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297624 | Use Nakayama's Lemma to show that [imath]I[/imath] is principal, generated by an idempotent.
Let [imath]I[/imath] be a finitely generated ideal in [imath]R[/imath], such that [imath]I^2 = I[/imath]. Using the fact there exists [imath]x\in R[/imath] such that [imath]e = 1 - x\in I[/imath] and [imath]xI = 0[/imath], use Nakayama's Lemma to show that [imath]I[/imath] is principal, generated by an idempotent. | 255885 | Constructing Idempotent Generator of Idempotent Ideal
Exercise 2.1 in Matsumura's Commutative Ring Theory reads as follows: "Let [imath]A[/imath] be a commutative ring and [imath]I[/imath] an ideal that is finitely generated and [imath]I=I^2[/imath]. Then [imath]I[/imath] is generated by an idempotent." In trying to solve it, i first followed a constructive approach, where e.g. for the case of two generators i tried to construct an idempotent generator. However, it seemed difficult. Then i realized that i could apply Nakayama's lemma to the [imath]A[/imath]-module [imath]I[/imath] and the existence of the idempotent generator follows. My question is: How could one go about finding this idempotent generator? Is there a systematic way? |
390103 | How to show that [imath]\exists~f\in C(X,\mathbb R)[/imath] such that [imath]f[/imath] is unbounded?
Let [imath]X[/imath] be a non-compact metric space. How to show that [imath]\exists~f\in C(X,\mathbb R)[/imath] such that [imath]f[/imath] is unbounded? | 114123 | Compactness of [imath](M, d)[/imath] in terms of continuous functions [imath]f:M\to \mathbb R[/imath]
Suppose [imath](M,d)[/imath] a metric space. I want to show that If every continuous real-valued function on [imath]M[/imath] attains a maximum, then [imath](M, d)[/imath] is compact. I was trying to do this by assuming [imath]M[/imath] non-compact and showing explicitly a function which does not attain a maximum. If [imath]M[/imath] is not bounded it's enough to take the distance from a fixed point as we can always find a sequence which "exits" every compact and pushes the distance at infinity. But what if [imath]M[/imath] is bounded? |
390474 | Lebesgue Integral Rudin Problem
Suppose {[imath]n_k[/imath]} is an increasing sequence of positive integers and E is the set of all x[imath]\in[/imath]([imath]-\pi, \pi[/imath]) at which {sin[imath]n_k x[/imath]} converges. Prove that [imath]m(E)=0[/imath]. Hint: For every A [imath]\subset[/imath] E, [imath]\int_A[/imath] sin[imath]n_k x dx \rightarrow 0[/imath] and 2[imath]\int_A [/imath](sin[imath]n_k x)^2dx[/imath] = [imath]\int_A (1-cos2n_k x)dx[/imath] [imath]\rightarrow m(A)[/imath] as [imath]n \rightarrow \infty[/imath] | 247799 | Prove the set of which sin(nx) converges has Lebesgue measure zero (from Baby Rudin Chapter 11)
I am trying to work through Rudin. This is a question from chapter 11: Suppose that [imath]\{n_k\}[/imath] is an increasing sequence of positive integers and [imath]E[/imath] is the set of all [imath]x[/imath] in [imath](-\pi,\pi)[/imath] at which [imath]\sin(n_kx)[/imath] converges. Prove that [imath]E[/imath] has Lebesgue measure zero. Hint: For every subset [imath]A[/imath] of [imath]E[/imath], [imath]\int_A \sin (n_kx) dx[/imath] tends to zero, and [imath]2\int_{A} \sin^2 (n_k x)dx[/imath] tends to the measure of [imath]A[/imath]. So, if I can use the hint, I'm pretty sure I can get the question. The thing is, I have no idea how to prove the hint. Using the hint, here's how I think you do the question: Define [imath]f[/imath] on [imath]E[/imath] as \begin{align*}f(x)= \lim_{k\rightarrow \infty} \sin(n_kx) \end{align*} Notice that [imath]\sin(n_kx)\leq 1[/imath], and [imath]1\in L[/imath] on [imath]A\subset E[/imath] for every measurable [imath]A[/imath] (since [imath]E[/imath] has finite measure), so Theorem [imath]11.32[/imath] in Rudin (the Lebesgue dominated convergence theorem) says \begin{align*} \int_A f(x) dx = \int_A \lim_{k\rightarrow \infty} \sin(n_kx) = \lim_{k\rightarrow \infty} \int_A \sin(n_kx) = 0 \end{align*} (by the first hint). But this was true for all [imath]A\subseteq E[/imath], so by one of the questions on the last assignment, [imath]f(x)=0[/imath] almost everywhere on [imath]E[/imath] and so [imath]f^2(x)=0[/imath] almost everywhere on [imath]E[/imath]. Using Theorem [imath]11.32[/imath] again, we get \begin{align*} \mu(A) = \lim_{k \rightarrow \infty} \int_A 2\sin^2(n_kx) = 2\int_A \lim_{k\rightarrow \infty} \sin^2(n_kx)= 2 \int_A f^2(x)=0 \end{align*} Therefore [imath]\mu(E)=0[/imath]. Does anybody know how to prove the hint? Thanks! |
390653 | The geometric interpretation
In the course of mathematical analysis, there was one problem that i excited to know more about it: What is the geometric interpretation of [imath] \int_a^b f(x)\,d(\alpha(x)) [/imath] and [imath]\alpha(x)[/imath] is function in [imath][a,b][/imath] | 378621 | Stieltjes Integral meaning.
Can anybody give a geometrical interpretation of the Stieltjes integral: [imath]\int_a^bf(\xi)\,d\alpha(\xi)[/imath] How would we calculate? [imath]\int_a^b \xi^3\,d\alpha(\xi)[/imath] for example. |
279332 | Inverse of a bijection f is equal to its derivative
Does there exist a differentiable bijection [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] such that [imath]f'(x) = f^{-1}(x)[/imath] ? | 763996 | Inverse Function Differential Equation
For the differential equation [imath]\frac{d}{dx}[y(x)]=y^{(-1)}(x)[/imath] where [imath]y^{(-1)}(x)[/imath] is the inverse of [imath]y(x)[/imath], find y(x). I gave up on finding the solution analytically pretty quickly and decided that a numerical approach might be more effective. But, I'm not sure that this problem can be solved, even numerically; an Euler or Runge-Kutta type method will not work because to find the value of [imath]y^{(-1)}(a)[/imath], one must first know the value of [imath]y(b)[/imath], where [imath]a[/imath] is not necessarily equal to [imath]b[/imath]. Sort of like trying to solve [imath]\frac{d}{dx}[y(x)]=y(x+1)[/imath], I don't know of any numerical approaches that can handle a problem of that type. If anyone has any ideas on how this might be solved (or proven unsolvable), they would be appreciated. Thanks! |
134400 | preservation of completeness under homeomorphism
Does homeomorphic metric spaces preserves completeness?I mean two metric space which are homeomorphic and one of them is complete[imath]\Rightarrow[/imath] another one is also complete? | 391071 | [imath]X \cong Y[/imath], [imath]X[/imath] complete [imath]\implies Y[/imath] complete?
Let [imath]\cong[/imath] denote the homeomorphic notation. Let [imath]X,Y[/imath] be metric spaces, and let [imath]X \cong Y[/imath]. If [imath]X[/imath] is a complete metric space does it imply [imath]Y[/imath] is also complete. |
391015 | [imath]AX=C[/imath]: An Inconsistent Linear Equation
Question: Let [imath]A \in M_{n\times n}(F)[/imath]. Suppose that the system of linear equations [imath]AX = B[/imath] has more than one solution. Prove that there is a column [imath]C \in F^n[/imath] such that the system of linear equations [imath]AX = C[/imath] is inconsistent. Attempt: Lemma: Let [imath]AX=C[/imath] be a system of linear equations. Then the system is consistent if and only if [imath]\text{dim Im}~(A)=\text{dim Im}~(A \mid C)[/imath]. Let [imath]Y[/imath] and [imath]Y'[/imath] be solutions to the system [imath]AX=B[/imath]. Then [imath] A(Y) = A(Y')=B,[/imath] or [imath] A(Y)-A(Y')=A(Y-Y')=0,[/imath] so [imath]\dim \text{Ker}~(A)\geq 1[/imath], which means [imath]\dim \text{Im}~(A)<n[/imath], or simply that [imath]\text{span}~(\{A^{(1)},\dots,A^{(n)}\})\neq F^n.[/imath] In other words, there exists at least one [imath]A^{(i)}[/imath] such that [imath]A^{(i)}=k_1A^{(1)}+\cdots +k_{i-1}A^{(i-1)}+k_{i+1}A^{(i+1)}+\cdots + k_nA^{(n)}\neq ce_i,[/imath] for some [imath]e_i\in \{e_1,\dots,e_n\}[/imath] the standard basis for [imath]F^n[/imath] with [imath]c,k_i\in F[/imath]. Thus, by the lemma, we clearly see that such an [imath]e_i[/imath] is the desired [imath]C[/imath] for which [imath]AX=C[/imath] has no solutions, namely [imath]\text{dim Im}~(A)\neq \text{dim Im}(A \mid e_i)=\text{dim Im}(A \mid C).[/imath] This is so because if [imath]\text{dim Im}(A)=m[/imath], then [imath]\text{dim Im}~(A \mid e_i)=m+1[/imath] for such an [imath]e_i[/imath] described above. Acknowledgements: @jkn | 390796 | Inconsistent System of Linear Equations
Let [imath]A ∈ M_{n\times n}(F)[/imath]. Suppose that the system of linear equations [imath]AX = B[/imath] has more than one solution. Prove that there is a column [imath]C ∈ F^n[/imath] such that the system of linear equations [imath]AX = C[/imath] is inconsistent. I thought it'd be clever to use this (underlined by my pencil): |
391293 | Solve a cubic polynomial given that one root is four times a second root?
How exactly would you solve the equation: "Solve the equation [imath]10x^3+23x^2+5x−2=0[/imath] given that one root is four times a second root." How would you go about solving this? Any help would be greatly appreciated. | 391250 | Solve a cubic polynomial (given one root is four times a second root)?
So, I've been stuck on a question for a long time now: "Solve the equation [imath]10x^3 + 23x^2 + 5x - 2 = 0[/imath] given that one root is four times a second root." How would you go about solving this? Any help would be greatly appreciated. |
390791 | Question on Coordenate Base of a Infinite Banach Space
Let [imath]X[/imath] be a infinite dimensional Banach space. If [imath]\mathcal{B}=\{e_i:i\in I\}[/imath] is a Hamel base for [imath]X[/imath]. How to show that only a finite number of the coordinate functionals [imath]e^{*}_i[/imath] will be continuous. My approach: Suppose that there is a [imath]I_0\subseteq I[/imath] infinite such that, for each [imath]i\in I_0[/imath], [imath]e^*_i[/imath] is continuous. Then, there exists a countable infinite number of [imath]i\in I_0[/imath] such that [imath]e^*_i[/imath] is continuous. But having trouble to derive an absurd from this fact. | 79184 | At most finitely many (Hamel) coordinate functionals are continuous - different proof
If [imath]X[/imath] is a vector space over [imath]\mathbb R[/imath] and [imath]B=\{x_i; i\in I\}[/imath] is a Hamel basis for [imath]X[/imath], then for each [imath]i\in I[/imath] we have a linear functional [imath]a_i(x)[/imath] which assigns to [imath]x[/imath] the [imath]i[/imath]-th coordinate, i.e., the functions [imath]a_i[/imath] are uniquely determined by the conditions that [imath]x=\sum_{i\in I} a_i(x)x_i,[/imath] where only finitely many summands are non-zero. If [imath]X[/imath] is a Banach space, then at most finitely many of them can be continuous. I have learned the following argument from comments in this question. Suppose that [imath]\{b_i; i\in\mathbb N\}[/imath] is an infinite subset of [imath]B[/imath] such that each [imath]f_{b_i}[/imath] is continuous. W.l.o.g. we may assume that [imath]\lVert{b_i}\rVert=1[/imath]. Let [imath]y:=\sum_{i=1}^\infty \frac1{2^i}b_i.[/imath] (Since [imath]X[/imath] is complete, the above sum converges.) We also denote [imath]y_n:=\sum_{i=1}^n \frac1{2^i}b_i[/imath]. Since [imath]y_n[/imath] converges to [imath]y[/imath], we have [imath]f_{b_k}(y)=\lim\limits_{n\to\infty} f_{b_k}(y_n)=\frac1{2^k}[/imath] for each [imath]k\in\mathbb N[/imath]. Thus the point [imath]x[/imath] has infinitely many non-zero coordinates, which contradicts the definition of Hamel basis. I have stumbled upon Exercise 4.3 in the book Christopher Heil: A Basis Theory Primer. Springer, New York, 2011. In this exercise we are working in an infinite-dimensional space [imath]X[/imath]. Basically the same notation as I mentioned above is introduced, [imath]a_i[/imath]'s are called coefficient functionals and then it goes as follows: (a) Show by example that it is possible for some particular functional [imath]a_i[/imath] to be continuous. (b) Show that [imath]a_i(x_j) = \delta_{ij}[/imath] for [imath]i,j\in I[/imath]. (c) Let [imath]J = \{i\in I : a_i \text{ is continuous}\}[/imath]. Show that [imath]\sup \{j\in J; \lVert a_j \rVert<+\infty\}[/imath] (d) Show that at most finitely many functionals [imath]a_i[/imath] can be continuous, i.e., [imath]J[/imath] is finite. (e) Give an example of an infinite-dimensional normed linear space that has a Hamel basis [imath]\{x_i; i\in I\}[/imath] such that each of the associated coefficient functionals [imath]a_i[/imath] for [imath]i\in I[/imath] is continuous. The part (c) can be shown easily using Banach-Steinhaus theorem (a.k.a. Uniform boundedness principle). But I guess that the author of the book has in mind a different proof for part (d) from what I sketched above, since (c) is an easy consequence of (d) -- so he would probably not be put the exercises in this order. (But maybe I was just trying to read to much between the lines.) Question: I was not able to find a proof od (d) which uses (c). Do you have some idea how to do this? NOTE: My question is not about the parts (a), (b), (e). I've included them just for the sake of the completeness, in order to include sufficient context for the question. |
391439 | What does [imath]x^\pi[/imath] mean?
I was just wondering, what does [imath]x^\pi[/imath] or for that matter, [imath]x[/imath] raised to any irrational number mean? For example, I want to represent [imath]x^2[/imath] then that would mean [imath]x * x[/imath] or if I want to do [imath]x^\frac{2}{3}[/imath] then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as [imath]\pi[/imath]? Thanks! | 132703 | What does [imath]2^x[/imath] really mean when [imath]x[/imath] is not an integer?
We all know that [imath]2^5[/imath] means [imath]2\times 2\times 2\times 2\times 2 = 32[/imath], but what does [imath]2^\pi[/imath] mean? How is it possible to calculate that without using a calculator? I am really curious about this, so please let me know what you think. |
391485 | Integral of [imath]\int \frac{x^4+2x+4}{x^4-1}dx[/imath]
I am trying to solve this integral and I need your suggestions. [imath]\int \frac{x^4+2x+4}{x^4-1}dx[/imath] Thanks | 384553 | Integral Of [imath]\frac{x^4+2x+4}{x^4-1}[/imath]
Any ideas how to solve it? [imath]\int\frac{x^4+2x+4}{x^4-1}dx[/imath] Thanks! |
391396 | For two disjoint compact subsets [imath]A[/imath] and [imath]B[/imath] of a metric space [imath](X,d)[/imath] show that [imath]d(A,B)>0.[/imath]
I was thinking about the following problem: For two disjoint compact subsets [imath]A[/imath] and [imath]B[/imath] of a metric space [imath](X,d)[/imath] show that [imath]d(A,B)>0.[/imath] I'm having doubt with my attemp. Please have a look and do comment: Let [imath]d(A,B)=\inf\{d(a,b):a\in A,b\in B\}=0.[/imath] Then [imath]\exists[/imath] sequences [imath]\{a_n\}\subset A,\{b_n\}\subset B[/imath] such that [imath]0\le d(a_n,b_n)<\dfrac{1}{n}~\forall~n.[/imath] Since [imath]A[/imath] is compact, [imath]\exists[/imath] a convergent subsequene [imath]\{a_{r_n}\}[/imath] of [imath]\{a_n\}[/imath] converging to some [imath]a\in A.[/imath] Then [imath]0\le d(a_{r_n},b_{r_n})<\dfrac{1}{r_n}~\forall~n.[/imath] Similarly since [imath]B[/imath] is compact, [imath]\exists[/imath] a convergent subsequene [imath]\{b_{r_{n_m}}\}[/imath] of [imath]\{b_{r_n}\}[/imath] converging to some [imath]b\in B.[/imath] Then $0\le d(a_{r_{n_m}},b_{r_{n_m}})[imath]<\dfrac{1}{r_{n_m}}~\forall~m.$ We note that $r_{n_m}\ge n_m\ge m>0[/imath]\implies0<\dfrac{1}{r_{n_m}}<\dfrac{1}{m}\to0$[imath]\implies\dfrac{1}{r_{n_m}}\to0.[/imath] Using the sqeezing lemma once again we can see that [imath]\exists[/imath] convergent sequences [imath]\{a_n\}\subset A[/imath] and [imath]\{b_n\}\subset B[/imath] such that [imath]a_n\to a\in A,b_n\to b\in B.[/imath] [imath]\exists[/imath] convergent sequences [imath]\{a_n\}\subset A[/imath] and [imath]\{b_n\}\subset B[/imath] such that [imath]a_n\to a\in A,b_n\to b\in B.[/imath] Then [imath]d(a_n,b_n)\to d(a,b)[/imath] Consequently, [imath]d(a,b)=0\implies a=b,[/imath] a contradiction to [imath]A\cap B=\emptyset.[/imath] | 48714 | A and B disjoint, A compact, and B closed implies there is positive distance between both sets
Claim: Let [imath]X[/imath] be a metric space. If [imath]A,B\in X[/imath] are disjoint, if A is compact, and if B is closed, then [imath]\exists \delta>0: |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B[/imath]. Proof. Assume the contrary. Let [imath]\alpha_n\in A,\beta_n\in B[/imath] be chosen such that [imath]|\alpha_n-\beta_n|\rightarrow0[/imath] as [imath]n\rightarrow \infty[/imath]. Since A is compact, there exists a convergent subsequence of [imath]\alpha_n\;(n\in\mathbb{N})[/imath], [imath]\alpha_{n_m}\;(m\in\mathbb{N})[/imath], which converges to [imath]\alpha\in A[/imath]. We have [imath]|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.[/imath] Hence [imath]\alpha[/imath] is a limit point of B and since B is closed [imath]\alpha\in B[/imath], contradiction. Is my proof correct? I feel as though I am missing something simple which would trivialize the proof. |
391516 | cardinality of a set with repeating elements?
What is the cardinality of a set which has repeating elements ? For example [imath]S = \{1,1,1,2,2\}[/imath] Is each individual element counted? Please quote a reference text if possible. | 155475 | [imath]\{1,1\}=\{1\}[/imath], origin of this convention
Is there any book that explicitly contain the convention that a representation of the set that contain repeated element is the same as the one without repeated elements? Like [imath]\{1,1,2,3\} = \{1,2,3\}[/imath]. I have looked over a few books and it didn't mention such thing. (Wikipedia has it, but it does not cite source). In my years learning mathematics in both US and Hungary, this convention is known and applied. However recently I noticed some Chinese students claim they have never seen this before, and I don't remember I saw it in any book either. I never found a book explicitly says what are the rules in how [imath]\{a_1,a_2,a_3,\ldots,a_n\}[/imath] specify a set. Some people believe it can only specify a set if [imath]a_i\neq a_j \Leftrightarrow i\neq j[/imath]. The convention shows that doesn't have to be satisfied. |
391643 | Proving the quadratic formula (for dummies)
I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from [imath]ax^2+bx+c=0[/imath] to [imath]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/imath]. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated. | 176439 | Proving Quadratic Formula
purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving [imath]ax^2 + bx + c = 0[/imath]." section. How does [imath]\dfrac{b}{2a}[/imath] become [imath]\dfrac{b^2}{4a^2}[/imath]? |
392027 | Find the limit [imath]\lim_{n\to\infty}\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}[/imath]
Find the limit [imath]\lim_{n\to\infty}\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}[/imath]. Remark:there are n times square root within [imath]n[/imath]. | 48503 | Nested Radicals: [imath]\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}[/imath]
Let [imath]a>0[/imath] . How we can find the limit of : [imath]\sqrt{a+\sqrt{2a+\sqrt{3a+\ldots}}}[/imath] Thanks in advance for your help |
391187 | Lebesgue Integration Proof
I have been trying to teach myself Lebesgue Integration and came across the following question in a textbook. Let [imath]g \subset [0,1][/imath] be bounded and measurable. Suppose that it satisifes the property [imath]\int_{[0,1]} g \chi_{[0,c)}\, d\mu = 0[/imath] for all [imath]c \in [0,1][/imath]. Show that [imath]g=0[/imath] a.e. | 86220 | showing that [imath]g=0[/imath] almost everywhere on [imath][0,1][/imath]
Let [imath]g \in L^1[0,1][/imath]. Suppose that given any pair of rationals [imath]0\leq p\lt q \leq 1[/imath], we have [imath]\int_p^q g(x) d\mu=0.[/imath] Please I would like help in showing that [imath]g=0[/imath] almost everywhere on [imath][0,1][/imath]. |
392204 | [imath]A^3 + A = 0[/imath] then [imath]rank (A) = 2[/imath]
Let [imath]A[/imath] be a [imath]3\times 3[/imath] non-zero real matrix and satisfies [imath]A^3 + A = 0[/imath]. Then prove that [imath]rank (A) = 2[/imath]. As [imath]A[/imath] is satisfying [imath]A^3 + A = 0[/imath], so [imath]0[/imath] is an eigen value of [imath]A[/imath].So [imath]\operatorname{rank} (A) < 3[/imath]. So [imath]\operatorname{rank} (A) = 0,1,\text{or}\, 2[/imath]. Clearly [imath]\operatorname{rank}(A) = 0[/imath] is not possible as [imath]A[/imath] is a non-zero matrix.How to show that [imath]\operatorname{rank} (A) = 1[/imath] is also not possible? | 390052 | [imath]A^3+A=0[/imath] We need to show [imath]\mathrm{rank}(A)=2[/imath]
Let [imath]A\ne 0[/imath] be a [imath]3\times 3[/imath] matrix with real entries such that [imath]A^3+A=0[/imath]. We need to show [imath]\mathrm{rank}(A)=2[/imath]. [imath]\det A(A^2+I)=0\Rightarrow\det A=0\Rightarrow \mathrm{rank}(A)<3[/imath], Suppose [imath]\mathrm{rank}(A)=1[/imath], Then I showed one matrix with rank [imath]1[/imath] which do not satisfies the given relation, is my answer is ok? Thank you for help and discussion |
392195 | Solve a cubic equation?
Need help with solving an equation: Solve the equation [imath]5x^3 - 24x^2 + 9x + 54 = 0[/imath] given that two of it's roots are equal. Any help would be greatly appreciated. Thanks! | 387486 | Solve a cubic polynomial?
I've been having trouble with this question: Solve the equation, [imath]5x^3 - 24x^2 + 9x + 54 = 0[/imath] given that two of its roots are equal. I've tried methods such as Vieta's formula and simultaneous equations, assuming the roots are: [imath]a[/imath], [imath]a[/imath], [imath]b[/imath], but I am still unsuccessful. Any help would be greatly appreciated. |
392189 | A question in Ring theory
R be a commutative ring with unity and it has exactly one maximal ideal. Then prove that the equation [imath]x^2 =x[/imath] has exactly two solutions. Show me the right way to solbe this one. | 391029 | If [imath]R[/imath] is a commutative ring with unity and [imath]R[/imath] has only one maximal ideal then show that the equation [imath]x^2=x[/imath] has only two solutions
If [imath]R[/imath] is a commutative ring with unity and [imath]R[/imath] has only one maximal ideal then show that the equation [imath]x^2=x[/imath] has only two solutions. I know that [imath]0[/imath] and [imath]1[/imath] are the solutions, but I can't proceed from there. |
392306 | Evaluating the improper integral [imath] \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\mathrm dx} [/imath]
I am trying to solve the following integral, but I don't have a solution, and the answer I am getting doesn't seem correct. So I am trying to integrate this: [imath] \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\,\mathrm dx} [/imath] To integrate this, I want to use a contour that looks like a pizza slice, out of a pie of radius R. One edge of this pizza slice is along the positive x-axis, if that makes sense. Since [imath] z^{10} + 1 [/imath] has 10 zeroes, the slice should only be one tenth of a whole circle. So let's call this contour [imath] C [/imath]. Then: [imath] \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = 2 \pi i\,\operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) [/imath] This is because this slice contains only one singularity. Furthermore: [imath] \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = \int_0^R{\frac{z^2}{z^{10} + 1}\,\mathrm dz} + \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} [/imath] And then, by the M-L Formula, we can say that [imath] \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} [/imath] goes to [imath]0[/imath] as [imath]R[/imath] goes to infinity. Evaluating [imath] 2 \pi i\ \operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) [/imath] I get [imath] \dfrac{\pi}{e^{i \pi/5}} [/imath]. Since this answer isn't real, I don't think this could be correct. What did I do wrong? | 110457 | Closed form for [imath] \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }[/imath]
I've been looking at [imath]\int\limits_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }[/imath] It seems that it always evaluates in terms of [imath]\sin X[/imath] and [imath]\pi[/imath], where [imath]X[/imath] is to be determined. For example: [imath]\displaystyle \int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}[/imath] [imath]\int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}[/imath] [imath]\int\limits_0^\infty {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}[/imath] So I guess there must be a closed form - the use of [imath]\Gamma(x)\Gamma(1-x)[/imath] first comess to my mind because of the [imath]\dfrac{{\pi x}}{{\sin \pi x}}[/imath] appearing. Note that the arguments are always the ratio of the exponents, like [imath]\dfrac{1}{4}[/imath], [imath]\dfrac{1}{3}[/imath] and [imath]\dfrac{2}{5}[/imath]. Is there any way of finding it? I'll work on it and update with any ideas. UPDATE: The integral reduces to finding [imath]\int\limits_{ - \infty }^\infty {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} [/imath] With [imath]a =\dfrac{n+1}{m}[/imath] which converges only if [imath]0 < a < 1[/imath] Using series I find the solution is [imath]\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} [/imath] Can this be put it terms of the Digamma Function or something of the sort? |
392245 | Do there exist some non-constant holomorphic functions such that the sum of the modulus of them is a constant
Do there exist some non-constant holomorphic functions [imath]f_1,f_2,\ldots,f_n[/imath]such that [imath]\sum_{k=1}^{n}\left|\,f_k\right|[/imath] is a constant? Can you give an example? Thanks very much | 96257 | Maximum of sum of finite modulus of analytic function.
Let [imath]f_1,f_2,\ldots,f_n [/imath] be analytic complex functions in domain [imath]D[/imath]. and [imath]f = \sum_{k=1}^n|f_k|[/imath] is not constant. Can I show the maximum of [imath]f[/imath] only appears on boundary of [imath]D\,[/imath]? |
392524 | Existence of a graph G?
The Q i am asked is as follows. Q:Let G=(V,E) be a graph with nine vertices such that each vertex has a degree 5 or 6. Show that G has at least 5 vertices of degree 6, or at least 6 vertices of degree 5. first lets bust the middle 5 of6 = 9-5 thus 4of 5 and 6 of 5 = 9-6 hence 3 of 6. so they are claiming that 4 vertex of degree 6 and 5 of degree 5 cannot be a graph. and anything else is fine. clearly [imath]4*6 + 5*5 = 24+25 = 49[/imath] thus [imath]E= 24.5[/imath] which i would like to see you show me half an edge? Now i could draw pictures of all the other graphs ( me being lazy i want to find a better way) or i could ask is the existence of an even sum here sufficient to prove that there exists a simple connected graph? i know that having half an edge is sufficient to disprove the existence but for a set number of vertices if there exists an even number of edges does a graph need to exist? EDIT: ( to downvote and comment on duplicate) My question is for a set number of vertices if there exists an even number of edges does a graph need to exist? | 388768 | Show that a graph with 9 vertices has at least five vertices of degree 6, or at least six vertices of degree 5.
Let [imath]G = (V, E)[/imath] be a graph with nine vertices, such that each vertex has degree [imath]5[/imath] or [imath]6[/imath]. Show that [imath]G[/imath] has at least five vertices of degree [imath]6[/imath], or at least six vertices of degree [imath]5[/imath]. My friend and I have been working on this question for the past two days and we have nothing. Any pointers or help? |
392728 | Prove [imath]\gcd\left((a^{2m}−1)/(a+1),a+1\right)=\gcd(a+1,2m)[/imath]
Show or prove that [imath] \gcd\left(\frac{a^{2m}−1}{a+1},a+1\right)=\gcd(a+1,2m), [/imath] and that [imath] \gcd\left(\frac{a^{2m+1}+1}{a+1},a+1\right)=\gcd(a+1,2m+1). [/imath] | 392649 | Proving two equations involving the greatest common divisor
Show or prove that [imath]\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),[/imath] and that [imath]\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).[/imath] |
393994 | Find the value of [imath]\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}[/imath]
I want to know an objective approach to solve these type of expression in a quick time Which of the expression equals to [imath]\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}[/imath] a)[imath]1-\tan\theta-\cot\theta[/imath] b)[imath]1+\tan\theta-\cot\theta[/imath] c)[imath]1-\tan\theta+\cot\theta[/imath] d)[imath]1+\tan\theta+\cot\theta[/imath] I've tried it several ways like taking LCM,change whole into [imath]\sin\theta[/imath] and [imath]\cos\theta[/imath].but I've stuck. | 385582 | Another trigonometric proof...?
...sigh..another problem how shall I prove the following? [imath] {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A[/imath] so what now? the following's what I've done: [imath]\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A[/imath] |
12989 | A smooth function's domain of being non-analytic
I am wondering how much a smooth function may be non-analytic, because in proofs, whilst there non-analytic smooth functions, it would suffice if a smooth function were analytic on only a "small set". More exactly: Let [imath]U \in \mathbb R^n[/imath] be open, [imath]C^\infty = C^\infty(U,\mathbb R)[/imath]. A smooth function in [imath]C^\infty[/imath] is analytic in [imath]a \in U[/imath], iff there exists [imath]\epsilon > 0[/imath], s.t. the function is equal to its own Taylor series in [imath]B_\epsilon(a)[/imath]. There exist smooth functions that are non-analytic, i.e. there exists [imath]f \in C^\infty, b \in U, \epsilon > 0[/imath] s.t. the function is not its taylor series at [imath]x[/imath] in [imath]B_\epsilon (b)[/imath]. Let [imath]A[/imath] be the union of all [imath]\epsilon[/imath]-Balls in [imath]U[/imath] where [imath]f[/imath] is analytic. By definition, [imath]A[/imath] is open. It's complement [imath]C = A^c[/imath]is the closed set of points where [imath]f[/imath] is non-analytic. Does [imath]C[/imath] have an interior? | 1288304 | an infinitely differentiable function,Taylor's formula does not hold
Give an infinitely differentiable function [imath]g:\mathbb{R}\mapsto\mathbb{R}[/imath] for which Taylor's formula does not hold, i.e, Taylor expansion of [imath]g[/imath] does not equal to [imath]g[/imath]. Prove that this function is indeed [imath]C^\infty[/imath] on [imath]\mathbb{R}[/imath]. Basically, I believe I need to look for a function [imath]g[/imath] whose Remainder term in the Taylor's expansion tends to infinity as [imath]n[/imath] tends to infinity. Is this right? though I cannot come up with such a function by myself. After browsing I came across the function [imath]f(x) = 0, x \le 0, f(x) = e^{-1/x}, x > 0[/imath] is this the right function? |
392179 | Why is this function continuous on [imath] x \in \mathbb{R} [/imath] where [imath] x < 1 [/imath]?
The function is [imath] \exp(\frac{1}{3}\text{log}[(z-1)(z-2)(z-3)]) [/imath] where [imath] \log[(z-1)(z-2)(z-3)] = \int_4^z{\frac{[(z-1)(z-2)(z-3)]'}{[(z-1)(z-2)(z-3)]}dz} + \log 6 [/imath] This function is clearly continuous on [imath] \mathbb{C} \backslash \{ x \in \mathbb{R} : x < 3 \} [/imath] since this is a simply connected domain and [imath] [(z-1)(z-2)(z-3)]' [/imath] and [imath] [(z-1)(z-2)(z-3)] [/imath] don't equal zero anywhere in the domain. I don't see how to to show this is continuous on the real axis for [imath] x \lt 1 [/imath] though. Any thoughts? | 391066 | Making a cube root function analytic on [imath]\mathbb{C}\backslash [1,3][/imath]
I am still not convinced by the post that the function[imath]\sqrt[3]{(z-1)(z-2)(z-3)}[/imath] can be defined so it is analytic on [imath]\mathbb{C}\backslash [1,3][/imath]. We define for each [imath]z\in \mathbb{C}\backslash (-\infty,3][/imath] the function [imath]f(z)=\int_4^z \frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}\,dz +\ln 6[/imath] and it claims that function [imath]\exp\left(\frac{1}{3}f(z)\right)[/imath] is continuous on [imath](-\infty,1)[/imath]. This is the part I don't understand. How do we prove that it is continuous on [imath](-\infty,1)[/imath]. I computed the integral from 4 to [imath]-4[/imath] along the upper semicircle of [imath]|z|=4[/imath], and the integral from 4 to [imath]-4[/imath] along the lower semicircle of [imath]|z|=4[/imath] and they are not equal. So I don't see how the function can be continuous at [imath]-4[/imath], for example. |
394168 | Evaluate the Sum [imath]\sum_{i=0}^\infty \frac {i^N} {4^i}[/imath]
[imath]\displaystyle\sum_{i=0}^\infty \frac {i^N} {4^i}[/imath] I'm supposed to evaluate this as I'm working through Data Structures and Algorithm Analysis in C++. I've solved similar problems, and after trying all of the algebra-fu manipulations I could think of, I'm stumped. My first efforts were, similar to how I solved the three previous problems in the series, to try doing things like setting it equal to S, dividing both sides by four, and subtracting the equations, in various ways in an attempt to cancel terms. However, I'm missing some step to make it happen. | 4317 | Generalizing [imath]\sum \limits_{n=1}^{\infty }n^{2}/x^{n}[/imath] to [imath]\sum \limits_{n=1}^{\infty }n^{p}/x^{n}[/imath]
For computing the present worth of an infinite sequence of equally spaced payments [imath](n^{2})[/imath] I had the need to evaluate [imath]\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x>1.[/imath] The method I used was based on the geometric series [imath]\displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x}[/imath] differentiating each side followed by a multiplication by [imath]x[/imath], differentiating a second time and multiplying again by [imath]x[/imath]. There is at least a second (more difficult) method that is to compute the series partial sums and letting [imath]n[/imath] go to infinity. Question: Is there a closed form for [imath]\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x>1,p\in\mathbb{Z}^{+}\quad ?[/imath] What is the sketch of its proof in case it exists? |
394429 | Why are the p-adic integers a linearly ordered group?
In a previous question, someone suggested the p-adic integers as an example of a non-archimedean linearly ordered group. I'm not sure why these are linearly ordered - specifically, it doesn't seem to me like there would be a total order. E.g. under the 3-adic norm I think [imath]1\leq 4[/imath] and [imath]4\leq 1[/imath] (since [imath]3^0[/imath] is the highest that divides both) yet [imath]1\not = 4[/imath] so it's not total. When you talk about the "p-adic" integers as an ordered group, is it implied that you "mod out" everything which isn't a power of [imath]p[/imath] so that the order is total? | 49990 | The p-adic numbers as an ordered group
So I understand that there is no order on the field of [imath]$p$[/imath]-adic numbers [imath]$\mathbb{Q}_p$[/imath] that makes it into an ordered field (i.e.) compatible with both addition and multiplication. Now, from the responses to a couple of my previous questions, [imath]$\mathbb{Q}_p$[/imath] is a divisible abelian group under addition (being a field of characteristic [imath]0[/imath]). [imath]$\mathbb{Q}_p$[/imath] is torsion-free. It admits an order compatible with the group operation (addition), since every torsion-free abelian group is orderable. My question is, can I write an explicit ordering of [imath]$\mathbb{Q}_p$[/imath] compatible with the group operation? By "explicit", I mean an ordering in which, given two [imath]$p$[/imath]-adic numbers, I can decide which is greater. P.S.: I was not sure how to classify this problem, so please feel free to change the tags. |
367249 | [imath]a^{37} \equiv a \pmod {1729}[/imath]
Use Euler theorem to show that [imath]a^{37} \equiv a \pmod {1729}[/imath] is true for any integer [imath]a[/imath]. I am a beginner in Number theory and stuck in this problem. Show me the right direction. | 340249 | For any integer [imath]a[/imath], [imath]a^{37} \equiv a \left( \text{mod } 1729\right)[/imath]
For any integer [imath]a[/imath], [imath]a^{37} \equiv a \left( \text{mod } 1729\right).[/imath] We're asked to use Euler's Theorem to prove this. What I've tried: [imath]\phi(1729)=\phi(7)\phi(13)\phi(19)=1296[/imath]. If [imath](a,n)=1[/imath] then [imath]a^{1296} \equiv 1 \left( \text{mod } 1729\right)[/imath]. I note that [imath]1296=36^2[/imath] and that something equivalent to what I need to prove is [imath]a(a^{36}-1)\equiv 0 \left( \text{mod } 1729\right).[/imath] I am now wondering what are the conditions for [imath]a^n \equiv b^n \left( \text{mod } m\right)[/imath] implying [imath]a \equiv b \left( \text{mod } m\right)[/imath] so then I can say [imath](a^{36})^{36} \equiv 1 (\text{mod } 1729)[/imath] implies [imath]a^{36} \equiv 1 (\text{mod }1729)[/imath] at which point the equivalent statement would be true. |
395677 | Last non zero digit of [imath]n![/imath]
What is the last non zero digit of [imath]100![/imath]? Is there a method to do the same for [imath]n![/imath]? All I know is that we can find the number of zeroes at the end using a certain formula.However I guess that's of no use over here. | 130352 | Last non Zero digit of a Factorial
I ran into a cool trick for last non zero digit of a factorial. This is actually a recurrent relation which states that: If [imath]D(N)[/imath] denotes the last non zero digit of factorial, then [imath]D(N)=4D\left(\left\lfloor{\frac N5}\right\rfloor\right)\cdot D(\mbox{Units digit of $N$}) \qquad \mbox{(If tens digit of $N$ is odd)}[/imath] [imath]D(N)=6D\left(\left\lfloor{\frac N5}\right\rfloor\right)\cdot D(\mbox{Units digit of $N$}) \qquad \mbox{(If tens digit of $N$ is even)}[/imath] Where [imath]\left\lfloor\cdots\right\rfloor[/imath] is greatest integer function. I was wondering, if anybody could explain why this works? |
395747 | Prove that if [imath]G[/imath] is abelian, then [imath]H = \{a \in G \mid a^2 = e\}[/imath] is subgroup of [imath]G[/imath]
Let [imath]G[/imath] be an abelian group. Prove that [imath]H = \{a \in G \mid a^2 = e\}[/imath] is subgroup of [imath]G[/imath], where [imath]e[/imath] is the neutral element of [imath]G[/imath]. I need some help to approach this question. | 146871 | If [imath]G[/imath] is abelian, then the set of all [imath]g \in G[/imath] such that [imath]g = g^{-1}[/imath] is a subgroup of [imath]G[/imath]
Prove that if [imath]G[/imath] is abelian then the set [imath]H[/imath] of all elements of [imath]G[/imath] that are their own inverses is a subgroup of [imath]G[/imath]. Naturally in an abelian group, [imath]ab = ba[/imath] for [imath]a, b \in G[/imath], however I'm not sure how to show the set elements that are their own inverses is a subgroup of [imath]G[/imath] using arbitrary elements. |
395760 | generalizing De Morgan's Laws
Show: [imath] B - \bigcup_{a \in A}F_{a} = \bigcap_{a \in A} (B - F_{a}) [/imath] and show [imath] B - \bigcap_{a \in A}F_{a} = \bigcup_{a \in A} (B - F_{a}) [/imath] I struggle with proofs. This is what I have for the first proof, going in one direction. If I am correct then the other direction is straight forward. [imath] \text{let } x \in B - \bigcup_{a \in A} F_{a} [/imath] then [imath] x \in B \text{ and } x\notin F_{a} \text{ for all } x \in A[/imath] then [imath] x \in \bigcap_{a \in A}(B - F_{a}) [/imath] | 207570 | Infinite DeMorgan laws
Let [imath]X[/imath] be a set and [imath]\{Y_\alpha\}[/imath] is infinite system of some subsets of [imath]X[/imath]. Is it true that: [imath]\bigcup_\alpha(X\setminus Y_\alpha)=X\setminus\bigcap_\alpha Y_\alpha,[/imath] [imath]\bigcap_\alpha(X\setminus Y_\alpha)=X\setminus\bigcup_\alpha Y_\alpha.[/imath] (infinite DeMorgan laws) Thanks a lot! |
395763 | Rationalizing quotients
I have [imath]\frac{\sqrt{10}}{\sqrt{5} - 2}[/imath] I have no idea what to do, I know that I can do some tricks with splitting square roots up but pulling out whole numbers like I know that [imath]\sqrt{27}[/imath] is just [imath]\sqrt{3*9}[/imath] so I can pull out a nine which becomes a 3. Here though I have no such options, what can I possibly do? | 249563 | Rationalizing the denominator of [imath]\frac {\sqrt{10}}{\sqrt{5} -2}[/imath]
I have the expression [imath]\frac {\sqrt{10}}{\sqrt{5} -2}[/imath] I can't figure out what to do from here, I can't seem to pull any numbers out of either of the square roots so it appears that it must remain as is. |
395715 | Structure Theorem for finitely generated modules over PIDs
Let [imath]A[/imath] be a real [imath]4 \times 4[/imath] matrix. Supose [imath]i,-i[/imath] are the eigenvalues of [imath]A[/imath]. Show that there exists an invertible matrix [imath]P[/imath] such that [imath]PAP^{-1}[/imath] is either [imath]\begin{pmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&1&0&-1\\ 0&0&1&0 \end{pmatrix}, \begin{pmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{pmatrix}.[/imath] My professor gives me some hints: The characteristic polynomial of [imath]A[/imath] is [imath]f_A (x)=(x^2 +1)^2[/imath] . Therefore, by structure theorem of modules over a PID, [imath]Im(A)[/imath] is isomorphic to [imath]\mathbb{R}[x]/((x^2 +1)^2 )[/imath] or [imath]\mathbb{R}[x]/(x^2 +1) \times \mathbb{R}[x]/(x^2 +1)[/imath] ... | 395299 | Two problems about Structure Theorem for finitely generated modules over PIDs
1) Let [imath]A[/imath] be a real 4 by 4 matrix. Supose [imath]i,-i[/imath] are the eigenvalues of [imath]A[/imath]. Show that there exists an invertible matrix [imath]P[/imath] such that [imath]PAP^{-1}[/imath] is either [imath]\begin{pmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&1&0&-1\\ 0&0&1&0 \end{pmatrix}, \begin{pmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{pmatrix}.[/imath] 2) Let [imath]A= \begin{pmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{pmatrix} [/imath]. Find an integer [imath]r>0[/imath] and positive integers [imath]1<d_1\mid d_2\mid\cdots\mid d_s[/imath] such that [imath]\mathbb{Z}^3/A\mathbb{Z}^3[/imath] is isomorphic to [imath]\mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_s\mathbb{Z} \times \mathbb{Z}^r[/imath]. I think both problems have something to do with the Structure Theorem for finitely generated modules over PIDs, but I am not familiar with that, would someone please help. Actually I don't even understand what is meant by [imath]\mathbb{Z}^3/A\mathbb{Z}^3[/imath]... |
395223 | Find the factorization of the polynomial as a product of irreducible
Find the factorization of the polynomial [imath]x^5-x^4+8x^3-8x^2+16x-16[/imath] as a product of irreducible on rings [imath]R[x][/imath] and [imath]C[x][/imath] Testing with the simplest possible root in this case, [imath]P(1)=0[/imath] Applying the schema of Ruffini [imath] \begin{array}{c|lcr} & 1 & -1 & 8 & -8 & 16 & -16 \\ 1 & & 1 & 0 & 8 & 0 & 16 \\ \hline & 1 & 0 & 8 &0 &16 & 0 \\ \end{array} [/imath] [imath]P(x)=(x-1).(x^4+8x^2+16)[/imath] [imath]P(x)=(x-1).(x^2+4)^2 \rightarrow Irreducible[/imath] [imath]in[/imath] [imath]R[x][/imath] So, I must to factorize this to make it irreducible in [imath]C[x][/imath] Applying Bhaskara in [imath](x^2+4)^2[/imath] [imath]a=1; b=0; c=4[/imath] [imath]\frac{\pm\sqrt{-4.1.4}}{2}=\frac{\pm\sqrt{-16}}{2}= R_1, R_2[/imath] [imath]R_1=-2i[/imath] [imath]R_2=2i[/imath] So (...) [imath]P(x)= (x-2i).(x+2i).(x-1) \rightarrow Irreducible [/imath] [imath]in [/imath] [imath]C[x][/imath] this is well done? | 395115 | Find the factorization of the polynomial as a product of irreducible on rings R[x] and C[x]
Find the factorization of the polynomial [imath]x^5-x^4+8x^3-8x^2+16x-16[/imath] as a product of irreducible on rings [imath]\Bbb R[x][/imath] and [imath]\Bbb C[x][/imath] Testing with the simplest possible root in this case, [imath]P(1) = 0[/imath] Applying the schema of Ruffini [imath] \begin{array}{c|lcr} & 1 & -1 & 8 & -8 & 16 & -16 \\ 1 & & 1 & 0 & 8 & 0 & 16 \\ \hline & 1 & 0 & 8 &0 &16 & 0 \\ \end{array} [/imath] [imath]P(x)=(x-1).(x^4+8x^2+16)[/imath] So, I need to make it irreducible in [imath]R[x][/imath], there's no more roots, but I don't know how to reduce it, I tried factoring: [imath]x^4+8.(x^2+2).(x-1)[/imath], but I can not keep working it. What I can do to solve it? |
396323 | analysis: limit of product of sequences
I would really appreciate help with this question: Show that if the limit [imath]\lim_{n \to \infty} {a_n} = 0[/imath] and sequence [imath]\{b_n\}[/imath] is bounded, then [imath]\lim_{n\to \infty} a_nb_n =0[/imath] thanks | 9649 | If [imath]b_n[/imath] is a bounded sequence and [imath]\lim a_n = 0[/imath], show that [imath]\lim(a_nb_n) = 0[/imath]
This is a real-analysis homework question so I of course have to be very precise and justify anything or any theorem I use. If [imath]b_n[/imath] is a bounded sequence and [imath]\lim(a_n) = 0[/imath], show that [imath]\lim(a_nb_n) = 0[/imath] Intuitively, since [imath]b_n[/imath] is bounded, then sup([imath]b_n[/imath]) is some finite number and therefore we can take an [imath]N[/imath] natural number as large as we need such that for all [imath]n\gt N[/imath] [imath]b_na_n[/imath] approaches [imath]0[/imath]. At first I thought to use the limit theorems, but since [imath]a_n[/imath] is not bounded, the general limit theorems do not reply. (I am referring to [imath]\lim(X + Y) = \lim X + \lim Y[/imath] for [imath]X,Y[/imath] sequences etc). I was thinking then to use the definition of the limit somehow to show that since [imath]b_n[/imath] is bounded we can take as intuitively stated above [imath]N[/imath] large enough to show the statement is true. I'm not sure how to proceed with this. Thank you for your replies in advance! |
396420 | If [imath]a^3=a[/imath] in a ring, prove: the ring is commutative
Let [imath]R[/imath] be a ring, not necessarily with a unit element. [imath]R[/imath] is not necessarily integral. If for any [imath]a \in R[/imath], [imath]a^3=a[/imath], prove: [imath]R[/imath] is commutative: Any [imath]a, b \in R[/imath], [imath]ab=ba[/imath]. My efforts on it: I can use the relation [imath](a+b)^3=a+b[/imath] and [imath](a-b)^3=a-b[/imath]. But the relation is cubic and I do not know how to reduce it to a relation about [imath]ab[/imath] and [imath]ba[/imath]. | 185086 | Name a ring of 2 by 2 matrices where [imath]a^3 = a[/imath] and a belonging to this ring?
I need an example of a ring consisting of 2 by 2 matrices where [imath]a^3=a[/imath] with [imath]a[/imath] belonging to this ring. If someone can list the elements I would be satisfied. What I'm trying to get at it is conceptualize why a ring [imath]R[/imath] is always commmuative when [imath]a^3=a[/imath]. I know of one such example and that is the factor ring [imath]\mathbb{Z}/3\mathbb{Z}.[/imath] Does anyone know how to prove this statement mathematically as well as giving me an example of a ring of 2 by 2 matrices? |
396421 | Method to partial fractions.
For Example: [imath]\frac{ax^2 + bx+c}{(dx+e)(fx^2+g)}\equiv\frac{A}{dx+e}+\frac{Bx+C}{fx^2+g}[/imath] and [imath]\frac{ax^4 + bx^3+cx^2+dx+e}{(x+f)(x^2+g)}\equiv Ax+B+\frac{C}{x+f}+\frac{Dx+E}{x^2+g}[/imath] How do you know how to format the right hand side, in the equation below in partial fractions? (what is the 'pattern'?) [imath]\frac{2ax}{(x-2a)(x^2+a^2)}[/imath] where a is a non zero constant Could you look at it as the top of the fraction being a simple derivative of the bottom? If so how would you know to add the Ax+B in the latter example? | 368665 | Derivation of the general forms of partial fractions
I'm learning about partial fractions, and I've been told of 3 types or "forms" that they can take (1) If the denominator of the fraction has linear factors: [imath]{5 \over {(x - 2)(x + 3)}} \equiv {A \over {x - 2}} + {B \over {x + 3}}[/imath] (2) If the denominator of the fraction has quadratic factors that don't factorise: [imath]{{2x + 3} \over {(x - 1)({x^2} + 4)}} \equiv {A \over {x - 1}} + {{Bx + C} \over {{x^2} + 4}}[/imath] (3) If the denominator has a factor that repeats: [imath]{{5x + 3} \over {(x - 2){{(x + 3)}^2}}} \equiv {A \over {x - 2}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}[/imath] I'd appreciate it if someone could explain to me how these equivalence relationships are derived, essentially how do they take the forms that they do? I sort of understand the first one, but the other two I don't. Thank you |
396388 | Infinite Series question
The first one, the effective resistance is [imath]2R[/imath], then [imath]5R/3[/imath] then [imath]13R/8[/imath] etc.... My job is to find the pattern/equation so I can find the total resistance when [imath]20[/imath] resistors are connected. Of course I can do the long way by using this formula I created [imath]R_n = (R_{n-1} \times R)/(R_{n-1} +R) +R[/imath]. Can you guys help me find an equation where I can just input the value of [imath]n[/imath] and simply get an answer. Thanks! The second question is: What would I get if I had an infinite number of resistance. Help is very much appreciated :) | 353261 | The limit of a recurrence relation (with resistors)
Background to problem (not too important): My proposed solution: The infinitely long element, , however complex, can be represented as a single resistor of resistance [imath]R[/imath]. Remembering the initial resistor near [imath]A[/imath], we know [imath]R_{AB}= r+R[/imath]. However, as this is an infinitely long element, it is equivalent to a resistor of resistance [imath]R[/imath] attached to the right of two resistors of resistance [imath]r[/imath] (the resistance [imath]R[/imath] is an intrinsic property of the element, so is unaffected by the fact that the further to the right it is, the lower the current passing through it). Thus, taking [imath]R[/imath] in series with [imath]r[/imath], then the result in parallel with [imath]r[/imath], then in series with [imath]r[/imath]: [imath]R_{AB}= r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}} \right )[/imath] On the second iteration (moving [imath]R[/imath] further to the right): [imath]R_{AB}= r+ \frac{1}{\frac{1}{r}+\frac{1}{r + \left (\frac{1}{\frac{1}{R+r}+\frac{1}{r}} \right )}}[/imath] Ad infinitum. I understand this may not be the fastest solution, but I'd like to know a little more about it nonetheless. The mathematics [imath]u_{1}=r+R[/imath] [imath]\large u_{n+1}=r+\frac{1}{\frac{1}{r}+\frac{1}{u_n}}[/imath] Does [imath]\lim_{n \rightarrow \infty} (u_n)[/imath] exist (important: is the limit a function of [imath]R[/imath]?), and, if so, what is it? First cases [imath]u_{2}=\frac{3r^2+2rR}{R+2r}[/imath] [imath]u_{3}=\frac{8r^2+5rR}{3R+5r}[/imath] [imath]u_{4}=\frac{21r^2+13rR}{8R+13r}[/imath] [imath]u_{4}=\frac{55r^2+34rR}{21R+34r}[/imath] [imath]\lim _{n \rightarrow \infty} (u_n)\stackrel{?}{=}\varphi r[/imath] Seems the doing of Fibonnaci. How does one take the limit of this (I assume it requires knowledge of knowledge of the form of [imath]f(n)=F_n[/imath]). Intuitively, why does Fibonnaci appear here? What are the rabbits in this case? Link to Wolfram's computation. |
17747 | Why a complete graph has [imath]\frac{n(n-1)}{2}[/imath] edges?
I'm studying graphs in algorithm and complexity, (but I'm not very good at math) as in title: Why a complete graph has [imath]\frac{n(n-1)}{2}[/imath] edges? And how this is related with combinatorics? | 199695 | How many edges are in a clique of n vertices?
I believe the answer is [imath]\frac12(n-1)^2[/imath], but I couldn't confirm by googling, and I'm not confident in my ability to derive the formula myself. |
396764 | What is [imath]2 - 1 + 1[/imath]?
[imath]2-1+1[/imath]; a fairly straightforward question, but I (well, not me, but Henry Reich) found something strange. Most people would evaluate it as [imath]2+(-1)+1 = 2[/imath]; however, this goes against the famed, and fairly standard B.E.D.M.A.S./P.E.D.M.A.S., which states that addition goes first, and then subtraction. If this is the case, then the answer is [imath]2 - (1 + 1) = 2-2 =0[/imath]. Which is the correct answer, and why is the conventional way (B.E.D.M.A.S./P.E.D.M.A.S.) so ambiguous? | 395985 | Order of operations (BODMAS)
[imath]40-20/2+15\times1.5\\\hspace{.1in}\\40-20/2+15\times1.5=\\ 40-10+22.5=7.5[/imath] I'm studying and this is from an example. In BODMAS, aren't addition and subtraction have same level? So, in the 3rd line, it should be from left to right, correct? |
396800 | Infinite Series Problem Using Residues
Show that [imath]\sum_{n=0}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\coth\pi a+\frac{1}{2a^2}, a>0[/imath] I know I must use summation theorem and I calculated the residue which is: [imath]Res\left(\frac{1}{z^2+a^2}, \pm ai\right)=-\frac{\pi}{2a}\coth\pi a[/imath] Now my question is: how do I get the last term [imath]+\frac{1}{2a^2}[/imath] after using the summation theorem? | 208317 | Show [imath]\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}[/imath]
How to show the following equality? [imath]\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}[/imath] |
396849 | How does [imath]\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1[/imath] result into 1+3+5...
How can [imath]\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1[/imath] result into 1+3+5...? Why odd numbers? | 136237 | Direct Proof that [imath]1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n[/imath]
Prove that for all integers [imath]n[/imath], [imath]n \geq 1[/imath], [imath]1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n[/imath] How would I go about proving this? |
396969 | Definition of Exact functors
As we know by definition, a functor for example [imath]T\colon R\textrm{-}\mathsf{Mod}\to\mathsf{Ab}[/imath] (from the category of [imath]R[/imath]-modules to the category of abelian groups) is "exact" precisely when for any short exact sequence: [imath]0 \to A \to B\to C\to 0[/imath] the sequence [imath]0\to T(A) \to T(B) \to T(C) \to 0[/imath] is also exact. My question is: Is this equivalent to the condition where these sequences do not begin with zero? In other words can we say [imath]T\colon R\textrm{-}\mathsf{Mod}\to\mathsf{Ab}[/imath] is exact precisely when for any exact sequence: [imath]A \to B \to C,[/imath] the sequence [imath]T(A) \to T(B) \to T(C)[/imath] is also exact? Obviously, the latter implies the exactness definition, but I am not yet convinced about the other direction. | 157783 | Equivalent definition of exactness of functor?
I'll use the following definition: (Def) A functor [imath]F[/imath] is exact if and only if it maps short exact sequences to short exact sequences. Now I'd like to prove the following (not entirely sure it's true but someone mentioned something like this to me some time ago): Claim: [imath]F[/imath] is exact if and only if it maps exact sequences [imath]M \to N \to P[/imath] to exact sequences [imath]F(M) \to F(N) \to F(P)[/imath] Proof: [imath]\Longleftarrow[/imath]: Let [imath]0 \to M \to N \to P \to 0[/imath] be exact. Then [imath]0 \to M \to N[/imath], [imath]M \to N \to P[/imath] and [imath]N \to P \to 0[/imath] are exact and hence [imath]0 \to F(M) \to F(N) [/imath], [imath]F(M) \to F(N) \to F(P)[/imath] and [imath]F(N) \to F(P) \to 0[/imath] are exact. Hence [imath]0 \to F(M) \to F(N) \to F(P) \to 0[/imath] is exact. [imath]\implies[/imath]: This is direction I'm stuck with. I am trying to do something like this: Given [imath]M \to N \to P[/imath] exact, we have that [imath]0 \to ker(f) \to M \to im(f) \to 0[/imath] is exact. Hence [imath]0 \to F(ker(f)) \to F(M) \to F(im(f)) \to 0[/imath] is exact. Then I want to do this again for the other side of the sequence and stick it back together after applying [imath]F[/imath] to get the desired short exact sequence. How does this work? Perhaps I need additional assumptions on [imath]F[/imath]? Thanks for your help. |
397813 | Multiplicatives
Let [imath]f: N \to N[/imath], [imath]f(2) = 3[/imath], and [imath]f(ab) = f(a)f(b)[/imath], that is, f is a multiplicative function. f is also strictly increasing. Show that no such function exists. Progress: Apparently, this is proven by contradiction. So I used [imath]f(2) = f(2 * 1) = f(2) * f(1)[/imath]. This gives me [imath]3 = 3 * f(1)[/imath], which means [imath]f(1) = 1[/imath]. I see no contradiction though. f is strictly increasing, not decreasing, so it makes sense that [imath]f(1) = 1[/imath] but [imath]f(2) = 3[/imath]. Help please? | 154793 | Nonexistence of a strongly multiplicative increasing function with [imath]f(2)=3[/imath]
Show that there does not exist a strictly increasing function [imath]f:\mathbb{N}\rightarrow\mathbb{N}[/imath] satisfying [imath]f(2)=3[/imath] [imath]f(mn)=f(m)f(n)\forall m,n\in\mathbb{N}[/imath] Progress: Assume the function exists. Let [imath]f(3)=k[/imath] Since [imath]2^3 < 3^2[/imath], [imath]3^2=f(2)^3=f(2^3)<f(3^2)=f(3)^2=k^2[/imath] so [imath]k>5[/imath] and since [imath]3^3 < 5^2[/imath], then [imath]k^3=f(3)^3=f(3^3)<f(2^5)=f(2)^5=3^5=243<343=7^3[/imath] so [imath]k<7[/imath] therefore [imath]k=6[/imath]. I've messed around with knowing [imath]f(3)=6[/imath] and [imath]f(2)=3[/imath] but I am stuck. |
4400 | Boy Born on a Tuesday - is it just a language trick?
The following probability question appeared in an earlier thread: I have two children. One is a boy born on a Tuesday. What is the probability I have two boys? The claim was that it is not actually a mathematical problem and it is only a language problem. If one wanted to restate this problem formally the obvious way would be like so: Definition: Sex is defined as an element of the set [imath]\\{\text{boy},\text{girl}\\}[/imath]. Definition: Birthday is defined as an element of the set [imath]\\{\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday},\text{Sunday}\\}[/imath] Definition: A Child is defined to be an ordered pair: (sex [imath]\times[/imath] birthday). Let [imath](x,y)[/imath] be a pair of children, Define an auxiliary predicate [imath]H(s,b) :\\!\\!\iff s = \text{boy} \text{ and } b = \text{Tuesday}[/imath]. Calculate [imath]P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y))[/imath] I don't see any other sensible way to formalize this question. To actually solve this problem now requires no thought (infact it is thinking which leads us to guess incorrect answers), we just compute [imath] \begin{align*} & P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y)) \\\\ =& \frac{P(x\text{ is a boy and }y\text{ is a boy and }(H(x)\text{ or }H(y)))} {P(H(x)\text{ or }H(y))} \\\\ =& \frac{P((x\text{ is a boy and }y\text{ is a boy and }H(x))\text{ or }(x\text{ is a boy and }y\text{ is a boy and }H(y)))} {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\ =& \frac{\begin{align*} &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday}) \\\\ + &P(x\text{ is a boy and }y\text{ is a boy and }y\text{ born on Tuesday}) \\\\ - &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday and }y\text{ born on Tuesday}) \\\\ \end{align*}} {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\ =& \frac{1/2 \cdot 1/2 \cdot 1/7 + 1/2 \cdot 1/2 \cdot 1/7 - 1/2 \cdot 1/2 \cdot 1/7 \cdot 1/7} {1/2 \cdot 1/7 + 1/2 \cdot 1/7 - 1/2 \cdot 1/7 \cdot 1/2 \cdot 1/7} \\\\ =& 13/27 \end{align*} [/imath] Now what I am wondering is, does this refute the claim that this puzzle is just a language problem or add to it? Was there a lot of room for misinterpreting the questions which I just missed? | 87436 | The Tuesday Birthday Problem - why does the probability change when the father specifies the birthday of a son?
I've most recently read about the Tuesday Boy Problem via twitter and I, as probably most other people, was sure that the probability has to be 1/2. After having read through a lot of solutions which were not identical at all, I've come to the conclusion that [imath]P = \frac{13}{27}[/imath] sounds the most reasonable. The corresponding argument was as follows: Say the older child is the boy born on Tuesday. Then, if the younger child is female, there are seven possibilities and analogous if the younger child is male. In case the younger child is the boy born on Tuesday, for an older daughter, again there are seven possibilities, but for an older son, there are only six, because the case that the older son was born on Tuesday has already been counted. Thus, [imath]P = \frac{6+7}{6+7+7+7} = \frac{13}{27}. [/imath] My first question is: Is this the correct solution? I've found other websites giving different solutions, however I could never agree with any of those. In case this is the correct solution: Why does the probability change when the father specifies the birthday of a son? (does it actually change? A lot of answers/posts stated that the statement does matter) What I mean is: It is clear that (in case he has a son) his son is born on some day of the week. I could replace Tuesday with any day of the week and the probability would be the same. Say the father would have stated: I have two children. (At least) One of them is a son. Then, without loss of generality, we could say that this son is born on a Tuesday and again we would have [imath]P = \frac{13}{27}[/imath]. But looking at an equivalent (?) problem, we get a completely different probability: Let's toss two coins consecutively and say heads is equivalent to "son" and tails to "daughter". Then, if we know that we tossed at least one heads, if it was the first tossed coin, the probability for another heads is 1/2. If the second coin was heads, then we can only take the case in account where the first coin was tails. So the probability for two heads is [imath]P = \frac{1}{3}[/imath]. So my second (actually third) question is: Where did I go wrong? Lastly I want to ask (as my knowledge in probability/statistics is limited to what I have been taught in high school) whether the argument which gives [imath]P=\frac{13}{27}[/imath] for the original Tuesday Birthday Problem already takes the possibility of twins in consideration. Can it be regarded as a special case of two boys both born on Tuesday (I ask because in every argument leading to this probability there was a distinction between older child = boy vs. younger child = boy)? Thank you very much in advance for any answer. |
397743 | Why is [imath]2\pi i \neq 0?[/imath]
We know that [imath]e^{\pi i} = -1[/imath] because of de Moivre's formula. ([imath]e^{\pi i} = \cos \pi + i\sin \pi = -1).[/imath] Suppose we square both sides and get [imath]e^{2\pi i} = 1[/imath](which you also get from de Moivre's formula), then shouldn't [imath]2\pi i=0[/imath]? What am I missing here? | 281528 | What is wrong with this fake proof [imath]e^i = 1[/imath]?
[imath]e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1[/imath] Obviously, one of my algebraic manipulations is not valid. |
398124 | show [imath]\sum_{n=0}^{\infty}{z^n\over n}[/imath] is convergent on the unit circle
I need to show [imath]\sum_{n=0}^{\infty}{z^n\over n}[/imath] is convergent on the unit circle except the point [imath]z=1[/imath], well at [imath]z=1[/imath] we get our known divergent harmonic series, but I am not able to show easily the other part. | 118686 | Determining precisely where [imath]\sum_{n=1}^\infty\frac{z^n}{n}[/imath] converges?
Inspired by the exponential series, I'm curious about where exactly the series [imath]\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}[/imath] for [imath]z\in\mathbb{C}[/imath] converges. I calculated [imath] \limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}} [/imath] and [imath] \lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1. [/imath] So the radius of convergence is [imath]1[/imath], so the series converges on all [imath]z[/imath] inside [imath]S^1[/imath]. But is there a way to tell for which [imath]z[/imath] on the unit circle the series converges? I know it converges for [imath]z=-1[/imath], but diverges for [imath]z=1[/imath], but I don't know about the rest of the circle. For what other [imath]z[/imath] does this series converge? Thanks. |
398144 | Open Cover / Real Analysis
I have the next question: Let [imath]K \subset [/imath] [imath]R^1[/imath] consist of [imath]0[/imath] and the numbers 1/[imath]n[/imath], for [imath]n=1,2,3,\ldots[/imath] Prove that [imath]K[/imath] is compact directly from the definition (without using Heine-Borel). I'm trying to understand compact sets so I would be grateful if someone could give me some examples of open covers and subcovers. Thank you! | 71333 | Proving that [imath]S=\{\frac{1}{n}:n\in\mathbb{Z}\}\cup\{0\}[/imath] is compact using the open cover definition
Let [imath]S=\{\frac{1}{n}:n\in\mathbb{Z}\}\cup\{0\}[/imath] be a subset of [imath]\mathbb{R}[/imath]. I have to prove using the open cover definition that this is compact. Could you help me, please? |
398158 | which of the followings are positive definite:
(CSIR-UGC NET Mathematical Sciences-2011) Suppose [imath]A,B[/imath] are [imath]n \times n[/imath] positive definite. Then which of the followings are positive definite: [imath]A+B[/imath] [imath]ABA^{*}[/imath] [imath]A^2+I[/imath] [imath]AB[/imath] | 158559 | which are positive definite matrix
Given that [imath]A,B[/imath] are positive definite matrix, are they also PD? [imath]A+B[/imath] [imath]AB[/imath] [imath]A^2 +I[/imath] [imath]ABA^{*}[/imath] [imath]x^TAx>0, x^TBx>0[/imath] so [imath]1[/imath], is correct, could you tell me about the 2, 3,4? |
398165 | [imath]n[/imath] divides [imath]\phi(a^n -1)[/imath] where [imath]a, n[/imath] are positive integer.
Let [imath]n[/imath] and [imath]$a$[/imath] be positive integers with [imath]$a > 1$[/imath]. I need to show that [imath]$n$[/imath] divides [imath]\phi(a^n -1)[/imath]. Here, [imath]$\phi$[/imath] denotes the Euler totient function. Could any one give me a hint? | 1956537 | Show that [imath]n[/imath] divides [imath]\phi(a^n -1)[/imath]
Show that [imath]n[/imath] divides [imath]\phi(a^n -1)[/imath] for any [imath]a>n[/imath] where [imath]\phi[/imath] is Euler's totient function. I would really appreciate a hint. |
396084 | question about idempotent operators
Let [imath]H[/imath] be a Hilbert space and [imath]\dim(H)=\infty[/imath]. If each [imath]T\in B(H)[/imath] is finite sum of idempotent operators? If each [imath]T\in B(H)[/imath] is infinite sum of idempotent operators? | 162214 | Every Hilbert space operator is a combination of projections
I am reading a paper on Hilbert space operators, in which the authors used a surprising result Every [imath]X\in\mathcal{B}(\mathcal{H})[/imath] is a finite linear combination of orthogonal projections. The author referred to a 1967 paper by Fillmore, Sums of operators of square zero. However, this paper is not online. I wonder whether someone has a hint on how this could be true since there are all kinds of operators while projections have such a regular and restricted form. Thanks! |
398382 | CRT and systems of modular equations
The formula I found: [imath]\sum_{i=1}^{k} a_{i}b_{i}b^{`}_{i} (\mod m_{i})[/imath] where: [imath]b_{i} = \frac{M}{m_{i}}[/imath] [imath]b_{i}^{`} = b_{i}^{-1} (mod m_{i})[/imath] And for example: [imath]x \equiv -7 \mod 13[/imath] [imath]x \equiv 39 \mod 15[/imath] [imath]M = 13*15 = 195[/imath] [imath]b_{1} = 13[/imath] [imath]b_{2} = 15[/imath] [imath]b_{1}^{`} = \frac{1}{13} mod 13? = \frac{1}{13}?[/imath] [imath]b_{2}^{`} = \frac{1}{15} mod 15? = \frac{1}{15}?[/imath] It would make [imath]x = (a_{1} mod 13) + (a_{2} mod 15) = -7 + 9 = 2[/imath], which is wrong. What am I missing? Maybe the formula is wrong? | 398327 | System of two simple modular equations
[imath]x \equiv -7 \mod 13[/imath] [imath]x \equiv 39 \mod 15[/imath] I need to find the smallest x for which these equations can be solved. I've been always doing this using Chinese Reminder Theorem, but it seems that it doesn't work here, I'm not sure why though. Let's see: [imath]N = 13*15 = 195[/imath] [imath]N_{1} = \frac{195}{13} = 13[/imath] [imath]N_{2} = \frac{195}{15} = 15[/imath] [imath]GCD(13, 13) = 1*13 + 0*13[/imath] [imath]GCD(15,15) = 1*15 + 0*15[/imath] [imath]x = 1 * (-7) * 13 + 1 * 39 * 15 = 480[/imath] The answer however is supposed to be [imath]x = 84[/imath]...What's wrong? |
398399 | About the property of [imath]m[/imath]: if [imath]n < m[/imath] is co-prime to [imath]m[/imath], then [imath]n[/imath] is prime
The number [imath]30[/imath] has a curious property: All numbers co-prime to it, which are between [imath]1[/imath] and [imath]30[/imath] (non-inclusive) are all prime numbers! I tried searching(limited search, of course) for numbers [imath]\gt 30[/imath] that have this property, but could not find any. Are there any such numbers [imath]\gt 30[/imath]? | 245702 | Let n>30. Then prove there exists a natural number [imath] 1< m\leq n[/imath] such that [imath](n,m)=1[/imath] and [imath]m[/imath] is not prime.
Let [imath]n>30[/imath]. Then prove there exists a natural number [imath]1<m\leq n[/imath] such that [imath](n,m)=1[/imath] and [imath]m[/imath] is not prime. [imath](m,n)[/imath] denotes the greatest common divisor of [imath]m[/imath] and [imath]n[/imath]. Thanks |
399183 | [imath]A\in\mathbb C^{2\times 2}[/imath] be such that [imath]A^3=A.[/imath] Then the number of such [imath]A[/imath] is infinite.
Verify: [imath]A\in\mathbb C^{2\times 2}[/imath] be such that [imath]A^3=A.[/imath] Then the number of such [imath]A[/imath] is infinite. My attempt: Choose a nonsingular [imath]P\in\mathbb C^{2\times 2}.[/imath] And let [imath]A=P\begin{pmatrix}1&0\\0&-1\end{pmatrix}P^{-1}.[/imath] Then [imath]A^3=A.[/imath] So infinite such [imath]A[/imath] can occur. Am I correct? | 243161 | Are there infinitely many [imath]A\in \mathbb{C}^{2 \times 2}[/imath] satisfying [imath]A^3 = A[/imath]?
Let [imath]A[/imath] be a [imath]2 × 2[/imath]-matrix with complex entries. The number of [imath]2 × 2[/imath]-matrices [imath]A[/imath] with complex entries satisfying the equation [imath]A^3 = A[/imath] is infinite. Is the above statement true? I know that [imath]0[/imath] and I are two solutions. But are there any more solutions? |
399178 | Prove that [imath]f(f(x))=x[/imath] has no roots .... [imath]f[/imath] having a general form
This problem gave me some headache, especially because [imath]f[/imath] have its own general form : let [imath]f(x) = ax^2 + bx + c[/imath]. Suppose that [imath]f(x) = x[/imath] has no real roots. Show that equation [imath]f(f(x))=x[/imath] has also no real roots. | 370151 | If [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is continuous and [imath]f(x)\neq x[/imath] for all [imath]x[/imath], must it be true that [imath]f(f(x))\neq x[/imath] for all [imath]x[/imath]?
Let [imath]f: \Bbb R → \Bbb R[/imath] be a continuous function such that [imath]f(x)=x[/imath] has no real solution . Then is it true that [imath]f(f(x))=x[/imath] also has no real solution ? |
399238 | About Uniform Convergence of [imath]\sum_{n=1}^\infty\frac{\sin nx}{n}[/imath] on [imath][0,2\pi][/imath]
Is [imath]\sum_{n=1}^\infty\dfrac{\sin nx}{n}[/imath] uniform convergent on [imath][0,2\pi][/imath]? I think it is not. However, I could not prove it by Cauchy's criterion. | 28830 | Does [imath]\sum{\frac{\sin{(nx)}}{n}}[/imath] converge uniformly for all [imath]x[/imath] in [imath][0,2\pi][/imath]
This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this. Given [imath]a_n[/imath] a decreasing sequence of positive numbers and suppose that [imath]\sum_{n=0}^{\infty}{a_n \sin{(nx)}}[/imath] Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that [imath]n a_n \to 0[/imath]. Clearly [imath]\frac{1}{n}[/imath] fits the description of [imath]a_n[/imath], and [imath]n \frac{1}{n} \to 1 \neq 0[/imath], so this would prove that there is a mistake in the problem if [imath]\sum{\frac{\sin{(nx)}}{n}}[/imath] converge uniformly for all [imath]x[/imath]. So my question is if [imath]\sum{\frac{\sin{(nx)}}{n}}[/imath] converge uniformly for every [imath]x[/imath]. (I know that the series converge uniformly for every x in [imath][\delta, 2\pi - \delta][/imath], for [imath]0 < \delta <2\pi[/imath] by using the Dirichlet criterion.) |
116976 | I want to prove the following limit [imath] \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 [/imath] using the definition of limit.
I want to prove the following limit \begin{equation} \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 \end{equation} using the definition of limit. Then: \begin{equation} \left| \frac{\sin x}{x}-1 \right|=\left|\frac{\sin x}{ x}\right| \left|1 -\frac{x}{\sin x} \right|\leq \left|1 -\frac{x}{\sin x} \right| \end{equation} Also \begin{equation} \left|\frac{x}{\sin x}-1 \right|\leq \left|\frac{1}{\cos x}-1 \right|=\left|\frac{\sin^{2} x}{\cos x(1+\cos x)}\right| \end{equation} noting that [imath]\cos x>0[/imath] in the neighbourhood [imath](-\delta_{\epsilon},\delta_{\epsilon})[/imath], I have: \begin{equation} \leq \left|\frac{\sin^{2} x}{\cos x}\right|=\left| \sin x \right| \left| \tan x \right|\leq \left| \tan x \right|\leq \tan \delta_{\epsilon} \leq \epsilon \end{equation} it is verified for [imath]\delta_{\epsilon}=arc\tan(\frac{\varepsilon}{2})[/imath] (for example) Can anyone "see" a simpler solution (using the definition of limit)? | 285313 | Prove [imath]\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1[/imath] with the epsilon-delta definition of limit.
It is well known that [imath]\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1[/imath] I know several proofs of this: the geometric proof shows that [imath]\cos(\theta)\leq\frac {\sin(\theta)}{\theta}\leq1[/imath] and using the Squeeze Theorem I conclude that [imath]\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1[/imath], other proof uses the Maclaurin series of [imath]\sin(x)[/imath]. My question is: is there a demonstration of this limit using the epsilon-delta definition of limit? |
320965 | Given [imath]f:S\to T[/imath] and [imath]g:T\to S[/imath], prove [imath]\exists A\subseteq S, B\subseteq T[/imath] such that [imath]f(A)=B[/imath] and [imath]g(T\setminus B)=S\setminus A[/imath]
The question: Let [imath]S[/imath] and [imath]T[/imath] be sets and let [imath]f: S \to T[/imath] and [imath]g: T \to S[/imath] be arbitrary functions. Prove that there is a subset [imath]A \subset S[/imath] and [imath]B \subset T[/imath] such that [imath]f \left( A \right) = B[/imath] and [imath]g \left( T \setminus B \right )= S \setminus A [/imath] My issue is that I read the problem as a statement, not a proof. I tried writing it out but I'm still stuck: [imath] \exists \left( A \subset S , B \subset T \right) \left( f \left( A \right) = B \wedge g \left( T \setminus B \right) = S \setminus A \right) [/imath] How do I go about proving this thing? | 22581 | Do there exist sets [imath]A\subseteq X[/imath] and [imath]B\subseteq Y[/imath] such that [imath]f(A)=B[/imath] and [imath]g(Y-B)=X-A[/imath]?
This is a little exercise I've been fiddling with for a while now. Let [imath]f\colon X\to Y[/imath] and [imath]g\colon Y\to X[/imath] be functions. I want to show that there are subsets [imath]A\subseteq X[/imath] and [imath]B\subseteq Y[/imath] such that [imath]f(A)=B[/imath] and [imath]g(Y-B)=X-A[/imath]. Of course, if [imath]f[/imath] is surjective, then taking [imath]A=X[/imath], one has [imath]f(A)=B=Y[/imath] and [imath]g(Y-Y)=g(\emptyset)=\emptyset=X-X=X-A[/imath], and you're done. So I suppose [imath]f[/imath] is not surjective. I tried to approach it by contradiction. It seems that for any [imath]A\subseteq X[/imath], there is obviously a [imath]B\subseteq Y[/imath] such that [imath]f(A)=B[/imath], so if the result is not the case, for each pair of subsets [imath]A[/imath] and [imath]f(A)[/imath], we must have [imath]g(Y-f(A))\neq X-A[/imath]. Then for every [imath]f(A)\subseteq Y[/imath], there exists a [imath]y\in Y-f(A)[/imath] such that [imath]g(y)\notin X-A[/imath]. This would imply [imath]g(y)\notin X \lor g(y)\in A[/imath], but since [imath]g(y)\in X[/imath], we must have [imath]g(y)\in A[/imath]. The only thing I can glean from this is that [imath]g[/imath] is surjective, since for any singleton [imath]\{x\}\subseteq X[/imath], we could then find a [imath]y\in Y-\{f(x)\}[/imath] such that [imath]g(y)=x[/imath]. But I don't quite see how to get a contradiction. What direction should I head? I attempted to apply the fact that a monotone function on power sets has a fixed point, but that only seems to apply then the function maps from a power set into itself. Thanks! |
399992 | Cross product as factor in dot product
Given there are two vectors [imath]w,v[/imath] with [imath]||w||=4[/imath] , [imath]||v||=1[/imath] and [imath]\phi=\frac{2\pi}{3}[/imath] How do you transform the following expression into a form in which it can be computed with the given information? [imath] < w \times 5v,v-3w> [/imath] The cross product as first factor is what puzzles me about this | 399306 | Computing cross product using norm and angle
Sorry for the weird title, if someone finds a better title for my problem be my guest to edit it ;) For [imath]\mathbf{v,w} [/imath] in R³ with [imath]\mathbf{||v||=1 ;||w||=4; \theta =\frac{2\pi}{3}}[/imath] Solve the following: [imath]\lVert (3v+w)\times(v-2w)\rVert[/imath] [imath]\langle w\times 5v, v-3w\rangle[/imath] I've already calculated [imath]\langle v,w\rangle = \frac{-1}{2}||v|| \cdot||w|| =-2[/imath] Also, I've transformed 1. into [imath]\lVert-7(v\times w)\rVert[/imath] But I'm a bit lost here. Next I masked [imath]w\times 5v [/imath] as [imath]x[/imath] so as 2. we now have [imath]\langle x,v-3w\rangle[/imath] This all seems to bring me not a bit closer to the solution, can someone suggest a valid approach? |
398743 | Sobolev spaces - about weak derivative
Let [imath]U[/imath] a bounded and open subset of [imath]R^n[/imath]. Let [imath]u \in H^{1}(U)[/imath] a bounded function , [imath]v \in H^{1}_{0}(U)[/imath] a non negative function. Consider [imath]\varphi : R \rightarrow R[/imath] a convex and smooth function.Is true that [imath]\varphi^{'}(u)v \in H^{1}_{0}(U)[/imath]? I am trying , but nothing works... Someone can give me a hint ? | 398686 | sobolev spaces - product of two functions
I am working in a exercise, to my solution works I need the following affirmation is true: Let [imath]\varphi : R \rightarrow R[/imath] a convex and smooth function. Let [imath]u \in H^{1}(U)[/imath] a bounded function and [imath]v \in H^{1}_{0}(U)[/imath] a non negative function. Then [imath]\varphi^{'}(u)v \in H^{1}_{0}(U)[/imath]. I am trying to use the definiton , but it's not working. Someone can give me a hint? Thanks in advance |
400081 | Proving [imath]f(x) = x^2 \sin(1/x)[/imath], [imath]f(0)=0[/imath] is differentiable at [imath]0[/imath], with derivative [imath]f'(0)= 0[/imath] at zero
I need a solution for this question. I've been trying out this question for days and I haven't been able to find out its solution yet. And some explanation would help too. Show that the function f defined by: [imath]f(x):= \begin{cases} x^2\sin(1/x) &:\text{if $x \ne 0$} \\ 0 &:\text{if $x=0$} \end{cases}[/imath] is differentiable at [imath]x=0[/imath], and that [imath]f'(0)=0[/imath]. | 232672 | Show that the function [imath]g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)[/imath] is everywhere differentiable and that [imath]g′(0) = 0[/imath]
Show that the function [imath]g(x) = x^2 \sin\left(\frac{1}{x}\right) ,(g(0) = 0)[/imath] is everywhere differentiable and that [imath]g′(0) = 0[/imath]. |
272114 | Prove that [imath]x = 2[/imath] is the unique solution to [imath]3^x + 4^x = 5^x[/imath] where [imath]x \in \mathbb{R}[/imath]
Yesterday, my uncle asked me this question: Prove that [imath]x = 2[/imath] is the unique solution to [imath]3^x + 4^x = 5^x[/imath] where [imath]x \in \mathbb{R}[/imath]. How can we do this? Note that this is not a diophantine equation since [imath]x \in \mathbb{R}[/imath] if you are thinking about Fermat's Last Theorem. | 61812 | Proving that [imath] 2 [/imath] is the only real solution of [imath] 3^x+4^x=5^x [/imath]
I would like to prove that the equation [imath] 3^x+4^x=5^x [/imath] has only one real solution ([imath]x=2[/imath]) I tried to study the function [imath] f(x)=5^x-4^x-3^x [/imath] (in order to use the intermediate value theorem) but I am not able to find the sign of [imath] f'(x)= \ln(5)\times5^x-\ln(4)\times4^x-\ln(3)\times3^x [/imath] and I can't see any other method to solve this exercise... |
259801 | Question about [imath]p(z,w)=\alpha_0(z)+\alpha_1(z)w+\cdots+\alpha_k(z)w^k[/imath]
Let [imath]p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k[/imath] ,where [imath]k \le 1[/imath] and [imath]p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k_0,\cdots, p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k_k[/imath] are non-constant polynomials in the complex variable [imath]z[/imath]. Then: [imath](z,w) \in C\times C:p(z,w)=0[/imath] is: Bounded with empty interior. Unbounded with empty interior. bounded with nonempty interior . unbounded with nonempty interior. How can I be able to solve this problem ? I have no idea at all. | 400545 | If [imath]p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k[/imath] are non constant polynomial.
[imath]p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k[/imath] where [imath]a_i(z)[/imath] are non constant polynomials in complex variables with [imath]k\ge 1[/imath]. I need know if [imath]\{(z,w):p(z,w)=0\}[/imath] which of these are true or false: [imath]1[/imath]. bounded with empty interior [imath]2[/imath]. unbounded with empty interior [imath]3[/imath]. bounded with non empty interior [imath]4[/imath]. unbounded with non empty interior I think only [imath]1[/imath] is true as the set consists of roots of a polynomial in two variables with some specific degree. Am I right? |
401381 | Cardinality of the set of surjective functions on [imath]\mathbb{N}[/imath]?
I know that the set of all surjective mappings of ℕ onto ℕ (lets name this set as F) should have cardinality |ℝ|. How to strictly prove that? From the fact that cardinality of every possible function is |ℝ|, |F| <= |ℝ|. Saw similar question on this site, but I need strictly defined function that shows that |F|>=|ℝ|. Thank you for your time. | 399503 | Is the set of surjective functions from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath] uncountable?
I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form [imath]\Bbb{N} \to \Bbb{N}[/imath] is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function [imath]p[/imath].The problem that I'm having is that I don't know in which way to change the diagonal elements so that [imath]p[/imath] should, but cannot be in the matrix. Can someone help me, or at least give me some tips? |
401727 | Group equals union of two subgroups
Suppose [imath]G=H\cup K[/imath], where [imath]H[/imath] and [imath]K[/imath] are subgroups. Show that either [imath]H=G[/imath] or [imath]K=G[/imath]. What I did: For finite [imath]G[/imath], if [imath]H\neq G[/imath] and [imath]K\neq G[/imath], then [imath]|H|,|K|\le |G|/2[/imath]. But they clearly share the identity element, so [imath]H\cup K\neq G[/imath]. How can I extend it to [imath]G[/imath] infinite? | 334405 | If a group is the union of two subgroups, is one subgroup the group itself?
"Let [imath]G[/imath] be a group, and suppose [imath]G=H \cup K[/imath], where [imath]H[/imath] and [imath]K[/imath] are subgroups. Show that either [imath]H=G[/imath] or [imath]K=G[/imath]." Let [imath]h \in H[/imath] and [imath]k \in K[/imath]. Then [imath]hk \in H[/imath] or [imath]hk \in K[/imath] (since every element of [imath]G[/imath] is in either [imath]H[/imath] or [imath]K[/imath]). If [imath]hk=h'[/imath] for some [imath]h' \in H[/imath], then [imath]k=h^{-1}h'[/imath], so [imath]k \in H[/imath]. If [imath]hk=k'[/imath] for some [imath]k' \in K[/imath], then [imath]h=k'k^{-1}[/imath] so that [imath]h \in K[/imath]. If for all [imath]h \in H[/imath] we have [imath]h \in K[/imath], or if for all [imath]k \in K[/imath] we have [imath]k \in H[/imath], then [imath]H \subseteq K[/imath] or [imath]K \subseteq H[/imath]. Then since [imath]G=H \cup K[/imath], we must have either [imath]H=G[/imath] or [imath]K=G[/imath]. I'm not sure if the first paragraph of my 'proof' implies the second. I've shown that for arbitrary [imath]h \in H[/imath], [imath]h \in H[/imath] and possibly [imath]h \in K[/imath], and similar for [imath]k \in K[/imath]. I don't know how to wrap it up (or perhaps this route won't lead anywhere at all). If this way won't work, I'd just like a hint on a new direction to take. Thanks. |
401709 | Let [imath]W=\{p(B):p\text{ be a polynomial with real coefiicient}\}[/imath]
Let [imath]W=\{p(B):p\text{ be a polynomial with real coefiicient}\}[/imath] and where [imath]B=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}[/imath] Then the dimension [imath]d[/imath] of the space [imath]W[/imath] satisfies 1.[imath]4\le d\le 6[/imath] 2.[imath]6\le d\le 9[/imath] 3.[imath]3\le d\le 8[/imath] 4.[imath]3\le d\le 4[/imath] I have calculated that [imath]B^3=I[/imath] so [imath]I,B,B^2[/imath] will be the basis vebtors in [imath]W[/imath] so I hope [imath]\dim=3[/imath] but I dont know why they have put bounds, It was question in an past examination. | 261913 | Finding the Dimension of a Matrix Polynomial: [imath]W[/imath] = { [imath]p(B)[/imath] : [imath]p[/imath] is a polynomial with real coefficients}
Let [imath]W = \{ p(B) : p \text{ is a polynomial with real coefficients}\}[/imath], where [imath]B= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{pmatrix}[/imath] Which of the following possibilities presents the tightest bounds on the dimension [imath]d[/imath] of the vector space [imath]W[/imath]? [imath]4 ≤ d ≤ 6[/imath] [imath]6 ≤ d ≤ 9[/imath] [imath]3 ≤ d ≤ 8[/imath] [imath]3 ≤ d ≤ 4[/imath] |
401763 | Non-self-mapping automorphism implies abelian
Suppose [imath]\sigma\in\text{Aut}(G)[/imath]. If [imath]\sigma^2=1[/imath] and [imath]x^{\sigma}\neq x[/imath] for [imath]1\neq x\in G[/imath], show that if [imath]G[/imath] is finite, it must be abelian. There's a hint to show that the set [imath]\{x^{-1}x^{\sigma}\mid x\in G\}[/imath] is the whole group [imath]G[/imath]. I have already proved this hint ([imath]x^{-1}x^{\sigma} = y^{-1}y^{\sigma}\rightarrow (xy^{-1})^{\sigma} = xy^{-1}\rightarrow x=y[/imath] since [imath]x^{\sigma}=x[/imath] only for [imath]x=1[/imath]), but I could not use it to solve the problem. | 302222 | The existence of a group automorphism with some properties implies commutativity.
Let [imath]G [/imath] be a finite group, [imath]T[/imath] be an automorphisom of [imath] G [/imath] st [imath] Tx = x \iff x=e [/imath]. Suppose further that [imath] T^2 =I [/imath]. Prove that [imath] G [/imath] is abelian. I was thinking if I show [imath] T aba^{-1} b^ {-1}=aba^ {-1}b^{-1} \forall a, b \in G[/imath]. But I was unable to show it. Please give me any hints about it. |
401773 | Let [imath]a_1,\dots,a_n[/imath] be real numbers, and set [imath]a_{ij} = a_ia_j[/imath].
Let [imath]a_1,\dots,a_n[/imath] be real numbers, and set [imath]a_{ij} = a_ia_j[/imath]. Consider the [imath]n \times n[/imath] matrix [imath]A=(a_{ij})[/imath]. Then It is possible to choose [imath]a_1.\dots,a_n[/imath] such that [imath]A[/imath] is non-singular matrix [imath]A[/imath] is positive definite if [imath](a_1,\dots,a_n)[/imath] is nonzero vector matrix [imath]A[/imath] is positive semi definite for all [imath](a_1,\dots,a_n)[/imath] for all [imath](a_1,\dots,a_n)[/imath], [imath]0[/imath] is an eigen value of [imath]A[/imath] I have calculated upto [imath]3\times 3[/imath] that determimant is [imath]0[/imath] but I have no idea how to conclude rigoriously. please help? | 257066 | Let [imath]a_{i,j} =a_ia_j[/imath] , [imath]1 ≤ i, j ≤ b[/imath], where [imath](a_1,a_2,\ldots,a_n)[/imath], are real numbers. Multiple choice question.
Let [imath]a_{i,j} =a_ia_j[/imath] , [imath]1 ≤ i, j ≤ b[/imath], where [imath](a_1,a_2,\ldots,a_n)[/imath], are real numbers. Let [imath]A =(a_{i,j})[/imath] be the [imath]n ×n[/imath] matrix. Then It is possible to choose [imath](a_1,a_2.,….,a_n)[/imath], so as to make the matrix A non singular The matrix [imath]A[/imath] is positive definite if [imath](a_1,a_2,\ldots,a_n)[/imath], is a non zero vector The matrix [imath]A[/imath] is positive semi definite for all [imath](a_1,a_2,\ldots,a_n)[/imath], For all [imath](a_1,a_2,\ldots,a_n)[/imath], zero is an eigenvalue of [imath]A[/imath]. For second order matrix I checked that (1) and (2) are not correct. But I am not sure about the others that they are true/false. Thanks to help me. |
401981 | Absolute summable sequence in normed space
I have studied that for a sequence of real numbers absolute summability implies summability. What can we say about the sequences [imath]\{x_k\}[/imath] in a normed space . If it is not true in general could anybody show me with the counter examples in normed space? Thanks for the help. | 120180 | [imath]\ell_1[/imath] and unconditional convergence
Thanks to the Riemann theorem we know that absolute convergence and unconditional convergence are the same for [imath]\mathbb{R}[/imath]. In all the Frechet spaces absolute convergence implies unconditional convergence. There are counterexamples for the converse implication in some spaces - for example in [imath]\ell_p[/imath] ([imath]p>1[/imath]) the series [imath]\sum_{n=1}^\infty \frac{1}{n}e_n[/imath] is unconditionally convergent but not absolutely convergent (where [imath]e_n[/imath] is [imath]1[/imath] for [imath]n[/imath] and [imath]0[/imath] otherwise). However it doesn't work for [imath]\ell_1[/imath]. Here is my question: Does unconditional convergence imply absolute convergence in [imath]\ell_1[/imath]? I can't come up with any argument or a counterexample. I'm particularly interested in the real case if it matters. |
401675 | Consider a matrix with integer entries such that [imath]a_{ii}=1[/imath] and [imath]a_{ij}=0[/imath] for [imath]i>j[/imath]
Consider a matrix with integer entries such that [imath]a_{ii}=1[/imath] and [imath]a_{ij}=0[/imath] for [imath]i>j[/imath] Then [imath]A^{-1}[/imath] exists and it has integers entries. [imath]A^{-1}[/imath] exists and it has some entries which are not integers. [imath]A^{-1}[/imath] exists and it is a polynomial function in [imath]A[/imath] with integers coefficient. [imath]A^{-1}[/imath] is a not a power of [imath]A[/imath] unless [imath]A=I[/imath] As [imath]\det A=1[/imath] so [imath]A[/imath] inverse exist and [imath]A^{-1}=\dfrac{\operatorname{Adj} A}{\det A}[/imath] clearly [imath]A^{-1}[/imath] will have all integer entries, [imath]2[/imath] is false,[imath]c_nA^n+\dots +I=0\Rightarrow A^{-1}=-c_nA^n+\dots+ c_{n-1}A[/imath] as [imath]\det A=c_0=1[/imath] so [imath]3[/imath] is true, say [imath]A^{k-1}=A^{-1}[/imath] then [imath]A^k-I=0[/imath] so minpoly must divide [imath]x^k-1[/imath] and as its charpoly is [imath](x-1)^n=0[/imath] so minpoly must divide [imath](x-1)^n[/imath] too but that forces minpoly is [imath](x-1)[/imath] and hence [imath]A=I[/imath]. is my all logics are okay? | 257951 | Consider a matrix [imath]A[/imath] with integer entries such that [imath]a_{ij}=0[/imath] for [imath]i>j[/imath] and [imath]a_{ii}=1[/imath] . then which of the followings are true?
Consider a matrix [imath]A=(a_{ij})_{ n ×n }[/imath] with integer entries such that [imath]a_{ij}=0[/imath] for [imath]i>j[/imath] and [imath]a_{ii}=1[/imath] for [imath]i=1,…,n[/imath]. then which of the followings are true? [imath]A^{-1}[/imath] exists and it has integer entries. [imath]A^{-1}[/imath] exists and it has some entries that are not integer. [imath]A^{-1}[/imath] is a polynomial of [imath]A[/imath] with integer coefficients. [imath]A^{-1}[/imath] is not a power of [imath]A[/imath] unless [imath]A[/imath] is the identity matrix. By the given conditions [imath]A[/imath] is the upper triangular matrix with diagonal elements [imath]1[/imath].so eigenvalues are [imath]1[/imath].so their product=determinant of [imath]A =1.[/imath] So 1 is true. Inverse of the identity matrix is itself with has all integer entries so 2 is false. But I have no idea about (3) and (4) can anyone help me please. |
402731 | Prove that [imath]5^n - 2^n[/imath] is divisible by [imath]3[/imath] for all nonnegative integers [imath]n[/imath] using mathematical induction
Using mathematical induction, prove for all integers n 1 that [imath]5^n - 2^n[/imath] is divisible by 3. Can someone help me with this? | 372976 | Why is every answer of [imath]5^k - 2^k[/imath] divisible by 3?
We have the formula [imath]5^k - 2^k[/imath] I have noticed that every answer you get from this formula is divisible by 3. At least, I think so. Why is this? Does it have to do with [imath]5-2=3[/imath]? |
298234 | Prove that none of [imath]\{11, 111, 1111,\dots \}[/imath] is the perfect square of an integer
Please help me with solving this : prove that none of [imath]\{11, 111, 1111 \ldots \}[/imath] is the square of any [imath]x\in\mathbb{Z}[/imath] (that is, there is no [imath]x\in\mathbb{Z}[/imath] such that [imath]x^2\in\{11, 111, 1111, \ldots\}[/imath]). | 812567 | To prove this sequence does not contain a perfect square
I have to prove that the sequence [imath]\{11,111,1111, \dots \} [/imath] doesn't contain any perfect square numbers. I can realize it but I am unable to prove it. Please help. |
200971 | How do you explain the appearance of a sine in the integral for calculating the surface area of a sphere?
Let's say I want to calculate the surface area of a sphere. For simplicity, let's just use the unit sphere. A naïve argument might go like this. Let's say I mark the north and south "poles" and draw half of a great circle, which has length [imath]\pi[/imath]. I could say that since I need to go all the way around the sphere, I need to multiply this by [imath]2\pi[/imath] (the circumference of the equator). Therefore, the surface area of the unit sphere is [imath]2\pi^2[/imath]. Now, as we all know it should be [imath]4\pi[/imath]. Let's say we do an integral, using the following parametrization: [imath] T(\theta, \phi) = \begin{pmatrix} \sin \phi \cos \theta \\ \sin \phi \sin \theta \\ \cos \phi \end{pmatrix}, [/imath] with [imath]0 \le \phi \le \pi[/imath] and [imath]0 \le \theta \le 2\pi[/imath]. If we work out all the formulas, we get that [imath]Area(S^2) = \int_0^{2\pi} \int_0^\pi \sin \phi\ \mathrm{d}\phi \ \mathrm{d}\theta = 2\pi \int_0^\pi \sin \phi\ \mathrm{d}\phi.[/imath] The [imath]2\pi[/imath] is there all right, but it multiples not [imath]\pi[/imath] but [imath]\int_0^\pi \sin \phi\ \mathrm{d}\phi[/imath], which equals [imath]2[/imath]. Where does this come from? In other words, why is it wrong to just multiply [imath]2\pi[/imath] by half the length of a great circle? It would be great if there was a geometric explanation, with as little calculus as possible involved. | 131735 | Surface Element in Spherical Coordinates
In spherical polars, [imath]x=r\cos(\phi)\sin(\theta)[/imath] [imath]y=r\sin(\phi)\sin(\theta)[/imath] [imath]z=r\cos(\theta)[/imath] I want to work out an integral over the surface of a sphere - ie [imath]r[/imath] constant. I'm able to derive through scale factors, ie [imath]\delta(s)^2=h_1^2\delta(\theta)^2+h_2^2\delta(\phi)^2[/imath] (note [imath]\delta(r)=0[/imath]), that: [imath]h_1=r\sin(\theta),h_2=r[/imath] [imath]dA=h_1h_2=r^2\sin(\theta)[/imath] I'm just wondering is there an "easier" way to do this (eg. Jacobian determinant when I'm varying all 3 variables). I know you can supposedly visualize a change of area on the surface of the sphere, but I'm not particularly good at doing that sadly. |
381720 | Find all integer solutions to [imath]x^2+4=y^3[/imath].
Find all integer solutions to [imath]x^2+4=y^3[/imath]. Some obvious solutions are [imath](x,y)=(\pm2,2)[/imath]. Are these the only ones? | 263622 | Integers that satisfy [imath]a^3= b^2 + 4[/imath]
Well, here's my question: Are there any integers, [imath]a[/imath] and [imath]b[/imath] that satisfy the equation [imath]b^2[/imath][imath]+4[/imath]=[imath]a^3[/imath], such that [imath]a[/imath] and [imath]b[/imath] are coprime? I've already found the case where [imath]b=11[/imath] and [imath]a =5[/imath], but other than that? And if there do exist other cases, how would I find them? And if not how would I prove so? Thanks in advance. :) |
398413 | Double summation including power and factorial
I am finding some trouble in computing the following sum: [imath]\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}[/imath] Could you please provide a result? Thanks in advance | 402698 | Cauchy product on exponential-looking power series
Original posting by dioxen here: Double summation including power and factorial I am finding some trouble in computing the following sum: [imath]\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}[/imath] Could you please provide a result? Thanks in advance After a failed attempt at getting to the binomial expansion, I tried to play around with it a little bit. I came up with two ideas of a solution, but can't see it through. Both involve applying the Cauchy product. \begin{equation} \Bigg(\sum_{n=0}^\infty a_n \Bigg) \Bigg(\sum_{n=0}^{\infty} b_n \Bigg) = \sum_{n=0}^{\infty} c_n \space \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k} \space \space \space \space \space \space \space \space \space \space \space \space (1) \end{equation} 1: In this problem, we cound treat [imath] c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!}=\frac{1}{n!}\sum_{k=0}^n\frac {(xy)^{n-k}}{(n-k)!}x^{k}=\frac{1}{n!}\sum_{k=0}^n a_k b_{ n-k}[/imath] So if we're thinking within the framework of the first two summations in (1), our [imath]a_n[/imath] and [imath]b_n[/imath] are as follows: [imath]a_n=x^n[/imath] and [imath] b_n = \frac {(xy)^{n}}{n!}\\[/imath] our only problem is the we're left with the [imath]\frac{1}{n!}[/imath] in front of the summation. If we thought up some way to get rid of it, we'd have a nice geometric series and [imath]e^{xy}[/imath]. But I can't think of a way...any leads here? 2: Alternatively, we could leave the [imath]\frac{x^n}{n!}[/imath] outside and treat it as a number (because it is, before we choose to sum it up over [imath]n[/imath]). [imath] c_n= \frac{x^n}{n!}\;\sum_{k=0}^n\frac {y^k}{k!} 1^{n-k}=\frac{x^n}{n!}\sum_{k=0}^n a_k b_{ n- k}[/imath] which gives us [imath]a_n=\frac{y^n}{n!}[/imath] and [imath]b_n=1[/imath]. Could we do something about [imath]\frac{x^n}{n!}[/imath]? [imath] \Bigg(\sum_{n=0}^\infty \frac{x^n}{n!} \Bigg) \Bigg(\sum_{n=0}^{\infty} \frac{x^n}{n!} \frac{y^n}{n!} \Bigg) [/imath] Does this even make sense? Because otherwise I can't think of a way this sum would be computable. |
403414 | Matrices AB, BA eigenspaces
Take two matrices [imath]A[/imath] [imath]n\times m[/imath] and [imath]B[/imath] [imath]m\times n[/imath]. They both have a nonzero eigenvalue [imath]\lambda[/imath]. How do you prove that the dimension of the eigenspaces of [imath]AB[/imath] and [imath]BA[/imath] corresponding to [imath]\lambda[/imath] are equal? i.e., [imath]\dim E_\lambda(AB) = \dim E_\lambda(BA)[/imath]. Attempt: If you take [imath]L[/imath] independent eigenvectors [imath]v_1, \ldots, v_n[/imath] of [imath]AB[/imath], they get [imath]Bv_1, \ldots, Bv_n[/imath] eigenvectors in [imath]BA[/imath]. This is apparently important in the proof, but I can't understand where to use this fact. Please help. | 403086 | Geometric Multiplicity, Eigenspace of Matrix
If [imath]A[/imath] is an [imath]r[/imath] x [imath]s[/imath] matrix and B is an [imath]s[/imath] x [imath]r[/imath] matrix. [imath]E_\mu(C)[/imath] is the eigenspace of square matrix C with eigenvalue [imath]\mu ≠ 0[/imath]. Proof: [imath]dimE_\mu(AB) = dimE_\mu(BA)[/imath] I imagine one would need to find the basis of eigenvectors for both spaces, but I'm not sure how... |
403383 | Doubt in [imath]\varepsilon[/imath]-[imath]\delta[/imath] proof of continuity of [imath]x^2[/imath]
I have one elementary doubt on the proof that [imath]f(x)=x^2[/imath] is continuous for every [imath]a \in \Bbb R[/imath]. The initial steps usually presented are: to deduce which [imath]\delta[/imath] will work we write: [imath]|x^2-a^2|=|x-a||x+a|[/imath] Now, we must find some bound for [imath]|x+a|[/imath], so we require that [imath]|x-a|<1[/imath] or some other positive real number. Here comes my first doubt, this trick appears very often, what's the intuition on making [imath]|x-a|<1[/imath]? After that we have that this implies [imath]x < 1 + a[/imath] and this implies that [imath]|x+a| < 1 + 2|a|[/imath]. Now it's obvious that if we set [imath]|x-a|< \varepsilon/(1+2|a|)[/imath] and this obviously will make [imath]|f(x)-f(a)|<\varepsilon[/imath]. However this was deduced under the hypothesis that [imath]|x-a|<1[/imath], so what if it were not true that [imath]|x-a|<1[/imath]? Also the end of the proof is normally saying that we must take [imath]\delta = \min\{1, \varepsilon/(1+2|a|)\}[/imath] However it happens that if the minimum is [imath]1[/imath], it seems to not work, because when we use this [imath]\delta[/imath] the [imath]\varepsilon[/imath] doesn't even appear. I think my problem is that I didn't really get the trick of making [imath]|x-a|<1[/imath]. Thanks very much in advance! | 209440 | How to show that [imath]f(x)=x^2[/imath] is continuous at [imath]x=1[/imath]?
How to show that [imath]f(x)=x^2[/imath] is continuous at [imath]x=1[/imath]? |
403518 | prove [imath]\inf S = -\sup (-S)[/imath]
Let [imath]-S[/imath] be the set [imath]\left\{ -s:s \in S \right\}[/imath] where [imath]-S[/imath] is the set that contains negatives of the members of [imath]S[/imath]. We want to prove that [imath]\inf(S) = -\sup(-S)[/imath] Here is how I proved it Let [imath]s_0= \sup(-S)[/imath]. That is for all [imath]-s_1\in -S[/imath] then [imath]-s_1 \leq s_0[/imath]. Multiplying both sides by [imath]-1[/imath] we get [imath]-s_0 \leq s_1[/imath] for all [imath]s_1 \in S[/imath]. So [imath]\inf(S)=-s_0=-\sup(-S)[/imath] It looks short and sweet. Not sure if its right though. | 392129 | Proof that [imath]\inf A = -\sup(-A)[/imath]
Let [imath]A[/imath] be a nonempty subset of real numbers which is bounded below. Let [imath]-A[/imath] be the set of of all numbers [imath]-x[/imath], where [imath]x[/imath] is in [imath]A[/imath]. Prove that [imath]\inf A = -\sup(-A)[/imath] So far this is what I have Let [imath]\alpha=\inf(A)[/imath], which allows us to say that [imath]\alpha \leq x[/imath] for all [imath]x \in A[/imath]. Therefore, we know that [imath]-\alpha \geq -x[/imath] for all [imath]x \in -A[/imath]. Therefore we know that [imath]-\alpha[/imath] is an upper bound of [imath]-A[/imath]. [imath]\ \ \ \ [/imath] Now let [imath]b[/imath] be the upper bound of [imath]-A[/imath]. There exists [imath]b \geq-x \implies-b \leq x[/imath] for all [imath]x \in A[/imath]. Hence, \begin{align} -b & \leq \alpha\\ -\alpha & \leq b\\ -\alpha & = - \inf(A) = \sup (-A) \end{align} By multiplying [imath]-1[/imath] on both sides, we get that [imath]\inf(A) = -\sup (-A)[/imath] Is my proof correct? |
403508 | Integrability of a function (Darboux)
Question: Let [imath]f: [a,b] \to \mathbb{R}[/imath] and assume [imath]0 \leq f(x) \leq B[/imath] for [imath]x\in [a,b][/imath]. Show that [imath]U(f^2,P) -L(f^2,P) \leq 2B\ ( U(f,P) - L(f,P) )[/imath] for all partitions [imath]P[/imath] of [imath][a,b][/imath]. Show that if [imath]f[/imath] is integrable on [imath][a,b][/imath], then so is [imath]f^2[/imath] [no positivity assumption here]. I found this hint buried in the latex file my prof gave me. (Hint: [imath]f(x)^2 -f(y)^2 = (f(x) + f(y)) \cdot (f(x) -f(y))[/imath].) It really wasn't easy to put it to good use... [imath]\begin{align} U(f^2,P) -L(f^2,P) &= (U(f,p) + L(f,p)) \cdot (U(f,p) - L(f,p)) \\ &\leq (|U(f,p)| + |L(f,p)|) \cdot (U(f,p) - L(f,p)) \end{align}[/imath] We know [imath]|U(f,p)|\geq |L(f,p)|[/imath], and since [imath]f(x) \leq B[/imath], [imath]|f(x)| \leq B[/imath] for all [imath]x \in [a,b][/imath] where [imath]B \in \mathbb {R}[/imath]. Knowing that [imath]|L(f,p)| \leq |U(f,p)|\leq |f(x)| \leq B[/imath], we get: [imath](B + B) \cdot (U(f,p) - L(f,p)) = 2B\ (U(f,P) - L(f,P))[/imath] Hence [imath]U(f^2,P)-L(f^2,P) \leq 2B\ (U(f,P) - L(f,P))[/imath]. Anyone agree with me? Part 2) Theorem: Let [imath]f:[a,b] \to [c,d][/imath] be Daraboux-integrable and [imath]g:[c,d] \to \mathbb {R}[/imath] be continuous. Then the composition [imath]g \circ f[/imath] is Daraboux-integrable. Let [imath]g=x^2[/imath], so that [imath]g \circ f=f^2[/imath]. Since [imath]g[/imath] is continous on [imath][c,d][/imath] for all [imath]c,d \in \mathbb {R}[/imath], [imath]f^{2}[/imath] is integrable as long as [imath]f[/imath] is integrable on [imath][a,b][/imath] by the theorem above. | 395051 | Upper and lower integration inequality
I would like to learn how to prove that the following inequality holds. Let [imath]F[/imath] be a bounded function on an interval [imath][a,b][/imath], so that there exists [imath]B\geq 0[/imath] such that [imath]|f(x)| \leq B[/imath] for every [imath]x\in [a,b][/imath]. Show that [imath][ U(f^2,P) -L(f^2,P) \leq 2B [ U(f,P) -L(f,P) ] ][/imath] for all partitions [imath]P[/imath] of [imath][a,b][/imath]. |
403879 | Show that if [imath]x\geq 0[/imath] and [imath]n[/imath] is a positive integer, then [imath]\sum_{k=0}^{n-1}\left\lfloor {x+\frac{k}{n}}\right\rfloor=\lfloor {nx}\rfloor[/imath]
I need help with this question, where [imath]\lfloor x\rfloor[/imath] means the floor function of [imath]x[/imath]. Show that if [imath]x\geq 0[/imath] and [imath]n[/imath] is a postive integer, then [imath]\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n} \right\rfloor =\lfloor nx\rfloor[/imath] | 184361 | Floor function properties: [imath][2x] = [x] + [ x + \frac12 ][/imath] and [imath][nx] = \sum_{k = 0}^{n - 1} [ x + \frac{k}{n} ] [/imath]
I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of [imath][x][/imath], so I'm going with this one: DEFINITION Given [imath]x\in \Bbb R[/imath], the integer part of [imath]x[/imath] is the unique [imath]z\in \Bbb Z[/imath] such that [imath]z\leq x < z+1[/imath] and we denote it by [imath][x][/imath]. Now he asks to prove some basic things about it, such as: if [imath]n\in \Bbb Z[/imath], then [imath][x+n]=[x]+n[/imath] So I proved it like this: Let [imath]z=[x+n][/imath] and [imath]z'=[x][/imath]. Then we have that [imath]z\leq x+n<z+1[/imath] [imath]z'\leq x<z'+1[/imath] Then [imath]z'+n\leq x+n<z'+n+1[/imath] But since [imath]z'[/imath] is an integer, so is [imath]z'+n[/imath]. Since [imath]z[/imath] is unique, it must be that [imath]z'+n=z[/imath]. However, this doesn't seem to get me anywhere to prove that [imath]\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right][/imath] in and in general that [imath]\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} [/imath] Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of [imath][x][/imath] in terms of [imath]x[/imath] to work this out. Another property is [imath][-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}[/imath] I argue: if [imath]x\in\Bbb Z[/imath], it is clear [imath][x]=x[/imath]. Then [imath]-[x]=-x[/imath], and [imath]-[x]\in \Bbb Z[/imath] so [imath][-[x]]=-[x]=[-x][/imath]. For the other, I guess one could say: [imath]n \leqslant x < n + 1 \Rightarrow - n - 1 < x \leqslant -n[/imath] and since [imath]x[/imath] is not an integer, this should be the same as [imath] - n - 1 \leqslant -x < -n[/imath] [imath] - n - 1 \leqslant -x < (-n-1)+1[/imath] So [imath][-x]=-[x]-1[/imath] |
404044 | Let N1 and N2 are normal subgroups in the finite group G. Is it true that if N1≃N1 then G∖N1≃G∖N2.?
Let [imath]N_1[/imath] and [imath]N_2[/imath] are normal subgroups in the finite group [imath]G[/imath]. Is it true that if [imath]N_1 \simeq N_2[/imath] then [imath]G/ N_1 \simeq G/ N_2[/imath]? | 7720 | Finite group with isomorphic normal subgroups and non-isomorphic quotients?
I know it is possible for a group [imath]G[/imath] to have normal subgroups [imath]H, K[/imath], such that [imath]H\cong K[/imath] but [imath]G/H\not\cong G/K[/imath], but I couldn't think of any examples with [imath]G[/imath] finite. What is an illustrative example? |
404227 | Big-O: Prove [imath]2^n[/imath] is [imath]O(n!)[/imath]
I am a little stuck trying to prove that [imath]2^n[/imath] is [imath]O(n!)[/imath]. Obviously, I can tell in a few ways that this is the case. For one, based on Big-[imath]O[/imath] hierarchy, the exponential is beneath the factorial in terms of complexity. I also substituted [imath]C = 2[/imath] and tried several values of [imath]n[/imath], all following the Big-[imath]O[/imath] definition of [imath]f(n) \le cg(n)[/imath]. However, my professor says neither of these is a sufficient justification and I should be able to use the definition alone. How can I do this? I know in the past it has been advantageous for me to compare the functions by converting one into a form that resembles the other, but I just can't figure out how to relate the factorial and [imath]2^n[/imath] functions. Can anyone point me in the right direction? | 393713 | How do I prove that [imath]2^n=O(n!)[/imath]?
How do I prove that [imath]2^n=O(n!)[/imath]? Is this a valid argument? 2<=2, 2<3, 2<4,.... 2<n if n>2 therefore 2.2.2....n times < 1.2. ... n so,2^n <n! Thanks in advance. |
404488 | units and irreducible elements over UFD
Pick out the correct statements from the following list: a. A homomorphic image of a UFD (unique factorization domain) is again a UFD. b. The element [imath]2 \in \mathbb Z[\sqrt{−5}][/imath] is irreducible in [imath]\mathbb Z[\sqrt{−5}][/imath]. c. Units of the ring [imath]\mathbb Z[\sqrt{−5}][/imath] are the units of [imath]\mathbb Z[/imath]. d. The element [imath]2[/imath] is a prime element in [imath]\mathbb Z[\sqrt{−5}][/imath]. | 221527 | some question related to [imath]\mathbb{Z}[\sqrt{-5}][/imath] and UFD
could any one give me examples/proofs/counter examples against or for of the followings? 1.Homomorphic image of a UFD is again a UFD 2.The element [imath]2\in\mathbb{Z}[\sqrt{-5}][/imath] is irreducible 3.Units of the ring [imath]\mathbb{Z}[\sqrt{-5}][/imath] are units of [imath]\mathbb{Z}[/imath] 4.[imath]2[/imath] is a prime element in [imath]\mathbb{Z}[\sqrt{-5}][/imath] for 1, I know that [imath]\mathbb{Z}[/imath] is a UFD, to show the statement is false from [imath]\mathbb{Z}[/imath] to where I should construct a homomorphism? for 2, In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units. 3 is true. 4 is false |
401323 | Accumulation points of accumulation points of accumulation points
Let [imath]A'[/imath] denote the set of accumulation points of [imath]A[/imath]. Find a subset [imath]A[/imath] of [imath]\Bbb R^2[/imath] such that [imath]A, A', A'', A'''[/imath] are all distinct. I can find a set [imath]A[/imath] such that [imath]A[/imath] and [imath]A'[/imath] are distinct, but not one where [imath]A,A',A'',A'''[/imath] are all distinct. | 1118036 | Example of subset of [imath]\mathbb{R}²[/imath] such that [imath]A\neq A'\neq A''\neq A'''[/imath]?
I am looking for a subset of [imath]\mathbb{R}²[/imath] such that [imath]A\neq A'\neq A''\neq A'''[/imath] (where [imath]A'[/imath] is the set of limit points of [imath]A[/imath]). I read it's possible but I don't even see how it could be ... I've tried a subset that has an isolated point (or several) and is open otherwise. The furthest I could go was picking [imath]\lbrace(\frac{1}{n},0):n\in\mathbb{Z}\rbrace \cup B((2,0),1)[/imath] because then [imath]A'[/imath] has an isolated point. I'm sure I'm missing a point here, could you help me? Thank you very much |
405030 | Question about new proof in number theory?
Very recently, Yitang "Tom" Zhang from the University of New Hampshire made a huge step toward proving the twin prime conjecture. He proved that there are infinitely many pairs of prime separated by a constant distance, and that distance is less that [imath]7 \cdot 10^7[/imath]. For example, in the special case of the twin prime conjecture, pairs of primes are separated by 2, such as 3 and 5, 5 and 7, 11 and 13, etc. Now people are saying that they might be able to use his methods to prove that the difference is lower than 16. My question is, does his proof limit the possibility of these pairs of primes to only one pair, or does it potentially allow for all the 7 million possibilities? Thanks! By the way this is where I found out about this: https://www.youtube.com/watch?v=vkMXdShDdtY | 398763 | Yitang Zhang: Prime Gaps
Has anybody read Yitang Zhang's paper on prime gaps? Wired reports "[imath]70[/imath] million" at most, but I was wondering if the number was actually more specific. *EDIT*[imath]^1[/imath]: Are there any experts here who can explain the proof? Is the outline in the annals the preprint or the full accepted paper? |
405124 | Restricted Partitions of [imath]n[/imath]
The original question was to find the number of ways to split an integer, [imath]n[/imath], into any number of partitions where each of the parts belong to the set [imath]\lbrace 1,3,4,9\rbrace[/imath]. Assuming I did this right I found the answer to be simply: The [imath]x^n[/imath] coefficient of [imath](1-x)^{-1}(1-x^3)^{-1}(1-x^4)^{-1}(1-x^9)^{-1}.[/imath] But now the restriction has been made that the number of parts in the composition that equal [imath]9[/imath] has to be less than or equal to the number of parts that equal [imath]4[/imath]. If someone could help me figure out the correct generating function I'd be really thankful! | 405046 | Counting number of solutions with restrictions
I want to count the number of non-negative integer solutions to an equation such as [imath]x+5y+8z=n[/imath] I can do this using generating functions; for example, the answer here is [imath][x^n]\frac{1}{(1-x)(1-x^5)(1-x^8)}[/imath] where [imath][x^n][/imath] is the coefficient of [imath]x^n[/imath]. But what if I add a restriction between variables such as [imath]y \le z[/imath]? I have no idea how to count the number of solutions with this additional constraint. Any suggestions? |
405164 | Is there any compelling reason why [imath]0[/imath] *shouldn't* be a natural number?
This seems to be (from what I've heard from various math people of various statures) a heated debate. One of my previous professors proclaimed very strongly that [imath]0[/imath] is not a natural number. Another recently said the same. But then there are people who say it should be. One very good logical reasoning given was that, if the natural numbers are supposed to be the "counting" numbers, then [imath]0[/imath] is useful in the sense that it denotes the lack of something. I do realize that the natural numbers were defined prior to the discovery of the concept of "zero", but that shouldn't be a reason why we can't formally agree that it should be a natural number. And for the cases where using [imath]0[/imath] breaks things, there's no reason why calculations can't explicitly omit [imath]0[/imath]. | 391840 | Should [imath]\mathbb{N}[/imath] contain [imath]0[/imath]?
This is a classical question, that has led to many a heated argument: Should the symbol [imath]\mathbb{N}[/imath] stand for [imath]0,1,2,3,\dots[/imath] or [imath]1,2,3,\dots[/imath]? It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician [imath]A[/imath] define [imath]\mathbb{N}[/imath] to be the positive integers including [imath]0[/imath], and mathematician [imath]B[/imath] to define [imath]\mathbb{N}[/imath] to be the positive integers excluding [imath]0[/imath]. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future. I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that [imath]0 \in \mathbb{N}[/imath] because: 1) morally, [imath]\mathbb{N}[/imath] is the cardinalities of finite sets 2) the empty set is a set with [imath]0[/imath] elements. However, recently I realised that this reasoning applies to [imath]\omega[/imath] rather than [imath]\mathbb{N}[/imath], and - much to my horror - I saw [imath]\omega[/imath] and [imath]\mathbb{N}[/imath] used side by side with the only distinction being that [imath]0 \in \omega[/imath] while [imath]0 \not \in \mathbb{N}[/imath]. For another example, [imath]\mathbb{N}[/imath] seems to be a much nicer semigroup if [imath]0 \not \in \mathbb{N}[/imath] (and in any case, adding [imath]0[/imath] to a semigroup is a more natural operation than removing it), which would an argument for taking [imath]0 \not \in \mathbb{N}[/imath]. The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for. Question: Does there exists a convincing argument for deciding if [imath]0 \in \mathbb{N}[/imath] ? (I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using [imath]\mathbb{Z}_+[/imath] (or even [imath]\mathbb{Z}_{>0}[/imath] and [imath]\mathbb{Z}_{\geq 0}[/imath]) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.) |
400745 | Transitivity of a square of a relation
Transitivity [imath]\def\p{\mathrel p}[/imath]If [imath]\p[/imath] is a relation on a set [imath]A[/imath], define [imath]\p^2[/imath] by [imath]a \mathrel{\p^2} b[/imath] if and only if there exists [imath]c[/imath] with [imath]a \p c[/imath] and [imath]c \p b[/imath]. If [imath]p[/imath] is reflexive/symmetric/transitive does [imath]p^2[/imath] have the same properties? So i understand and have proven the reflexive and symmetric properties (see also this question: Reflexivity, Transitivity, Symmertry of the square of an relation), however is this correct for transitivty. Suppose [imath]\p[/imath] is transitive Let [imath]a,b,c[/imath] [imath]\in[/imath] A WTS - a [imath]\p^2[/imath] b, b [imath]\p^2[/imath] c, therefore a [imath]\p^2[/imath] c So if p is transitive that means : [imath]a \p b[/imath] [imath]b \p c[/imath] [imath]a \p c[/imath] For b [imath]\p^2[/imath] c we need [imath]a \p c[/imath] and [imath]c \p c[/imath] for it to be related So since p is transitive we can say that [imath]a \p c[/imath] and that [imath]c \p c[/imath] since its reflexive, so then we have our condition satisfied so therefor we can say that if a [imath]\p^2[/imath] b and b [imath]\p^2[/imath] c Then a [imath]\p^2[/imath] c since condition [imath]a \p c[/imath], [imath]c \p c[/imath] is satisfied because p is transitive qed am i on the right track here? | 400015 | Reflexivity, Transitivity, Symmertry of the square of an relation
[imath]\def\p{\mathrel p}[/imath]If [imath]\p[/imath] is a relation on a set [imath]A[/imath], define [imath]\p^2[/imath] by [imath]a \mathrel{\p^2} b[/imath] if and only if there exists [imath]c[/imath] with [imath]a \p c[/imath] and [imath]c \p b[/imath]. If [imath]p[/imath] is reflexive/symmetric/transitive does [imath]p^2[/imath] have the same properties? I'm not even sure how to start this, I assume I would need to use the [imath]a[/imath] related to [imath]c[/imath], [imath]c[/imath] related to [imath]b[/imath] somehow? |
405332 | How to calculate: [imath]\sum_{n=1}^{\infty} n a^n[/imath]
I've tried to calculate this sum: [imath]\sum_{n=1}^{\infty} n a^n[/imath] The point of this is to try to work out the "mean" term in an exponentially decaying average. I've done the following: [imath]\text{let }x = \sum_{n=1}^{\infty} n a^n[/imath] [imath]x = a + a \sum_{n=1}^{\infty} (n+1) a^n[/imath] [imath]x = a + a (\sum_{n=1}^{\infty} n a^n + \sum_{n=1}^{\infty} a^n)[/imath] [imath]x = a + a (x + \sum_{n=1}^{\infty} a^n)[/imath] [imath]x = a + ax + a\sum_{n=1}^{\infty} a^n[/imath] [imath](1-a)x = a + a\sum_{n=1}^{\infty} a^n[/imath] Lets try to work out the [imath]\sum_{n=1}^{\infty} a^n[/imath] part: [imath]let y = \sum_{n=1}^{\infty} a^n[/imath] [imath]y = a + a \sum_{n=1}^{\infty} a^n[/imath] [imath]y = a + ay[/imath] [imath]y - ay = a[/imath] [imath]y(1-a) = a[/imath] [imath]y = a/(1-a)[/imath] Substitute y back in: [imath](1-a)x = a + a*(a/(1-a))[/imath] [imath](1-a)^2 x = a(1-a) + a^2[/imath] [imath](1-a)^2 x = a - a^2 + a^2[/imath] [imath](1-a)^2 x = a[/imath] [imath]x = a/(1-a)^2[/imath] Is this right, and if so is there a shorter way? Edit: To actually calculate the "mean" term of a exponential moving average we need to keep in mind that terms are weighted at the level of [imath](1-a)[/imath]. i.e. for [imath]a=1[/imath] there is no decay, for [imath]a=0[/imath] only the most recent term counts. So the above result we need to multiply by [imath](1-a)[/imath] to get the result: Exponential moving average "mean term" = [imath]a/(1-a)[/imath] This gives the results, for [imath]a=0[/imath], the mean term is the "0th term" (none other are used) whereas for [imath]a=0.5[/imath] the mean term is the "1st term" (i.e. after the current term). | 1244297 | Show the closed form of the sum [imath]\sum_{i=0}^{n-1} i x^i[/imath]
Can anybody help me to show that when [imath]x\neq 1[/imath] [imath]\large \sum_{i=0}^{n-1} i\, x^i = \frac{1-n\, x^{n-1}+(n-1)\,x^n}{(1-x)^2}[/imath] |
405601 | what methods to solve the eqution [imath]y'(x)=y(x+1)[/imath]
I want to solve : [imath]y'(x)=y(x+1)[/imath] ! I don't know how to solve it ! form what tried I know that y must be nonlinear function also it can't be polynomial! if we replace the 1 with [imath]\frac{\pi}{2}[/imath] then [imath]y=sin(x)[/imath] must satisfy the equation ! if : [imath]y'(x)=\frac {y(x+1)}{e}[/imath] then [imath]y=e^x[/imath] is a solution , so how we solve these equations ? | 45392 | Differential equations that are also functional
I was toying with equations of the type [imath]f(x+\alpha)=f'(x)[/imath] where [imath]f[/imath] is a real function. For example if [imath]\alpha=\frac{\pi}{2}[/imath] then the solutions include the function [imath]f_{\lambda,\mu}(x)=\lambda cos(x+\mu)[/imath]. Are there more solutions? On the other hand, if I want to solve the equation for any [imath]\alpha[/imath], I can assume a solution of the form [imath]f(x)=e^{\lambda x}[/imath], and find [imath]\lambda[/imath] as a complex number that enables me to solve the equation... I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator [imath]\phi: f(x)\mapsto f(x+1)[/imath] adds another one? Is there some litterature about this kind of equations? Please satisfy my curiosity if you can... Thanks! |
405677 | Question on cardinal arithmetic
How do I go about showing: [imath]\aleph_{\omega}^{\aleph_{1}}[/imath]=[imath]\aleph_{\omega}^{\aleph_0}[/imath].[imath]2^{\aleph_1}[/imath] My first thought was to apply Hausdorff's formula directly, but that doesn't quite work. The book provides a hint using products and sums that I don't quite understand. Can you please help? The question is from Introduction to Set Theory by Hrbacek and Jech. | 216817 | Why is [imath]{\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}[/imath]?
I am supposed to prove that [imath]{\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}[/imath] , but I really have no idea how to start or what to do. I thought I could use the following fact: [imath]{2}^{\aleph_1}= {\aleph_1}^{\aleph_1}[/imath], because of the infiniteness of [imath]{\aleph_1}[/imath]. I hope someone will show me how this works. Thanks in advance! |
405966 | If I have three points, is there an easy way to tell if they are collinear?
Points [imath](a,b)[/imath], [imath](m,n)[/imath], and [imath](x,y)[/imath] are selected at random. What is the quickest/easiest way to tell if they are collinear? At first I thought it was a matter of comparing slopes but that doesn't appear to be enough. | 38338 | Methods for showing three points in [imath]\mathbb{R}^2[/imath] are colinear (or not)
A common question is to prove/disprove that three points in [imath]\mathbb{R}^2[/imath] are colinear. For, example Show that [imath](-1, 8)[/imath], [imath](1, -2)[/imath] and [imath](2, 1)[/imath] lie on a common line. What are some methods one could use to answer questions of this nature? |
406230 | How many different ways to combine $2?
Suppose there are these coins: 1c, 5c, 10c, 25c, \[imath]1 (100c) and \[/imath]2 (200c). How many different ways can \[imath]2 be made using any number of coins?[/imath] I figure we need to use generating functions, so let C(x)$ be the generating function: [imath] C(x)=\frac{1}{(1-x)(1-x^{5})(1-x^{10})(1-x^{25})(1-x^{100})(1-x^{200})} [/imath] How to compute (like an algorithm) the coefficient of $ x^{200} [imath]?[/imath] This question is similar to number of ways to make 2.00 but no computing the solution there. | 397502 | number of ways to make $2.00
How many different ways can you make $2.00 using only 1 cent, 5 cent, 10 cent, and 25 cent pieces, and 1 and 2 dollar bills (there are 100 cents in a dollar)? I have worked out an equation: [imath]p + 5n + 10d + 25q + 100l + 200t = 200[/imath] Where all variables are non-negative integers. But I don't know what to do from here to figure out how many different sets of non-negative integers there are that satisfy this equation. What do I do next? |
396167 | question about Riemann zeta [imath]\zeta (0)[/imath]
i know that [imath]\zeta (m)=\sum_{n=1}^\infty n^{-m}[/imath] so [imath]\zeta (0)=\sum_{n=1}^\infty n^0=1+1+1+1+1+1+\cdots=\infty [/imath] but actually [imath]\zeta (0)=-0.5[/imath] where is the wrong please help thanks for all | 79105 | How to find [imath]\zeta(0)=\frac{-1}{2}[/imath] by definition?
I would like to know how we can find the following result: [imath]\zeta(0)=-\frac12[/imath] Is there a way, using the definition, [imath]\zeta(s)=\sum_{i=1}^{\infty}i^{-s}[/imath] to find this? |
287866 | "Fat" Cantor Set
So the standard Cantor set has an outer measure equal to [imath]$0$[/imath], but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of [imath]$1$[/imath]. I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of [imath]$0$[/imath] as well... Are there other constraints that need to be made in order to accomplish this? | 1040486 | example for Lebesgue measure
Construct a closed, uncountable, perfect, nowhere dense subset of [imath][0,1][/imath] which has Lebesgue measure [imath]\frac{1}{2}[/imath]. (Hint: Find the Cantor subset of [imath][0, 1][/imath] with Lebesgue measure [imath]\frac{1}{2}[/imath]) |
405470 | Help with Cartesian product subsets
I want to prove that if [imath]A \subseteq C\,[/imath] and [imath]\,B \subseteq D,\,[/imath] then [imath]\,A \times B \subseteq C \times D.[/imath] I know that [imath]A \subseteq C \iff a \in A \rightarrow a \in C[/imath] and that [imath]B\subseteq D\iff b \in B \rightarrow b \in D[/imath] I also know that [imath]A \times B = \{(a, b)\mid a\in A, b\in B\}[/imath] and that [imath]C\times D = \{(c, d) \mid c \in C, d \in D\}[/imath]. So keep that in mind, how do I connect the dots? | 319975 | If [imath]A \subseteq C[/imath] and [imath]B \subseteq D[/imath] then [imath]A \times B \subseteq C \times D[/imath]
Show that: if [imath]A \subseteq C\,[/imath] and [imath]\,B \subseteq D,\,[/imath] then [imath]\,A \times B \subseteq C \times D.[/imath] Can anyone help me with this? |
406597 | Proving by induction that [imath]n^{n+1} > (n+1)^n[/imath] for [imath]n \ge 3[/imath]
Prove the following inequality by mathematical induction: [imath]n^{n+1}>{(n+1)}^n \qquad (n\geq3)[/imath] Obviously it holds for [imath]n=3[/imath]. Assume [imath]P(n)[/imath] holds, then [imath]LHS=n^{n+1}[/imath] | 77935 | Prove by induction that for all [imath]n \geq 3[/imath]: [imath]n^{n+1} > (n+1)^n[/imath]
I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples. Now he showed me an exercise he was having trouble with, which states that one should prove [imath]n^{n+1} > (n+1)^n[/imath] for all [imath]n \geq 3[/imath]. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the [imath](n+2)^{n+1}[/imath], primarely the number [imath]2[/imath] is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school. Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem? Thanks for answers in advance. |
406750 | Representing a real valued function as a sum of odd and even functions
With [imath]f(x)[/imath] being a real valued function we can write it as a sum of an odd function [imath]m(x)[/imath] and an even function [imath]n(x)[/imath]: [imath]f(x)=m(x)+n(x)[/imath] Write an equation for [imath]f(-x)[/imath] in terms of [imath]m(x)[/imath] and [imath]n(x)[/imath]: My attempt using the properties - even function if: [imath]f(x)=f(-x)[/imath] and odd function if: [imath]-f(x)=f(-x)[/imath] [imath]f(x)=m(x)+n(x) \implies f(-x)=m(-x)+n(-x) \implies f(-x)= -m(x)+n(x) \implies f(-x)=n(x)-m(x)[/imath] I think that is correct but then I need to find equations for both [imath]m(x)[/imath] and [imath]n(x)[/imath] in terms of [imath]f(x)[/imath] and [imath]f(-x)[/imath] so a suggestion on how to tackle that would be great. | 5274 | How do I divide a function into even and odd sections?
While working on a proof showing that all functions limited to the domain of real numbers can be expressed as a sum of their odd and even components, I stumbled into a troublesome roadblock; namely, I had no clue how one divides the function into these even and odd parts. Looking up a solution for the proof, I found these general formulas for the even and odd parts of a function [imath]f(n)[/imath]: [imath]\begin{align*} f_e(n)&\overset{\Delta}{=}\frac{f(n)+f(-n)}{2}\\ f_o(n)&\overset{\Delta}{=}\frac{f(n)-f(-n)}{2} \end{align*}[/imath] While I understand that in an even function [imath]f(n) = f(-n)[/imath] and that in an odd function [imath]f(-n) = -f(n)[/imath], I still don't get how these general formulas for the even and odd parts were obtained. Can someone guide me through the logic? |
406872 | Divisibility and factors
1) Can factors be negative? Please prove your opinion. 2)If prime factorization is given to you, how will you find out how many composite factors are there? Not the factors, just how many. For 2), my teacher gave me an easy method: let x = [imath]a^{a_1} . b^{b_1}. c^{c_1}..[/imath] Then your answer is : [imath](a_1 +1)(b_1 + 1)(c_1 +1)...[/imath] But again, he didnt prove it, and I didnt have any time left in the class to ask him, so please help me prove it... | 89485 | Prove a property of divisor function
Let [imath]n[/imath] be a positive natural number whose prime factorization is [imath]n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}[/imath], where [imath]p_i[/imath] are natural distinct prime numbers, and [imath]a_i[/imath] are positive natural numbers. Using induction to show that the number of divisors of [imath]n[/imath] is [imath](a_1+1)(a_2+1)\cdots(a_k+1)[/imath] |