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105990 | Showing that [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}\backslash\mathbb{Q}[/imath] are equinumerous using Cantor-Bernstein
I need to prove that [imath]\mathbb{R}\backslash\mathbb{Q} \sim \mathbb{R} [/imath] Using Cantor-Bernstein, need to show an injection from [imath]\mathbb{R}\backslash\mathbb{Q}[/imath] to [imath]\mathbb{R}[/imath] and from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}\backslash\mathbb{Q}[/imath]. [imath]\mathbb{R}\backslash\mathbb{Q}[/imath] is a subset of [imath]\mathbb{R}[/imath] so only need to show injection from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}\backslash\mathbb{Q}[/imath] to complete the proof. Possible injection: [imath]f:\mathbb{R}\to \mathbb{R}\backslash\mathbb{Q}[/imath] defined as [imath]f(x) = \pi x[/imath] if [imath]x[/imath] is not a multiple of [imath]\pi[/imath]; otherwise [imath]f(x) = \sqrt{2} x[/imath]. Not sure if [imath]f[/imath] actually is an injection... | 2263217 | Showing that irrational numbers have the same cardinality as real numbers
So I did this but I'm not sure if it is an actual proof because it is based on a hypothesis, and also I'm not sure if I understand the Continuum Hypothesis correctly. Because [imath]\boldsymbol{I} \subseteq \boldsymbol{R}[/imath], l can have either the same or less cardinality than [imath]\boldsymbol{R}[/imath]. By the Continuum Hypothesis, l can not have a lower cardinality than [imath]\boldsymbol{R}[/imath], so it must have the same cardinality. Is this correct, and/or is there a more rigorous method? Also, can I use the notation [imath]\operatorname{card} A < \operatorname{card} B[/imath]? |
24873 | Elementary proof that [imath]\mathbb{R}^n[/imath] is not homeomorphic to [imath]\mathbb{R}^m[/imath]
It is very elementary to show that [imath]\mathbb{R}[/imath] isn't homeomorphic to [imath]\mathbb{R}^m[/imath] for [imath]m>1[/imath]: subtract a point and use the fact that connectedness is a homeomorphism invariant. Along similar lines, you can show that [imath]\mathbb{R^2}[/imath] isn't homeomorphic to [imath]\mathbb{R}^m[/imath] for [imath]m>2[/imath] by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary. However, the general result that [imath]\mathbb{R^n}[/imath] isn't homeomorphic to [imath]\mathbb{R^m}[/imath] for [imath]n\neq m[/imath], though intuitively obvious, is usually proved using sophisticated results from algebraic topology, such as invariance of domain or extensions of the Jordan curve theorem. Is there a more elementary proof of this fact? If not, is there intuition for why a proof is so difficult? | 814100 | Elementary proof that [imath]Gl_n(\mathbb R)[/imath] and [imath]Gl_m(R)[/imath] are homeomorphic iff [imath]n=m[/imath]
Prove that [imath]Gl_n(\mathbb R)[/imath] and [imath]Gl_m(R)[/imath] are homeomorphic iff [imath]n=m[/imath] Since [imath]Gl_n(\mathbb R)[/imath] is homeomorphic to an open subset of [imath]\mathbb R^{n^2}[/imath], this boils down to proving that two open subsets of [imath]\mathbb R^n[/imath] and [imath]\mathbb R^m[/imath] are homeomorphic iff [imath]n=m[/imath]. This can be done via homology, of which I know absolutely nothing. Do you know a proof that doesn't cover things an undergraduate isn't supposed to know ? EDIT: I'll close the question, since it's a duplicate of Elementary proof that [imath]\mathbb{R}^n[/imath] is not homeomorphic to [imath]\mathbb{R}^m[/imath] |
121076 | Proving [imath](1 + 1/n)^{n+1} \gt e[/imath]
I'm trying to prove that [imath] \left(1 + \frac{1}{n}\right)^{n+1} > e [/imath] It seems that the definition of [imath]e[/imath] is going to be important here but I can't work out what to do with the limit in the resulting inequality. | 894933 | Why is sequence [imath](1+\frac{1}{n})^{n+1}[/imath] descending?
I was studying the proof of [imath]e[/imath] number when I noticed something: Why is the sequence [imath](1+\frac{1}{n})^{n+1}[/imath] descending? It starts ascending with grater n but in one moment it starts descending? Why is that? On what [imath]n[/imath] this happens? Edit: Let me try to rephrase the question... How come that sequence [imath](1+\frac{1}{n})^n[/imath] rises, but the sequence [imath](1+\frac{1}{n})^{n+1}[/imath] falls, just because of the one more 1+1/n multiplication (power)? |
105680 | eigenvalues and eigenvectors of [imath]vv^T[/imath]
Given a column vector [imath]v[/imath] in [imath]\mathbb{R}^n[/imath], what are the eigenvalues of matrix [imath]vv^T[/imath] and associated eigenvectors? PS: not homework even though it may look like so. | 660872 | Eigenvalues and eigenvectors of [imath]v v^T[/imath] matrix
If [imath]v[/imath] is a column vector, then how many non-zero eigenvalues does the matrix [imath]vv^T[/imath] have? What are the eigenvalues? What are the corresponding eigenvectors? What are the eigenvectors corresponding to the zero eigenvalues? [Note: This is not homework] |
28589 | Sharp interpolation inequality for Lebesgue spaces
Suppose [imath]f\in L^{p}(\mathbb{R}^{n}) \cap L^{q}(\mathbb{R}^{n})[/imath]. How can I prove that for any [imath]p \lt r \lt q[/imath], [imath] \lVert f \rVert_{r} \leq (\lVert f \rVert_{p})^{(1/r-1/q)/(1/p-1/q)} (\lVert f \rVert_{q})^{(1/r-1/q)/(1/p-1/q)}\:? [/imath] I tried using Hölder to show that [imath]f[/imath] is in [imath]L^{r}[/imath] but I'm completely lost on how to go about it... | 1529859 | [imath]L^p[/imath] spaces inclusion
Show that for any measurable function [imath]f[/imath] in a measure space, we have: [imath] ||f||_p \leq \max\{||f||_r ,||f||_s \} [/imath] whenever [imath]0<r<p<s[/imath]. Now by splitting the integrals into parts where [imath]f>1[/imath] and [imath]f\leq 1[/imath], we could easily show that [imath] ||f||_p \leq ||f||_r+||f||_s \ [/imath] But I have no idea how to show the stronger inequality as above. |
88257 | Uniform continuity and boundedness
I have come across a proof which I understand almost completely, except for one part: THEOREM: If [imath]f[/imath] is uniformly continuous on a bounded interval [imath]I[/imath], then [imath]f[/imath] is also bounded on [imath]I[/imath]. PROOF: In this case we assume that [imath]I[/imath] is of the form [imath](a,b), (a,b], [a,b)[/imath], or [imath][a,b][/imath], with [imath]a,b \in \mathbb{R}[/imath]. Fix an [imath]\epsilon > 0[/imath], for instance [imath]\epsilon = 1[/imath]. Since [imath]f[/imath] is uniformly continuous, there is a [imath]\delta > 0[/imath] such that: [imath]|f(x_1) - f(x_2)| < \epsilon = 1[/imath] when [imath]x_1, x_2 \in I[/imath] and [imath]|x_1 - x_2| < \delta[/imath] Divide [imath]I[/imath] into [imath]N[/imath] intervals, [imath]I_1, . . ., I_N[/imath], where [imath]N[/imath] is chosen so that [imath]\frac{b-a}{N} < \delta[/imath]. Let [imath]z_i[/imath] be the center point of [imath]I_i[/imath]. For each [imath]i[/imath] and [imath]x \in I_i[/imath], [imath]|x - z_i| < \delta[/imath], and then we have: [imath]|f(x)| = |f(x) - f(z_i) + f(z_i)| \leq |f(x) - f(z_i)| + |f(z_i)| \leq 1 + |f(z_i)|[/imath]. Then for [imath]x \in I_i[/imath], [imath]|f(x)| \leq 1 + \max_{1 \leq i \leq N}\{|f(z_i)|\}[/imath]. Let [imath]M = \max_{1 \leq i \leq N}\{|f(z_i)|\}[/imath]. Then [imath]|f(x)| \leq 1 + M[/imath] QED OK, so the one thing I am a bit unsure of here, is when we write: Let [imath]M = \max_{1 \leq i \leq N}\{|f(z_i)|\}[/imath]. How is it that we know for sure that each [imath]|f(z_i)|[/imath] is also bounded? I see how the presence of a maximum value completes the proof, but why is it not possible that we have an [imath]|f(z_i)|[/imath] which is unbounded? If anyone could explain this to me I would greatly appreciate it! | 1565279 | A uniformly continuous function is bounded
Suppose that the function [imath]f:[0,1)\rightarrow \mathbb{R}[/imath] is uniformly continuous on [imath][0,1).[/imath] Prove that the function [imath]f[/imath] is bounded. (i.e. that the range[imath](f)[/imath] is a bounded set) |
9916 | [imath]H[/imath] is a subgroup of index [imath]2[/imath] in finite group [imath]G[/imath], then every left coset of [imath]H[/imath] is a right coset as well
I know this means that there are two cosets. I also know that one must be H itself. This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then [imath]aH = bH[/imath] and [imath]Ha = Hb[/imath], since each element in G acting on the subgroup must produce the same set. This means [imath]aH = Ha[/imath] I believe. From there [imath]H = aHa^{-1}[/imath], which given the information I am not sure if this good or how to use it. Any help would good. | 2667688 | If [imath][G:H]=2[/imath], prove that [imath]gH=Hg[/imath].
Question. If [imath][G:H]=2[/imath], prove that [imath]gH=Hg[/imath]. Should I use set equality? Or is it because since there are only two distinct left cosets and I use logical reasoning from there? By the way, would one of the left cosets be [imath]eH[/imath]? I am still new to this coset topic, and so I have no real concrete work on how to go about this proof... |
70976 | Why is [imath]\mathbb{Z}[\sqrt{-n}], n\ge 3[/imath] not a UFD?
I'm considering the ring [imath]\mathbb{Z}[\sqrt{-n}][/imath], where [imath]n\ge 3[/imath] and square free. I want to see why it's not a UFD. I defined a norm for the ring by [imath]|a+b\sqrt{-n}|=a^2+nb^2[/imath]. Using this I was able to show that [imath]2[/imath], [imath]\sqrt{-n}[/imath] and [imath]1+\sqrt{-n}[/imath] are all irreducible. Is there someway to conclude that [imath]\mathbb{Z}[\sqrt{-n}][/imath] is not a UFD based on this? Thanks. | 318543 | For which [imath]d<0[/imath] is [imath]\mathbb Z[\sqrt{d}][/imath] an Euclidean Domain?
I know that for [imath]d=-1, -2[/imath] the ring [imath]\mathbb Z[\sqrt{d}][/imath] is an Euclidean Domain. I believe that it is not an Euclidean Domain for and [imath]d \leq-3[/imath]. I have been able to prove it for a handful of examples (like [imath]d=-3[/imath]), showing that the resulting ring is not a UFD, but am not sure how to prove the claim in general. (And whether the resulting rings are not UFDs in the general case or merely not Euclidean.) |
3271 | How to evaluate Riemann Zeta function
How do I evaluate this function for given [imath]s[/imath]? [imath]\zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots[/imath] | 1988430 | Faster Convergence for the Smaller Values of the Riemann Zeta Function
I have a C++ program that uses the equation [imath]\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}[/imath] to calculate the Riemann zeta function. This equation converges fast for larger values, like 183, but converges much slower for smaller values, like 2. For example, calculating the value of [imath]\zeta(2)[/imath] took an hour to be accurate to 5 digits, but one second for [imath]\zeta(183)[/imath] to be accurate to 100 digits. Are there any equations for calculating the Riemann zeta function that are faster for calculating smaller values? Because I am coding in C++, I cannot use [imath]\int[/imath] (without implementing external libraries, which is not really an option here). |
42609 | Strong and weak convergence in [imath]\ell^1[/imath]
Let [imath]\ell^1[/imath] be the space of absolutely summable real or complex sequences. Let us say that a sequence [imath](x_1, x_2, \ldots)[/imath] of vectors in [imath]\ell^1[/imath] converges weakly to [imath]x \in \ell^1[/imath] if for every bounded linear functional [imath]\varphi \in (\ell^1)^*[/imath], [imath]\varphi(x_n) \rightarrow \varphi(x)[/imath] as [imath]n \to \infty[/imath]. How may I show that weak convergence, in this sense, is the same as the usual convergence-in-norm? It's clear the weak convergence implies pointwise convergence, but that's not good enough to conclude strong convergence... By linearity, it suffices to prove that if [imath]\varphi(x_n) \longrightarrow 0[/imath] for every [imath]\varphi \in (\ell^1)^*[/imath], then [imath]\| x_n \| \longrightarrow 0[/imath]. Let [imath]x_n(k)[/imath] be the [imath]k[/imath]-th component of the vector [imath]x_n[/imath]. Then, [imath]x_n(k) \longrightarrow 0[/imath] for every [imath]k[/imath], so [imath]\sup_n |x_n(k)| < \infty[/imath] for each [imath]k[/imath], and this implies [imath]\lim_{N \to \infty} \lim_{n \to \infty} \sum_{k=1}^{N} |x_n(k)| = 0[/imath] This is almost what I want, but the limits are the wrong way around. The obvious next thing to try is to construct some clever functional, or even a family of clever functionals, but I can't think of anything useful here. I can see that pointwise convergence alone is not good enough — if [imath]x_n[/imath] is the standard basis vector, then [imath]x_n \longrightarrow 0[/imath] pointwise, but [imath]\| x_n \| = 1[/imath] for all [imath]n[/imath]. The fact that it doesn't converge strongly can be detected by the linear functional [imath]\varphi(x_n) = \sum_k x_n(k)[/imath], but I'm at a loss as to how to generalise this. | 365115 | weakly convergent sequence in [imath]l^1[/imath]
Prove that every weakly convergent sequence in [imath]l^1[/imath] converges. By Riesz on [imath]L^p[/imath] spaces, every linear functional [imath]L\in (l^1 )^*[/imath], is [imath]L(x) = \langle u,x\rangle[/imath] for some [imath]u\in l^{\infty}[/imath] (which is to say that [imath]|u_i|\leq C[/imath] for some [imath]C[/imath] and all [imath]i\geq 1[/imath] (here [imath]x[/imath] and [imath]u[/imath] are sequences of complex numbers). In particular, we can easily deduce that if [imath]x^i \rightarrow x[/imath] weakly then [imath]x^i_j \rightarrow x_j[/imath] for every [imath]j[/imath], but we really need to show that [imath]\sum_0^{\infty}|x^i_j-x_j| \rightarrow 0[/imath] as [imath]i\rightarrow \infty[/imath], which requires more. I'm really not sure how to proceed with this. Hints? |
536 | Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1
How does one prove that a simple (steps of length [imath]1[/imath] in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in [imath]1[/imath] or [imath]2[/imath] dimensions returns to the origin with probability [imath]1[/imath]? Edit: note that while returning to the origin is guaranteed [imath](p = 1)[/imath] in [imath]1[/imath] and [imath]2[/imath] dimensions, it is not guaranteed in higher dimensions; this means that something in a correct justification for the [imath]1[/imath]- or [imath]2[/imath]-d case must fail to extend to [imath]3[/imath]-d (or fail when the probability for each direction drops from [imath]\frac14[/imath] to [imath]\frac16[/imath]). | 359115 | Find the probability that he go back to his original grid after finite moves.
Assume that Tom is on a grid paper of infinite size, and he go left, right, forward or backward randomly in each move. Find the probability that he go back to his original grid after [imath]n[/imath] moves. I wrote a program and find that if I gave him [imath]10^4, 10^5, 10^6, 10^7[/imath] chances to moves, the required probability will be around 74%, 78%, 81%, 84%. How can I find out the probability that he go back to his original grid after [imath]n[/imath] moves mathematically? Thanks. |
300179 | Finding all invertible matrices [imath]A[/imath] where [imath]A = A^{-1}[/imath] and [imath]A^{-1} = A^T[/imath]
Finding all invertible [imath]A[/imath], a [imath]2\times 2[/imath] matrix that satisfies [imath]A = A^{-1}[/imath] and [imath]A^{-1} = A^T[/imath]. Hint: The identity [imath]\cos^2t + \sin^2t = 1[/imath] may be useful. I have no idea how to start this. Any help would be much appreciated. | 300067 | About finding [imath]2\times 2[/imath] matrices that are their own inverses
They ask me to find all invertible matrices [imath]A[/imath] of the form: [imath]\begin{bmatrix}a & b\\ c&d \end{bmatrix}[/imath] and satisfying [imath]A=A^{-1}[/imath] and [imath]A^t=A^{-1}[/imath]. I find that rather complex; does it have anything to do with orthogonality? Not sure. Any help please?? |
104083 | An open ball is an open set
Prove that for any [imath]x_0 \in X[/imath] and any [imath]r>0[/imath], the open ball [imath]B_r(x_o)[/imath] is open. My attempt: Let [imath]y\in B_r(x_0)[/imath]. By definition, [imath]d(y,x_0)<r[/imath]. I want to show there exists an [imath]r_1\in\mathbb{R^+}[/imath] s.t. [imath]B_{r_{1}}(y)\subseteq B_r(x_0)[/imath]. Let [imath]a\in B_{r_{1}}(y)[/imath]. Then, [imath]d(a,y)<r_1[/imath]. For [imath]a\in B_{r}(x_0)[/imath], [imath]d(a,x_0)<r[/imath]. I want to show [imath]d(a,y)<r_1[/imath] implies [imath]d(a,x_0)<r[/imath]. By triangle inequality, [imath]d(a,y)\leq d(a,x_0) + d(y,x_0) \rightarrow[/imath] [imath]d(a,y)<r_1\leq d(a,x_0)+d(y,x_0)<2r...[/imath] I'm a little stuck after this point. | 473213 | Prove [imath]B(x,\delta)[/imath] is open.
Prove [imath]B(x,\delta)[/imath] is open. What this question is asking me to prove...? I don't understand nor have a clue to approach the question... |
243049 | How do I convince someone that [imath]1+1=2[/imath] may not necessarily be true?
Me and my friend were arguing over this "fact" that we all know and hold dear. However, I do know that [imath]1+1=2[/imath] is an axiom. That is why I beg to differ. Neither of us have the required mathematical knowledge to convince each other. And that is why, we decided to turn to stackexchange for help. What would be stack's opinion? | 1155238 | Why are the rules of logic universally applicable?
We can imagine physical constants to be different in a different universe or even not be constant in our own universe. We can imagine and simulate different physical and information-theoretical laws, e.g. regarding entropy or correlation of force and distance. However, [imath]1 + 1 = 2[/imath] as we usually define addition over natural, integer, fractional and complex numbers will always and provably hold, regardless of context and numeral system (of course, the representation may change, but not the semantics). Why is this? Why can't [imath]1 + 1 = 3[/imath] given the same axioms? And no, Peano's axioms do not define logic. This is a common misunderstanding. In fact, they are based on logic concepts. Why are mathematical laws completely invariant such that logic is even possible and sort of inevitable? From what I understand is that logic seems to be the only solution that works (and everything else breaks down at some point), because it is possible to conclude everything from a false statement (see also In classical logic, why is (p⇒q) True if both p and q are False?). But then again, this leaves us right at where we left ("Why is it the only solution?"). Also, I don't see how Peano explains structures like this: http://www.nature.com/news/two-hundred-terabyte-maths-proof-is-largest-ever-1.19990 |
76097 | What is the limit of [imath]n \sin (2 \pi \cdot e \cdot n!)[/imath] as [imath]n[/imath] goes to infinity?
I tried and got this [imath]e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}[/imath] [imath]n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m[/imath] where [imath]m[/imath] is an integer. [imath]\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0[/imath] Is it correct? | 374137 | A rather ugly limit
Evaluate [imath]\lim_{n \rightarrow \infty} n \sin (2\pi e n!).[/imath] I wanna ask what's wrong with my method: Define [imath]C_n= n \cos (2\pi e n!)[/imath] and [imath]S_n=n \sin (2\pi e n!)[/imath], then [imath]C_n+iS_n=ne^{i2\pi en!}=n1^{en!}=n[/imath], comparing the imaginary parts yeilds [imath]S_n=0[/imath]. Therefore the limit equals [imath]0[/imath]. Also, I am asking for a way to solve the problem, thank you. |
157372 | Proving [imath]\frac{\sin x}{x} =\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right) \left(1-\frac{x^2}{3^2\pi^2}\right)\cdots[/imath]
How to prove the following product? [imath]\frac{\sin(x)}{x}= \left(1+\frac{x}{\pi}\right) \left(1-\frac{x}{\pi}\right) \left(1+\frac{x}{2\pi}\right) \left(1-\frac{x}{2\pi}\right) \left(1+\frac{x}{3\pi}\right) \left(1-\frac{x}{3\pi}\right)\cdots[/imath] | 447011 | How to prove that [imath]\frac{\sin \pi x}{\pi x}=\prod_{n=1}^{\infty}(1-\frac{x^2}{n^2})[/imath]
How to prove that [imath]\frac{\sin \pi x}{\pi x}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)[/imath] I tried it with the Taylor series of [imath]\sin(x)[/imath] but I failed. Is there any help? |
77769 | Equal simple field extensions?
I have a question about simple field extensions. For a field [imath]F[/imath], if [imath][F(a):F][/imath] is odd, then why is [imath]F(a)=F(a^2)[/imath]? | 1393510 | Prove that [imath]E=F[\alpha^2][/imath]
Let [imath]E=F[\alpha][/imath], [imath]\alpha[/imath] is algebraic over [imath]F[/imath] and [imath][E:F][/imath] is odd. Prove that [imath]E=F[\alpha^2][/imath]. Now clearly [imath][F[\alpha^2]:F]|[E:F][/imath] so [imath][F[\alpha^2]:F][/imath] is also odd. But how can we show that [imath]E=F[\alpha^2][/imath]? |
66963 | Largest integer that can't be represented as a non-negative linear combination of [imath]m, n = mn - m - n[/imath]? Why?
This seemingly simple question has really stumped me: How do I prove that the largest integer that can't be represented with a non-negative linear combination of the integers [imath]m, n[/imath] is [imath]mn - m - n[/imath], assuming they are coprime? I got as far as this, but now I can't figure out where to go: [imath]mx + ny = k[/imath], where [imath]k = mn - m - n + c[/imath], for some [imath]c > 0[/imath] [imath]\Rightarrow m(x + 1) + n(y + 1) = mn + c[/imath] If I could only prove this must have a non-negative solution for [imath]x[/imath] and [imath]y[/imath], I'd be done... but I'm kind of stuck. Any ideas? | 1979138 | Natural numbers and linear combinations
Prove that every natural number [imath]n\geqslant 12[/imath] can be written as a linear combination of [imath]4[/imath] and [imath]5[/imath] with non negative integer coefficients. I proceeded by induction. For [imath]n=12[/imath] we have [imath]12=3\cdot 4 +0\cdot 5[/imath] Let [imath]n>12[/imath] Suppose that the hypothesis is valid for every [imath]k\in \mathbb{N}[/imath] with [imath]12<k<n[/imath] Now by doing some calculations we see that for [imath]n=13,14,\ldots,24[/imath] can be written as a linear combination of [imath]4[/imath] and [imath]5[/imath] with non negative integer coefficients. Suppose that [imath]n>24[/imath] then [imath]n=12+(12+k)[/imath] where [imath]12<12+k<n[/imath] and by induction hypothesis [imath]12+k=4x+5y[/imath] , [imath]x,y\geqslant0[/imath] We conclude that [imath]n=12+4x+5y[/imath]. Is this proof right or does it have a logical mistake? My question is a proof verification and about a possible logical mistake in using the strong induction. It's not an exercise that I want someone to solve for me. |
57292 | For every matrix [imath]A\in M_{2}( \mathbb{C}) [/imath] there's [imath]X\in M_{2}( \mathbb{C})[/imath] such that [imath]X^2=A[/imath]?
True\False? For every matrix [imath]A\in M_{2}( \mathbb{C}) [/imath] there's [imath]X\in M_{2}( \mathbb{C})[/imath] such that [imath]X^2=A[/imath]. I know that every complexed matrix has a Jordan form matrix [imath]J[/imath] such that [imath]P^{-1}CP=J[/imath], But it's not diagonalizable for sure. Thanks | 670793 | Jordan canonical form of 2X2 matrix
I got this problem: Is it true or not that for every matrix [imath]A\in M_2( \mathbb C)[/imath] there is a matrix [imath]X\in M_2( \mathbb C)[/imath] such that [imath]X^2=A[/imath]? prove the answer. I know that we can pass to Jordan form, and I proved that this claim is true for all the Jordan forms except [imath] \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} [/imath] any suggestions what to do with this matrix? and also, there is a general version of this problem? thanks for helpers! |
24378 | Prove that any two consecutive terms of the Fibonacci sequence are relatively prime
Prove that any two consecutive terms of the Fibonacci sequence are relatively prime. My attempt: We have [imath]f_1 = 1, f_2 = 1, f_3 = 2, \dots[/imath], so obviously [imath]\gcd(f_1, f_2) = 1[/imath]. Suppose that [imath]\gcd(f_n, f_{n+1}) = 1[/imath]; we will show that [imath]\gcd(f_{n+1}, f_{n+2}) = 1[/imath]. Consider [imath]\gcd(f_{n+1}, f_{n+2}) = \gcd(f_{n+1}, f_{n+1} + f_n)[/imath] because [imath]f_{n+2} = f_{n+1} + f_n.[/imath] Then [imath]\gcd(f_{n+1}, f_{n+1} + f_n) = \gcd(f_{n+1}, f_{n}) = 1[/imath] (gcd property). Hence, [imath]\gcd(f_n, f_{n+1}) = 1[/imath] for all [imath]n > 0[/imath]. Am I on the right track? Any feedback would be greatly appreciated. Thanks, | 1669760 | Greatest Common Divisor with Fibonacci Numbers
Prove that for all integers [imath]n\geq 0[/imath]: [imath]\gcd(F_{n+1},F_n)=1[/imath] I am extremely lost. Please can some provide some hint or direction? Thank you so very much |
26918 | Why is [imath]\pi[/imath] = 3.14... instead of 6.28...?
Inspired by a paper (from 2001) entitled Pi is Wrong: Why is [imath]\pi[/imath] = 3.14... instead of 6.28... ? Setting [imath]\pi[/imath] = 6.28 would seem to simplify many equations and constants in math and physics. Is there an intuitive reason we relate the circumference of a circle to its diameter instead of its radius, or was it an arbitrary choice that's left us with multiplicative baggage? | 1801513 | Why does convention not recognize [imath]2\pi[/imath] as the fundamental quantity?
It looks as though my question might turn out to be a duplicate. If so, it does not need an answer, after all, thanks. ORIGINAL QUESTION Why does convention recognize [imath]\pi\approx 3.14[/imath], rather than [imath]2\pi\approx 6.28[/imath], as the fundamental quantity? You have some contour integrals that come out to [imath]\sqrt{\pi}[/imath] or [imath]1/\sqrt{\pi}[/imath], so that's kind of neat; but I don't know that that arises more often than, say, [imath]1/\sqrt{2\pi}[/imath], as in the Fourier transform. Anyway, all that special-function action is so advanced that it misses what might seem to some to be the main point: [imath]2\pi[/imath] is a circle. How much more fundamental can you get than that? But mathematicians are smart people (and I am just a building-construction engineer), so I do not doubt that a good reason exists. I just do not know what the reason is. Hence the question. Why wasn't some other symbol defined, [imath]\kappa\approx 6.28[/imath]? |
270393 | How to prove Boole’s inequality
I am trying to prove Boole’s inequality [imath]P\left(\ \bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty P(A_i).[/imath] I can show it of any finite [imath]n[/imath] using induction. What to do for [imath]\infty[/imath] ? | 1285823 | prove [imath]p (\bigcup A_i) \leq \sum p (A_i) [/imath]
Prove [imath]p (\bigcup A_i) \leq \sum p (A_i) [/imath] Now i doubt that this has not been proven already as it is a common question, but i am typing latex on my tablet because i spilled water on my computer so i am having trouble finding the question in the archives. if somebody could post a link. Thanks. In case it has not been proven, i am thinking of using induction, but it just doesnt feel like the right thing to do to prove this question. |
66118 | How can I compute [imath]\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} [/imath]?
This sum is difficult. How can I compute it, without using calculus? [imath]\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}[/imath] If someone can explain some technique to do it, I'd appreciate it. Or advice using a telescopic sum, I think with a telescopic could go, but do not know how to assemble it. | 389116 | Proof that [imath]\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1} = \frac{2^{m+1}-1}{m+1}[/imath]
Recently I needed to compute [imath]E[\frac{1}{X+1}][/imath] where [imath]X\sim Bin(m, \frac 1 2)[/imath]. While expanding, I came across the sum [imath]\sum_{k=0}^m \binom{m}{k}\frac{1}{k+1}[/imath], which I was unable to solve. Plugging into Mathematica gives [imath]\frac{2^{m+1}-1}{m+1}[/imath], but I don't see how to derive this result. Can anyone give a proof of this identity or an alternate way to compute [imath]E[\frac{1}{X+1}][/imath]? |
44126 | Primes of the form [imath]n^2+1[/imath] - hard?
I met a student that is trying to prove for fun that there are infinitely many primes of the form [imath]n^2+1[/imath]. I tried to tell him it's a hard problem, but I lack references. Is there a paper/book covering the problem? Is this problem really hard or I remember incorrectly? | 1765164 | Are there infinitely many primes such that...
Are there infinitely many primes [imath]p[/imath], such that, [imath]p-1=n^2[/imath] where [imath]n\in \mathbb{N}[/imath]. As [imath]p[/imath] must be [imath]odd[/imath], we can say that [imath]p[/imath] must be of the form [imath]4k+1[/imath]. But I do not know how to proceed. |
21199 | Is [imath]\frac{\textrm{d}y}{\textrm{d}x}[/imath] not a ratio?
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative [imath]$dy/dx$[/imath] is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula [imath]$dy = f'(x)dx$[/imath] we are able to plug in values for [imath]$dx$[/imath] and calculate a [imath]$dy$[/imath] (differential). Then if we rearrange we get [imath]$dy/dx$[/imath] which could be seen as a ratio. I wonder if the author says this because [imath]$dx$[/imath] is an independent variable, and [imath]$dy$[/imath] is a dependent variable, for [imath]$dy/dx$[/imath] to be a ratio both variables need to be independent.. maybe? | 367174 | [imath]dy\over dx[/imath] is one things but why in integration we can treat it as 2 different terms
when i am learning differentiation, my lectuer tell us that the deriative [imath]dy\over dx[/imath] is one things, it is not the ration between dy and dx. However when i learn about integrating, sometime we need to do substitution, like integrating [imath]\int_{0}^{1}2xdx[/imath] when substituting [imath]y=2x[/imath], we can substitute [imath]dy=2dx[/imath], but why in this case it can be treated as 2 different terms instead of 1 term?? |
298536 | Let [imath]A[/imath] and [imath]B[/imath] be two finite subgroups of a group [imath]G[/imath]. Show that [imath]|AB|=\frac{|A| \times |B|}{|A \cap B|}[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be two finite subgroups of a group [imath]G[/imath]. Show that [imath]|AB|=\frac{|A| \times |B|}{|A \cap B|}[/imath] I have no idea how to start. Anyone can help ? I think of divisibility to prove this, but I got nowhere. | 168942 | Order of a product of subgroups. Prove that [imath]o(HK) = \frac{o(H)o(K)}{o(H \cap K)}[/imath].
Let [imath]H[/imath], [imath]K[/imath] be subgroups of [imath]G[/imath]. Prove that [imath]o(HK) = \frac{o(H)o(K)}{o(H \cap K)}[/imath]. I need this theorem to prove something. |
3633 | On the functional square root of [imath]x^2+1[/imath]
There are some math quizzes like: find a function [imath]\phi:\mathbb{R}\rightarrow\mathbb{R}[/imath] such that [imath]\phi(\phi(x)) = f(x) \equiv x^2 + 1.[/imath] If such [imath]\phi[/imath] exists (it does in this example), [imath]\phi[/imath] can be viewed as a "square root" of [imath]f[/imath] in the sense of function composition because [imath]\phi\circ\phi = f[/imath]. Is there a general theory on the mathematical properties of this kind of square roots? (For instance, for what [imath]f[/imath] will a real analytic [imath]\phi[/imath] exist?) | 459621 | Any chance for a solution of [imath]f[f(x)]=x^2+1[/imath]?
Not sure if this is a closed or open question. But the question is suppose [imath]f[f(x)]=x^2+1[/imath] then what is [imath]f(x)[/imath]? Though the question does not refer to the domain let us suppose it is defined on [imath]\Bbb R[/imath]. Thanks. |
108288 | Infinite tetration, convergence radius
I got this problem from my teacher as a optional challenge. I am open about this being a given problem, however it is not homework. The problem is stated as follows. Assume we have an infinite tetration as follows [imath]x^{x^{x^{.^{.^.}}}} \, = \, a[/imath] With a given [imath]a[/imath] find [imath]x[/imath]. The next part of the problem was to discuss the convergence radius of a. If a is too big or too small the tetration does not converge. Below is my humble stab at the problem. My friend said you would have to treat the tetration as a infinnite serie, and therefore coul not perform algebraic manipulations on it before it is know wheter it converges or diverges. However my attempt is to first do some algebraic steps, then discuss the convergence radius. I) Initial discussion At the start it is obvious that the tetration converges when [imath]a=1[/imath] (just set [imath]x=1[/imath]) Now after some computer hardwork it seems that the tetration fails to converge when a is roughly larger than 3. II) Algebraic manipulation [imath] x^{x^{x^{.^{.^.}}}} \, = \, a[/imath] This is the same as [imath] x^a \, = \, a[/imath] [imath] \log_a(x^a) \, = \, \log_a(a)[/imath] [imath] \log_a(x) \, = \, \frac{1}{a}[/imath] [imath] x \, = \, a^{\frac{1}{a}}[/imath] Now, if we let [imath]a=2[/imath] then [imath]x = sqrt{2}[/imath]. After some more computational work, this seems to be correct. Which makes me belive this fomula is correct. III) Discsussion about convergence By looking at the derivative of [imath] \displaystyle \large a^{\frac{1}{a}} [/imath] we see that the maxima occurs when [imath]a=e[/imath]. Which also seems to correspond with the inital computational work. Now I think, that the minima of [imath]\displaystyle \large a^{1/a}[/imath] is zero, by looking at its graph. And study its derivative and endpoints. So that my "guess" or work shows that a converges when [imath] a \in [0 \, , \, 1/e] [/imath] VI) My questions Can my algebraic manipulations be justified? They seem rather sketchy taking the a`th logarithm and so on . (Although they seem to "magicaly" give out the right answer) By looking at wikipedia it seems that the tetration converge when [imath] a \in \left[ 1/e \, , \, e \right] [/imath] This is almost what I have, why is my lower bound worng? How can I find the correct lower bound? | 362407 | Values for which tetration to infinite heights (i.e., [imath]x^{x^{x^{x^{.^{.^{.}}}}}}[/imath]) converges
I was reading Evaluating tetration to infinite heights (e.g., [imath]2^{2^{2^{2^{.^{.^.}}}}}[/imath]). For what values of [imath]x[/imath] does tetration to infinite heights (i.e., [imath]x^{x^{x^{x^{.^{.^{.}}}}}}[/imath]) converge? |
181928 | [imath]gHg^{-1}\subset H[/imath] whenever [imath]Ha\not = Hb[/imath] implies [imath]aH\not =bH[/imath]
If [imath]H[/imath] be a subgroup of a group [imath]G[/imath] such that [imath]Ha \not=Hb[/imath] implies that [imath]aH\not=bH[/imath].Then how can I show that [imath]gHg^{-1}\subset H[/imath] [imath]\forall[/imath] [imath]g\in G[/imath]? I do not think I have made any progress.However, this is what I have done: For any [imath]h\in H[/imath],[imath](ghg^{-1})(gh^{-1}g^{-1})=e[/imath] where [imath]e[/imath] is the identity element of the group [imath]G[/imath]. And for [imath]h,k\in H[/imath], [imath](ghg^{-1})(gkg^{-1})=g(hk)g^{-1}[/imath] which is in [imath]gHg^{-1}[/imath] as [imath]hk\in H[/imath]. Thus [imath]gHg^{-1}[/imath] is a group. I do not think I am getting anywhere. | 1578104 | Showing that [imath]H[/imath] is normal
This question has been bugging me for a while. Suppose that [imath]H[/imath] is a subgroup of [imath]G[/imath] such that [imath]Ha\not=Hb[/imath] implies that [imath]aH\not =bH[/imath]. I need to show that [imath]gHg^{-1}\subset H[/imath].(I don't mean a proper subset, just subset) Now, take [imath]x\in gHg^{-1}[/imath]. Suppose it is not in [imath]H[/imath]. Then [imath]x=ghg^{-1}[/imath] for some [imath]h[/imath]. As [imath]x\not \in H[/imath], [imath]Hx\not=H[/imath] whih implies [imath]xH\not=H[/imath] but this doesn't seem to be a terribly bright application of the given condition.(I could have got it using [imath]x\not \in H[/imath]) Can someone give me a hint to the solution? |
182316 | Is the set of irrationals separable as a subspace of the real line?
I am trying to find an example of a separable Hausdorff space which has a non-separable subspace. This led me to ask the question in the title: is the set of irrationals, regarded as a subspace of the real line, separable or non-separable? A space is separable if it contains a countable dense subset. A subset [imath]A[/imath] of a space [imath]X[/imath] is dense in [imath]X[/imath] if [imath]\bar{A}=X[/imath]. It's easy to come up with a dense set and a countable set in [imath]\mathbb{R}\setminus\mathbb{Q}[/imath], since (trivially) [imath]\overline{\mathbb{R}\setminus\mathbb{Q}}=\mathbb{R}\setminus\mathbb{Q}[/imath] in the subspace topology, and as a countable set we can take something like [imath]\{k\pi \mid k \in \mathbb{Z}\}[/imath]. But is there a subset that is both dense AND countable? And of course, how do we prove the result? | 2633669 | Is [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] separable?
Let [imath](X,d)[/imath] be a metric space. We say that [imath]X[/imath] is separable if [imath]X[/imath] contains a countable dense subset. We say that [imath]X[/imath] is second countable if [imath]X[/imath] has a countable base. The following fact is well-known. Fact: If [imath]X[/imath] is a separable metric space, then any subset [imath]Y[/imath] of [imath]X[/imath] is also separable. One main idea to prove above fact is that subset of a second countable space is again second countable. Question: Let [imath]X=\mathbb{R}[/imath] and [imath]d[/imath] be euclidean norm(that is, absolute value [imath]|\cdot|[/imath]). Is [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] separable? If we use the above fact, then [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] should be separable as [imath]\mathbb{R}[/imath] contains [imath]\mathbb{Q},[/imath] which is a countable dense subset of [imath]\mathbb{R},[/imath] and [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] is a subset of [imath]\mathbb{R}.[/imath] However, I do not see immediately a countable dense subset of [imath]\mathbb{R}\setminus \mathbb{Q},[/imath] as [imath]\mathbb{R}\setminus\mathbb{Q}[/imath] is uncountable, though it is dense in [imath]\mathbb{R}.[/imath] Any hint would be appreciated. |
46168 | Cantor-Bernstein-like theorem: If [imath]f\colon A\to B[/imath] is injection and [imath]g\colon A\to B[/imath] is surjective, can we prove there is a bijection as well?
I've been trying to find this proof: If there exists [imath]f \colon A\to B[/imath] injective and [imath]g \colon A \to B[/imath] surjective, prove there exists [imath]h \colon A \to B[/imath] bijective. I thought of using cardinality, but I think it's possible to prove that without using it. Anyone knows how to? | 314442 | Can a surjection and injection exist but not a bijection?
If I there exists an injection [imath]\phi: S_1 \to S_2[/imath] and a surjection [imath]\tau: S_1 \to S_2[/imath], does there necessarily exist a bijection between sets [imath]S_1[/imath] and [imath]S_2[/imath]? I'd like this to be true, but I don't see a way to construct a bijection directly from [imath]\phi[/imath] and [imath]\tau[/imath]. |
160157 | Integration of [imath]\int\frac{1}{x^{4}+1}\mathrm dx[/imath]
I don't know how to integrate [imath]\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx[/imath]. Do I have to use trigonometric substitution? | 1394710 | How to compute [imath]\int_0^1 \frac{1}{1+x^4}\;dx[/imath]?
I am having problems trying to compute [imath]\int_0^1 \frac{1}{1+x^4}\;dx[/imath] Wolfram alpha gives an answer [imath]\frac{\pi + 2 \coth^{-1}(\sqrt{2})}{4 \sqrt{2}}[/imath] |
57177 | Two problems on number theory
I need some ideas (preferable some tricks) for solving these two problems: Find the largest number [imath]n[/imath] such that [imath](2004!)![/imath] is divisible by [imath]((n!)!)![/imath] For which integer [imath]n[/imath] is [imath]2^8 + 2^{11} + 2^n[/imath] a perfect square? For the second one the suggested solution is like this : [imath] 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12[/imath] But I can't understand the approach,any ideas? | 642564 | If [imath]n,k\in\mathbb N[/imath], solve [imath]2^8+2^{11}+2^n=k^2[/imath].
If [imath]n,k\in\mathbb N[/imath], solve [imath]2^8+2^{11}+2^n=k^2[/imath] It's hard for me to find an idea. Some help would be great. Thanks. |
14625 | How can a structure have infinite length and infinite surface area, but have finite volume?
Consider the curve [imath]\frac{1}{x}[/imath] where [imath]x \geq 1[/imath]. Rotate this curve around the x-axis. One Dimension - Clearly this structure is infinitely long. Two Dimensions - Surface Area = [imath]2\pi\int_∞^1\frac{1}{x}dx = 2\pi(\ln ∞ - \ln 1) = ∞[/imath] Three Dimensions - Volume = [imath]\pi\int_∞^1{x}^{-2}dx = \pi(-\frac{1}{∞} + \frac{1}{1}) = \pi[/imath] So this structure has infinite length and infinite surface area. However it has finite volume, which just does not make sense. Even more interesting, the "walls" of this structure are infinitely thin. Since the volume is finite, we could fill this structure with a finite amount of paint. To fill the structure the paint would need to cover the complete surface area of the inside of this structure. Since the "walls" are infinitely thin, why would a finite amount of paint not be able to cover the outside of the "walls" too? Please help me make sense of this whole thing. | 1637785 | How can revolving an infinite area have a finite volume
The area of the region bounded by [imath]f(x) = \frac{1}{x}[/imath], [imath]y = 0[/imath], and [imath]x = 1[/imath] is [imath] A = \int_1^{+\infty} f(x) \, \textrm{d}x = \lim_{b \to +\infty} \int_1^b \frac{\textrm{d}x}{x} = \lim_{b \to +\infty} [\ln b - \ln 1] = +\infty [/imath] The volume of the solid formed by revolving the same region as above is [imath] V = \pi \int_1^{+\infty} [f(x)]^2 \, \textrm{d}x = \lim_{b \to +\infty} \pi \int_1^b \frac{\textrm{d}x}{x^2} = \pi \lim_{b \to +\infty} [-\frac{1}{b} + 1] = \pi [/imath] I understand that [imath]y = \frac{1}{x}[/imath] and [imath]y = \frac{1}{x^2}[/imath] are different, so it makes perfect algebraic sense that the area under both curves will be different; one being divergent and the other convergent in this question. But how can it make geometric sense that revolving an infinite area results in a finite volume? |
86263 | Characteristic of a field is [imath]0[/imath] or prime
I'm trying to prove that the characteristic of any field [imath]F[/imath] is either [imath]0[/imath] or a prime number, but I have no idea what to do. Help? | 573170 | Characteristic of a field [imath]F[/imath] is prime
If Char[imath]F[/imath] [imath]\neq 0[/imath], then Char[imath]F[/imath] must be prime number. MY try: If Char[imath]F[/imath][imath] = nk [/imath] for integers [imath]n[/imath] and [imath]k[/imath], then by definition, [imath]nk = 0 \implies n = 0[/imath] or [imath]k = 0[/imath] which implies Char[imath]F=0[/imath] which is a contradiction. Is this correct? |
287957 | When is [imath](p - 1)! + 1[/imath] a power of [imath]p[/imath]?
A friend asked me this question: If [imath]p[/imath] is a prime, prove that [imath](p - 1)! + 1[/imath] is a power of [imath]p[/imath] if and only if [imath]p = 2, 3[/imath] or [imath]5[/imath]. Clearly one direction is obvious, namely that [imath]p=2,3,5[/imath] implies [imath](p - 1)! + 1[/imath] is a power of [imath]p[/imath]. The other direction is not clear to me. Since by Wilson's theorem [imath]p[/imath] divides [imath](p - 1)! + 1[/imath] so we need to show that if there are no other prime factors then [imath]p=2,3,5[/imath]. Can someone give me a hint for establishing this? Thanks | 2092823 | Prove for [imath]k>5[/imath] prime, [imath](k-1)!+1[/imath] has at least [imath]2[/imath] **different** prime divisors.
Let [imath]k\in\mathbb{N},~k>5[/imath] be a prime number. Prove [imath](k-1)!+1[/imath] has at least two different prime divisors. I am thinking on Wilson's Theorem, but I don't know how to use it. |
141293 | continuous functions on [imath]\mathbb R[/imath] such that [imath]g(x+y)=g(x)g(y)[/imath]
Let [imath]g[/imath] be a function on [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath] which is not identically zero and which satisfies the equation [imath]g(x+y)=g(x)g(y)[/imath] for [imath]x[/imath],[imath]y[/imath] in [imath]\mathbb R[/imath]. [imath]g(0)=1[/imath]. If [imath]a=g(1)[/imath],then [imath]a>0[/imath] and [imath]g(r)>a^r[/imath] for all [imath]r[/imath] in [imath]\mathbb Q[/imath]. Show that the function is strictly increasing if [imath]g(1)[/imath] is greater than [imath]1[/imath], constant if [imath]g(1)[/imath] is equal to [imath]1[/imath] or strictly decreasing if [imath]g(1)[/imath] is between zero and one, when [imath]g[/imath] is continuous. | 1628538 | Non-linear functional form satisfying [imath]g(x-y) = \frac{g(x)}{g(y)}[/imath].
Is there a general non-linear mapping function family, which obeys this general rule: [imath]g(x-y) = \frac{g(x)}{g(y)}.[/imath] I am looking at identifying a classification of admissible kinetics relaxation functions. |
91665 | Proving [imath]\mathbb{N}^k[/imath] is countable
Prove that [imath]\mathbb{N}^k[/imath] is countable for every [imath]k \in \mathbb{N}[/imath]. I am told that we can go about this inductively. Let [imath]P(n)[/imath] be the statement: “[imath]\mathbb{N}^n[/imath] is countable” [imath]\forall n \in \mathbb{N}[/imath]. Base Case: [imath]\mathbb{N}^1 = \mathbb{N}[/imath] is countable by definition, so [imath]\checkmark[/imath] Inductive Step: [imath]\mathbb{N}^{k+1}[/imath] [imath]“=”[/imath] [imath]\mathbb{N}^k \times \mathbb{N}[/imath] We know that [imath](A,B)[/imath] countable [imath]\implies[/imath] [imath]A \times B[/imath] is countable. I am stuck on the part where I have to prove the rest, but I know that, for example, [imath](1,2,7) \in \mathbb{N}^3 \notin \mathbb{N}^2 \times \mathbb{N}[/imath] but instead [imath]((1,2),7) \in \mathbb{N}^2 \times \mathbb{N}[/imath]. So how would I go about proving the statement. | 1008629 | A question about countable sets.
let [imath]k[/imath] be an integer, and consider the set [imath]S[/imath] of functions from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath] where [imath]g(x) =0[/imath] for all [imath]x>k[/imath]. I want to show [imath]S[/imath] is countable. This is what I have so far, but I am not quite sure if I am going about it correctly, or what to do next. So our functions will look like this [imath]A_1=\{(1,a_1),(2,a_2),(3,a_3),(4,a_4),....,(k,a_k), (k+1,0), (k+2,0)....\}[/imath] It seems from here I should be able to show that there are only countable ways of doing this, but I am not quite sure how to say that. I think the best way would to come up with a bijection from the naturals to [imath]S[/imath] but for the life of me I can't come up with one. I would love some help. Thanks in advance. |
223140 | Let [imath]L_p[/imath] be the complete, separable space with [imath]p>0[/imath].
Let [imath]L_p[/imath] be the complete, separable space with [imath]p>0[/imath]. [imath]\mathbf{J}=\{I = (r,s] \}[/imath] where [imath]r[/imath] and [imath]s[/imath] are rational numbers. [imath]\mathbf{A}[/imath] is the algebra generated by [imath]\mathbf{J}[/imath], with [imath]\mathbf{S}=\operatorname{span}(\mathbf{A})[/imath]. a). Try to verify that [imath]\mathbf{S}[/imath] is dense in [imath]L_p[/imath] space with respect to [imath]L_p[/imath] metric. b). Try to verify that for any [imath]p>0[/imath], [imath]L_p[/imath] is complete. | 610442 | Proving that [imath]L^{p}\left(\mathbb{R},\mu\right) [/imath] is separable for [imath]p\in\left[1,\infty\right) [/imath]
EDIT: As someone in the comments noted the claim is not true for all measure spaces so I have edited the question. I want to show that [imath]L^{p}\left(\mathbb{R},\mu\right)[/imath] (where [imath]\mu[/imath] is the Lebesgue measure) is separable for all [imath]p\in\left[1,\infty\right)[/imath]. I already know that the set [imath]\mathcal{S}[/imath] of simple functions [imath]s[/imath] such that [imath]\mu\left(x\,|\, s\left(x\right)\neq0\right)<\infty[/imath] is dense. So it would suffice to find a countable dense subset of [imath]\mathcal{S}[/imath]. Of course the immediate line of thought is to look at simple functions with rational values but suppose that [imath]s[/imath] receives the rational values [imath]q_{1},...,q_{n}[/imath] on the sets [imath]A_{n}=\left\{ x\,|\, s\left(x\right)=q_{n}\right\}[/imath] there is an uncountable number of ways to define such functions. I would appreciate if someone could help me with constructing a countable dense set of [imath]\mathcal{S}[/imath] Thanks in advance! |
106786 | [imath]A^m\hookrightarrow A^n[/imath] implies [imath]m\leq n[/imath] for a ring [imath]A\neq 0[/imath]
I'm trying to prove that if [imath]A\neq 0[/imath] is a commutative ring and there is an injective [imath]A[/imath]-module homomorphism [imath]A^m\hookrightarrow A^n[/imath] then [imath]m\leq n[/imath] must necessarily hold. This is exercise 2.11 from Atiyah and MacDonald's Introduction to Commutative Algebra, but unfortunately all the solutions available online are either very sparse with regard to this question or seem to use (to my surprise) a generalized version of Cramer's rule. Is there perhaps some other "cleaner" approach to solving this problem? | 310166 | [imath]R[/imath]-linear injection
If [imath]f: R^n\rightarrow R^m[/imath] is an injective map, which is also [imath]R[/imath]-linear, where [imath]R[/imath] is a commutative ring with unity. Is it true that [imath]n[/imath] has to be less than or equal to [imath]m[/imath] always? |
169797 | Write down the sum of sum of sum of digits of [imath]4444^{4444}[/imath]
Let [imath]A = 4444^{4444}[/imath]; Then sum of digits of [imath]A = B[/imath]; Then sum of digits of [imath]B = C[/imath]; Then sum of digits of [imath]C = D[/imath]; Find [imath]D[/imath]. What should be the approach here? | 744811 | If [imath]d(n)[/imath] is the sum of digits of n, find [imath]d(d(d(n)))[/imath] of [imath]n=4444^{4444}[/imath]
If [imath]d(n)[/imath] is the sum of digits of n, find [imath]d(d(d(n)))[/imath] of [imath]n=4444^{4444}[/imath]. My attempt: [imath]4444^{4444}<10000^{4444}[/imath] Now, [imath]\max d(10000^{4444})<9\times 17776[/imath] Again, [imath]\max d(159984)\le 45[/imath] Again [imath]\max d(45)\le 12[/imath] So, [imath]\max d(d(d(n)))\le 12[/imath]. Now, what am I supposed to do? Please help. |
232219 | Extreme points of unit ball in [imath]C(X)[/imath]
Let [imath]X[/imath] be a compact Hausdorff space and [imath]C(X)[/imath] be the space of continuos functions in sup-norm. I read in Douglas' Banach algebra techniques in operator theory that the followings are equivalent: 1)[imath]f\in C(X)[/imath] is an extreme point of the unit ball; and 2)[imath]|f(x)|=1[/imath] for all [imath]x\in X[/imath]. It is easy to show that 2) implies 1). However, I am unable to show the converse. I could show that [imath]f[/imath] is extreme implies [imath]\|f\|=1[/imath] but this is far from what we need. My guess: if 2) is false. We try to construct a nonnegative function [imath]r[/imath] on [imath]X[/imath] which is strictly positive when [imath]|f(x)|\neq 1[/imath] and meanwhile \begin{equation} |(1+r)f|\le 1 \end{equation}on [imath]X[/imath]. Then we can decompose \begin{equation} f=\frac{1}{2}(1+r)f+\frac{1}{2}(1-r)f. \end{equation} If [imath]|f|[/imath] is bounded away from [imath]0[/imath], then [imath]r=-1+1/|f|[/imath] would be a good choice, but I cannot rule out the case when [imath]f[/imath] vanishes at certain points. Can somebody help? Thanks! Also a related problem, Douglas then says these extreme points of the unit ball spans the entire [imath]C(X)[/imath]. Can somebody also give a hint on this? Thanks! | 1176556 | Can continuous functions with range in the unit circle be the average of two different continuous functions with the same range?
Suppose [imath]f:[0,1]\to {\Bbb C}[/imath] is a continous function such that [imath]|f(x)|=1[/imath] for all [imath]x\in[0,1][/imath]. Does there exist different continuous functions [imath]f_1,f_2:[0,1]\to{\Bbb C}[/imath] such that [imath]f=(f_1+f_2)/2[/imath] and [imath]\|f_i\|_\infty=1[/imath] ? I don't know how to get started. Could anyone come up with some hints? |
76743 | Limit of [imath]{a_n}^{1/n}[/imath] is equal to [imath]\lim_{n\to\infty} a_{n+1}/a_n[/imath]
Today my lecturer put up on the board that: If [imath]\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}[/imath] exists and [imath]a_n>0[/imath] then [imath]\displaystyle \limsup\limits_{n\to\infty}\left(a_n^{\frac{1}{n}}\right)=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}[/imath] however I am not sure why this is true, can somebody give me a hint or something as to how to go about proving this. thanks for any help | 561076 | Is [imath]\lim_{n \rightarrow \infty} a_{n+1}/a_n=L \implies \lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L[/imath] true? If not, is there a counter example?
We were told, in recitation class, about a test for sequences convergence (not series) Which goes as follows: if [imath]\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=L[/imath] then [imath]\lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L[/imath]. In a previous question I asked: This limit: [imath]\lim_{n \rightarrow \infty} \sqrt [n] {nk \choose n}[/imath]. I was told that this fact is not true. My question is, can anyone think of a counter example for it? Because, If yes, Then I would like to let my tutor know about it, but, I don't want to doubt him befor I am sure of it. Thank you! |
15875 | When is a function satisfying the Cauchy-Riemann equations holomorphic?
It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if [imath]f[/imath] is continuously differentiable as a function from [imath]U \subset \mathbb{R}^2[/imath] to [imath]\mathbb{R}^2[/imath] and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or Wikipedia). Why is the [imath]C^1[/imath] condition necessary? I don't see where this comes in to the proof below. Assume that [imath]u(x,y)[/imath] and [imath]v(x,y)[/imath] are continuously differentiable and satisfy the Cauchy-Riemann equations. Let [imath]h=h_1 + h_2i[/imath]. Then \begin{equation*} u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|) \end{equation*} and \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|). \end{equation*} Multiplying the second equation by [imath]i[/imath] and adding the two together gives \begin{align*} (u+iv)(z+h)-(u+iv)(z) &= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\ &= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|). \end{align*} Now dividing by [imath]h[/imath] gives us the desired result. Does there exist a differentiable but not [imath]C^1[/imath] function [imath]f: U \rightarrow \mathbb{R}^2[/imath] which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function? | 2088282 | If Cauchy-Riemann equations are satisfied everywhere does that mean the function is analytic?
Let [imath]f(z)=\frac{z^{5}}{|z|^4}[/imath] for ([imath]z\neq 0[/imath]) and [imath]f(z)=0[/imath] for [imath]z=0[/imath] then [imath]u[/imath] and [imath]v[/imath] satisfy Cauchy-Riemann equations everywhere but [imath]f(z)[/imath] is not differentiable at [imath]z=0[/imath] which means its not analytical because for a function to be analytic it must be differnetiable everywhere. I am confused because my teacher said that if [imath]u[/imath] and [imath]v[/imath] satisfy Cauchy-Riemann equations then [imath]f(z)[/imath] is analytic but this example proves that this is false. Am I misunderstanding something or is my teacher wrong? |
124130 | Sum of two closed sets in [imath]\mathbb R[/imath] is closed?
Is there a counterexample for the claim in the question subject, that a sum of two closed sets in [imath]\mathbb R[/imath] is closed? If not, how can we prove it? (By sum of sets [imath]X+Y[/imath] I mean the set of all sums [imath]x+y[/imath] where [imath]x[/imath] is in [imath]X[/imath] and [imath]y[/imath] is in [imath]Y[/imath]) Thanks! | 906440 | If [imath]A[/imath] and [imath]B[/imath] are closed subsets of the set of real numbers, then is [imath]A+B[/imath] closed?
Let [imath]A[/imath] and [imath]B[/imath] be two closed subsets of the set of real numbers. Define [imath]A+B=\{a+b\in\mathbb{R}:a\in A ,b\in B\}[/imath]. Is it true that [imath]A+B[/imath] is closed in [imath]\mathbb{R}[/imath]? If not, could you give a counter-example? |
82682 | Proof of Convergence: Babylonian Method [imath]x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})[/imath]
a) Let [imath]a>0[/imath] and the sequence [imath]x_n[/imath] fulfills [imath]x_1>0[/imath] and [imath]x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})[/imath] for [imath]n \in \mathbb N[/imath]. Show that [imath]x_n \rightarrow \sqrt a[/imath] when [imath]n\rightarrow \infty[/imath]. I have done it in two ways, but I guess I'm not allowed to use the first one and the second one is incomplete. Can someone please help me? We already know [imath]x_n \rightarrow \sqrt a[/imath], so we do another step of the iteration and see that [imath]x_{n+1} = \sqrt a[/imath]. Using limit, [imath]x_n \rightarrow x, x_{n+1} \rightarrow x[/imath] (this is the part I think it's incomplete, don't I have to show [imath]x_{n+1} \rightarrow x[/imath], how?), we have that [imath]x = \frac x 2 (1 + \frac a {x^2}) \Rightarrow 1 = a/x^2 \Rightarrow x = \sqrt a[/imath] b) Let the sequence [imath]x_n[/imath] be defined as [imath]x_{n+1}= 1 + \frac 1 {x_n} (n \in \mathbb N), x_1=1[/imath]. Show that it converges and calculate it's limit. "Tip: Show that sequences [imath]x_{2n}[/imath] and [imath]x_{2n+1}[/imath] monotone convergent to the limit." I didn't understand the tip, how can this help me? Does it make a difference if the number is odd or even? Thanks in advance! | 436547 | Recurrence relation: [imath]x_{n+1} = \frac 12 x_n + \frac 1 {x_n},[/imath]
[imath]x_{n+1} = \frac 12 x_n + \frac 1 {x_n}, x_0 \neq 0[/imath] [imath] a = \frac a2 + \frac 1a \Rightarrow a = \frac {a^2 + 2} {2a} \Rightarrow 2a^2 = a^2 + 2 \Rightarrow a^2 = 2 \Rightarrow a = \pm \sqrt 2[/imath] If [imath]x_0 > 0[/imath], then any subsequent terms will be positive too. If [imath]x_0 < 0[/imath] , then - negative. Hence, if [imath]x_0 > 0[/imath], then [imath]\lim_{n \rightarrow \infty} x_n = \sqrt 2[/imath]. If [imath]x_0 < 0[/imath], then [imath]\lim_{n \rightarrow \infty} x_n = - \sqrt 2[/imath] Is it complete solution? I know that it's correct answer, but may be I missed something in proof? |
141521 | Examples of perfect sets.
Let [imath]0\lt a\lt 1[/imath]. Can I get examples of of subsets of [imath][0,1][/imath] that are perfect sets, contains no intervals and has measure [imath]1-a[/imath]. Well, I know by construction the Cantor set is perfect, contains no intervals. However, it has measure [imath]0[/imath]. So it's no good. | 406453 | An example of a generalized Cantor set with positive Lebesgue measure
I want to know if there exist a set [imath] X\subset \mathbb R[/imath] such that [imath]X[/imath] is [imath]i)[/imath] Perfect [imath]ii)[/imath] Compact [imath]iii)[/imath] Has empty interior [imath]iv)[/imath] Totally disconnected [imath]v)[/imath] Is not countable But [imath]X[/imath] has positive Lebesgue measure. The sets that are defined with the above properties are called generalized Cantor sets. Please could you tell me how to construct an explicit example? |
303760 | Is [imath]\sum \sin^2(k)/k[/imath] Convergent?
A student recently used the series [imath]\displaystyle\sum_{k=1}^\infty\frac{\sin^2k}{k}[/imath] as an example of a divergent series whose terms tend to [imath]0[/imath]. However, I'm having trouble convincing myself that this series does in fact converge. Anyone have any ideas? | 273559 | Convergence of [imath]\sum_{n=1}^\infty \frac{\sin^2(n)}{n}[/imath]
Does the series [imath] \sum_{n=1}^\infty \frac{\sin^2(n)}{n} [/imath] converge? I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance. |
238997 | Prove the divergence of the sequence [imath]\left\{ \sin(n) \right\}_{n=1}^{\infty}[/imath].
I am looking for nice ways of proving the divergence of the sequence [imath]\left\{x_n\right\}_{n=1}^{\infty}[/imath] defined by [imath]x_n=\sin{(n)}.[/imath] One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals [imath]I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)[/imath] and one where they lie in [imath]J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).[/imath] If [imath]{x_n}[/imath] converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval [imath]\left[ \frac{1}{2}, 1\right][/imath] while the second has its values in [imath]\left[-1,-\frac{1}{2}\right][/imath]. Contradiction. Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea? | 1430155 | Divergence of [imath]\sin(n)[/imath]
Using only the standard analytic properties of the sine function, but no particular definition of it, what's a neat little proof that the sequence [imath]\sin(n)[/imath] diverges? |
90746 | If [imath]Y[/imath] is connected, why is [imath]A\cup Y[/imath] connected in this case?
If [imath](X,\mathcal{T})[/imath] is a connected space, and [imath]Y[/imath] a connected subset, and [imath]X\setminus Y=A\cup B[/imath] for separated sets [imath]A[/imath] and [imath]B[/imath], then why is [imath]A\cup Y[/imath] connected as well? Thank you kindly. | 2019419 | prove the connectedness of a set
Suppose that [imath]Y\subset X[/imath] are connected spaces and suppose that [imath]A[/imath] and [imath]B[/imath] are separation of [imath]X-Y[/imath]. Prove that the union [imath]Y\cup A[/imath] is connected. I want to prove by contradiction.Firstly I assume there exists a separation of [imath]Y\cup A[/imath],say [imath]C\cup D[/imath] .Then since [imath]Y[/imath] is connected,then [imath]Y[/imath] must belong to [imath]C[/imath] or [imath]D[/imath]. Whithout loss of generality,we can suppose [imath]Y\subset C[/imath].I guess [imath]C\cup B[/imath] and [imath]D\cup A[/imath] should be the separation of [imath]X[/imath].But I don't know how to prove it in formal mathematics. |
41878 | Why is the even root of a number always positive?
Let [imath]n \in \mathbb N[/imath] be a natural number and [imath]a \in \mathbb R[/imath] be a real number. The [imath]n[/imath]-th root of the number [imath]a[/imath] is defined as follows: Case I: [imath]n[/imath] is an odd number. In this case the [imath]n^{\text{th}}[/imath] root of [imath]a[/imath] is defined to be that number [imath]b \in \mathbb R[/imath] such that [imath]b^n = a[/imath]. Case II: [imath]n[/imath] is an even number. In this case the [imath]n^{\text{th}}[/imath] root of [imath]a[/imath] is defined to be that number [imath]b \geq 0[/imath] such that [imath]b^n = a[/imath]. Why is it that when [imath]n[/imath] is even, we only consider [imath]b \geq 0[/imath]. For example, both [imath]+2[/imath] and [imath]-2[/imath] squared equal [imath]4[/imath], but when we say the square root of [imath]4[/imath] is [imath]2[/imath]. Is there a reason for this? | 954153 | why square root of a positive number is positive?
We have [imath](+3)^2=(-3)^2=9[/imath]. But why do we define [imath]\sqrt 9=+3?[/imath] Why [imath]\sqrt9=-3[/imath] is false? Thank you |
91318 | Proving the Cantor Pairing Function Bijective
How would you prove the Cantor Pairing Function bijective? I only know how to prove a bijection by showing (1) If [imath]f(x) = f(y)[/imath], then [imath]x=y[/imath] and (2) There exists an [imath]x[/imath] such that [imath]f(x) = y[/imath] How would you show that for a function like the Cantor pairing function? | 1364774 | Prove that [imath]C(m,n)=\frac{(m+n)(m+n+1)}{2}+m[/imath] is a bijection from [imath]\mathbb{N}^2[/imath] to [imath]\mathbb{N}[/imath]
I've been stuck for a couple of hours on how to prove that [imath]C(m,n)=\frac{(m+n)(m+n+1)}{2}+m[/imath] is a bijection from [imath]\mathbb{N}^2[/imath] to [imath]\mathbb{N}[/imath]. I read in another question that in order to prove that it is injective, you have to show: [imath]m+n<m'+n'\Rightarrow C(m,n)<C(m',n')[/imath] From where [imath]C(m,n)=C(m',n')\Rightarrow m+n=m'+n'[/imath], and then [imath]m=m',n=n'[/imath]. However, I failed to show the first implication, and I also couldn't understand how to go from [imath]m+n=m'+n'[/imath] to [imath]m=m',n=n'[/imath]. Would someone be kind enough to give me some tips on how to proceed? |
113674 | Proof of a simple property of real, constant functions.
I recently came across the following theorem: [imath] \forall x_1, x_2 \in \mathbb{R},\textrm{function, } f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto y; \ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2 \implies f \textrm{ is constant.}\ \mathbf{(1)} [/imath] I've been trying for some time, but the proof of [imath]\mathbf{(1)}[/imath] remains as elusive as ever. I've made two major attempts, the second of which I'll outline here. Though, I would be glad to detail the first as well if requested, I won't now since I think it's mostly wrong. But for the second, this is what I have so far: If, [imath]\forall x_1,\ x_2,\ |f(x_1) - f(x_2)| \leq (x_1-x_2)^2[/imath], then [imath]f[/imath] is continuous. This is so as [imath]f[/imath] is defined for all reals, [imath](\forall x \in \mathbb{R})\ f[/imath] has finite limits, and each of those limits equals [imath]f(x)[/imath]. Assume [imath]f[/imath] wasn't constant, then [imath]\exists x_1,\ x_2 \ni x_1 \neq x_2 \implies |f(x_1)- f(x_2)| > 0[/imath]. Since [imath]f[/imath] is continuous, there exist an infinity of such pairs, [imath]x_1[/imath] and [imath]x_2[/imath]. For all such [imath]x_1[/imath] and [imath]x_2[/imath], we may construct a set, [imath]S[/imath], consitsting of [imath]f(x_1)[/imath] and [imath]f(x_2)[/imath] (not as pairs); since f is defined for all [imath]x,\ S[/imath] is "absolutely" bounded and as such has a least upper bound and and greatest lower bound, which we will denote as [imath]\alpha_1\ = f(a_1)[/imath] and [imath]\alpha_2 = f(a_2)[/imath] respectively. To show [imath]f[/imath] is constant, it will suffice to show that [imath]\alpha_1 = \alpha_2[/imath]. Does anyone see how the proof could be completed? Or even, do you think there might be a better approach? Thank you all in advance. | 334917 | Suppose that the function [imath]f[/imath] satisfies |[imath]f(x)-f(t)[/imath]| [imath]\le (x-t)^2[/imath] for each [imath]x[/imath], [imath]t[/imath] on the reals. Prove that [imath]f[/imath] must be a constant function.
Suppose that the function [imath]f[/imath] satisfies |[imath]f(x)-f(t)[/imath]| [imath]\le (x-t)^2[/imath] for each [imath]x[/imath], [imath]t[/imath] on the reals. Prove that [imath]f[/imath] must be a constant function. I believe I'm close as I can get to the answer without actually being able to put it together, and it's very frustrating. |
306477 | Limit [imath](x,y)\to (0,0) (x^2 \cdot y^2)/(x^3 + y^3)[/imath]
I'm not sure how to solve this problem. As [imath](x,y)\rightarrow(0,0)[/imath], [imath]\frac{x^2\cdot y^2}{x^3+y^3} = L[/imath] With [imath]y=ax[/imath], I found [imath](a²x)/(a³+1)[/imath] so [imath]a[/imath] cannot be -1. Does this prove the that limit does not exist? I tried [imath]y³= A(x)-x³[/imath] and got nothing. Does the limit exist? Sorry for any mistakes, English is not my native language. | 220453 | Limit of [imath]\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y^2}{x^3+y^3}[/imath]
I want to know whether [imath]\lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^3+y^3}[/imath] exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example, [imath]f(x,mx^2) = \dfrac{m^2x^3}{1+m^3x^3}[/imath] But that doesn't show that the limit is 0. |
52651 | What is the name of the vertical bar in [imath](x^2+1)\vert_{x = 4}[/imath] or [imath]\left.\left(\frac{x^3}{3}+x+c\right) \right\vert_0^4[/imath]?
I've always wanted to know what the name of the vertical bar in these examples was: [imath]f(x)=(x^2+1)\vert_{x = 4}[/imath] (I know this means evaluate [imath]x[/imath] at [imath]4[/imath]) [imath]\int_0^4 (x^2+1) \,dx = \left.\left(\frac{x^3}{3}+x+c\right) \right\vert_0^4[/imath] (and I know this means that you would then evaluate at [imath]x=0[/imath] and [imath]x=4[/imath], then subtract [imath]F(4)-F(0)[/imath] if finding the net signed area) I know it seems trivial, but it's something I can't really seem to find when I go googling and the question came up in my calc class last night and no one seemed to know. Also, for bonus internets; What is the name of the horizontal bar in [imath]\frac{x^3}{3}[/imath]? Is that called an obelus? | 295612 | What does the long '|' mean?
I have often seen the long '|' symbol with a subscript expression afterward. What does this mean in mathematics? Here is an example I found from Wikipedia: [imath]\large\left.\frac{dy}{dx}\right|_{x=c} \;\;= \;\;\;\;\left.\frac{dy}{du}\right|_{u=g(c)}\cdot \;\;\;\;\left.\frac{du}{dx}\right|_{x=c}[/imath] |
298067 | Show that [imath]\mathcal{O}^+_K[/imath] contains [imath]\mathcal{O}_K[/imath], and that the discriminant [imath]\Delta(K)[/imath] is the index [imath][\mathcal{O}^+_K : O_K][/imath].
Let [imath]K[/imath] be a number field, let [imath]\mathcal{O}_K[/imath] be its ring of integers, and let [imath]B = \{b_1,\ldots,b_d\}[/imath] be a subset of [imath]K[/imath] of cardinality [imath]d[/imath] such that [imath]\mathcal{O}_K = \mathbb{Z}b_1+\cdots+\mathbb{Z}b_d[/imath]. Define the trace-dual basis [imath]B^+[/imath] by the property [imath]\mathrm{Tr}(b_ib'_j)=\delta_{i,j}[/imath] (Kronecker delta). i) Show that the [imath]\mathbb{Z}[/imath]-span [imath]\mathbb{Z}b'_1+\cdots+ \mathbb{Z}b'_d[/imath] of [imath]B^+[/imath] does not depend on the choice of [imath]B[/imath]. This abelian group is called the trace dual of [imath]O_K[/imath], and we denote it by [imath]O^+_K[/imath]. ii) Show that [imath]\mathcal{O}^+_K[/imath] contains [imath]\mathcal{O}_K[/imath], and that the discriminant [imath]\Delta(K)[/imath] is the index [imath][\mathcal{O}^+_K : O_K][/imath]. | 297197 | Show that the [imath]\mathbb{Z}[/imath]-span [imath]\mathbb{Z}b'_1+\cdots+ \mathbb{Z}b'_d[/imath] of [imath]B^+[/imath] does not depend on the choice of [imath]B[/imath]
Let [imath]K[/imath] be a number field, let [imath]\mathcal{O}_K[/imath] be its ring of integers, and let [imath]B = \{b_1,\ldots,b_d\}[/imath] be a subset of [imath]K[/imath] of cardinality [imath]d[/imath] such that [imath]\mathcal{O}_K = \mathbb{Z}b_1+\cdots+\mathbb{Z}b_d[/imath]. Define the trace-dual basis [imath]B^+[/imath] by the property [imath]\mathrm{Tr}(b_ib'_j)=\delta_{i,j}[/imath] (Kronecker delta). i) Show that the [imath]\mathbb{Z}[/imath]-span [imath]\mathbb{Z}b'_1+\cdots+ \mathbb{Z}b'_d[/imath] of [imath]B^+[/imath] does not depend on the choice of [imath]B[/imath]. This abelian group is called the trace dual of [imath]O_K[/imath], and we denote it by [imath]O^+_K[/imath]. ii) Show that [imath]\mathcal{O}^+_K[/imath] contains [imath]\mathcal{O}_K[/imath], and that the discriminant [imath]\Delta(K)[/imath] is the index [imath][\mathcal{O}^+_K : O_K][/imath]. |
43314 | Failure of isomorphisms on stalks to arise from an isomorphism of sheaves
It is well-known (Hartshorne 2.1.1) that if [imath]F[/imath] and [imath]G[/imath] are sheaves on a space [imath]X[/imath], then [imath]\phi:F\rightarrow G[/imath] is an isomorphism if and only if the induced stalk map [imath]\phi_p:F_p\rightarrow G_p[/imath] is an isomorphism for all [imath]p\in X[/imath]. However, if we have a collection of isomorphisms [imath]\{\psi_p:F_p\rightarrow G_p\}_{p\in X}[/imath], this does not guarantee that [imath]F[/imath] and [imath]G[/imath] are isomorphic, because the [imath]\psi_p[/imath] might not be related to each other, i.e. there might not be a sheaf map [imath]\psi:F\rightarrow G[/imath] such that [imath]\psi_p[/imath] is the induced stalk map for all [imath]p\in X[/imath]. However, I was recently making this point to someone and was unable to think of a good example of non-isomorphic [imath]F[/imath] and [imath]G[/imath] having isomorphisms [imath]\psi_p:F_p\rightarrow G_p[/imath]. I'm sure I knew one at some point, but I'm blanking on it now. Can someone provide an illustrative example, e.g. an example that occurs in some natural or basic problem, or one that captures the essential pattern of any example where this issue arises, or one where it is clear that [imath]F[/imath] and [imath]G[/imath] could not be isomorphic? | 596622 | Isomorphism of sheaves
Let [imath]X[/imath] be a scheme and [imath]\mathscr{F}[/imath], [imath]\mathscr{G}[/imath] sheaves on [imath]X[/imath]. Suppose we know that for every [imath]x\in X[/imath] the stalks are isomorphic: [imath] \mathscr{F}_x \cong \mathscr{G}_x [/imath] Can we conclude immediately from here that [imath]\mathscr{F}\cong \mathscr{G}[/imath], or do we first need to find a morphism of sheaves [imath]f:\mathscr{F}\to \mathscr{G}[/imath] such that [imath]f_x[/imath] induces the above isomorphisms on the stalks for every [imath]x[/imath] ? Remark: Proposition 1.1 (page 63) of Hartshorne says that: A morphism of sheaves [imath]f : \mathscr{F} \to \mathscr{G}[/imath] on a topological space [imath]X[/imath] is an isomorphism iff the induced map on stalks [imath]f_x : \mathscr{F}_x \to \mathscr{G}_x[/imath] is an isomorphism for every [imath]x\in X[/imath]. Extra question: What can we say, instead, if we have isomorphisms between every fiber of [imath]\mathscr{F}[/imath] and [imath]\mathscr{G}[/imath] ? |
61755 | Is there a way to solve for an unknown in a factorial?
I don't want to do this through trial and error, and the best way I have found so far was to start dividing from 1. [imath]n! = \text {a really big number}[/imath] Ex. [imath]n! = 9999999[/imath] Is there a way to approximate n or solve for n through a formula of some sort? Update (Here is my attempt): Stirling's Approximation: [imath]n! \approx \sqrt{2 \pi n} \left( \dfrac{n}{e} \right ) ^ n[/imath] So taking the log: [imath](2\cdot \pi\cdot n)^{1/2} \cdot n^n \cdot e^{-n}[/imath] [imath](2\cdot \pi)^{1/2} \cdot n^{1/2} \cdot n^n \cdot e^{-n}[/imath] [imath].5\log (2\pi) + .5\log n + n\log n \cdot -n \log e[/imath] [imath].5\log (2\pi) + \log n(.5+n) - n[/imath] Now to solve for n: [imath].5\log (2\pi) + \log n(.5+n) - n = r[/imath] [imath]\log n(.5+n) - n = r - .5 \log (2\pi)[/imath] Now I am a little caught up here. | 1966379 | What is the reverse operation of a factorial?
Is there a mathematical function [imath]f[/imath] that takes [imath]n![/imath] as an input, and returns [imath]n[/imath]? For example, [imath]f(720) = 6[/imath] |
159707 | Is there any geometric way to characterize [imath]e[/imath]?
Let me explain it better: after this question, I've been looking for a way to put famous constants in the real line in a geometrical way -- just for fun. Putting [imath]\sqrt2[/imath] is really easy: constructing a [imath]45^\circ[/imath]-[imath]90^\circ[/imath]-[imath]45^\circ[/imath] triangle with unitary sides will make me have an idea of what [imath]\sqrt2[/imath] is. Extending this to [imath]\sqrt5[/imath], [imath]\sqrt{13}[/imath], and other algebraic numbers is easy using Trigonometry; however, it turned difficult working with some transcendental constants. Constructing [imath]\pi[/imath] is easy using circumferences; but I couldn't figure out how I should work with [imath]e[/imath]. Looking at made me realize that [imath]e[/imath] is the point [imath]\omega[/imath] such that [imath]\displaystyle\int_1^{\omega}\frac{1}{x}dx = 1[/imath]. However, I don't have any other ideas. And I keep asking myself: Is there any way to "see" [imath]e[/imath] geometrically? And more: is it true that one can build any real number geometrically? Any help will be appreciated. Thanks. | 308462 | Is there a simple geometrical description of [imath]e[/imath]?
Of course I am not looking for a definition through [imath]\int_1^e{1\over x} \, \mathrm{d}x=1[/imath] or that slope of [imath]a^x[/imath] at [imath]x=0[/imath] is [imath]1[/imath] when [imath]a=e[/imath]. I am looking for something understandable by a kid who has begun comprehending [imath]\pi[/imath] as the ratio of circumference to diameter of a circle. Or perhaps by one who is a couple of years older. (And of course there is no reason to suspect that every mathematical constant has a simple geometrical description. But a "definition" that might not be suitable for a calculus text could be suitable for introduction to laymen.) Edit 1: One area I would find interesting would be a definition that uses the entire hyperbola [imath]xy=1[/imath] instead of pieces of it. Of course both area (between hyperbola and asymptotes) and the length, measured in the usual fashion, will be infinity. I have attempted projecting the shape onto a sphere to see if I get a number similar to [imath]e[/imath]; but with no luck. |
118536 | Prove the map has a fixed point
Assume [imath]K[/imath] is a compact metric space with metric [imath]\rho[/imath] and [imath]A[/imath] is a map from [imath]K[/imath] to [imath]K[/imath] such that [imath]\rho (Ax,Ay) < \rho(x,y)[/imath] for [imath]x\neq y[/imath]. Prove A have a unique fixed point in [imath]K[/imath]. The uniqueness is easy. My problem is to show that there a exist fixed point. [imath]K[/imath] is compact, so every sequence has convergent subsequence. Construct a sequence [imath]{x_n}[/imath] by [imath]x_{n+1}=Ax_{n}[/imath],[imath]\{x_n\}[/imath] has a convergent subsequence [imath]\{ x_{n_k}\}[/imath], but how to show there is a fixed point using [imath]\rho (Ax,Ay) < \rho(x,y)[/imath]? | 2791153 | Existence of unique fixed point in compact Metric space
Let [imath](X,d)[/imath] be compact. Show: for a map [imath]f[/imath] that when [imath]\forall x, y \in X[/imath] with [imath]x\neq y[/imath] [imath]d(f(x),f(y))<d(x,y)[/imath] is fulfilled. Then [imath]f[/imath] has a unique fixed point. |
24429 | on the boundary of analytic functions
Suppose I have a function [imath]f[/imath] that is analytic on the unit disk [imath]D = \{ z \in \mathbb{C} : |z| < 1 \}[/imath] that is also continuous up to [imath]\bar{D}[/imath]. If [imath]f[/imath] is identically zero on some segment of of the boundary (e.g. [imath]\{ e^{it}, 0 \leq t \leq \pi/2 \}[/imath] ), is it then true that [imath]f[/imath] is identically zero on the entire boundary? I know that analytic functions that are zero at an accumulation point inside the domain of analyticity are identically zero throughout the entire domain, but I don't know what (if anything) can be said if something similar occurs on the boundary of the domain. | 407070 | Show that [imath]f(z)=0[/imath].
Suppose that [imath]f:\mathbb{C}\rightarrow\mathbb{C}[/imath] is analytic on the open unit disc and continuous on the closed unit disc. Assume that [imath]f(z)=0[/imath] on an arc of the circle [imath]\{z\in\mathbb{C}:|z|=1\}[/imath]. Show that [imath]f(z)=0[/imath]. |
41667 | Fibonacci, tribonacci and other similar sequences
I know the sequence called the Fibonacci sequence; it's defined like: [imath]\begin{align*} F_0&=0\\ F_1&=1\\ F_2&=F_0+F_1\\ &\vdots\\ Fn&=F_{n-1} + F_{n-2}\end{align*}[/imath] And we know that there's Binet formula for computing [imath]n[/imath]-th element in the sequence. However, I'm trying to find something totally different. We know that [imath]K=2[/imath] for the Fibonacci sequence; let's call [imath]K[/imath] the number of previous elements to get the [imath]n[/imath]-th element. For example, [imath]\begin{align*} K=2&\Rightarrow F_n= F_{n-1} + F_{n-2},\\ K=3&\Rightarrow F_n= F_{n-1} + F_{n-2} + F_{n-3},\\ \end{align*}[/imath] and so on. How to compute the [imath]n[/imath]-th element for given [imath]K[/imath]? I couldn't find any formula for [imath]K > 2[/imath]. Thanks for any help. | 827565 | How to find the nth term of tribonacci series
I want to know the nth term of tibonacci series, given by the recurrence relation [imath] a_{n + 3} = a_{n + 2} + a_{n + 1} + a_n [/imath] with [imath]a_1 = 1, a_2 = 2, a_3 = 4[/imath], so the first few terms are [imath] 1,2,4,7,13,24,44, \ldots [/imath] I am more interested in derivation of how the nth term is calculated rather than direct comming up with black box formula. |
301771 | Prove that if [imath]n[/imath] is a composite, then [imath]2^n-1[/imath] is composite.
Not sure if I'm doing this correctly but this is what I've done: Assume that [imath]n[/imath] is composite and suppose [imath]2^n-1[/imath] is a prime for [imath]n \gt 2[/imath]. Then, [imath]2^n-1 = 2k[/imath] for some [imath]k \in \Bbb Z [/imath], [imath]\forall n[/imath]. But this would be a contradiction since [imath]n \gt 2[/imath]. I'm not sure if this a correct proof. EDIT: I realized that it's to prove [imath]2^n - 1[/imath] NOT [imath]2^{n-1}[/imath] so my attempt above is completely wrong. | 2237162 | Showing [imath]2^n -1[/imath] is composite when n is composite
I'd like to show that if [imath]n[/imath] is a composite number, then so is [imath]2^n-1[/imath]. When [imath]n[/imath] is an even number, this is simple. Let [imath]n[/imath] = 2k. Then, [imath]2^n -1 = (2^k)^2- 1 = (2^k -1)(2^k + 1)[/imath]. For the case where [imath]3|n[/imath], this is also not too hard. For [imath]n = 6[/imath], we have [imath]2^6 -1 = 63 = 7 \cdot 9[/imath]. Assume [imath]7|(2^{3k} -1)[/imath]. Then, [imath]2^{3(k +1)} -1 = 7\cdot2^{3k} + (2^{3k} -1) [/imath] which is divisible by 7. So, we can show that this case also leads to a composite number via induction. However, I can't seem to generalize it for an arbitrary [imath]n[/imath]. Any hints to the right direction is much appreciated. |
624 | Balance chemical equations without trial and error?
In my AP chemistry class, I often have to balance chemical equations like the following: [imath] \mathrm{Al} + \text O_2 \to \mathrm{Al}_2 \mathrm O_3 [/imath] The goal is to make both side of the arrow have the same amount of atoms by adding compounds in the equation to each side. A solution: [imath] 4 \mathrm{Al} + 3 \mathrm{ O_2} \to 2 \mathrm{Al}_2 \mathrm{ O_3} [/imath] When the subscripts become really large, or there are a lot of atoms involved, trial and error is impossible unless performed by a computer. What if some chemical equation can not be balanced? (Do such equations exist?) I tried one for a long time only to realize the problem was wrong. My teacher said trial and error is the only way. Are there other methods? | 2694438 | Solve system of equations with no constant terms
I am trying to balance a chemical equation algebraically and have found the resulting system of equations: [imath]\begin{align*} 3a+8b+d&=12e+4f+2g\\ a&=e\\ 4a+4c+3d&=4e+3f+g+3h\\ 2b+d&=3e+f\\ c&=h \end{align*}[/imath] I am able to reduce this to a system of three equations: [imath]\begin{align*} 8b+d&=9e+4f+2g\\ c+3d&=3f+g\\ 2b+d&=3e+f \end{align*}[/imath] Now, I recognize that there are more variables than there are equations, and so there is no single solution to this equation. However, I need a solution with reasonably low numbers and only integer values for all the variables ([imath]a,b,c,d,e,f,g,h[/imath] or, by substitution, [imath]b,c,d,e,f,g[/imath]). A matrix solution won't work since (in standard [imath]AX=B[/imath] form) the only elements in [imath]B[/imath] are zero and therefore all the variables are zero. As a side note, the reaction is: [imath]H_3PO_4 + (NH_4)_2 + MoO_4 + HNO_3 \longrightarrow (NH_4)_3PO_4 + MoO_3 + NH_4NO_3 + H_2O[/imath] If there is a better way to go about this (besides inspection; I have been trying that for hours) please let me know. Any ideas? |
45976 | Connectivity, Path Connectivity and Differentiability
I have two questions which pertain to differentiability, connectivity and path connectivity. Ocasionally, I will encounter an author who defines connectivity in the following way: An open subset [imath]U[/imath] of [imath]\mathbb{R}^n[/imath] is said to be connected if and only if given two points [imath]a[/imath] and [imath]b[/imath] of [imath]U[/imath] there exists a differentiable mapping [imath]\phi: \mathbb{R} \rightarrow U[/imath] such that [imath]\phi(0) = a[/imath] and [imath]\phi(1) = b[/imath]. This particular example is from Edward's Advanced Calculus of Several Variables p 84. Now, this is obviously not the standard definition we learn from topology which has nothing to do with differentiability but rather whether there exists two nonempty open subsets that comprise a separation. It also seems to me that what the author is really defining what it means for a space to be "smoothly path connected", which of course implies connectivity and, it seems to me, considerably more. My first question is: Is "smoothly path connected", as defined above actually equivalent to "connected", in the topological sense, in [imath]\mathbb{R}^n[/imath]? Next, in Vector Calculus by Baxandall and Liebeck on p 150 the authors state the existence of a continuous path [imath]\alpha[/imath] from the closed interval [imath][0,1][/imath] to an open subset [imath]D[/imath] of [imath]\mathbb{R}^n[/imath] where [imath]\alpha (0) = a \in D[/imath] and [imath]\alpha(1) = b \in D[/imath] guarantees the existence of a differentiable path with the same properties. This claim is stated without proof. My second question is: Can someone provide a reference to a proof of the above claim or explain why it is so? | 2398824 | Smooth path connecting two points in a manifold
I received the following question. It's also in Chapter 2 of Spivak's book on Differential Geometry, Volume 1. Given a smooth connected manifold [imath]M[/imath], prove that between any two points [imath]p,q \in M[/imath], there exists a smooth curve [imath]c : [0,1] \to M[/imath] such that [imath]c(0) = p, c(1) = q[/imath]. Furthermore, [imath]c[/imath] can be chosen to be one-one. How I propose to do this problem is as follows : I would like to find a common chart(open neighbourhood homeomorphic to [imath]\mathbb R^n[/imath]) containing [imath]p[/imath] and [imath]q[/imath]. This is done as follows : take a chart [imath]C_p[/imath] containing [imath]p[/imath], and take a path connecting [imath]p[/imath] and [imath]q[/imath] (This must exist by the definition of a manifold as a second countable Hausdorff space, but I'm unable to say why). Around this path, consider a contractible neighbourhood [imath]N[/imath] (Can I do this?). Now, [imath]N \cup C_p[/imath] is a chart containing [imath]p[/imath] and [imath]q[/imath]. Therefore, since this neighbourhood is homeomorphic to [imath]\mathbb R^n[/imath], I can find a 1-1 smooth path connecting the images of [imath]p[/imath] and [imath]q[/imath] in [imath]\mathbb R^n[/imath] (this is a standard property of [imath]\mathbb R^n[/imath], easily provable), and pulling this back into [imath]M[/imath] gives me the [imath]c[/imath] I desire, which is also [imath]1-1[/imath]. Is this correct? EDIT : This question is different from a proposed duplicate, given in the comment below, since here I am attempting the question in a way different from how it been done in the duplicate, and furthermore I am asking for a critique of my attempts, not the answer itself. When I am confident, I will answer this question myself and close it. |
32689 | Isomorphism between [imath]I_G/I_G^2[/imath] and [imath]G/G'[/imath]
Ok, this has been bugging me for a while, and I'm sure there's something obvious I'm missing. The references I've looked at for this result in an effort to resolve the issue didn't address it. [imath]G[/imath] is a group, [imath]\mathbb{Z}[G][/imath] its integral group ring, [imath]I_G[/imath] the augmentation ideal (i.e. the kernel of the map [imath]\mathbb{Z}[G]\rightarrow\mathbb{Z}[/imath] sending all group elements to 1), and [imath]G/G'[/imath] is the abelianization of [imath]G[/imath]. I'm attempting to show that [imath]I_G/I_G^2[/imath] is isomorphic to [imath]G/G'[/imath]. It's straightforward to show that [imath](g-1)+(h-1)\equiv(gh-1)\bmod I_G^2\hskip0.5in(*)[/imath] and thus every equivalence class mod [imath]I_G^2[/imath] is equal to some [imath](g-1)+I_G^2[/imath]. Now all we have to do is define a map [imath]\phi:G/G'\rightarrow I_G/I_G^2[/imath] and show that it has an inverse [imath]\psi:I_G/I_G^2\rightarrow G/G'[/imath]. The definitions are obvious enough: [imath]\phi(gG')=(g-1)+I_G^2[/imath] [imath]\psi((g-1)+I_G^2)=gG'[/imath] and starred equation shows that these are homomorphisms. But! My problem is showing that these maps are well-defined. For example, if [imath](g-1)+I_G^2=(h-1)+I_G^2[/imath], i.e. [imath](g-1)-(h-1)\equiv (gh^{-1}-1)\equiv0\bmod I_G^2,[/imath] we need to show that [imath]\psi((g-1)+I_G^2)=gG'=hG'=\psi((h-1)+I_G^2),[/imath] i.e. [imath]gh^{-1}\in G'[/imath]. If we have [imath]gh^{-1}\in G'[/imath], i.e. [imath]gh^{-1}[/imath] equals some [imath]\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}[/imath], then [imath]gh^{-1}-1=\left(\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}\right)-1,[/imath] and, unwinding using the starred equation, [imath]\left(\prod_{a,b\in G} (aba^{-1}b^{-1})^{n_{a,b}}\right)-1\equiv \sum n_{a,b}(aba^{-1}b^{-1}-1)\equiv[/imath] [imath]\sum n_{a,b}\left[(ab-1)-(a-1)-(b-1)\right]=\sum n_{a,b}(a-1)(b-1)\equiv0\bmod I_G^2[/imath] but for some reason I can't make this work the other way. I'm sure I'm being silly; someone please point out where. | 275249 | Quotient of ideal in group ring is isomorphic to abelianization
Let [imath]G[/imath] be a group and [imath]\mathbb Z G[/imath] the group ring over the integers. Let [imath]I[/imath] be the ideal of elements [imath]\sum_{g\in G} n_g g[/imath] with [imath]\sum_{g\in G} n_g = 0[/imath]. I am trying to prove that [imath]I/I^2[/imath] is isomorphic (as an additive abelian group) to the abelianization [imath]G/[G,G][/imath]. I have tried showing that it satisfies the universal property of the abelianization, but this does not seem to work. I also tried constructing an explicit isomorphism, but could not find one. Any ideas? |
54507 | [imath]|e^a-e^b| \leq |a-b|[/imath] for complex numbers with non-positive real parts
Came across this problem on an old qualifying exam: Let [imath]a[/imath] and [imath]b[/imath] be complex numbers whose real parts are negative or 0. Prove the inequality [imath]|e^a-e^b| \leq |a-b|[/imath]. If [imath]f(z)=e^z[/imath] and [imath]z=x+iy[/imath], then [imath]|f'(z)|=e^x\leq 1[/imath] given that [imath]x \leq 0[/imath]. I played around with the limit definition of the derivative, but wasn't able to get anywhere. Not sure what else to try; a hint would be very helpful! | 608571 | [imath]|\mathrm{e}^{z}- \mathrm{e}^{w}| \leq |z-w|[/imath] for [imath]\{z \in \mathbb{C} : \mathrm{Re}z \leq 0 \}[/imath]
I'm looking at this pretty easy problem, but I need the final argument for the mean-value property that I'm looking for. The problem goes like this. Prove that if [imath]V = \{z \in \mathbb{C} : \mathrm{Re} z \leq 0 \}[/imath], then [imath]|\mathrm{e}^{z}- \mathrm{e}^{w}| \leq |z-w|[/imath] for [imath]z,w \in V[/imath]. So my idea is obviously to apply the mean-value theorem for [imath]f(z) = \mathrm{e}^z[/imath] and bound it by [imath]\left| \frac{f(z) - f(w)}{z-w}\right|\leq 1.[/imath] But how do I argue that [imath]\frac{f(z) - f(w)}{z-w} \leq f'(0)?[/imath] This would finish the problem immediately. |
3776 | Limit of a particular variety of infinite product/series
I was musing about a particular limit, [imath]L = \prod\limits_{n > 0} \bigl(1 - 2^{-n}\bigr)[/imath]: we may bound 0.288 < L < 0.308, which we may show by taking the logarithm: [imath]\ln(L) = \ln \bigl( \frac{315}{1024}\bigr) + \sum\limits_{n > 4} \ln\bigl(1 - 2^{-n}\bigr) > \ln\bigl(\frac{315}{1024}\bigr) - \sum\limits_{n > 4} 2^{-n} =\; \ln\bigl(\frac{315}{1024} \cdot \mathrm e^{-1/16}\bigr)[/imath]. I was wondering if this type of infinite product (or the corresponding sum of logarithms) has a name, and whether there are techniques for obtaining a closed form expression for the limit. | 2838489 | How to compute [imath]\prod_{1\le i\le n} \left(1-\frac{1}{2^i}\right)[/imath]
Could you please tell me how to compute [imath]\prod_{1\le i\le n} \left(1-\frac{1}{2^i}\right)[/imath]? Especially, I would like to know its limit when [imath]n\to \infty[/imath]. Thanks |
56770 | What does recursive cosine sequence converge to?
I have a sequence defined as follow: [imath]a_0 = 1, a_n=\cos\left(a_{n-1}\right)[/imath]. I want to count [imath]\lim_{n\rightarrow\infty} a_n[/imath] - it definitely does have limit by looking at the graph, the first few numbers of the limit are 0.7390851, but I have no idea, if that number is related to some other real number ([imath]\pi[/imath] or something like that). The sequence is from this site, but they don't provide actual result for their own sequence. | 1542884 | Let [imath]x_0=0[/imath]. Define [imath]x_{n+1}=\cos x_n[/imath] for every [imath]n\ge 0[/imath]
Let [imath]x_0=0[/imath]. Define [imath]x_{n+1}=\cos x_n[/imath] for every [imath]n\ge0[/imath]. Then A) [imath]\{x_n\}[/imath] is increasing and convergent B) [imath]\{x_n\}[/imath] is decreasing and convergent C) [imath]\{x_n\}[/imath] is convergent and [imath]x_{2n}\lt\lim_{m\to\infty} x_m\lt x_{2n+1} [/imath] for every [imath]n\in\Bbb N[/imath] D) [imath]\{x_n\}[/imath] is not convergent Attempt: [imath]x_0=0,x_1=1[/imath], and [imath]0\le x_n\le1[/imath] for [imath]n\gt 1[/imath] So this sequence is non constant and bounded, Hence convergent But, I am unable to pick the right option. |
50006 | Different ways to prove there are infinitely many primes?
This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways. Proof 1 : Euclid's. If there are finitely many primes then [imath]p_1 p_2 ... p_n + 1[/imath] is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes. Proof 2 : Consider the sequence [imath]a_n = 2^{2^n} + 1[/imath]. We have that [imath] 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), [/imath] so that for [imath]m < n[/imath], [imath](2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1[/imath]. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them. Proof 3 : (Note : I particularly like this one.) Define a topology on [imath]\mathbb Z[/imath] in the following way : a set [imath]\mathscr N[/imath] of integers is said to be open if for every [imath]n \in \mathscr N[/imath] there is an arithmetic progression [imath]\mathscr A[/imath] such that [imath]n \in \mathscr A \subseteq \mathscr N[/imath]. This can easily be proven to define a topology on [imath]\mathbb Z[/imath]. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set [imath] \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z [/imath] should be open and closed, but by the fundamental theorem of arithmetic, its complement in [imath]\mathbb Z[/imath] is the set [imath]\{ -1, 1 \}[/imath], which is not open, thus giving a contradiction. Proof 4 : Let [imath]a,b[/imath] be coprime integers and [imath]c > 0[/imath]. There exists [imath]x[/imath] such that [imath](a+bx, c) = 1[/imath]. To see this, choose [imath]x[/imath] such that [imath]a+bx \not\equiv 0 \, \mathrm{mod}[/imath] [imath]p_i[/imath] for all primes [imath]p_i[/imath] dividing [imath]c[/imath]. If [imath]a \equiv 0 \, \mathrm{mod}[/imath] [imath]p_i[/imath], since [imath]a[/imath] and [imath]b[/imath] are coprime, [imath]b[/imath] has an inverse mod [imath]p_i[/imath], call it [imath]\overline{b}[/imath]. Choosing [imath]x \equiv \overline{b} \, \mathrm{mod}[/imath] [imath]p_i[/imath], you are done. If [imath]a \not\equiv 0 \, \mathrm{mod}[/imath] [imath]p_i[/imath], then choosing [imath]x \equiv 0 \, \mathrm{mod}[/imath] [imath]p_i[/imath] works fine. Find [imath]x[/imath] using the Chinese Remainder Theorem. Now assuming there are finitely many primes, let [imath]c[/imath] be the product of all of them. Our construction generates an integer coprime to [imath]c[/imath], giving a contradiction to the fundamental theorem of arithmetic. Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...) Do you have any other nice proofs? | 1675288 | How many ways are there to prove that there is no largest prime?
Is there any other proof by which I can show that there is no largest prime? I saw an example where it is proved with contradiction.(Idea is basically that of Euclid's proof) Imagine that the largest prime prime is [imath]13[/imath].So, total number of primes we know are-[imath]2,3,5,7,11,13[/imath]. Now,if I do [imath](2\times3\times5\times7\times11\times13)+1=30031[/imath].So, we can see that [imath]30031[/imath] is not divisible by [imath]2,3,5,7,11,13[/imath] as they leave remainder [imath]1[/imath]. Also,as it is formed by multiplying only primes it does not have any other composite factors.We also see that [imath]30031=59\times 509[/imath].Which are again two primes.Thus,[imath]13[/imath] is not the largest prime. What are the other ways to prove that there is no largest prime? Thanks for any proof!! |
148558 | Density of the set [imath]S=\{m/2^n| n\in\mathbb{N}, m\in\mathbb{Z}\}[/imath] on [imath]\mathbb{R}[/imath]?
Let [imath]S=\{\frac{m}{2^n}| n\in\mathbb{N}, m\in\mathbb{Z}\}[/imath], is [imath]S[/imath] a dense set on [imath]\mathbb{R}[/imath]? | 1807464 | Denseness of set S in R
Consider the set [imath]S= \{\frac{p}{2^n} : p,n \in \mathbb{Z}\}[/imath]. Is it dense in [imath]\mathbb{R}[/imath] ?? To me intuitively it seems to be dense but I cannot come up with any analytic proof or disprove . |
113121 | lim sup inequality [imath]\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n [/imath]
I´m not sure how to start with this proof, how can I do it? [imath] \limsup ( a_n b_n ) \leqslant \limsup a_n \limsup b_n [/imath] I also have to prove, if [imath] \lim a_n [/imath] exists then: [imath] \limsup ( a_n b_n ) = \limsup a_n \limsup b_n [/imath] Help please, it´s not a homework I want to learn. | 506198 | Proof for limit superior's property: [imath]\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n[/imath]
Let [imath]a_n,b_n>0[/imath] for all [imath]n\in\mathbb N[/imath]. Prove that [imath]\limsup (a_n b_n ) \leq \limsup a_n \cdot \limsup b_n[/imath] I know that [imath]\limsup (a_n+b_n ) \leq \limsup a_n + \limsup b_n[/imath]. But I don't know how to use this to get what I want to show. |
55468 | How to prove that exponential grows faster than polynomial?
In other words, how to prove: For all real constants [imath]a[/imath] and [imath]b[/imath] such that [imath]a > 1[/imath], [imath]\lim_{n\rightarrow\infty}\frac{n^b}{a^n} = 0[/imath] I know the definition of limit but I feel that it's not enough to prove this theorem. | 1046432 | Proving that [imath]\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0[/imath]
I need to prove that [imath]\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0[/imath] where [imath]k\in \mathbb{N}[/imath]. All I can think of is to use something like L'Hopital's rule but I suppose there must be a another simpler way. I would much appreciate if someone could give me a hint. Thanks |
1196 | Proof that [imath]n^3+2n[/imath] is divisible by [imath]3[/imath]
I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs! Problem: For any natural number [imath]n , n^3 + 2n[/imath] is divisible by [imath]3.[/imath] This makes sense Proof: Basis Step: If [imath]n = 0,[/imath] then [imath]n^3 + 2n = 0^3 +[/imath] [imath]2 \times 0 = 0.[/imath] So it is divisible by [imath]3.[/imath] Induction: Assume that for an arbitrary natural number [imath]n[/imath], [imath]n^3+ 2n[/imath] is divisible by [imath]3.[/imath] Induction Hypothesis: To prove this for [imath]n+1,[/imath] first try to express [imath]( n + 1 )^3 + 2( n + 1 )[/imath] in terms of [imath]n^3 + 2n[/imath] and use the induction hypothesis. Got it [imath]( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}[/imath] [imath] = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying and regrouping}\}[/imath] [imath] = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out the 3}\}[/imath] which is divisible by [imath]3[/imath], because [imath](n^3 + 2n )[/imath] is divisible by [imath]3[/imath] by the induction hypothesis. What? Can someone explain that last part? I don't see how you can claim [imath](n^3+ 2n ) + 3( n^2 + n + 1 )[/imath] is divisible by [imath]3.[/imath] | 1368373 | Prove by induction that [imath]\frac{n^3}{3}+\frac{2n}{3}[/imath] is an integer.
The question that I am working on is: Prove that [imath]\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N[/imath] The method that I think would be will work for this question is that I say that [imath]3|(n^3+2n)[/imath] and prove that.Would this be a good way to do this question? So far I have done the following: 1) Base Case n = 1 [imath]3|n^3+2n = 3|3 \checkmark[/imath] 2) Assume, [imath]n^3+2n[/imath] is divisible by [imath]3[/imath] for [imath]n =k, k \in \mathbb N[/imath] [imath]k^3+2k = 3m , m \in \mathbb N[/imath] Let [imath]n = k + 1[/imath] ; Then: [imath](k+1)^3 + 2(k+1)[/imath] = [imath]k^3+2k+3k^2+3k+3[/imath] Since [imath]k^3+2k = 3m[/imath] [imath]3m + 3k^2+3k+3[/imath] Now, I'm stuck I don't know what else to do further. |
305976 | [imath]\int_{0}^{\infty} [x]e^{-x}dx[/imath]
could any one tell me how to solve this integration? It was a question from a prev year of my institute qual: Thank you for help. [imath]\int_{0}^{\infty} [x]e^{-x}dx[/imath] | 305242 | Stuck on this integral involving exp and the floor function
Here is the integral [imath]\int_0^\infty \lfloor x \rfloor e^{-x}dx[/imath] Here is what I have so far: [imath]I = \sum_{n=0}^\infty \int_n^{n+1} n e^{-x}dx[/imath] [imath] = \sum_{n=0}^\infty -ne^{-n-1} + ne^{-n}[/imath] [imath] = \sum_{n=0}^\infty -ne^{-n-1} + ne^{-n}[/imath] [imath] = \sum_{n=0}^\infty ne^{-n}(1 -e^{-1})[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty ne^{-n}[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty \sum_{k=0}^{\infty} \frac{-(-n)(-n)^k}{k!}[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty \sum_{k=0}^{\infty} \frac{-(-n)^{k+1}}{k!}[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty \sum_{k=0}^{\infty} \frac{-(-n)^{k+1}}{(k+1)!}(k+1)[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty \sum_{k=0}^{\infty} \frac{-k(-n)^{k+1}}{(k+1)!} + \frac{-(-n)^{k+1}}{(k+1)!}[/imath] [imath] = (1 -e^{-1})\sum_{n=0}^\infty (1-e^{-n})[\sum_{k=0}^{\infty} \frac{-k(-n)^{k+1}}{(k+1)!}][/imath] At this point I gave up. |
84870 | How to show that a set of discontinuous points of an increasing function is at most countable
I would like to prove the following: Let [imath]g[/imath] be a monotone increasing function on [imath][0,1][/imath]. Then the set of points where [imath]g[/imath] is not continuous is at most countable. My attempt: Let [imath]g(x^-)~,g(x^+)[/imath] denote the left and right hand limits of [imath]g[/imath] respectively. Let [imath]A[/imath] be the set of points where [imath]g[/imath] is not continuous. Then for any [imath]x\in A[/imath], there is a rational, say, [imath]f(x)[/imath] such that [imath]g(x^-)\lt f(x)\lt g(x^+)[/imath]. For [imath]x_1\lt x_2[/imath], we have that [imath]g(x_1^+)\leq g(x_2^-)[/imath]. Thus [imath]f(x_1)\neq f(x_2)[/imath] if [imath]x_1\neq x_2[/imath]. This shows an injection between [imath]A[/imath] and a subset of the rationals. Since the rationals are countable, [imath]A[/imath] is countable, being a subset of a countable set. Is my work okay? Are there better/cleaner ways of approaching it? | 1488477 | How to see that this set is countable?
Let [imath]f : \mathbb{R}\to \mathbb{R}[/imath] be a monotone function and consider the set [imath]S[/imath] of all points [imath]a\in \mathbb{R}[/imath] such that [imath]\lim_{x\to a^-}f(x)\neq \lim_{x\to a^+}f(x).[/imath] I want to show that this set is countable, but I'm not finding any way to do it. I thought one way would be to show that this set is discrete, so that it is countable. For that to happen, I would need to show that there is no convergent sequence [imath](x_n)[/imath] of points of [imath]S[/imath] all different than [imath]a[/imath] whose limit is [imath]a[/imath]. My idea then is to suppose there is such a sequence [imath](x_n)[/imath]. Then [imath]x_n\in S[/imath] and [imath]\lim x_n = a[/imath] with [imath]x_n\neq a[/imath]. In that case we have [imath]\lim_{x\to x_n^-}f(x)\neq \lim_{x\to x_n^+}f(x)[/imath]. Intuition says this should contradict [imath]f[/imath] being monotone, but I'm not finding a way to prove it. So how can I finish this proof? Is my strategy correct? If so how do I finish it? If not, how should I prove this result? |
221078 | Poisson Distribution of sum of two random independent variables [imath]X[/imath], [imath]Y[/imath]
[imath]X \sim \mathcal{P}( \lambda) [/imath] and [imath]Y \sim \mathcal{P}( \mu)[/imath] meaning that [imath]X[/imath] and [imath]Y[/imath] are Poisson distributions. What is the probability distribution law of [imath]X + Y[/imath]. I know it is [imath]X+Y \sim \mathcal{P}( \lambda + \mu)[/imath] but I don't understand how to derive it. | 2290242 | How can I prove that [imath]X+Y[/imath] is a Poisson process with parameter [imath]\lambda_X+\lambda_Y[/imath]?
For 2 independent Poisson processes [imath]X,Y[/imath], with parameters [imath]\lambda_X, \lambda_Y[/imath] respectively, how can I prove that [imath]X+Y[/imath] is a Poisson process with parameter [imath]\lambda_X+\lambda_Y[/imath]? To do this, I suppose you would use the rule [imath] P(X+Y = k) = \sum_{n=0}^k P(X = n) \cdot P(Y = k-n) \\ = \sum_{n=0}^k\frac{\lambda_X^n e^{\lambda_X}}{n!} \cdot \frac{\lambda_Y^n e^{\lambda_Y}}{(k-n)!} [/imath] However, I am unsure of how to continue from here. Can anyone prompt me in the right direction? |
112043 | Continuous Functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{Q}[/imath]
The following is not a homework problem. I am doing it for self study. Prove that any continuous function from [imath]\mathbb{R}[/imath] to [imath]\mathbb{Q}[/imath] is constant. Here is my proof: Let [imath]f:\mathbb{R}\rightarrow \mathbb{Q}[/imath] be such a function. We first show that [imath]\mathbb{Q}[/imath] is disconnected. Let [imath]p[/imath] be an irrational number. Then we can write [imath]\mathbb{Q}[/imath] as [imath](-\infty,p)\cup (p,\infty)[/imath]. [imath]\mathbb{Q}[/imath] is also totally disconnected, as any open subset of [imath]\mathbb{Q}[/imath] must contain two rational numbers and there is always an irrational number between the two which we can use to create a disconnection. Therefore the only connected subsets of [imath]\mathbb{Q}[/imath] are the singleton sets [imath]\{q\}[/imath], where [imath]q \in \mathbb{Q}[/imath], which are closed. We will show that [imath]f(\mathbb{R})[/imath] must be connected. Assume not, then [imath]f(\mathbb{R})=U\cup V[/imath], where [imath]U[/imath] and [imath]V[/imath] are nonempty open subsets of [imath]\mathbb{Q}[/imath] such that [imath]U \cap V= \emptyset[/imath]. This implies \begin{align*} \mathbb{R}&=f^{-1}(U \cup V)\\ &=f^{-1}(U)\cup f^{-1}(V), \end{align*} where [imath]f^{-1}(U),f^{-1}(V)[/imath] are nonempty, nonintersecting subsets of [imath]\mathbb{R}[/imath]. This, however, is a contradiction, as [imath]\mathbb{R}[/imath] is connected. Therefore no such set [imath]U[/imath] and [imath]V[/imath] can exist. As we have already shown, the only connected subsets of [imath]\mathbb{Q}[/imath] are the singleton sets [imath]\{q\}[/imath], so [imath]f(\mathbb{R})[/imath] must be such a set. Something about this does not seem quite satisfactory, as if I am missing something. Could anyone tell me a flaw in my logic? Also, is there a more satisfactory way to prove this theorem? | 1224547 | Suppose that [imath]f : \mathbb{R}\rightarrow \mathbb{R}[/imath] is continuous and that [imath]f(x)\in \mathbb{Q}[/imath] for all [imath]x \in \mathbb{R}[/imath]. Prove [imath]f[/imath] is constant.
I am really stuck on this proving this statement, so could someone please help me get through this. Thank You. P.S. This can be proved through elementary analysis results instead of going into topology. |
75623 | How slow/fast can [imath]L^p[/imath] norm grow?
This is actually an exercise in Rudin's Real and Complex Analysis, [imath]L^p[/imath] spaces chapter. Could anyone help me out? Thanks in advance. Motivation: It's well known that if we have a function [imath]f[/imath] which belongs to [imath]L^p(0,1)[/imath] for all [imath]p\ge 1[/imath]. Then [imath]\lim_{p\rightarrow \infty}\|f\|_p=\|f\|_{\infty}[/imath] (moreover, [imath]\|f\|_p[/imath] is increasing in [imath]p[/imath]). This is true even if [imath]\|f\|_{\infty}=\infty[/imath]. Question: How slow (fast) can [imath]\|f\|_p[/imath] grow when [imath]\|f\|_{\infty}=\infty[/imath]? More precisely, given any positive increasing function [imath]\Phi[/imath] with [imath]\lim_{p\rightarrow \infty}\Phi(p)=\infty[/imath], can we always find a function [imath]f[/imath] which belongs to [imath]L^p(0,1)[/imath] for all [imath]p\ge 1[/imath], and [imath]\|f\|_{\infty}=\infty[/imath], such that [imath]\|f\|_p\le (\ge)\Phi(p)[/imath] for large [imath]p[/imath]? | 2390878 | Can [imath]||f||_p[/imath] tend toward infinity arbitrarily slowly?
Suppose [imath]f[/imath] is Lebesgue measurable on [imath](0,1)[/imath], and not essentially bounded. Can [imath]||f||_p[/imath] tend to [imath]\infty[/imath] arbitrarily slowly? I am really lost on this problem. I assume the answer is that it can tend to [imath]\infty[/imath] arbitrarlily slowly but showing that is really difficult. I have tried letting [imath]f=\sum\limits_{k=1}^{\infty}a_k 1_{E_k}[/imath] where [imath]E_i[/imath] is a disjoint collection of subset of [imath](0,1)[/imath], and working from there but I am at a loss here. Any guidence? Arbitrarilly slowly- We say that [imath]||f||_p[/imath] tend to infinity arbitrarilly slowly if for every positive function [imath]\Phi(p)[/imath] on [imath](0,\infty)[/imath] with [imath]\lim_{p\to \infty}\Phi(p)=\infty[/imath] there exists a function [imath]f[/imath] on [imath](0,1)[/imath] such that [imath]||f||_p\leq \Phi(p)[/imath] for all large enough [imath]p[/imath]. |
253394 | If [imath]P \leq G[/imath], [imath]Q\leq G[/imath], are [imath]P\cap Q[/imath] and [imath]P\cup Q[/imath] subgroups of [imath]G[/imath]?
[imath]P[/imath] and [imath]Q[/imath] are subgroups of a group [imath]G[/imath]. How can we prove that [imath]P\cap Q[/imath] is a subgroup of [imath]G[/imath]? Is [imath]P \cup Q[/imath] a subgroup of [imath]G[/imath]? Reference: Fraleigh p. 59 Question 5.54 in A First Course in Abstract Algebra. | 283794 | Prove - intersection of subgroups is a subgroup.
Since I have some similar question to this one, I'll be happy to understand how to solve it. [imath]G[/imath] is a group and [imath]A,B\le G[/imath]. I need to show that [imath]A \cap B \le G[/imath] (If [imath]A, B,[/imath] are subgroups of [imath]G[/imath], I need to show that [imath]A \cap B[/imath] is a subgroup of [imath]G[/imath], too.) |
169020 | Fermat's theorem on sums of two squares
There's Fermat's theorem on sums of two squares. As the prime numbers that are [imath]1\bmod4[/imath] can be divided into the sum of two squares, will the squared numbers be unique? For example, [imath]41=4^2+5^2[/imath] and the squared numbers will be 4 and 5. | 529617 | show the p is a sum of two squares - number theory
If [imath]p[/imath] is a the sum of two squares are integers [imath]a[/imath] and [imath]b[/imath] s.t. [imath]p=a^2 + b^2[/imath] then [imath]p=1[/imath] mod [imath]4[/imath]. I need help proving that. |
108191 | Prove that [imath]f[/imath] continuous and [imath]\int_a^\infty |f(x)|\;dx[/imath] finite imply [imath]\lim\limits_{ x \to \infty } f(x)=0[/imath]
I'd love your help proving the following claim: If [imath]f[/imath] is continuous and [imath]\int_a^\infty |f(x)|\;dx[/imath] is finite then [imath]\lim\limits_{ x \to \infty } f(x)=0[/imath]. Here the counter example of all these sine functions won't do, since their infinite integral is not absolute converges. Any hints for a proof? | 1527496 | [imath]f:[a,\infty)\to\mathbb R[/imath] be continuous such that [imath]\int_a^\infty f(x)dx[/imath] exists finitely , then [imath]\lim_{x \to \infty}f(x)=0[/imath] ?
Let [imath]f:[a,\infty)\to\mathbb R[/imath] be a continuous function such that [imath]\int_a^\infty f(x)dx[/imath] exists finitely , then is it true that [imath]\lim_{x \to \infty}f(x)[/imath] exists and is equal to [imath]0[/imath] ? If not , then what if we restrict the range of [imath]f[/imath] to be non-negative only ? |
125774 | How to expand [imath]\cos nx[/imath] with [imath]\cos x[/imath]?
Multiple Angle Identities: How to expand [imath]\cos nx[/imath] with [imath]\cos x[/imath], such as [imath]\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1[/imath] See a list of trigonometric identities in english/ chinese | 1017707 | [imath]\cos(nx)=Q_n(\cos(x))[/imath] for polynomial [imath]Q_n[/imath] of degree [imath]n[/imath]
Are there any proofs of this equality online? I'm just looking for something very simply that I can self-verify. My textbook uses the result without a proof, and I want to see what a proof would look like here. |
256737 | Intersection of cyclic subgroups: [imath](x^m) \cap (x^n) = (x^{lcm(m,n)})[/imath]
This group theory problem has stumped me. I want to prove that if [imath]G=(x)[/imath] is a finite cyclic group that [imath](x^n) \cap (x^m) = (x^{\operatorname{lcm}(m,n)})[/imath] for all integers [imath]m[/imath] and [imath]n[/imath], where [imath](x)[/imath] is the group generated by [imath]x[/imath]. Thoughts? | 132251 | Show [imath]\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath]
So I want to show that [imath]\langle a^m \rangle \cap \langle a^n \rangle = \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath]. My approach to this problem was to show a double containment, i.e. to show that [imath]\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath] and [imath] \langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle[/imath]. I would like to see a full proof for this, specifically [imath]\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath]. (I tried it with the approach of breaking it down into to cases; [imath]a[/imath] has infinite order and [imath]a[/imath] has finite order, the latter of which i would appreciate the most help on.) My approach to solving the whole problem: (I would appreciate any feedback on anything that is wrong, or a different approach to the proof.) To show the easier containment, [imath]\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle[/imath], I did the following: Let [imath]l = \operatorname{lcm}(m, n)[/imath]. Let [imath]j \in \langle a^l\rangle[/imath], so [imath]j = (a^l)^k = a^{lk}[/imath] for some [imath]k \in \mathbb Z[/imath]. Since [imath]l[/imath] is a multiple of [imath]m[/imath] and [imath]n[/imath] by definition, we can say [imath]l = ms = nt[/imath] for some [imath]s, t \in \mathbb Z[/imath]. Now [imath]j = a^{kl} = a^{kms} = (a^m)^{ks} \in \langle a^m\rangle[/imath]. Similarly, [imath]j = a^{kl} = a^{knt} = (a^n)^{kt} \in \langle a^n\rangle[/imath]. Now, since [imath]j \in \langle a^m\rangle[/imath] and [imath]j \in \langle a^n\rangle[/imath], it follows that [imath]j \in \langle a^m \rangle \cap \langle a^n \rangle[/imath]. Thus, by definition, [imath]\langle a^{\operatorname{lcm}(m, n)}\rangle \subseteq \langle a^m \rangle \cap \langle a^n \rangle[/imath]. For the second containment, the one which I'm having more problems with, I attempted the following: Case in which [imath]a[/imath] is infinite: Suppose that [imath]\vert a \vert = \infty[/imath]. Let [imath]c \in \langle a^m \rangle \cap \langle a^n \rangle[/imath]. Then [imath]c = a^{mx} = a^{ny}[/imath] where [imath]x, y \in \mathbb Z[/imath]. It follows that [imath]a^{mx - ny} = e[/imath] and so [imath]mx = ny[/imath] because if [imath]mx > ny[/imath] then the difference would not be zero, and we would have an element that was finite, contradicting our hypothesis. And since [imath]mx = ny[/imath] we know [imath]\operatorname{lcm}(mx, ny) = mx = ny[/imath] and [imath]\operatorname{lcm}(mx, ny)[/imath] is a multiple of [imath]\operatorname{lcm}(m, n)[/imath]. Hence [imath]c \in \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath] and thus [imath]\langle a^m \rangle \cap \langle a^n \rangle \subseteq \langle a^{\operatorname{lcm}(m, n)}\rangle[/imath]. Case in which [imath]a[/imath] is finite: I tried starting it out the same as the previous case, but i could never reach my conclusion :( |
80364 | Crafty solutions to the following limit
The following problem came up at dinner, I know some ways to solve it but they are quite ugly and as some wise man said: There is no place in the world for ugly mathematics. These methods are using l'Hôpital, but that becomes quite hideous very quickly or by using series expansions. So I'm looking for slick solutions to the following problem: Compute [imath]\displaystyle \lim_{x \to 0} \frac{\sin(\tan x) - \tan(\sin x)}{\arcsin(\arctan x) - \arctan(\arcsin x)}[/imath]. I'm curious what you guys will make of this. | 1110550 | Prove that the limit is equal to one.
How to prove that [imath]\lim_{x\rightarrow 0 }\frac{\sin\big(\tan(x)\big)- \tan\big(\sin(x)\big)}{\arcsin\big(\arctan(x)\big)-\arctan\big(\arcsin(x)\big) } = 1[/imath] First terms in Taylor expansion of the numerator and denominator is [imath]-\frac{x^7}{30}[/imath]. So the direct proof is not beautiful and hard. |
137727 | Evaluate [imath]\sum\limits_{k=0}^n \binom{n}{k}[/imath] combinatorially
Please help me to evaluate combinatorially the following sum: [imath]\sum_{k=0}^n \binom{n}{k}[/imath] Thank you. | 1771661 | Story proof for [imath]\sum_{k=0}^n {n \choose k} = 2^n[/imath]
I found a solution online that uses the Binomial Theorem. Is it possible to prove this without using that theorem? |
209672 | Show that [imath]n[/imath] lines separate the plane into $\frac{n^2+n+2}{2}$ regions
Show that [imath]n[/imath] lines separate the plane into [imath]$\frac{n^2+n+2}{2}$[/imath] regions if no two of these lines are parallel and no three pass through a common point. I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0. But I really have no idea how to begin the inductive step. How do we know what k+1 we're supposed to arrive at? Thanks! | 1186344 | Prove simple induction
Suppose we we have n straight lines on the plane such that no two of them are parallel and no three of them go through the same point. Prove that the number of different regions that are created by these lines (regions that are bounded between line segments and/or those that are unbounded) is exactly [imath]\frac{n(n + 1)}{2} + 1[/imath] for all n ≥ 1. |
119636 | Formula for calculating [imath]\sum_{n=0}^{m}nr^n[/imath]
I want to know the general formula for [imath]\sum_{n=0}^{m}nr^n[/imath] for some constant r and how it is derived. For example, when r = 2, the formula is given by: [imath]\sum_{n=0}^{m}n2^n = 2(m2^m - 2^m +1)[/imath] according to http://www.wolframalpha.com/input/?i=partial+sum+of+n+2%5En Thanks! | 612219 | How to find [imath]\sum_{k=1}^{n}{3^kk}[/imath]
I need hint to find this sum [imath]\sum_{k=1}^{n}{3^kk}[/imath] [imath]\sum_{k=1}^{n}{3^k}[/imath] can be easily calculated as it is sum of geometric progression. |
78675 | [imath]G[/imath] modulo [imath]N[/imath] is a cyclic group when [imath]G[/imath] is cyclic
If G is a cyclic group and N is a subgroup of G, show that G/N (or GmodN) is a cyclic group. What I have so far: Since N is a subgroup of the cyclic group G, G/N is a cyclic group. I think I'm missing details. What suggestions do you have? | 1051684 | show that a factor group is cyclic
Prove that if G is cyclic and N is a subgroup then, G/N is cyclic. Here what I have so far. Proof: I know that G is cyclic, that means G= < g > = {[imath]g^{n}[/imath] | n [imath]\in[/imath] Z} and N < G. Each element of G can be written as [imath]g^{k}[/imath]. [imath]g^{k}[/imath]N= [imath](gN)^{k}[/imath]. would [imath](gN)^{k}[/imath] be my generator of G/N. |
85217 | Why is this series of square root of twos equal [imath]\pi[/imath]?
Wikipedia claims this but only cites an offline proof: [imath]\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi[/imath] for [imath]n[/imath] square roots and one minus sign. The formula is not the "usual" one, like Taylor series or something like that, so I can't easily prove it. I wrote a little script to calculate it and it's clearly visible, but that's not a proof. | 1769785 | I discovered a sequence that should converge to [imath]\pi[/imath] , but how to prove that it really converges to [imath]\pi[/imath]?
So yesterday I took paper and pencil and did some constructions of geometrical nature and I discovered a sequence that should converge to [imath]\pi[/imath]. It goes like this: Define [imath]a_1=\sqrt{2}[/imath] and for every other [imath]n \in \mathbb N[/imath] define [imath]a_n=\sqrt{2+a_{n-1}}[/imath]. Define [imath]S_n=2^n \cdot \sqrt{2-a_{n-1}}[/imath]. Then we should have [imath]\lim_{n \to \infty}S_n=\pi[/imath]. This formula (more correctly, sequence) can be seen on this page and is labeled with number (66) so I did not discover anything new. I found this sequence but I did not rigorously prove that it converges to [imath]\pi[/imath], the story is that I saw a pattern for the first few natural numbers and then defined the expression for general [imath]n[/imath] and when I did that I found the sequence on the above linked page. My question would be: In which way would you prove that this sequence converges to [imath]\pi[/imath]? |
134577 | How do you integrate [imath]\int \frac{1}{a + \cos x} dx[/imath]?
How do you integrate [imath]\int \frac{1}{a + \cos x} dx[/imath]? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer. Thanks! | 1323388 | Integral using contour integration
Here is the integral I want to evaluate: [imath]\int_{0}^{2\pi} \frac{dx}{a+b \cos x }, \quad a>b >0[/imath] Apparently there are limitations as to what values [imath]a, b[/imath] are supposed to take but let us not concern about this. Since using the sub [imath]u =\tan \frac{x}{2}[/imath] (the Weiersstrass sub) results that the integral is [imath]0[/imath] (as it should, since it is not [imath]1-1[/imath] function in this interval) I got down down the way of contour integration. Hence: [imath]\begin{aligned} \int_{0}^{2\pi}\frac{dx}{a+ b\cos x} &\overset{x=i \ln u}{=\! =\! =\!} \oint \limits_{|z|=1} \frac{dz}{iz \left [ a + \frac{b}{2}\left ( z+z^{-1} \right ) \right ]} \\ &= \frac{1}{i} \oint \limits_{|z|=1} \frac{dz}{za + \frac{bz^2}{2}+\frac{b}{2}}\\ &=\frac{2}{i} \oint \limits_{|z|=1} \frac{dz}{bz^2 +b +2za} \\ &= \frac{2}{i} 2\pi i \sum_\text{residues} f(z)\\ &= \frac{4\pi}{1+ \sqrt{1-8ab}+2a} \end{aligned}[/imath] However, judging by intuition this must not be the result. Because this one restricts the integral too much. What i mean is, that for [imath]a=6, b=3[/imath] we have that: [imath]\int_0^{2\pi} \frac{dx}{6+3\cos x}= \frac{2\pi}{3\sqrt{3}}[/imath] My formula cannot derive the result because then radical would be negative. What am I doing wrong here? |
8389 | A normal subgroup intersects the center of the [imath]p[/imath]-group nontrivially
If [imath]G[/imath] is a finite [imath]p[/imath]-group with a nontrivial normal subgroup [imath]H[/imath], then the intersection of [imath]H[/imath] and the center of [imath]G[/imath] is not trivial. | 502491 | Normal subgroup in a [imath]p[/imath]-group
Let [imath]G[/imath] be a [imath]p[/imath]-group, and [imath]H[/imath] is a normal subgroup of [imath]G[/imath] with [imath]|H| = p[/imath]. Prove that [imath]H \leq Z(G)[/imath]. More general, if [imath]K[/imath] is a normal subgroup of [imath]G[/imath], then [imath]K\cap Z(G) \neq \{e\}[/imath], with [imath]e[/imath] is the unity of [imath]G[/imath], where [imath]Z(G)[/imath] is the center of [imath]G[/imath], [imath]Z(G) = \{x\in G: xa = ax,\forall\; a\in G\}[/imath]. |
204645 | Why (directly!) does every number divide 9, 99, 999, ... or 10, 100, 1000, ..., or their product?
A curiosity that's been bugging me. More precisely: Given any integers [imath]b\geq 1[/imath] and [imath]n\geq 2[/imath], there exist integers [imath]0\leq k, l\leq b-1[/imath] such that [imath]b[/imath] divides [imath]n^l(n^k - 1)[/imath] exactly. The question in the title is obviously the case when n = 10. This serves as a motivating example: if we take b = 7 and n = 10, then k = 6, l = 0 works (uniquely), and if we calculate [imath]\dfrac{n^k - 1}{b}[/imath], we see that it turns out to be 142857 - the repeating part of the decimal expansion of 1/7. A (very sketchy, but correct!) sketch proof, which I've included for completeness: Notice that [imath]\dfrac{1}{99\dots 9} = 0.00\dots 01\; 00\dots 01\; 00\dots 01\dots[/imath]. Notice that [imath]1/7[/imath] must have either a repeating or a terminating decimal expansion: just perform the long division, and at each stage you will get remainders of 0 (so it terminates) or 1, 2, ..., 6 (so some of these will cycle in some order). It turns out the decimal expansion is repeating, and the 'repeating part' is 142857. This has length (k=)6. [imath]\dfrac{1}{10^6 - 1} = 0.000001\;000001\dots[/imath], so [imath]\dfrac{142857}{10^6 - 1} = 0.142857\;142857\dots = 1/7[/imath], and so [imath]7\times 142857 = 10^6 - 1[/imath]. And we can do the same thing with [imath]\dfrac{1}{10\dots 0} = 0.00\dots 01[/imath] and terminating decimals. And the same proof of course holds in my more general setting. But this (using the long division algorithm after spotting an unwieldy decimal expansion) feels a little artificial to me, and the statement is sufficiently general that I'm sure there must be a direct proof that I'm missing. Of course, the [imath]n^k[/imath] part is easy, but the [imath]n^k-1[/imath] part has me a little stumped. My question is: is there a direct proof of the latter part? Thanks! | 665462 | Does every prime [imath]p \neq 2, 5[/imath] divide at least one of [imath]\{9, 99, 999, 9999, \dots\}[/imath]?
I was thinking of decimal expressions for fractions, and figured that a fraction of the form [imath]\frac{1}{p}[/imath] must be expressed as a repeating decimal if [imath]p[/imath] doesn't divide [imath]100[/imath]. Thus, [imath]\frac{p}{p}[/imath] in decimal would equal [imath]0.\overline{999\dots}[/imath] for some number of [imath]9[/imath]s, thus there must be some amount of [imath]9[/imath]s such that [imath]p | 999...[/imath] in order for a decimal representation of [imath]\frac{1}{p}[/imath] to be possible. Furthermore, the question could be rephrased to "an infinite number of" since if it divides [imath]999\dots[/imath] where there are [imath]k[/imath] nines, it also divides when there are [imath]2k, 3k, \dots[/imath] nines. Is this reasoning correct? If so, this is how I thought about proving it: We can reduce the set to [imath]\{1, 11, 111, 1111, \dots\}[/imath] since [imath]p=3[/imath] obviously works. Let [imath]a_k = 111\dots[/imath] where there are [imath]k[/imath] ones. This satisfies the recursion [imath]a_k = 10a_{k-1} + 1[/imath] But I'm unsure what to do past this point (tried looking at modular cases or something but wasn't able to get anywhere). Am I on the right track at all? Is this "conjecture" even correct? |
2158 | Division of Factorials
I have a partition of a positive integer [imath](p)[/imath]. How can I prove that the factorial of [imath]p[/imath] can always be divided by the product of the factorials of the parts? As a quick example [imath]\frac{9!}{(2!3!4!)} = 1260[/imath] (no remainder), where [imath]9=2+3+4[/imath]. I can nearly see it by looking at factors, but I can't see a way to guarantee it. | 2455964 | Factorial of a sum of n numbers divisible by product of factorial of these numbers
Let [imath]a_1, a_2, a_3,...,a_n[/imath] be [imath]n[/imath] positive integers and if [imath]a=a_1+a_2+a_3+...+a_n[/imath]. Prove that [imath]a_1!a_2!a_3!...a_n! | a![/imath] Factorial of a sum of [imath]n[/imath] numbers is divisible by product of factorial of these numbers. |
12328 | RSA: Fast factorization of N if d and e are known
I stumbled across this paragraph in a paper: Hence, user b cannot decrypt C directly. But using e and d , user b can quickly factor N. How is it possible to speedup the prime factorization when knowing e (public key) and d (private key)? For clarification: RSA provides us with these equations: [imath]n = pq[/imath] [imath]\phi = (p-1)(q-1)[/imath] [imath]gcd(e, \phi) = 1[/imath] [imath]de = 1\pmod{\phi}[/imath] In order to determine [imath]p[/imath] and [imath]q[/imath] an attacker has to factor n which is not feasible. However the paper stated that it is easy to reconstruct [imath]p[/imath] and [imath]q[/imath] when a person knows both (his) private and public keys. | 1839263 | On equivalence of RSA and factoring
Suppose we are given a number "[imath]A[/imath]" which is multiple of [imath]\phi(n)[/imath]. One can assume factorization to be hard. So you cannot find exact value of [imath]\phi(n)[/imath] from [imath]A[/imath]. Clearly using this we can crack RSA as for any encryption exponent '[imath]e[/imath]' we will find its inverse modulo [imath]A[/imath] and that will also we inverse of of '[imath]e[/imath]' modulo [imath]\phi(n)[/imath]. But will [imath]A[/imath] factor [imath]n[/imath] ??. Its looks to me as a natural question on the line of thought of equivalence of RSA and factoring |
185818 | Enestrom-Kakeya Theorem
The Enestrom-Kakeya theorem states that all roots of the polynomial: [imath]p(z):=\sum_{k=0}^n a_kz^k[/imath] lie outside the open unit disk if the sequence [imath](a_k)[/imath] is positive and decreasing. A proof can be found in Marden- Geometry of Polynomials. I was wondering if the following idea of mine constitutes a valid proof as well: [imath](1-z)p(z)=a_0+\sum_{k=1}^n (a_k-a_{k-1})z^k-a_nz^{n+1}[/imath] so [imath]|(1-w)p(w)|\geq a_0-\sum_{k=1}^n (a_{k-1}-a_k)|w|^k -a_n > a_0-\sum_{k=1}^n (a_{k-1}-a_k) -a_n=0[/imath] for all [imath]|w|<1.[/imath] (Using the reverse triangle inequality. The step at which strict inequality is introduced is only valid if some [imath]a_{k-1}-a_k[/imath] is nonzero. If this is not the case, our polynomial trivially has all roots on the unit circle.) | 188039 | Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.
Let [imath]P(z)=a_0+a_1z+\cdots+a_nz^n[/imath] be a polynomial whose coefficients satisfy [imath]0<a_0<a_1<\cdots<a_n.[/imath] I want to show that the roots of [imath]P[/imath] live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice [imath]f(z)=a_nz^n, g(z)=[/imath] (the rest). Any ideas? |
296899 | What is the free [imath]\mathbb{Z}[/imath] module on [imath]M \times N[/imath].
If [imath]M[/imath] and [imath]N[/imath] are left [imath]R[/imath] and right [imath]R[/imath] are modules, what is the free [imath]\mathbb{Z}[/imath] module [imath]F_\mathbb{Z}(M \times N)[/imath]. I don't quite see why this wouldn't be [imath]M \times N[/imath], according to the definition. | 296962 | What is the free abelian group on [imath]M \times N[/imath] where [imath]M,N[/imath] are modules.
The free abelian group (equivalently the free [imath]\mathbb{Z}[/imath]-module) [imath]F(M \times N)[/imath] is defined as the set of all linear combinations of elements of [imath]M \times N[/imath], [imath]F(M \times N) = \{n_1x_1 + \cdots +n_kx_k : n_i \in \mathbb{Z}, x_i \in M \times N, k \in \mathbb{Z}_{\ge0}\}[/imath] Wouldn't these linear combinations already be in [imath]M \times N[/imath]? How is the free abelian group different from [imath]M \times N[/imath]? |
72844 | Is the pointwise maximum of two Riemann integrable functions Riemann integrable?
Let [imath]f,g[/imath] be Riemann integrable functions, prove that the function [imath] h(x) [/imath] defined by [imath] h\left( x \right) = \max \left\{ {f\left( x \right),g\left( x \right)} \right\} [/imath] is also Riemann integrable. | 760056 | How to prove that [imath]\max\{f,g\}[/imath] is Riemann integrable?
If f(x) and g(x) are Riemann integrable in [a,b], why [imath]h(x)=\max\{f(x),g(x)\}[/imath] is still Riemann integrable in [a,b]? Or maybe it is wrong? |
32737 | Prove [imath]\gcd(a+b, a-b) = 1[/imath] or [imath]2\,[/imath] if [imath]\,\gcd(a,b) = 1[/imath]
I want to show that for [imath]\gcd(a,b) = 1[/imath] [imath]a,b \in Z[/imath] [imath]\gcd(a+b, a-b) = 1[/imath] or [imath]\gcd(a+b, a-b) = 2[/imath] holds. I think the first step should look something like this: [imath]d = \gcd(a+b, a-b) = \gcd(2a, a-b)[/imath] From here I tried to proceed with two cases. 1: [imath]a-b[/imath] is even, which leads to [imath]\gcd(a+b, a-b) = 2[/imath] 2: [imath]a-b[/imath] is odd, which leads to [imath]\gcd(a+b, a-b) = 1[/imath] My main problem I think is, that I do not know how I should include [imath]\gcd(a,b) = 1[/imath] in the proof. Any help is appreciated. Thx in advance. Cherio Woltan | 2180356 | How to show that greatest common factor of [imath](m+n[/imath] and [imath] m-n)\in\mathbb{N}[/imath] is equal to [imath]1[/imath] or [imath]2[/imath]?
We have [imath]m,n \in \mathbb{N}[/imath] and [imath]gcd(m,n) = 1.[/imath] How to prove that [imath]gcd(m+n,m-n) = 1[/imath] or [imath]2[/imath]? |
160248 | Evaluating [imath]\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}[/imath]
I'm supposed to calculate: [imath]\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}[/imath] By using W|A, i may guess that the limit is [imath]\frac{1}{2}[/imath] that is a pretty interesting and nice result. I wonder in which ways we may approach it. | 334224 | Central value of the partial exponential function
I need help calculating the central value of the partial exponential function : [imath]\lim_{n \to \infty} e^{-n} \sum^n_{k=0} \frac{n^k}{k!}[/imath] fd |