a set of positive Lebesgue measure, show that [imath]A + A[/imath] contains an interval. I can't quite see how to modify the Steinhaus proof though. | 426217 | [imath]E+F[/imath] contains an interval
Let [imath]E[/imath] and [imath]F[/imath] be measurable sets with [imath]m(E),m(F)>0[/imath]. Prove that [imath]E+F[/imath] contains an interval. This is a part of an exam preparation, I would appreciate a hint. Thanks! |
279388 | There are at least three mutually non-isomorphic rings with [imath]4[/imath] elements?
Is the following statement is true? There are at least three mutually non-isomorphic rings with [imath]4[/imath] elements. I have no idea or counterexample at the moment. Please help. So far I know about that a group of order [imath]4[/imath] is abelian and there are two non isomorphic groups of order [imath]4[/imath] like [imath]K_4(non cyclic)[/imath] and [imath]\mathbb Z_4(cyclic)[/imath]. | 2812808 | How many rings are there of order 4 up to isomorphism
How many rings of order 4 can exist upto isomorphism there are only two groups [imath]Z_4[/imath] and [imath]K_4[/imath] upto isomorphism where [imath]Z_4[/imath] is abelian and [imath]K_4[/imath] is non-abelian.To be a ring that must be a abelian group under one binary operation.Can it be say that [imath]Z_4[/imath] is only ring of order 4 upto isomorphism |
302609 | A question about convergence in [imath]L^p[/imath].
Let [imath]E[/imath] be measurable and [imath]1 \le p \le \infty[/imath]. Suppose [imath]\{f_n\}_{n \in \mathbb{N}}[/imath] all measurable and [imath]\{f_n\}_{n \in \mathbb{N}} \to f[/imath] pointwise a.e. [imath]E[/imath]. For [imath]p[/imath] as above, I want to show that: [imath]\{f_n\}_{n \in \mathbb{N}} \to f \in L^p(E)[/imath] if there is [imath]\theta >0[/imath] such that [imath]\{f_n\}_{n \in \mathbb{N}}[/imath] belongs to and is bounded as a subset of [imath]L^{p + \theta}(E)[/imath]. One of the things that give me trouble is that I cannot understand the mathematical meaning of "belongs to and is bounded as a subset of" in this context. | 283920 | Pointwise a.e. convergence implies strong convergence?
Let [imath] 1 \leq p_1 < p_2 < \infty[/imath], and suppose that [imath]f_n[/imath] is a sequence of functions in [imath]L^{p_1}[a,b][/imath] such that [imath]f_n \to f[/imath] pointwise a.e. on [imath][a,b][/imath]. Suppose in addition that [imath] ||f_n||_{p_2} \leq 1[/imath] for every [imath]n[/imath], where [imath]|| \cdot ||_{p_2}[/imath] denotes the [imath]L^{p_2}[/imath] norm, then how can we show that [imath]f_n \to f[/imath] strongly in [imath]L^{p_1}[/imath] ? What I have tried: Since [imath] 1 < p_2 < \infty[/imath] and [imath]\{f_n\}[/imath] has bounded [imath]L^{p_2}[/imath] norm, it follows that [imath]f_n \to f[/imath] weakly in [imath]L^{p_2}[/imath]. Now since [imath][a,b][/imath] has finite measure, if [imath]f_n \to f[/imath] strongly in [imath]L^{p_2}[/imath], then [imath]f_n \to f[/imath] strongly in [imath]L^{p_1}[/imath] also, so this is what I am trying to show, although I do not know whether we actually do have strong convergence in [imath]L^{p_2}[/imath]. Given [imath]f_n \to f[/imath] weakly in [imath]L^{p_2}[/imath], we have strong convergence in [imath]L^{p_2}[/imath] if and only if the norms converge, i.e. [imath]||f_n||_{p_2} \to ||f||_{p_2}[/imath]. But how can we argue that we do have convergence of the norms in this case? Maybe using the condition that [imath] ||f_n||_{p_2} \leq 1[/imath] ? |
226786 | Let [imath]f(z)[/imath] be entire function. Show that if [imath]f(z)[/imath] is real when [imath]|z| = 1[/imath], then [imath]f(z)[/imath] must be a constant function using Maximum Modulus theorem
Let [imath]f(z)[/imath] be entire function. Consider the functions [imath]e^{if(z)}[/imath] and [imath]e^{−if(z)}[/imath] and applying the Maximum Modulus Theorem, show that if [imath]f(z)[/imath] is real when [imath]|z| = 1[/imath], then [imath]f(z)[/imath] must be a constant function. (We take [imath]f(z)=u(z)+iv(z)[/imath]) I am confused as so far I have [imath]|g(z)|=|e^{if(z)}|=|e^{-v(z)}|[/imath] and then since [imath]f(z)[/imath] is real, [imath]f(z)=u(z)[/imath] and [imath]v(z)=0[/imath] so I assumed it would follow that [imath]|g(z)|=|e^{v(z)}|=1[/imath]. Similarly, [imath]|g(z)|=|e^{-if(z)}|=|e^{v(z)}|=1[/imath]. Using Liouville I assumed one could say that both [imath]g(z)[/imath] and [imath]h(z)[/imath] are bounded entire functions, they are constant and so it follows that [imath]v(z)[/imath] is constant, meaning that both its partial derivatives are equal to 0 and, due to Cauchy Riemann, both of the partial derivatives of [imath]u(z)[/imath] are equal to zero. It would then follow that [imath]f(z)[/imath] is constant. I don't know how to go about the question using the Maximum Modulus Theorem, also I feel I am overlooking the importance of [imath]|z|=1[/imath] perhaps? Any help would be much appreciated!! | 549417 | Prove that [imath]f[/imath] is constant if [imath]f[/imath] is real when [imath]|z|=1[/imath]
Let [imath]f[/imath] be a holomorphic function in [imath]\mathbb{C}[/imath]. Prove that if [imath]f[/imath] is real when [imath]|z|=1[/imath], then [imath]f[/imath] must be a constant function. I honestly do not know how to do this problem, consider using Schwarz's lemma, but do not get anywhere. |
84206 | How to calculate the determinant of all-ones matrix minus the identity?
How do I calculate the determinant of the following [imath]n\times n[/imath] matrices [imath]$$\begin {bmatrix} 0 & 1 & \ldots & 1 \\ 1 & 0 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & ... & 0 \end {bmatrix}$$[/imath] and the same matrix but one of columns replaced only with [imath]$1$[/imath]s? In the above matrix all off-diagonal elements are [imath]$1$[/imath] and diagonal elements are [imath]$0$[/imath]. | 346489 | Finding determinant of a simple matrix
Can someone please explain how to compute the determinant of [imath]J_n - I_n[/imath] where [imath]j_n[/imath] it a matrix of ones? E.g. for [imath]n=5[/imath] we get the following matrix [imath]\left(\begin{array}{ccccc} 0 & 1 & 1 &1 &1 \\ 1 & 0 & 1 &1 &1 \\ 1 & 1 & 0 &1 &1 \\ 1 & 1 & 1 &0 &1 \\ 1 & 1 & 1 &1 &0 \end{array} \right)[/imath] By computing it for the first few [imath]n[/imath]'s it seems to be [imath](-1)^{n+1}(n-1)[/imath] but I couldn't prove it. |
165816 | Computing determinant of a specific matrix.
How to calculate the determinant of [imath] A=(a_{i,j})_{n \times n}=\left( \begin{array}{ccccc} a&b&b& \cdots & b\\ b& a& b& \cdots& b\\ \vdots& \vdots& \vdots& \ddots&\cdots\\ b&b&b & \cdots&a \end{array} \right)? [/imath] | 388125 | Simplest way to calculate a determinant
The big [imath]1[/imath]'s here just mean that the lower and upper triangular entries are all [imath]1[/imath]'s. The trace entries are all zero. The matrix is for a general [imath]n\times n[/imath] matrix of this form. I'm trying to deduce a simpler relation. [imath]\operatorname{det}~\begin{pmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 &\cdots & 1\\ 1 & 1 & 0 &\cdots & 1\\ \vdots & \vdots & \vdots & \ddots & 1\\ 1 & 1 & 1 & \cdots & 0\end{pmatrix}[/imath] |
85448 | Why does the median minimize [imath]E(|X-c|)[/imath]?
Suppose [imath]X[/imath] is a real-valued random variable and let [imath]P_X[/imath] denote the distribution of [imath]X[/imath]. Then [imath] E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). [/imath] The medians of [imath]X[/imath] are defined as any number [imath]m \in \mathbb{R}[/imath] such that [imath]P(X \leq m) \geq \frac{1}{2}[/imath] and [imath]P(X \geq m) \geq \frac{1}{2}[/imath]. Why do the medians solve [imath] \min_{c \in \mathbb{R}} E(|X-c|) \, ? [/imath] | 1910145 | Prove: [imath]m[/imath] is a median of [imath]X[/imath] iff the function [imath]g(a)=\mathbb E(|X-a|)[/imath] is minimized at [imath]a=m[/imath]
Ok, so here is the question: Suppose [imath]X[/imath] has density [imath]f[/imath] and [imath]\mathbb E|X|<\infty[/imath]. Prove: [imath]m[/imath] is a median of [imath]X[/imath] iff the function [imath]g(a)=\mathbb E(|X-a|)[/imath] is minimized at [imath]a=m[/imath]. Recall [imath]m[/imath] is a median of [imath]X[/imath] iff [imath]P(X \leq m)\geq 1/2[/imath] and [imath]P(X \geq m)\geq 1/2[/imath]. For discrete case, I think I can just rearrange the data as order statistics. However, for continuous case, I tried to manipulte the integrals of the expectation but didn't achieve the conclusion. |
139393 | How to show that [imath]\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0[/imath]
Evaluate the integral: [imath] \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx[/imath] The answer is [imath]0,[/imath] but I am unable to get it. There is some symmetry I can not see. | 619333 | show that [imath]\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0[/imath]
show that [imath]I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0[/imath] I find this is Nice equalition! My try: let [imath]\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}[/imath] so [imath]dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt[/imath] so [imath]I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx[/imath] By parts,we have [imath]\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt[/imath] so [imath]I=0[/imath] this problem maybe have more other nice methods!Thank you |
61458 | If [imath]f[/imath] is measurable and [imath]fg[/imath] is in [imath]L^1[/imath] for all [imath]g \in L^q[/imath], must [imath]f \in L^p[/imath]?
Let [imath]f[/imath] be a measurable function on a measure space [imath]X[/imath] and suppose that [imath]fg \in L^1[/imath] for all [imath]g\in L^q[/imath]. Must [imath]f[/imath] be in [imath]L^p[/imath], for [imath]p[/imath] the conjugate of [imath]q[/imath]? If we assume that [imath]\|fg\|_1 \leq C\|g\|_q[/imath] for some constant [imath]C[/imath], this follows from the Riesz Representation theorem. But what if we aren't given that such a [imath]C[/imath] exists? | 453699 | Lp Space conjugate function
Suppose [imath]1<p,q<\infty[/imath], [imath]\dfrac{1}{p} + \dfrac{1}{q}=1[/imath], [imath](X,\Sigma,\mu[/imath] is [imath]\sigma[/imath]-finite measure space and [imath]g[/imath] is a measurable function such that [imath]fg \in L^1(X)[/imath] for every [imath]f \in L^p(X)[/imath]. Prove that [imath]g \in L^q(X)[/imath]. I have no idea where to start this question. It seems to me a lot like a close question to the Riesz Representation Theorem, or an application of it. Any help is much appreciated. I'm just refreshing many questions I haven't looked at in quite some time for preparation of my Analysis Qualifying Exams. |
1096 | Where is the flaw in this "proof" that 1=2? (Derivative of repeated addition)
Consider the following: [imath]1 = 1^2[/imath] [imath]2 + 2 = 2^2[/imath] [imath]3 + 3 + 3 = 3^2[/imath] Therefore, [imath]\underbrace{x + x + x + \ldots + x}_{x \textrm{ times}}= x^2[/imath] Take the derivative of lhs and rhs and we get: [imath]\underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = 2x[/imath] Which simplifies to: [imath]x = 2x[/imath] and hence [imath]1 = 2[/imath]. Clearly something is wrong but I am unable pinpoint my mistake. | 304636 | Funny Proof of [imath]2=1[/imath]
We know [imath]x.x=x^2[/imath](Consider [imath]x\ne 0[/imath]) [imath]x.x[/imath] is adding [imath]x[/imath] [imath]x[/imath] times. So we have [imath]x+x+\dots+x[/imath]([imath]x[/imath] times [imath]x[/imath] is added)=[imath]x^2[/imath]......(1) Differentiating both sides of (1) we get, [imath]1+1+\dots+1[/imath](x times)=[imath]2.x[/imath] [imath]\Rightarrow x=2x[/imath] [imath]1=2[/imath] Proved. Find out the mistake in this proof. |
141705 | Result of the product [imath]0.9 \times 0.99 \times 0.999 \times ...[/imath]
My question has two parts: How can I nicely define the infinite sequence [imath]0.9,\ 0.99,\ 0.999,\ \dots[/imath]? One option would be the recursive definition below; is there a nicer way to do this? Maybe put it in a form that makes the second question easier to answer. [imath]s_{i+1} = s_i + 9\cdot10^{-i-2},\ s_0 = 0.9[/imath] Edit: Suggested by Kirthi Raman: [imath](s_i)_{i\ge1} = 1 - 10^{-i}[/imath] Once I have the sequence, what would be the limit of the infinite product below? I find the question interesting since [imath]0.999... = 1[/imath], so the product should converge (I think), but to what? What is the "last number" before [imath]1[/imath] (I know there is no such thing) that would contribute to the product? [imath]\prod_{i=1}^{\infty} s_i[/imath] | 360356 | Evaluating [imath]\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots[/imath]
[imath]\displaystyle\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots=?[/imath] Usually, product of infinite many numbers which are less than 1, is 0. But How about this time? Thank you. |
227681 | How to find chromatic number of the [imath]n[/imath]-dimensional hypercube [imath]Q_n[/imath]?
How to find chromatic number the [imath]n[/imath]-dimensional hypercube [imath]Q_n[/imath]? I know [imath]\chi(Q_2)[/imath]=2 , [imath]\chi(Q_3)[/imath]=2 , [imath]\chi(Q_4)[/imath]=4 | 1941390 | Graph theory problem: n-dimensional cube
Let [imath]Q_n[/imath] be the [imath]n-dimensional[/imath] cube graph: Its vertices are all the [imath]n-tuples[/imath] of [imath]0[/imath] and [imath]1[/imath] with two vertices being adjacent if they dier in precisely one position.For example, in [imath]Q_3[/imath], the vertices [imath](1,0,0)[/imath] and [imath](1,0,1)[/imath] are adjacent because they differ only in the third position.Show that [imath]Q_n[/imath]is bipartite. Can anybody help me with this question please? |
9402 | Calculating the integral $\int_0^\infty \frac{\cos x}{1+x^2}\, \mathrm{d}x$ without using complex analysis
Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? [imath]$$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$[/imath] | 1103920 | Integration of [imath]\int_{-\infty}^{\infty}\frac{\cos x}{a^2+x^2}dx[/imath]
I'm trying to find the integral [imath]\int_{-\infty}^{\infty}\frac{\cos x}{a^2+x^2}dx[/imath] Wolfram alpha says this is [imath]\frac{\pi e^{-a}}{a}[/imath] But how do you get this result? I tried using partial integration and some substitutions but I'm not getting there... |
122898 | Why are the solutions of polynomial equations so unconstrained over the quaternions?
An [imath]n[/imath]th-degree polynomial has at most [imath]n[/imath] distinct zeroes in the complex numbers. But it may have an uncountable set of zeroes in the quaternions. For example, [imath]x^2+1[/imath] has two zeroes in [imath]\mathbb C[/imath], but in [imath]\mathbb H[/imath], [imath]{\bf i}\cos x + {\bf j}\sin x[/imath] is a distinct zero of this polynomial for every [imath]x[/imath] in [imath][0, 2\pi)[/imath], and obviously there are many other zeroes. What is it about [imath]\mathbb H[/imath] that makes its behavior in this regard to be so different from the behavior of [imath]\mathbb R[/imath] and [imath]\mathbb C[/imath]? Is it simply because [imath]\mathbb H[/imath] is four-dimensional rather than two-dimensional? Are there any theorems that say when a ring will behave like [imath]\mathbb H[/imath] and when it will behave like [imath]\mathbb C[/imath]? Do all polynomials behave like this in [imath]\mathbb H[/imath]? Or is this one unusual? | 2028288 | Why don't quaternions contradict the Fundamental Theorem of Algebra?
I don't pretend to know anything much about the Fundamental Theorem of Algebra (FTA), but I do know what it states: for any polynomial with degree [imath]n[/imath], there are exactly [imath]n[/imath] solutions (roots). Well, when it comes to quaternions, apparently [imath]i^2=j^2=k^2=-1[/imath], but [imath]i\ne j\ne k\ne i[/imath]. So now, we have apparently found three solutions to the second-degree polynomial [imath]x^2=-1[/imath]. I'm not aware of the justification of the FTA, nor I am I aware of Hamilton's justification for quaternions. However, I know a contradiction when I see one. What am I missing here? |
72284 | Proof of [imath](\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}[/imath]
I've just started to learn about the tensor product and I want to show: [imath](\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.[/imath] Can you tell me if my proof is right: [imath]\mathbb{Z}/m\mathbb{Z}[/imath] and [imath]\mathbb{Z} / n \mathbb{Z}[/imath] are both finite free [imath]\mathbb{Z}[/imath]-modules with the basis consisting of one single element [imath]\{ 1 \}[/imath]. So [imath](\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})[/imath] has the basis [imath]\{ 1 \otimes 1 \}[/imath]. Therefore, any element in [imath](\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})[/imath] is of the form [imath](ab) 1 \otimes 1[/imath] and any element in [imath]\mathbb{Z}/ \gcd(m,n)\mathbb{Z}[/imath] is of the form [imath]k 1 = k[/imath] where [imath]k \in \{ 0, \dots , \gcd(n,m) \}[/imath]. I would like to construct an isomorphism that maps [imath]ab[/imath] to some [imath]k[/imath]. Let this map be [imath]ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)[/imath]. This is a homomorphism between modules: it maps [imath]0[/imath] to [imath]0[/imath] because it maps the empty sum to the empty sum. It also fulfills [imath]f(a + b) = f(a) + f(b)[/imath] because there is only one element, [imath]a = 1[/imath]. It is surjective. So all I need to show is that it is injective. But that is clear too because if [imath]ab \equiv 0 \bmod \gcd(m,n)[/imath] then both [imath]a \equiv 0 \bmod n[/imath] and [imath]b \equiv 0 \bmod m[/imath] so the kernel is trivial. Many thanks for your help!! | 967872 | Using the universal property of tensor product to show that [imath](\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}[/imath]
In the algebra lecture i need to solve the following exercise Use the universal property of the tensor product to show that [imath](\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}[/imath] Here is my job: I tried to follow the example seen in the lecture where we showed that [imath](\Bbb{Z}/2\Bbb{Z}) \otimes (\Bbb{Z}/3\Bbb{Z})=0[/imath]. In that example the professor listed all the generators of [imath](\Bbb{Z}/2\Bbb{Z}) \otimes (\Bbb{Z}/3\Bbb{Z})[/imath] and then showed that every generator is actually zero just by using the properties of the tensors. Namely [imath]\begin{array}{|cc|}\hline 0\otimes n = 0\otimes n + 0\otimes n &\implies 0\otimes n =0 \\\hline 1\otimes0=1\otimes 0 + 1 \otimes 0 &\implies 1\otimes 0 =0 \\\hline 1\otimes 2 =2(1\otimes 1) = 0\otimes1&\implies1\otimes 2=0 \\\hline 1\otimes1 =1\otimes 4 = 1\otimes2 + 1\otimes 2 & \implies 1\otimes 1 =0\\\hline \end{array}[/imath] for [imath]n= 0,1,2[/imath]. Now to show that [imath](\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}[/imath] I tried to list all the generator and try to show that they are equal to [imath]0[/imath] or [imath]1[/imath]. We have [imath]4\cdot 6=24[/imath] generators and i noticed that if a generator has a even integer than it is equal [imath]0[/imath]. The problem is that i can not show that some generators are equal [imath]1[/imath]. Indeed the generators equal [imath]0[/imath] are [imath]\begin{array}{|cc|}\hline 0\otimes n=0\otimes n + 0\otimes n& \implies 0\otimes n =0 \,\,\,n\in \{0,1,2,3,4,5\} \\\hline n\otimes 0 = n\otimes 0 + n \otimes 0&\implies n \otimes 0 =0 \,\,\, n \in \{0,1,2,3\}\\\hline \end{array}[/imath] and \begin{align*} 1\otimes 2=0 && 1\otimes 4=0 && 1\otimes 6=0 \\ 2\otimes n=0 && 3\otimes 2=0 && 3\otimes 4=0 \\ 3\otimes 6=0 \end{align*} for [imath]n \in \{0,1,2,3,4,5\}[/imath] The other generator can be brought back at the case [imath]1\otimes 1[/imath] and therefore i wish i could show [imath]1\otimes 1[/imath] can be considered as [imath]1[/imath], but i'm not able to. I know that i should use the universal property of the tensor product but i don't really know how to use it, therefore i think that there must be another procedure to solve the exercise. Is what i've done so far correct? |
62171 | Proving [imath]1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2[/imath] using induction
How can I prove that [imath]1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2[/imath] for all [imath]n \in \mathbb{N}[/imath]? I am looking for a proof using mathematical induction. Thanks | 384534 | Prove summation using induction
[imath]\sum\limits_{i=1}^n i^3 = \left(\frac{n(n + 1)}{2}\right)^2[/imath] My basis step is [imath]P(1)[/imath] sets the [imath]LHS = RHS = 1[/imath]. For the inductive step, I assume [imath]n = k[/imath] holds for [imath]k+1[/imath]. On the [imath]RHS[/imath]: [imath]\left(\frac{(k + 1)((k + 1) + 1)}{2}\right)^2[/imath] But I don't know how to convert the summation into something that can evaluated algebraically. Disclaimer: this is a question from an exam review sheet. |
122821 | [imath]\mathbb{A}^{2}[/imath] not isomorphic to affine space minus the origin
Why is the affine space [imath]\mathbb{A}^{2}[/imath] not isomorphic to [imath]\mathbb{A}^{2}[/imath] minus the origin? | 2028423 | show [imath]\mathbf{A}^2\backslash\{(0,0)\}[/imath] is not isomorphic to [imath]\mathbf{A}^2[/imath]
If we have [imath]\mathscr{O}(\mathbf{A}^2\backslash\{(0,0)\})=k[x,y][/imath], now how can we show that [imath]\mathbf{A}^2\backslash\{(0,0)\}[/imath] is not isomorphic to [imath]\mathbf{A}^2[/imath]? |
48989 | How to prove [imath]\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))[/imath]?
How to prove [imath]\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))[/imath]? | 874771 | Rank of product of two matrices
I want to show that [imath]\text {rank} ( AB)\le \min(\text{rank} B, \text{rank} A) [/imath] and when the equality occurs? Please help me with this problem by giving hints, solving, or suggesting a book. Thanks |
199990 | Prime ideals in [imath]\mathbb{Z}[x][/imath]
Am I right that all prime ideals in [imath]\mathbb{Z}[x][/imath] has the form [imath]p\mathbb{Z}[x][/imath] for some prime [imath]p\in\mathbb{Z}[/imath]? Thanks a lot! | 174595 | Classification of prime ideals of [imath]\mathbb{Z}[X][/imath]
Let [imath]\mathbb{Z}[X][/imath] be the ring of polynomials in one variable over [imath]\Bbb Z[/imath]. My question: Is every prime ideal of [imath]\mathbb{Z}[X][/imath] one of following types? If yes, how would you prove this? [imath](0)[/imath]. [imath](f(X))[/imath], where [imath]f(X)[/imath] is an irreducible polynomial. [imath](p)[/imath], where [imath]p[/imath] is a prime number. [imath](p, f(X))[/imath], where [imath]p[/imath] is a prime number and [imath]f(X)[/imath] is an irreducible polynomial modulo [imath]p[/imath]. |
46317 | What kind of "symmetry" is the symmetric group about?
There are two concepts which are very similar literally in abstract algebra: symmetric group and symmetry group. By definition, the symmetric group on a set is the group consisting of all bijections of the set (all one-to-one and onto functions) from the set to itself with function composition as the group operation. When the set is finite, the group is sometimes denoted as [imath]S_n[/imath]. The Dihedral group [imath]D_n[/imath], which is a special case of the symmetry group, has a very strong geometric intuition about symmetry as the picture shows. I know nothing about the relation between these two concepts but the fact that [imath]D_3[/imath] and [imath]S_3[/imath] are actually the same. For me, symmetric group is more about "permutations". And actually its subgroups are also called permutation groups. Here are my questions: What's the relation between these two concepts: "symmetric group" and "symmetry group"? What kind of "symmetry" is the symmetric group about? Where is the name "symmetric group" from? | 316272 | What is symmetric about the symmetric group?
In my abstract algebra reader many times they use the word "symmetry". But I just found out that I'm not quite sure what they mean with it. Let [imath]X=\{1,...,n\}[/imath]. The set [imath]S_n[/imath] of bijections [imath]X\to X[/imath] under composition with identity [imath]X\to X :x\mapsto x[/imath] is called the symmetric group on [imath]n[/imath] symbols. If I think about symmetry I think about figures, as something I can visualize. However I'm not sure if or how I should visualize the symmetric group of [imath]n[/imath] symbols as something symmetric. Edit: Thanks for all the answers! They all contributed to be finally satisfied with calling [imath]S_n[/imath] symmetric :) Edit2: This helped me a lot as well: http://en.wikipedia.org/wiki/Frucht%27s_theorem Frucht's theorem says that every finite group is the symmetry group of some graph. So every finite abstract group is actually the symmetries of some explicit object. |
242034 | Is any Banach space a dual space?
Let [imath]X[/imath] be a Banach space. Is there always a normed vector space [imath]Y[/imath] such that [imath]X[/imath] and [imath]Y^*[/imath] are isometric or isomorphic as topological vector spaces (that is, there exists a linear homeomorphism between [imath]X[/imath] and [imath]Y^*[/imath])? | 885793 | Is every Banach space isometrically isomorphic to the dual of a normed space?
I have the following question: Given a Banach space [imath]V[/imath] over the field [imath]\mathbb{K}[/imath], is it true that there is always a normed space [imath]W[/imath] over the field [imath]\mathbb{K}[/imath], such that [imath]V[/imath] and [imath]W'=\mathscr{L}(W,\mathbb{K})[/imath] are isometrically isomorphic (with [imath]\mathbb{K}\in \{\mathbb{R},\ \mathbb{C}\}[/imath])? Thanks for the help! |
238171 | Prove that if [imath]g^2=e[/imath] for all [imath]g[/imath] in [imath]G[/imath] then [imath]G[/imath] is Abelian.
Prove that if [imath]g^2=e[/imath] for all [imath]g[/imath] in [imath]G[/imath] then [imath]G[/imath] is Abelian. This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it. (Please note that [imath]e[/imath] in the question is the group's identity.) Here's my attempt though... First I understand Abelian means that if [imath]g_1[/imath] and [imath]g_2[/imath] are elements of a group [imath]G[/imath] then they are Abelian if [imath]g_1g_2=g_2g_1[/imath]... So, I begin by trying to play around with the elements of the group based on their definition... [imath](g_2g_1)^r=e[/imath] [imath](g_2g_1g_2g_2^{-1})^r=e[/imath] [imath](g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e[/imath] I assume that the [imath]g_2^{-1}[/imath]'s and the [imath]g_2[/imath]'s cancel out so that we end up with something like, [imath]g_2(g_1g_2)^rg_2^{-1}=e[/imath] [imath]g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2[/imath] Then ultimately... [imath]g_1g_2=e[/imath] I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way. Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra. | 302120 | Let [imath](G,\circ)[/imath] be a group in which [imath]x\circ x[/imath] is the neutral element for all [imath]x \in G[/imath]
Prove that [imath]x=x^{-1}[/imath] for all [imath]x \in G[/imath] and that G is commutative. I have no idea where to start. I know that commutative will mean that for [imath]x,y \in G[/imath] [imath]x\circ y = y \circ x[/imath] but I don't know how to prove that. |
55358 | Euler's Constant: The asymptotic behavior of [imath]\left(\sum\limits_{j=1}^{N} \frac{1}{j}\right) - \log(N)[/imath]
I want to show that there exists a constant [imath]\gamma\in\mathbb{R}[/imath] such that [imath] \sum_{j=1}^N \frac1{j} = \log(N)+\gamma+O(1/N). [/imath] I know how to prove that the Euler-Mascheroni constant exists (which I believe [imath]\gamma[/imath] to be), but I am having trouble with the big-[imath]O[/imath] notation and the subsequent bounding. I've considered [imath] \left|\left(\sum_{j=1}^N \frac1{j}\right) - \log(N)-\gamma\right|\le |K/N| [/imath] for some [imath]K[/imath], and I was approaching this by trying to show the that the left side of the inequality decays faster, but so far am stuck. Any advice for this type of problem, or analogous ones, would be appreciated. Thanks! | 572206 | Convergence using Riemann integrability
For each positive integer n, let [imath]\gamma_n = 1+\frac12+ \cdots + \frac1n - \int_1^n \frac1x \, dx[/imath]. Prove that the sequence [imath]\{\gamma_n\}[/imath] converges. |
64371 | Showing group with [imath]p^2[/imath] elements is Abelian
I have a group [imath]G[/imath] with [imath]p^2[/imath] elements, where [imath]p[/imath] is a prime number. Some (potentially) useful preliminary information I have is that there are exactly [imath]p+1[/imath] subgroups with [imath]p[/imath] elements, and with that I was able to show [imath]G[/imath] has a normal subgroup [imath]N[/imath] with [imath]p[/imath] elements. My problem is showing that [imath]G[/imath] is abelian, and I would be glad if someone could show me how. I had two potential approaches in mind and I would prefer if one of these were used (especially the second one). First: The center [imath]Z(G)[/imath] is a normal subgroup of [imath]G[/imath] so by Langrange's theorem, if [imath]Z(G)[/imath] has anything other than the identity, it's size is either [imath]p[/imath] or [imath]p^2[/imath]. If [imath]p^2[/imath] then [imath]Z(G)=G[/imath] and we are done. If [imath]Z(G)=p[/imath] then the quotient group of [imath]G[/imath] factored out by [imath]Z(G)[/imath] has [imath]p[/imath] elements, so it is cylic and I can prove from there that this implies [imath]G[/imath] is abelian. So can we show theres something other than the identity in the center of [imath]G[/imath]? Second: I list out the elements of some other subgroup [imath]H[/imath] with [imath]p[/imath] elements such that the intersection of [imath]H[/imath] and [imath]N[/imath] is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let [imath]N[/imath] be generated by [imath]a[/imath] and [imath]H[/imath] be generated by [imath]b[/imath]. We can show [imath]NK= G[/imath], i.e every element in G can be wrriten like [imath]a^k b^l [/imath]. So for this method, we just need to show [imath]ab=ba[/imath] (remember, these are not general elements in the set, but the generators of [imath]N[/imath] and [imath]H[/imath]). Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine. | 432227 | Group of order [imath]p^2[/imath] is abelian.
Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well. Let [imath]|G| = p^2[/imath] for some prime [imath]p[/imath]. Consider [imath]x \in G[/imath]. So, [imath]|x| = p, [/imath] or [imath] p^2[/imath]. If it is the latter, then, we are done. So, if [imath]|x|= p[/imath], then, let, [imath]H = [h| h=x^i, i \in Z] \Rightarrow |H|= p[/imath]. As [imath]p[/imath] is the smallest prime dividing [imath]|G|[/imath], [imath]H[/imath] is a normal subgroup. Now, consider some [imath]y \in G[/imath], but not in [imath]H \Rightarrow y^p= e[/imath]. Then, its easy to show that [imath]G = \langle x,y \rangle[/imath]. Now, consider a homomorphism [imath]f:G \rightarrow \operatorname{Aut}(H), g \to c_g [/imath], wherein [imath]c_g[/imath] represents conjugacy by [imath]g[/imath]. So, for example, [imath]c_g(x) = gx g^{-1}[/imath]. Then, it is clear that, [imath]H \subset \operatorname{Ker}(f) [/imath], as it is cyclic and therefore abelian. So, we need only look at [imath]G/H[/imath]. Basically, I am asking for a hint as to why [imath]c_y(x) = yxy^{-1} = x = c_e(x)[/imath], where [imath]x[/imath] and [imath]y[/imath] are the generators of [imath]H[/imath] and [imath]G/H[/imath], respectively? Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that [imath]\operatorname{Im}(f)[/imath] is trivial and therefore that [imath]G[/imath] is abelian. Thank You! |
304461 | [imath]X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})[/imath]. what is the value of X?
[imath]X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})[/imath] [imath]\tan(90-\theta)=\cot\theta=\frac{1}{\tan\theta}[/imath] | 75825 | A "fast" way for computing [imath] \prod \limits_{i=1}^{45}(1+\tan i^\circ) [/imath]?
Which is the fastest paper-pencil approach to compute the product [imath] \prod \limits_{i=1}^{45}(1+\tan i^\circ) [/imath] |
7511 | Functions that are their Own nth Derivatives for Real [imath]n[/imath]
Consider (non-trivial) functions that are their own nth derivatives. For instance [imath]\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x[/imath] [imath]\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-x} = e^{-x}[/imath] [imath]\frac{\mathrm{d}^3}{\mathrm{d}x^3} e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2}) = e^{\frac{-x}{2}}\sin(\frac{\sqrt{3}x}{2})[/imath] [imath]\frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin x = \sin x[/imath] [imath]\cdots[/imath] Let [imath]f_n(x)[/imath] be the function that is it's own nth derivative. I believe (but I'm not sure) for nonnegative integer [imath]n[/imath], this function can be written as the following infinite polynomial: [imath]f_n(x) = 1 + \cos(\frac{2\pi}{n})x + \cos(\frac{4\pi}{n})\frac{x^2}{2!} + \cos(\frac{6\pi}{n})\frac{x^3}{3!} + \cdots + \cos(\frac{2t\pi}{n})\frac{x^t}{t!} + \cdots[/imath] Is there some sense in which this function can be extended to real n using fractional derivatives? Would it then be possible to graph [imath]z(n, x) = f_n(x)[/imath], and would this function be smooth and continuous on both [imath]n[/imath] and [imath]x[/imath] axes? Or would it have many discontinuities? | 1826552 | Repeating/"Periodic" Derivatives?
We know that [imath]Ce^x[/imath] and [imath]0[/imath] are the two functions whose first derivative is equal to itself, but what about derivatives of a higher order? For example, the second derivative of [imath]e^{-x}[/imath] is equal to itself, but not the first, and the fourth derivative of [imath]sin(x)[/imath] is equal to itself. In short, are there other examples of functions whose nth derivative is equal to itself, where [imath]n>1[/imath]? Thank you kindly! |
299005 | How do you find the Maximal interval of existence of a differential equation?
I have a really simple differential equation: [imath]\frac{dx}{dt} = x^2.t[/imath] with initial value [imath]x(0) = x_0[/imath]. Determine the maximal interval where it exists, depending on [imath]x_0[/imath] The maximal interval needs to be written in the form [imath](t^-,t^+)[/imath], but I don't understand how you determine [imath]t^+[/imath] and [imath]t^-[/imath]? please could someone explain if there is a method to do this? Thanks | 298974 | solution of first order differential equation and maximal interval
Find the solution of [imath]x' = x^2t[/imath] with initial value [imath]x(0) = x_{0}[/imath]. Determine the maximal interval where it exists, depending on [imath]x_{0}[/imath] Please help me find the maximal interval! |
231150 | Prove [imath]f(S \cup T) = f(S) \cup f(T)[/imath]
[imath]f(S \cup T) = f(S) \cup f(T)[/imath] [imath]f(S)[/imath] encompasses all [imath]x[/imath] that is in [imath]S[/imath] [imath]f(T)[/imath] encompasses all [imath]x[/imath] that is in [imath]T[/imath] Thus the domain being the same, both the LHS and RHS map to the same [imath]y[/imath], since the function [imath]f[/imath] is the same for both. Can you post the solution? | 553674 | How to I prove the equivalence of [imath]f(S_1 \cup S_2)[/imath] and [imath]f(S_1) \cup f(S_2)[/imath] (Discrete Mathematics)
If [imath]S_1[/imath] and [imath]S_2[/imath] are both subsets of some arbitrary set [imath]A[/imath], then how do I prove that [imath]f(S_1\cup S_2) = f(S_1) \cup f(S_2)[/imath] for ALL cases I understand that it is true, but I don't know how to prove it. |
175251 | How to show that this set is compact in [imath]\ell^2[/imath]
Let [imath](a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})[/imath] be a fixed sequence. Consider the subspace [imath]C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.[/imath] According to the book [Dunford and Schwartz, Linear operators part I, page 453] [imath]C[/imath] is compact in the [imath]\ell^2[/imath]-norm, but there is no proof. How can I show that [imath]C[/imath] is indeed compact in [imath]\ell^2[/imath] ? | 2967114 | Problem 24. Introduction to Hilbert spaces. Stein
Let [imath](e_k)_{k=1}^{\infty}[/imath] denote an orthonormal set in a Hilbert space [imath]H[/imath]. If [imath](c_k)_{k=1}^{\infty}[/imath] is a sequence of positive real numbers such that [imath]\sum c_k^2<\infty[/imath], then the set [imath]A=\left\{ \sum_{k=1}^{\infty} a_ke_k: |a_k|\leq c_k\right\}[/imath] is compact in [imath]H[/imath]. Some hint? Thanks in advance. |
86644 | Determinant of a specially structured matrix ([imath]a[/imath]'s on the diagonal, all other entries equal to [imath]b[/imath])
I have the following [imath]n\times n[/imath] matrix: [imath]A=\begin{bmatrix} a & b & \ldots & b\\ b & a & \ldots & b\\ \vdots & \vdots & \ddots & \vdots\\ b & b & \ldots & a\end{bmatrix}[/imath] where [imath]0 < b < a[/imath]. I am interested in the expression for the determinant [imath]\det[A][/imath] in terms of [imath]a[/imath], [imath]b[/imath] and [imath]n[/imath]. This seems like a trivial problem, as the matrix [imath]A[/imath] has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out. Any help would be appreciated. | 757320 | Prove determinant of [imath]n \times n[/imath] matrix is [imath](a+(n-1)b)(a-b)^{n-1}[/imath]?
Prove [imath]\det(A)[/imath] is [imath](a+(n-1)b)(a-b)^{n-1}[/imath] where [imath]A[/imath] is [imath]n \times n[/imath] matrix with [imath]a[/imath]'s on diagonal and all other elements [imath]b[/imath], off diagonal. |
168118 | Determining eigenvalues, eigenvectors of [imath]A\in \mathbb{R}^{n\times n}(n\geq 2)[/imath].
Let [imath]a[/imath] and [imath]b[/imath] be distinct nonzero real numbers and let [imath]A\in \mathbb{R}^{n\times n}(n\geq 2)[/imath] with each diagonal entry equal to [imath]a[/imath] and each off-diagonal entry equal to [imath]b[/imath]. Determine all eigenvalues and eigenvectors of [imath]A[/imath] together with their algebraic multiplicities. Is [imath]A[/imath] diagonalizable? Why? Determine the minimal polynomial of [imath]A[/imath]. My idea: Considering a [imath]4\times 4[/imath] matrix for simplicity: [imath]\left( \begin{array}{ccccc} a& b&b & b \\ b &a&b &b \\ b & b&a& b\\ b&b&b&a \end{array} \right)[/imath] now we perform row operations to transform this matrix into a simple one like this: [imath]\left( \begin{array}{ccccc} a-b& b-a& 0 & 0 \\ 0 &a-b& b-a&0 \\ 0 & 0&a-b & b-a\\ b&b&b&a \end{array} \right)[/imath] Then find the eigen values and from there eigen vectors? I think eigen values are [imath]a\pm b[/imath] but not sure how to prove it. Thanks Marvis, your answer is very helpful. Can you also comment on my ideas about the rest of the problem: to find eigen vectors we need [imath]x[/imath] such that [imath](A-\lambda I)x=0.[/imath] So, [imath]\lambda =a-b[/imath] gives me that [imath](A-\lambda I)=(b)[/imath] a matrix with all entries [imath]b[/imath], thus eigen vector in this case is a [imath]n\times 1[/imath] vector with first entry [imath]n-1[/imath] and the rest of the entries as 1's. So, using the eigen vectors I can find an invertible matrix [imath]P[/imath] and find out if [imath]P^{-1}AP=D[/imath] where [imath]D[/imath] is a diagonal matrix? Then I can get minimal polynomial from the diagonal matrix. | 382799 | a problem on solving a determinant equation
Let [imath]a[/imath] be a real number. What is the number of distinct real roots of the following [imath]\left| \begin{array}{ccc} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{array} \right|=0.[/imath] I guess some method is there other than calculating the determinant explicitly. |
144526 | Is it wrong to tell children that [imath]1/0 =[/imath] NaN is incorrect, and should be [imath]∞[/imath]?
I was on the tube and overheard a dad questioning his kids about maths. The children were probably about 11 or 12 years old. After several more mundane questions he asked his daughter what [imath]1/0[/imath] evaluated to. She stated that it had no answer. He asked who told her that and she said her teacher. He then stated that her teacher had "taught it wrong" and it was actually [imath]∞[/imath]. I thought the Dad's statement was a little irresponsible. Does that seem like reasonable attitude? I suppose this question is partly about morality. | 1382013 | Is [imath]\frac00=\infty[/imath]? And what is [imath]\frac10[/imath]? Are they same? Does it hold true for any constant [imath]a[/imath] in [imath]\frac{a}0[/imath]
I know that [imath]\lim_{x\to0}\frac{x}{x}=[/imath] 1. But in my text book, it is written that it is [imath]\infty[/imath] and even [imath]\frac10=\infty[/imath]. But how is it possible? And are they both same? What is the difference between indeterminate and undefined? And if they are [imath]\infty[/imath] then what is the reason? And why can't we define something for these numbers? Now see that book says when [imath]x=0[/imath], [imath]t=\infty[/imath]. If it is wrong, why did they just write it? |
296366 | Finding degree of the extension
Is it true that the degree of extension [imath]\mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},\dotsc,\sqrt {p_n}) / \mathbb Q[/imath] is [imath]2^n[/imath] where [imath]p_n[/imath] is the [imath]n[/imath]th prime number. If so, how to prove this? My idea is to consider the chain of extensions [imath]\mathbb Q\subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt{2},\sqrt{3}) \subset \dotsb \subset \mathbb Q(\sqrt {2},\sqrt {3},\sqrt {5},...,\sqrt {p_n})[/imath] and using transitivity. I am having problems in finding degrees of intermediate extensions. Please help me. | 113689 | Proving that [imath]\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n[/imath] for distinct primes [imath]p_i[/imath].
I have read the following theorem: If [imath]p_1,p_2,\dots,p_n[/imath] are distinct prime numbers, then[imath]\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.[/imath] I have tried to prove a more general statement but I have a problem at one point. (I still don't know how to prove the theorem above, too, because I don't know how not to use linear independence, which I do in the more general statement below.) Could you please help me overcome the obstacle I've encountered? I will post the intended proof and make it clear where I'm having trouble. I want to prove the following statement: Let [imath]n\geq 1[/imath]. The set [imath]B_n:=\left\{\sqrt {p_1^{\epsilon_1}}\sqrt {p_2^{\epsilon_2}}\cdots\sqrt {p_n^{\epsilon_n}}\,|\,(\epsilon_1,\epsilon_2,\cdots,\epsilon_n)\in\{0,1\}^n\right\}[/imath] has [imath]2^n[/imath] elements and is a [imath]\mathbb Q-[/imath]basis of [imath]\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right].[/imath] The proof will be by induction. For [imath]n=1,[/imath] we have [imath]B_n=\left\{1,\sqrt {p_1}\right\}.[/imath] It is clear that [imath]\sqrt{p_1}\neq 1,[/imath] so the set has [imath]2=2^1[/imath] elements. It is the basis of [imath]\mathbb Q[\sqrt{p_1}][/imath] because the minimal polynomial of [imath]\sqrt {p_1}[/imath] over [imath]\mathbb Q[/imath] has degree [imath]2,[/imath] and there is a theorem that [imath]K[a][/imath] has [imath]a^0,\cdots,a^{d-1}[/imath] as a basis, where [imath]d[/imath] is the degree of the minimal polynomial of [imath]a[/imath] over [imath]K[/imath]. Suppose the statement is true for [imath]n-1[/imath], where [imath]n\geq 2.[/imath] We have [imath] \left(B_n=B_{n-1}\cup\sqrt{p_n}B_{n-1}\right)\text { and } \left(B_{n-1}\cap\sqrt{p_n}B_{n-1}=\emptyset\right), [/imath] which is easy to see. It is also easy to see that [imath]\operatorname{card}(B_{n-1})=\operatorname{card}(\sqrt{p_n}B_{n-1}),[/imath] and therefore [imath] \operatorname{card}B_{n}=2^n. [/imath] Let [imath] \sum_{x\in B_{n}}q_xx=0 [/imath] for some [imath]\{q_x\}_{x\in B_n}\subset\mathbb Q.[/imath] Let [imath]p(x):=\sqrt{p_n}x[/imath] for all [imath]x\in B_{n-1}.[/imath] We have [imath] \sum_{x\in B_{n}}q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in \sqrt{p_n}B_{n-1}} q_xx=\sum_{x\in B_{n-1}} q_xx+\sum_{x\in B_{n-1}} q_{p(x)}\sqrt{p_n}x. [/imath] Therefore [imath] \sum_{x\in B_{n-1}} q_xx=-\sqrt{p_n}\sum_{x\in B_{n-1}} q_{p(x)}x,\tag1 [/imath] and we can make the following division iff [imath]q_{p(x)}\neq 0[/imath] for all [imath]x\in B_{n-1}[/imath] (because [imath]B_{n-1}[/imath] is linearly indepentent over [imath]\mathbb Q[/imath]): [imath] \sqrt{p_n}=-\frac{\sum_{x\in B_{n-1}} q_xx}{\sum_{x\in B_{n-1}} q_{p(x)}x}, [/imath] The right-hand side belongs to [imath]\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right],[/imath] so we have [imath] \sqrt{p_n}\in \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]. [/imath] Therefore we can write [imath]\sqrt{p_n}[/imath] uniquely in the basis [imath]B_{n-1}[/imath]. [imath] \sqrt{p_n}=\sum_{y\in B_{n-1}}c_yy [/imath] for some [imath]\{c_y\}_{y\in B_{n-1}}\subset \mathbb Q.[/imath] After squaring this equation we will obtain [imath] p_n=\sum_{y\in B_{n-1}}c_y^2y^2+2\sum_{y,z\in B_{n-1}}c_yc_zyz. [/imath] The last sum must be zero because it is not in [imath]\mathbb Q[/imath] and because after reducing it, we obtain a representation of [imath]p_n[/imath] in the basis [imath]B_{n-1},[/imath] which is unique. Thus [imath]p_n=\sum_{y\in B_{n-1}}c_y^2y^2.[/imath] Unfortunately, I can't prove that [imath]c_yc_z[/imath] is always zero. This was my first thought, but clearly there's trouble with the possibility of reductions in [imath] \sum_{y,z\in B_{n-1}}c_yc_zyz. [/imath] Different pairs [imath]y,z[/imath] may yield the same element of [imath]B_{n-1}[/imath] in the product [imath]yz.[/imath] This happens for example when [imath]y=\sqrt 5\sqrt 3,[/imath] [imath]z=\sqrt 5\sqrt 2,[/imath] and [imath]y'= \sqrt 11\sqrt 2,[/imath] [imath]z'=\sqrt 11\sqrt 3[/imath]. If it were true that [imath]c_yc_z[/imath] is always zero, I would be able to continue my proof as follows. We would have only one [imath]y_0[/imath] such that [imath]c_{y_0}\neq 0[/imath] and we'd get [imath]p_n=c_{y_0}^2y_0^2.[/imath] Let [imath]c_{y_0}=\frac kl[/imath]. We can write [imath]l^2p_n=k^2y_0^2.[/imath] But [imath]y_0^2[/imath] is the product of some primes different from [imath]p_n[/imath]. Therefore the greatest power of [imath]p_n[/imath] that divides the right-hand side is even. However, the greatest power of [imath]p_n[/imath] that divides the left-hand side is odd. A contradiction. The contradiction proves that [imath]q_{p(x)}=0[/imath] for all [imath]x\in B_{n-1}.[/imath] Hence [imath](1)[/imath] gives us that [imath] \sum_{x\in B_{n-1}} q_xx=0 [/imath] and linear independence of [imath]B_{n-1}[/imath] gives us that [imath]q_x=0[/imath] for all [imath]x\in B_{n-1}.[/imath] This gives us that [imath]B_n[/imath] is linearly independent. It generates the whole [imath]\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right][/imath] because [imath] \mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_n\right]=\left(\mathbb Q\left[\sqrt p_1,\sqrt p_2,\cdots,\sqrt p_{n-1}\right]\right)\left[\sqrt{p_n}\right]. [/imath] This would end the proof. |
69317 | Construct a monotone function which has countably many discontinuities
I read in a textbook, which had seemed to have other dubious errors, that one may construct a monotone function with discontinuities at every point in a countable set [imath]C \subset [a,b][/imath] by enumerating the points as [imath]c_1, c_2, \dots[/imath] and defining [imath]f(x) = \sum_{c_n < x}2^{-n}[/imath]. However, if seems that if we let [imath][a,b] = [0,1], C = \mathbb{Q} \cap [0,1][/imath], then [imath]f(x)[/imath] is constant [imath]1[/imath] everywhere except 0, an apparent counterexample. So my question is: how does one construct a monotonic function which has discontinuities precisely on a countable set [imath]C[/imath]? Further, are there any relatively easy-to-visualize constructions? | 1209351 | Is there a non-decreasing function that is discontinuous at every rational point?
A well-known theorem is that if [imath]f:[a,b]\to\mathbb{R}[/imath] is non-decreasing, then [imath]f[/imath] as at most countably many discontinuities. This led me think of the following question. Question: Is there a non-decreasing function [imath]f:[0,1]\to\mathbb{R}[/imath] such that [imath]f[/imath] is discontinuous at every rational point? Such a function would definitively be geometrically counter-intuitive. Intuitively, I would think that if it is discontinues at every rational point, then it will be discontinues everywhere, and hence would not exist by the aforementioned theorem. But I cannot think of any proof or disproof of these claims. |
60284 | How to show that [imath]\det(AB) =\det(A) \det(B)[/imath]?
Given two square matrices [imath]A[/imath] and [imath]B[/imath], how do you show that [imath]\det(AB) = \det(A)\det(B)[/imath] where [imath]\det(\cdot)[/imath] is the determinant of the matrix? | 1454173 | Show that [imath][(BA)a,(BA)b,(BA)c] = det(B) [Aa,Ab,Ac][/imath]
I am basically on my way to proving [imath]det(AB) = det(A)det(B)[/imath] but I need this step in my proof. We have [imath][(BA)a,(BA)b,(BA)c] = (BA)a \cdot (BA)b \times (BA)c[/imath] but I am failing to see how this simplifies to [imath]det(B) [Aa,Ab,Ac][/imath]. |
83380 | I have a problem understanding the proof of Rencontres numbers (Derangements)
I understand the whole concept of Rencontres numbers but I can't understand how to prove this equation [imath]D_{n,0}=\left[\frac{n!}{e}\right][/imath] where [imath][\cdot][/imath] denotes the rounding function (i.e., [imath][x][/imath] is the integer nearest to [imath]x[/imath]). This equation that I wrote comes from solving the following recursion, but I don't understand how exactly the author calculated this recursion. [imath]$$\begin {align*} D_{n+2,0} & =(n+1)(D_{n+1,0}+D_{n,0}) \\ D_{0,0} & = 1 \\ D_{1,0} & = 0 \end {align*} $$[/imath] | 166378 | De-arrangement in permutation and combination
This article talks about de-arrangement in permutation combination. Funda 1: De-arrangement If [imath]n[/imath] distinct items are arranged in a row, then the number of ways they can be rearranged such that none of them occupies its original position is, [imath]n! \left(\frac{1}{0!} – \frac{1}{1!} + \frac{1}{2!} – \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!}\right).[/imath] Note: De-arrangement of 1 object is not possible. [imath]\mathrm{Dearr}(2) = 1[/imath]; [imath]\mathrm{Dearr}(3) = 2[/imath]; [imath]\mathrm{Dearr}(4) =12 – 4 + 1 = 9[/imath]; [imath]\mathrm{Dearr}(5) = 60 – 20 + 5 – 1 = 44[/imath]. I am not able to understand the logic behind the equation. I searched in the internet, but could not find any links to this particular topic. Can anyone explain the logic behind this equation or point me to some link that does it ? |
20475 | Proving Pascal's Rule : [imath]{{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}[/imath] when [imath]1\leq r\leq n[/imath]
I'm trying to prove that [imath]{n \choose r}[/imath] is equal to [imath]{{n-1} \choose {r-1}}+{{n-1} \choose r}[/imath] when [imath]1\leq r\leq n[/imath]. I suppose I could use the counting rules in probability, perhaps combination= [imath]{{n} \choose {r}}=\frac{n!}{r!(n-r!)}[/imath]. I want to see an actual proof behind this equation. Does anyone have any ideas? | 392595 | What's the intuition behind this equality involving combinatorics?
What is the intuition behind [imath] \binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} [/imath] ? I can't grasp why picking a group of [imath]k[/imath] out of [imath]n[/imath] bijects to first picking a group of [imath]k-1[/imath] out of [imath]n-1[/imath] and then a group of [imath]k[/imath] out of [imath]n-1[/imath]. |
4581 | Why does [imath]K \leadsto K(X)[/imath] preserve the degree of field extensions?
The following is a problem in an algebra textbook, probably a well-known fact, but I just don't know how to Google it. Let [imath]K/k[/imath] be a finite field extension. Then [imath]K(X)/k(X)[/imath] is also finite with the same degree as [imath]K/k[/imath]. Obviously if [imath]v_1,...,v_n[/imath] is a [imath]k[/imath]-Basis of [imath]K[/imath], any polynomial in [imath]K(X)[/imath] can be written as a [imath]k(X)[/imath]-linear combination of [imath]v_1,...,v_n[/imath], but I have no idea what to do with a nontrivial denominator. Is there perhaps a more elegant way of proving this? | 141480 | When a field extension [imath]E\subset F[/imath] has degree [imath]n[/imath], can I find the degree of the extension [imath]E(x)\subset F(x)?[/imath]
This is not a problem I've found stated anywhere, so I'm not sure how much generality I should assume. I will try to ask my question in such a way that answers on different levels of generality could be possible. I'm also not sure that this question isn't trivial. Let [imath]E\subset F[/imath] be a fields (both can be finite if needed), and let [imath]n[/imath] be the degree of the field extension ([imath]n[/imath] can be finite if needed). Can we find the degree of the extension [imath]E(x)\subset F(x)[/imath] of rational function fields? Say [imath]E=\mathbb F_2[/imath] and [imath]F=\mathbb F_4[/imath]. Then [imath](F:E)=2.[/imath] I can take [imath]\{1,\xi\}[/imath] to be an [imath]E[/imath]-basis of [imath]F[/imath]. Now let [imath]f\in F(x),[/imath] [imath]f(x)=\frac{a_kx^k+\cdots +a_0 } {b_lx^l+\cdots+b_0 }[/imath] for [imath]a_0,\ldots a_k,b_0,\ldots,b_l\in F[/imath] I can write [imath]a_0,\ldots a_k,b_0,\ldots,b_l[/imath] in the basis: [imath]\begin{eqnarray}&a_i&=p_i\xi+q_i\\&b_j&=r_j\xi+s_j\end{eqnarray}[/imath] But all I get is [imath]f(x)=\frac{p_k\xi x^k+\cdots+p_0+q_kx^k+\cdots+q_0} { r_k\xi x^k+\cdots+r_0+s_kx^k+\cdots+s_0},[/imath] and I have no idea what to do with this. On the other hand, my intuition is that the degree of the extension of rational function fields should only depend on the degree of the extension [imath]E\subset F,[/imath] even regardless of any finiteness conditions. |
600 | Circular permutations with indistinguishable objects
Given n distinct objects, there are [imath]n![/imath] permutations of the objects and [imath]n!/n[/imath] "circular permutations" of the objects (orientation of the circle matters, but there is no starting point, so [imath]1234[/imath] and [imath]2341[/imath] are the same, but [imath]4321[/imath] is different). Given [imath]n[/imath] objects of [imath]k[/imath] types (where the objects within each type are indistinguishable), [imath]r_i[/imath] of the [imath]i^{th}[/imath] type, there are \begin{equation*} \frac{n!}{r_1!r_2!\cdots r_k!} \end{equation*} permutations. How many circular permutations are there of such a set? | 710090 | How many ways to make a bracelet with [imath]n[/imath] white balls and [imath]m[/imath] black balls?
Okay so I have [imath]n[/imath] black ball and [imath]m[/imath] white balls. How many bracelets can I make using all the balls? At first I thought there are [imath](n+m-1)![/imath] if all the balls were different. So we can divide into groups of [imath]n!\cdot m![/imath] that look the same if you permute the black balls between them and also for the white balls. However if I had [imath]6[/imath] white balls and [imath]2[/imath] black this would yield [imath]3.5[/imath] so clearly this can't be correct. How can I count this? Is there a generalized version form for [imath]k[/imath] colors? |
144703 | Existence of irreducible polynomials over finite field
Let [imath]F[/imath] be a finite field. How do we prove that for each [imath]n \in \mathbb{N}[/imath] there is an irreducible polynomial of degree [imath]n[/imath]? One can assume that [imath]F = \mathbb{F}_{p^m}[/imath] where [imath]p[/imath] is prime. If [imath]n \ge |F|[/imath] then I can construct an irreducible polynomial, namely [imath] p(x) = 1 + \prod_{j=1}^{|F|} ( x - a_j )[/imath] where [imath]a_j[/imath] are all the field elements. It is clear that [imath]p(x)[/imath] has no roots in [imath]F[/imath]. This trick doesn't work for [imath]n < |F|[/imath]. A counter-example: Let [imath]F = \mathbb{F}_3[/imath] and [imath]p(x) = 1 + (x-1)(x-2)[/imath], then [imath]p(1) = 1[/imath], [imath]p(2) = 1[/imath], [imath]p(0) = 1 + (-1)(-2) = 1 + 2 = 3 = 0 \pmod 3[/imath]. I know there is a way to count them using the Möbius function [imath]\mu(n)[/imath] but I want a proof without it that just shows existence. | 1395169 | How to show there are irreducible polynomials of any degree over [imath]\Bbb Z _p[/imath]?
How do I show there are irreducible polynomials of any degree in [imath]\mathbb{Z}_p[x][/imath], with [imath]p[/imath] prime? I tried counting the number of reducible polynomials of any degree but that turned out to be hard... Any help? |
18576 | Summation involving a factorial: [imath]1 + \sum_{j=1}^{n} j!j[/imath]
[imath]1 + \sum_{j=1}^{n} j!j[/imath] I want to find a formula for the above and then prove it by induction. The answer according to Wolfram is [imath](n+1)!-1[/imath], however I have no idea how to get there. Any hints or ideas on how I should tackle this one? | 912593 | Finding a formula for a given series.
I'm having trouble figuring out how to evaluate: [imath]\sum _{j = 1}^{n} j!j[/imath] I've tried plugging in numbers and looking for a pattern and I've also tried to find a general form like: [imath]n=1:[/imath] [imath] 1*1 = 1 = n^2[/imath] [imath]n =2:[/imath] [imath] 1*1 + 1*2*2 = 5 = (n-1)^2 + (n-1)(n)^2[/imath] Could anyone point me in the right direction? |
6501 | Is there a known well ordering of the reals?
So, from what I understand, the axiom of choice is equivalent to the claim that every set can be well ordered. A set is well ordered by a relation, [imath]R[/imath] , if every subset has a least element. My question is: Has anyone constructed a well ordering on the reals? First, I was going to ask this question about the rationals, but then I realised that if you pick your favourite bijection between rationals and the integers, this determines a well ordering on the rationals through the natural well order on [imath]\mathbb{Z}[/imath] . So it's not the denseness of the reals that makes it hard to well order them. So is it just the size of [imath]\mathbb{R}[/imath] that makes it difficult to find a well order for it? Why should that be? To reiterate: Is there a known well order on the Reals? If there is, does a similar construction work for larger cardinalities? Is there a largest cardinality for which the construction works? | 792320 | Finding a linear order in [imath]\mathbb R[/imath]
I'm just wondering if there is a linear order in [imath]\mathbb R[/imath] where every subset has minimal element? It's easy to find such orders in [imath]\mathbb N [/imath] and [imath] \mathbb Z[/imath] but I can't think of one in [imath]\mathbb R[/imath] |
121513 | Can the product [imath]AB[/imath] be computed using only [imath]+, -,[/imath] and reciprocal operators?
Can the product of [imath]A, B[/imath] be computed using only [imath]+, -,[/imath] and reciprocal operators using a calculator? You can use calculator's memory function (multiply and divide are broken though). Additional: I should have mentioned earlier, in addition to the 3 operators, the numberpad of the calculator can be used so yes 1 can be used. | 2624661 | How to multiply on a calculator which only allows add, subtract and reciprocal
Assume we have a calculator with the following flaw. The only operations can be done by it are [imath]+[/imath] and [imath]-[/imath] and [imath]\dfrac{1}{x}[/imath] i.e. you only can add or subtract two numbers and also calculate the reciprocal but you can't multiply or divide. Using this calculator how can you multiply two numbers? I twiddled with lots of formulas but I got nowhere. Sorry if I can't add any further information or idea. I appreciate any solution on this.... |
260422 | Proving a Ring is commutative if every element is idempotent
I had this question on a final exam. I was wondering if anyone knows a proof for it. Let R be a ring not necessarily with unity ([imath]1 \neq 0[/imath] and [imath]x \in R[/imath] so that [imath]x*1=x=1*x[/imath]) and let R have the property that every element of R is idempotent that is [imath]\forall x \in R\quad x*x=x[/imath]. Prove that [imath]R[/imath] is commutative. Not to put my whole answer up because I know I was wrong. I will put the part where I had trouble going to the next step. My argument went like this: let [imath]a,b \in R[/imath] then [imath]ab\in R[/imath] cause [imath]R[/imath] is a ring. Therefore [imath]abab=ab[/imath]. Now [imath]aabb=ab[/imath] as well, so [imath]abab=aabb\ \Rightarrow abab-aabb=0 \Rightarrow\ a(ba-ab)b=0\ \because\text{ distributive property of $R$}.[/imath] Now because R is not a division ring then the element [imath]a[/imath] could be a divisor of zero or [imath]b[/imath] could be as well so you cannot just assume [imath]ba-ab=0[/imath]. This is where I am stuck. | 10274 | How to show that every Boolean ring is commutative?
A ring [imath]R[/imath] is a Boolean ring provided that [imath]a^2=a[/imath] for every [imath]a \in R[/imath]. How can we show that every Boolean ring is commutative? |
301971 | Let [imath]X[/imath] be any random variable. Prove that [imath]\mathbf{E}[X^2] < \infty[/imath] iff [imath]\dots[/imath]
Let [imath]X[/imath] be any random variable. Prove that [imath]\mathbf{E}[X^2] < \infty[/imath] iff [imath]\sum_{n=1}^{\infty}n \, \mathbf{P}(\mid X \mid > n) < \infty[/imath]. | 242376 | A question on measure space and measurable function
Let [imath](X,\mu)[/imath] be a measure space, and [imath]\mu(X)< + \infty[/imath]. Let [imath]f[/imath] be a nonnegative measurable function on [imath]X[/imath] and [imath]E_n = \{x \in X|f(x) \geq n \}[/imath]. Then [imath]\int_X f^2d \mu < + \infty[/imath] if and only if [imath]\sum_{n=1}^{\infty} n \mu(E_n) <+\infty[/imath]. Do I have to use simple functions that converge to [imath]f[/imath]? Thanks a lot. |
87970 | Proof involving the logarithmic integral
Another exercise from Apostol's book, this time we're supposed to prove [imath]\mathrm{Li}(x)=\frac{x}{\log x}+\int_2^x \frac{dt}{\log^2t}-\frac{2}{\log 2}.[/imath] which is easy to do via integration by parts. But then he goes on to write, [...] and that, more generally, [imath]\mathrm{Li}(x)=\frac{x}{\log x} \left(1+ \sum_{k=1}^{n-1} \frac{k!}{\log^k x} \right)+n! \int_2^x \frac{dt}{\log^{n+1}t}+C_n,[/imath] where [imath]C_n[/imath] is independent of [imath]x[/imath]. My question may come across as stupid, but what's going on here? Is this a comment, or should I be able to prove this as well? If so, I'd very much like some small hint on how to approach this! Thanks! | 347729 | Approximation of [imath]\mathrm{Li}(x) = \int\limits_{0}^x \frac{dt}{\ln t}[/imath]
I am reading about the Riemann hypothesis, and the article mentioned the Li function: [imath]\mathrm{Li}(x) = \int\limits_{0}^x \frac{dt}{\ln t}[/imath] They said that this function can be approximated: [imath]\mathrm{Li}(x) \approx \frac{x}{\ln x} + \frac{x}{\ln^2(x)} + \frac{2x}{\ln^3(x)}+\cdots ,[/imath] thus is a better approximation for [imath]\pi(x)[/imath](the number of prime numbers less than or equal to [imath]x[/imath]). I could not figure out how to prove this approximation formula. Could anyone help me please? |
30282 | Does [imath]k+\aleph_0=\mathfrak{c}[/imath] imply [imath]k=\mathfrak{c}[/imath] without the Axiom of Choice?
I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of Cardinal Arithmetic, which states that for [imath]\kappa[/imath] and [imath]\lambda[/imath] cardinal numbers, the larger infinite and the smaller nonzero, then [imath]\kappa+\lambda=\kappa\cdot\lambda=\max(\kappa,\lambda)[/imath]. I was playing around with the equation [imath]k+\aleph_0=\mathfrak{c}[/imath] for some cardinal [imath]k[/imath]. From the above, it follows that [imath]\mathfrak{c}=k+\aleph_0=\max(k,\aleph_0)[/imath], which implies [imath]k=\mathfrak{c}[/imath]. I'm curious, can we still show [imath]k=\mathfrak{c}[/imath] without the Axiom of Choice? Is it maybe possible to bound [imath]\mathfrak{c}-\aleph_0[/imath] above and below by [imath]\mathfrak{c}[/imath]? But then I'm not quite sure such algebraic manipulations even mean anything, or work like that here. Certainly normal arithmetic does not! Thanks. | 1310105 | Proving [imath]|\mathbb{R}-\mathbb{S}|=2^{\aleph_0}[/imath] when [imath]\mathbb{S}\subset R[/imath] is countable
I wish to prove that [imath]|\mathbb{R}-\mathbb{S}|=2^{\aleph_0}[/imath] when [imath]\mathbb{S}\subset \mathbb{R}[/imath] is countable. I want to say that [imath]|\mathbb{R}-\mathbb{S}|= |\mathbb{R}|-|\mathbb{S}|[/imath] but we haven't studied yet what subtraction of cardinals means (I can guess, though). How could I prove this using only basic cardinal properties? |