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434144 | Combinatorially showing [imath]\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0[/imath]
I am trying to show that [imath]\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0[/imath]. I found that using stirling's approximation, I can get: [imath] \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to \infty}{\frac{(2n)!}{n!^2*2^{2n}}}= \lim_{n\to \infty}{\frac{(\frac{2n}{e})^{2n}\sqrt{2\pi (2n)}}{((\frac{n}{e})^n\sqrt{2\pi n})^2*2^{2n}}} =\lim_{n\to \infty}{\frac{2\sqrt{\pi n}}{2\pi n}} =0 [/imath] But this seems inelegant where there should be a more elegant, combinatorial proof. Is there an easier way? | 909678 | What is [imath]\lim_{n\to \infty}\frac{2n \choose {n}}{4^n}[/imath]?
What is the result of the following limit? [imath]\lim_{n\to \infty}\frac{2n \choose {n}}{4^n}[/imath] since [imath]\sum_{k=0}^{2n}{2n \choose {k}}=2^{2n}=4^n[/imath] then [imath]\frac{4^n}{2n+1}\leq{2n \choose {n}}\leq 4^n[/imath] and limit is clealy [imath]\in [0,1][/imath], but what is it exactly? |
476354 | Evaluating [imath]\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}[/imath]
What is the value of [imath]\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}[/imath]? | 112161 | Showing [imath] \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}[/imath]
I would like to show that: [imath] \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} [/imath] We have: [imath] \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} [/imath] I wanted to use the fact that [imath]\arctan(\sqrt{3})=\frac{\pi}{3} [/imath] but [imath]\arctan(x)[/imath] can only be written as a power series when [imath] -1\leq x \leq1[/imath]... |
868066 | Stuck trying to prove that [imath]e^{-x^{-2}}[/imath] is [imath]C^{\infty}[/imath]
This is Spivak's Calculus on Manifolds ex. 2-25, he says Define [imath]f:\mathbb{R}\to \mathbb{R}[/imath] by [imath]f(x) = \left\lbrace \begin{array}{l} e^{-x^{-2}} &\text{ if } x \neq 0\\ 0 &\text{ otherwise } \\ \end{array} \right.[/imath]. Show that [imath]f[/imath] is [imath]C^{\infty}[/imath] anf [imath]f^{(i)}(0)=0\; \forall i[/imath]. I haven't found a way to get a general solution. I only managed to prove: [imath](1)[/imath] [imath]f[/imath] is continuous at [imath]0[/imath]. This follows from [imath]f(0)=0[/imath] and [imath]\lim_{x\to0} f(x)=\lim_{x\to 0}e^{-x^{-2}}=\lim_{x\to 0}\displaystyle\frac{1}{e^{1/x^2}}=0[/imath] since [imath]\displaystyle\frac{1}{x^2}\to+\infty[/imath] if [imath]x\to 0[/imath] then [imath]e^{1/x^2}\to+\infty[/imath] if [imath]x\to 0[/imath], which means that [imath]\displaystyle\frac{1}{e^{1/x^2}}\to 0[/imath] if [imath]x\to 0[/imath] . [imath](2)[/imath][imath]f'(0)=0[/imath]. I got this by calculating the limit [imath]f'(0)=\lim_{h\to 0}\displaystyle\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\displaystyle\frac{f(h)}{h}=\lim_{h\to 0}\displaystyle\frac{e^{-h^{-2}}}{h}=\lim_{h\to 0}\displaystyle\frac{1/h}{e^{h^{-2}}}[/imath]. The last one follows from L'Hôpital: [imath]\lim_{h\to 0}\displaystyle\frac{\frac{d}{dh}(1/h)}{\frac{d}{dh}(e^{h^{-2}})}=\lim_{h\to 0}\displaystyle\frac{(-1/h^2)}{(e^{h^{-2}})(-2h^{-3})}=\lim_{h\to 0}\displaystyle\frac{h}{2e^{h^{-2}}}=\displaystyle\frac{1}{2}\left(\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}\right)\left(\lim_{h\to 0} h\right) = 0[/imath]. [imath](3)[/imath] [imath]f''(0)=0[/imath] I found that [imath]\displaystyle\frac{df}{dx} = \left\lbrace \begin{array}{l} 2x^{-3}e^{-x^{-2}} &\text{ if } x \neq 0\\ 0 &\text{ otherwise } \\ \end{array} \right.[/imath], then I can get [imath]f''(0)[/imath] computing the limit [imath]\lim_{h\to 0}\displaystyle\frac{f'(0+h)-f'(0)}{h} = \lim_{h\to 0}\displaystyle\frac{f'(h)}{h} = \lim_{h\to 0}\displaystyle\frac{2e^{-h^{-2}}}{h^4}= 2\lim_{h\to 0}\displaystyle\frac{1/h^4}{e^{h^{-2}}}=-2\lim_{h\to 0}\displaystyle\frac{-4h^{-5}}{-2h^{-3}e^{h^{-2}}}\\=4\lim_{h\to 0}\displaystyle\frac{1}{h^2e^{h^{-2}}}=4\lim_{h\to 0}\displaystyle\frac{h^{-2}}{e^{h^{-2}}})=-4\lim_{h\to 0}\displaystyle\frac{-2h^{-3}}{e^{h^{-2}}(-2h^{-3})}=-4\lim_{h\to 0}\displaystyle\frac{1}{e^{h^{-2}}}=0[/imath], after using L'Hôpital twice. But still I don't have a clue to prove that [imath]f^{(i)}(0)=0[/imath] for every [imath]i[/imath]. One of the ideas I managed was to find a general form for [imath]f^{(i)}[/imath] but after I got [imath]f'[/imath] doesn't seem possible to do so considering that I would have to apply the product rule each time. | 476195 | Infinitely differentiable functions: how to prove that [imath]e^\frac{1}{x^2-1}[/imath] has derivative of any order?
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a function given by [imath]f(x)=\left\{\begin{matrix} \exp\left(\frac{1}{x^2-1}\right) & \text{if }|x|<1\\ 0 & \text{if }|x|\geq 1 \end{matrix}\right.[/imath] I would like to prove that [imath]f\in C^\infty[/imath], that is, [imath]f\in C^k[/imath] for all [imath]k\in \mathbb{N}[/imath]. I think that it can be done by induction on [imath]k[/imath]. If [imath]|x|>1[/imath], the problem is trivial. On other points, the base case is the simplest and the only that I'm be able to do. Can someone help me? Thanks. |
867967 | Why Limit of [imath]0/x[/imath] is [imath]0[/imath], if [imath]x[/imath] approaches [imath]0[/imath]?
It might be a silly question, but to me it is not obvious why the following expression holds: [imath] \lim\limits_{x\rightarrow 0}\frac{0}{x}=0 ? [/imath] | 747904 | Limit of [imath]0/x[/imath] as x goes to 0
What is the limit of [imath]0/x[/imath] as x goes to [imath]0[/imath], without using L'Hopital's rule? Clearly it should be [imath]0[/imath], but I'm not sure how it is any different from something like the divergence of [imath]1/r^2[/imath] which yields a dirac delta. |
868357 | If [imath]f_{n}\rightharpoonup \bar{f}[/imath] and [imath]f_{n}(x) \rightarrow f(x)[/imath] pointwise a.e., then is [imath]\bar{f} = f[/imath] a.e.?
Suppose [imath]f_{n}[/imath] is a sequence of functions in [imath]L^{p}(\mathbb{R}^{d})[/imath] such that [imath]\|f_{n}\|_{L^{p}} \leq 1[/imath] for all [imath]n[/imath] and [imath]f_{n}(x) \rightarrow f(x)[/imath] pointwise almost everywhere as [imath]n \rightarrow \infty[/imath]. If I also know that [imath]f_{n}[/imath] converges weakly in [imath]L^{p}[/imath] to an [imath]\bar{f}[/imath], then is [imath]\bar{f} = f[/imath] a.e.? | 803344 | Pointwise a.e. convergence and weak convergence in Lp
I'm trying to prove the following Theorem: Let [imath]\{f_n\}_{n\in\mathbb N}\subset L^p(\Omega)[/imath], [imath]f_n \to f[/imath] in [imath]L^p(\Omega)[/imath] ([imath]\Omega\subset\mathbb{R}^n[/imath] is open and bounded, [imath]1\leq p \leq \infty[/imath]) and [imath]f_n \to \hat{f}[/imath] almost everywhere. Then [imath]f=\hat{f}[/imath] almost everywhere. Ideas for the proof: We should prove that [imath]\int (f-\hat{f}) =0[/imath] We can write [imath]\int (f-\hat{f}) = \int (f-f_n)+\int (f_n-\hat{f})[/imath] Then, by weak convergence, the first integral tends to [imath]0[/imath] (because [imath]1\in L^\infty\subseteq L^p[/imath]). The problem is the limit of the second integral. We can write [imath] \int (f_n-\hat{f})=\int_{\Omega \cap E} (f_n-\hat{f}) + \int_{\Omega\cap E^c} (f_n-\hat{f})[/imath] where [imath]E=\{x: f_n(x) \not \to \hat{f}(x)\}[/imath]. Then [imath]|E|=0[/imath] and the first integral vanishes. My question is: What should I do with the second integral? Is it enough if I use the Lebesgue Dominated Convergence Theorem, since by Hölder inequality I can easily dominate [imath]\{f_n\}[/imath] in [imath]L^1[/imath]? |
868434 | How to differentiate [imath]\lim\limits_{n\to\infty}\underbrace{x^{x^{x^{...}}}}_{n\text{ times}}[/imath]?
Let [imath]f(x)=\lim\limits_{n\to\infty}\underbrace{x^{x^{x^{...}}}}_{n\text{ times}}[/imath] Is it possible to find [imath]f'(x)[/imath]. If yes, please show all steps. | 138498 | Derivative of [imath]x^{x^{\cdot^{\cdot}}}[/imath]?
The infinite tetration is defined as [imath]f(x)=x^{x^{\cdot^{\cdot}}}[/imath] This function is defined for [imath]e^{-e} \leq x \leq e^{e-1}[/imath]. (Wikipedia image) Can one determine the derivative of this function? |
842855 | Let [imath]f ∈ L_1([0,1])[/imath] be a function such that [imath]\int_E f(x)dx = 0[/imath] for any measurable set [imath]E ⊂ [0,1][/imath] of Lebesgue measure [imath]0.99.[/imath]
Let [imath]f ∈ L_1([0,1])[/imath] be a function such that [imath]\int_E f(x)dx = 0[/imath] for any measurable set [imath]E ⊂ [0,1][/imath] of Lebesgue measure [imath]0.99.[/imath] Prove that [imath]f = 0[/imath] a.e. Not sure how to start this question. Any suggestions? | 651565 | If [imath]f[/imath] is Lebesgue integrable on [imath][0,2][/imath] and [imath]\int_E fdx=0[/imath] for all measurable set E such that [imath]m(E)=\pi/2[/imath]. Prove or disprove that [imath]f=0[/imath] a.e.
Let [imath]f[/imath] be a Lebesgue integrable function on [imath][0,2][/imath]. If [imath]\int_E fdx=0[/imath] for all measurable set [imath]E[/imath], such that [imath]m(E)=\pi/2[/imath]. Is [imath]f=0[/imath] a.e. Prove or disprove I could not figure out anything. Can a function be very oscillatory so that on every interval its integral is zero? Any hint and approach are welcome. |
868682 | Continuous function which has only rational values.
If [imath]f:[a,b]\to\mathbb{R}[/imath] is a continuous function and [imath]f(x)\in\mathbb{Q}[/imath] for all [imath]x\in[a,b][/imath] then what can say about [imath]f[/imath]? My try: I think f should be constant, if it is not constant then it contradicts the continuity. Can anyone prove that f is constant? | 521675 | Is a rational-valued continuous function [imath]f\colon[0,1]\to\mathbb{R}[/imath] constant?
Let [imath]f\colon[0,1]\to\mathbb{R}[/imath] be continuous such that [imath]f(x)\in\mathbb{Q}[/imath] for any [imath]x\in[0,1][/imath]. Intuitively I feel that [imath]f[/imath] is constant, since [imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath]. How can I formally write this down? |
869299 | Localization and Direct limit
Let [imath] A [/imath] be a ring. Let [imath] I [/imath] be a preordered set, filtering. Let [imath] \Sigma [/imath] a multiplicative subset of [imath] A [/imath]. Suppose for any given [imath] i \in I [/imath] a multiplicative subset [imath] S_i [/imath] of [imath] A [/imath] contained in [imath] \Sigma [/imath]. Make the following assumptions: [imath] \Sigma = \displaystyle \bigcup_i S_i [/imath] For [imath] (i, j) \in I^2 [/imath] with [imath] i \leq j [/imath], elements [imath] S_i [/imath] become invertible in [imath] S_ {j}^{-1} A [/imath]. [imath] \mathcal{D} = ((S_{i}^{-1} A)_{i \in I} \ , \ (S_{i}^{-1} A \to S_{j}^{-1} A)_{i \leq j}) [/imath] is a commutative diagram in the category of filter [imath] A [/imath] - algebras. Question : Show that [imath] \displaystyle \lim_{\longrightarrow} \mathcal{D} \simeq \Sigma^{-1} A [/imath]. To do that, we need to show that : [imath] 1) [/imath] exist morphisms [imath] \pi_i: S_{i}^{-1} A \to \Sigma^{-1} A [/imath] such that [imath] \pi_{i + p} f_{i, i + p } = \pi_{i} [/imath]. [imath] 2) [/imath] for any group [imath] B [/imath] equipped with morphisms [imath] \pi_i: S_{i}^{-1} A \to B [/imath] such that [imath] p_{i + p} f_{i, i + p} = p_{i} [/imath] there exists a unique morphism [imath] g: \Sigma^{-1} A \to B [/imath] such that [imath] p_i = g \circ \pi_i [/imath]. But, i d'ont know how to do it. Thank you in advance for your help. | 863513 | Localization and direct limit
Let [imath] A [/imath] be a ring. Let [imath] I [/imath] be a preordered set, filtering. Let [imath] \Sigma [/imath] a multiplicative subset of [imath] A [/imath]. Suppose for any given [imath] i \in I [/imath] a multiplicative subset [imath] S_i [/imath] of [imath] A [/imath] contained in [imath] \Sigma [/imath]. Make the following assumptions: [imath] \Sigma = \displaystyle \bigcup_i S_i [/imath] For [imath] (i, j) \in I^2 [/imath] with [imath] i \leq j [/imath], elements [imath] S_i [/imath] become invertible in [imath] S_ {j}^{-1} A [/imath]. [imath] \mathcal{D} = ((S_{i}^{-1} A)_{i \in I} \ , \ (S_{i}^{-1} A \to S_{j}^{-1} A)_{i \leq j}) [/imath] is a commutative diagram in the category of filter [imath] A [/imath] - algebras. Question : Show that [imath] \displaystyle \lim_{\longrightarrow} \mathcal{D} \simeq \Sigma^{-1} A [/imath]. Thank you in advance for your help. |
869300 | [imath]i^{-1} F[/imath] a sheaf if and only if [imath]\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y[/imath] is an isomorphism
Let [imath]X[/imath] be a topological space containing two closed points [imath]x,y[/imath] and let [imath]i : \{x,y\} \to X[/imath] denote the inclusion map. Notice that [imath]\{x,y\}[/imath] carries the discrete topology. Let [imath]F[/imath] be a sheaf on [imath]X[/imath]. Then [imath]i^{-1} F[/imath] is a presheaf on [imath]\{x,y\}[/imath] which is given by [imath](i^{-1} F)(\emptyset)=1[/imath] (the terminal set), [imath](i^{-1} F)(\{x\}) = F_x[/imath] (the stalk at [imath]x[/imath]), [imath](i^{-1} F)(\{y\})=F_y[/imath] (the stalk at [imath]y[/imath]) and [imath](i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).[/imath] Why is, [imath]i^{-1} F[/imath] a sheaf if and only if the canonical map [imath]\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.[/imath] is an isomorphism ?. Thanks a lot to all of you. | 869025 | [imath]i^{-1} F[/imath] a sheaf if and only if [imath]\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y[/imath] is an isomorphism.
Let [imath]X[/imath] be a topological space containing two closed points [imath]x,y[/imath] and let [imath]i : \{x,y\} \to X[/imath] denote the inclusion map. Notice that [imath]\{x,y\}[/imath] carries the discrete topology. Let [imath]F[/imath] be a sheaf on [imath]X[/imath]. Then [imath]i^{-1} F[/imath] is a presheaf on [imath]\{x,y\}[/imath] which is given by [imath](i^{-1} F)(\emptyset)=1[/imath] (the terminal set), [imath](i^{-1} F)(\{x\}) = F_x[/imath] (the stalk at [imath]x[/imath]), [imath](i^{-1} F)(\{y\})=F_y[/imath] (the stalk at [imath]y[/imath]) and [imath](i^{-1} F)(\{x,y\}) = \varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U).[/imath] Why is, [imath]i^{-1} F[/imath] a sheaf if and only if the canonical map [imath]\varinjlim_{ U \subseteq X \text{ open}, ~ x,y \in U } F(U) \to F_x \times F_y.[/imath] is an isomorphism ?. Thanks a lot to all of you. |
869700 | Closed form for the sum: [imath]\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}[/imath]
I tried using partial fractions to compute the sum of the series [imath] \sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)} [/imath] Another technique is to turn this series into a definite integral of 0 to 1. but do not know how to do. Thanks for any help. | 864609 | What is the closed form for [imath]\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}[/imath]?
A while ago, I started to look at expressions of the following form: [imath] S_p:=\sum_{n=1}^\infty \frac1n - \frac1{n+1/p}, [/imath] where [imath]p[/imath] is prime, because otherwise things get too complicated for me at the moment. What I found so far is the following: [imath] S_p= p -\log(2p) - \frac12 \pi \cot\left(\frac\pi p\right) + T_p, [/imath] where [imath]T_p[/imath] contains [imath]\frac{p-1}2[/imath] terms [imath]t_{p,k}[/imath] of the form: \begin{cases} \pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right) \\ \pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\sin\left(\frac{k\pi}{2p}\right)\right)\\ \pm 2 \sin\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right) \\ \pm 2 \cos\left(\frac{k\pi}{2p}\right) \log\left(\cos\left(\frac{k\pi}{2p}\right)\right) \end{cases} An example can be found here. How does the closed form for [imath]\sum_{n=1}^\infty \frac1n - \frac1{a+n}[/imath] look like? (EDIT: where [imath]a=1/p[/imath]) |
869866 | [imath]a\le b[/imath] iff there exist [imath]\left|A\right|=a, \left|B\right|=b[/imath] and [imath]A\subseteq B[/imath]
[imath]a\le b[/imath] iff there exist [imath]\left|A\right|=a, \left|B\right|=b[/imath] and [imath]A\subseteq B[/imath] My Proof: [imath](\Leftarrow)[/imath] [imath]A\subseteq B[/imath] implies immediately that [imath]\left|A\right|\le\left|B\right|[/imath]. Hence, [imath]a\le b[/imath]. [imath](\Rightarrow)[/imath] [imath]a\le b[/imath]. Lets assume that [imath]b[/imath] is finite, [imath]b=n\in\mathbb{N}[/imath]. Then, there is a finite set [imath]B=\left\{0,...,n-1\right\}[/imath] such that [imath]\left|B\right|=n[/imath]. [imath]a[/imath] is also finite, [imath]a=m\in \mathbb{N}[/imath]. We are allowed to reduce [imath]B[/imath] to [imath]A=\{0,...,n-m+1\}[/imath] because we are able to do this with a finite number of steps. We have that [imath]\left|A\right|=a, \left|B\right|=b[/imath] and [imath]A\subseteq B[/imath]. Assuming that [imath]b\ge \aleph_0[/imath]. Then there is (Can I just claim that?) [imath]\left|B\right| = b[/imath]. If [imath]a[/imath] is finite then... If [imath]a[/imath] is infinite then ... Can you help me complete the proof? Also, I'd be glad to get a review for what I did so far. Thanks. | 859038 | A question about cardinal numbers in ZF set theory.
It is well known that cardinal numbers and the relations between them can be defined in ZF set theory (using the notion of "rank"), without the need of additional axioms. Can the following statement be proved from just the axioms of [imath]\sf ZF[/imath]? " Let [imath]k,\ell[/imath] be cardinal numbers and let [imath]k<\ell[/imath]. If [imath]A[/imath] is any set whose cardinal number is [imath]k[/imath], then there always exists a set [imath]B[/imath] whose cardinal number is [imath]\ell[/imath] and which contains [imath]A[/imath] as a subset". |
869722 | How to find this integral [imath]\int_{0}^{1}\ln\ln\bigl(1/x+\sqrt{(1/x^2)-1}\,\bigr)dx[/imath]
How do I compute this integral ? [imath]I=\int_{0}^{1}\ln{\left(\ln{\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)}\right)}dx[/imath] In the math chatroom someone suggests setting [imath]x=\operatorname{sech}(t)[/imath] and that the result immediately follows. I don't agree with it because [imath]\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}=\frac{e^t+e^{-t}}{2}+\sqrt{\cosh^2{t}-1}=e^t[/imath] and [imath]dx=\frac{e^t}{(e^{2t}+1)^2}dt[/imath] so [imath]I=\int_{0}^{\infty}\ln(t)\frac{e^{t}}{(e^{2t}+1)^2}dt[/imath] Thanks for your help. | 390640 | Closed form for [imath]\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx[/imath]
Please help me to find a closed form for the following integral: [imath]\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.[/imath] I was told it could be calculated in a closed form. |
869776 | Disconnecting using totally disconnected sets
Let [imath]X[/imath] be [imath][0,1]^2[/imath] and [imath]S\subset X[/imath] a totally disconnected subset. Is it true that [imath]S^c[/imath] is always connected? If it is false, what can we say when [imath]X=[0,1]^n[/imath]? | 31667 | Complement of a totally disconnected compact subset of the plane
Let [imath]E \subset \mathbb{C}[/imath] be compact and totally disconnected. Is there an elementary way to prove that [imath]\mathbb{C} \setminus E[/imath] is connected? |
869216 | Exercise 7.10 Atiyah, [imath]M[x] [/imath] is a noetherian [imath]A[x] [/imath]-module
The exercise is: Let [imath]M[/imath] be a noetherian [imath]A[/imath]-module. Then [imath]M[x] [/imath] is a noetherian [imath]A[x] [/imath] module. The action of [imath]A[x] [/imath] on [imath]M[x] [/imath] is the obvious one. In a previous exercise it was shown that [imath]M[x] \cong A[x] \otimes_A M[/imath] How to proceed ? | 616152 | Show that [imath]M[x][/imath] is a Noetherian [imath]A[x][/imath]-module.
This is a question from Atiyah and Macdonald, Introduction to Commutative Algebra. Problem: Let [imath]M[/imath] be a Noetherian [imath]A[/imath]-module. Show that [imath]M[x][/imath] is a Noetherian [imath]A[x][/imath]-module. Solution: So, I can solve the problem with an extra assumption. That is, if we assume that [imath]M[/imath] is faithful (i.e., [imath]Ann(M)=0[/imath]). In this case, it follows that [imath]A[/imath] is necessarily Noetherian as well. Hence, by Hilbert's Basis Theorem, it follows that [imath]A[x][/imath] is Noetherian as well. It can easily be shown that [imath]M[x]\cong A[x]\bigotimes_A M.[/imath] I can also show that the tensor product of two Noetherian modules is Noetherian, hence the result. I am wondering though, does this result hold without this extra assumption? I guess, I'm not also sure whether the ring [imath]A[/imath] is always necessarily Noetherian if we do not require that our module [imath]M[/imath] be faithful? All I can show is that if [imath]M[/imath] is Noetherian as an [imath]A[/imath]-module then [imath]A/Ann(M)[/imath] is necessarily Noetherian as a ring. Thanks!! |
870420 | Is there a positive integer [imath]N[/imath] that fits?
Is there a positive integer [imath]N[/imath] such that the equation [imath]x^2 + y^2 = N[/imath] has at least [imath]2005[/imath] solutions in non-negative integers [imath]x[/imath] and [imath]y[/imath]? If so, prove it. If not, prove it! | 17496 | Number of integer solutions of [imath]x^2 + y^2 = k[/imath]
I'm looking for some help disproving an answer provided on a StackOverflow question I posted about computing the number of double square combinations for a given integer. The original question is from the Facebook Hacker Cup Source: Facebook Hacker Cup Qualification Round 2011 A double-square number is an integer [imath]X[/imath] which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because [imath]10 = 3^2 + 1^2[/imath]. Given [imath]X[/imath], how can we determine the number of ways in which it can be written as the sum of two squares? For example, [imath]10[/imath] can only be written as [imath]3^2 + 1^2[/imath] (we don't count [imath]1^2 + 3^2[/imath] as being different). On the other hand, [imath]25[/imath] can be written as [imath]5^2 + 0^2[/imath] or as [imath]4^2 + 3^2[/imath]. You need to solve this problem for [imath]0 \leq X \leq 2,147,483,647[/imath]. Examples: [imath]10 \Rightarrow 1[/imath] [imath]25 \Rightarrow 2[/imath] [imath]3 \Rightarrow 0[/imath] [imath]0 \Rightarrow 1[/imath] [imath]1 \Rightarrow 1[/imath] In response to my original question about optimizing this for F#, I got the following response which I'm unable to confirm solves the given problem correctly. Source: StackOverflow answer by Alexandre C. Again, the number of integer solutions of [imath]x^2 + y^2 = k[/imath] is four times the number of prime divisors of [imath]k[/imath] which are equal to [imath]1 \bmod 4[/imath]. Knowing this, writing a program which gives the number of solutions is easy: compute prime numbers up to [imath]46341[/imath] once and for all. Given [imath]k[/imath], compute the prime divisors of [imath]k[/imath] by using the above list (test up to [imath]\sqrt{k}[/imath]). Count the ones which are equal to [imath]1 \bmod 4[/imath], and sum. Multiply answer by [imath]4[/imath]. When I go through this algorithm for [imath]25[/imath], I get [imath]8[/imath] which is not correct. For each prime factor (pf) of [imath]25[/imath] ([imath]5[/imath], [imath]5[/imath] are prime factors of [imath]25[/imath]) if pf % [imath]4[/imath] = [imath]1[/imath] (true for both [imath]5[/imath]'s), add [imath]1[/imath] to count return [imath]4[/imath] * count (count would be [imath]2[/imath] here). So for [imath]25[/imath], this would be [imath]8[/imath] |
870524 | Is the sequences[imath]\{S_n\}[/imath] convergent?
Let [imath]S_n=e^{-n}\sum_{k=0}^n\frac{n^k}{k!}[/imath] Is the sequences[imath]\{S_n\}[/imath] convergent? The following is my answer,but this is not correct. please give some hints. For all [imath]x\in\mathbb{R}[/imath], [imath]\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{x^k}{k!}=e^x.[/imath] then [imath]\lim_{n\rightarrow\infty}e^{-n}\sum_{k=0}^n\frac{n^k}{k!}=1.[/imath] | 136996 | Partial sums of exponential series
What is known about [imath]f(k)=\sum_{n=0}^{k-1} \frac{k^n}{n!}[/imath] for large [imath]k[/imath]? Obviously it is is a partial sum of the series for [imath]e^k[/imath] -- but this partial sum doesn't reach close to [imath]e^k[/imath] itself because we're cutting off the series right at the largest terms. In the full series, the [imath](k+i-1)[/imath]th term is always at least as large as the [imath](k-i)[/imath]th term for [imath]1\le i\le k[/imath], so [imath]f(k)< e^k/2[/imath]. Can we estimate more precisely how much smaller than [imath]e^k[/imath] the function is? It would look very nice and pleasing if, say, [imath]f(k)\sim e^{k-1}[/imath] for large [imath]k[/imath], but I have no real evidence for that hypothesis. (Inspired by this question and my answer thereto). |
870545 | Why the Ito isometry implies this equality?
If [imath]{\rm Cov}[dW_t,dB_t]=\rho \, dt[/imath] then why [imath]\mathbb{Cov} \left( \int_0^t \sigma_{1}(s) \mathrm{d} W_s, \int_0^t \sigma_{2}(u) \mathrm{d} B_u \right)[/imath] [imath]\stackrel{\text{Ito isometry}}{=} \rho \int_0^t \sigma_1(s) \sigma_2(s) \, \mathrm{ds}[/imath] where [imath]\sigma_{1s}[/imath] and [imath]\sigma_{2s}[/imath] are two deterministic functions of [imath]t[/imath]? I do not understand why the [imath]\rho[/imath] emerges. I mean from the application of the Ito isometry the answear should not only be [imath]\int_0^t \sigma_1(s) \sigma_2(s) \, \mathrm{d}s[/imath]? | 870352 | Expected value of correlated stochastic integrals
I do not understand the following result: Suppose [imath]dz_\chi[/imath] and [imath] dz_\xi[/imath] are correlated increments of standard Brownian motion with [imath]dz_\chi dz_\xi=\rho dt[/imath] you have the following expectation giving the following result: [imath]\mathbb{E} [\int_{t}^{T}e^{\kappa s} dz_\chi \int_{t}^{T}dz_\xi] = \rho \int_{t}^{T}e^{\kappa s} ds[/imath] I really do not get it. This equation is justified by the Itô's isometry. In the book that I am reading the author just passes by saying that is easy to see, but if someone could show more explicit the steps that are taking to get this equation. |
222209 | Convergence in probability and almost surely
Let [imath]X_n[/imath] be a sequence of independent random variable which converges in probability to [imath]X[/imath]. Prove [imath]X[/imath] is a constant. Can someone give me a hint how I should go about proving this? I tried proving this by contradiction by saying [imath]X[/imath] taking 2 different values, but this can still still happen because [imath]X[/imath] is only almost surely constant. | 860149 | Sequence of independent RV
Let [imath](X_n)[/imath] be a sequence of independent RVs that converges in probability to some RV [imath]X[/imath]. Show that [imath]X[/imath] is constant a.e. My attempt So from definition of convergence I have: [imath]\forall\epsilon>0 \:\:\:\mathbb{P}(|X_n-X|>\epsilon)\longrightarrow0[/imath] I think that there is some useful theorem that I don't know. Can anyone help? I mean just give me some tip, not necessarily full answer. |
870911 | Flaw in the proof that a set is countable
Q: Let S be the set containing all sequences of 0's and 1's. i.e [imath]S = \{(a_1,a_2,a_3,a_4,\ldots) : a_i = 0 \text{ or } 1\}[/imath] Show that S is countable. Proof(Flawed) : Let [imath]A_i[/imath] be the set of all sequences of length i . Clearly there are finite( [imath]2^i[/imath]) such sequences of length i. Therefore, S = [imath]\bigcup_1^\infty A_i[/imath], a collection of countable set is also countable. P.S: Actually S is uncountable and can be proved using a similar technique to Cantor's Diagonalization method. Where is the flaw in the above mentioned proof? | 338094 | why is the set of all binary sequences not countable?
What is wrong with this reasoning: The union (finite or infinite) of countable sets is countable. The set of all binary sequences is the infinite union of the sets [imath]S_n[/imath] of all the binary sequences of length n, which are finite, hence countable. |
486401 | Induction proof [imath]2n+1<2^n[/imath]
I struggle to proof that: [imath]2n+1<2^n[/imath] By using induction. The base case is for [imath]n\ge3[/imath]. Any help will be appreciated! | 351151 | Induction to prove [imath]2n + 3 < 2^n[/imath]
I am having trouble and was wondering if someone could go over the steps slowly to show that: [imath]2n + 3 < 2^n \ \text{for} \ n \geq 4[/imath] Any help would be amazing! |
870195 | On an existence of a real-valued measurable function on a non-atomic probability measure space
Suppose that [imath](X,E,μ)[/imath] is a non-atomic probability measure space. Let [imath]\xi :X \to \mathbb{R}[/imath] be a random variable. A Borel measure [imath]\mu_{\xi}[/imath] in [imath]\mathbb{R}[/imath] defined by [imath]\mu_{\xi}(X)=\mu(\xi^{-1}(X))[/imath] for [imath]X \in B(\mathbb{R})[/imath] is called a Borel probability measure on [imath]\mathbb{R}[/imath] defined by [imath]\xi[/imath]. Question: Does there exists a measurable function [imath]\xi :X \to \mathbb{R}[/imath] such that a Borel probability measure [imath]\mu_{\xi}[/imath] is non-atomic? | 31962 | How to split an integral exactly in two parts
This question is a by-product of a conversation with Theo Buehler in comments to this answer. Let's settle definitions. Definition Let [imath](\Omega, \mathcal{F}, \mu)[/imath] be a measure space. We say that [imath]X\in \mathcal{F}[/imath] is an atom if [imath]\mu(X) > 0[/imath] and its only subset of strictly positive measure is [imath]X[/imath] itself. [imath]\Omega[/imath] is said to be non-atomic if no atoms exist. So in a non-atomic measure space we can always split a measurable subset into smaller ones. Now the question is: have we got some control over this splitting? Can we split something in exactly two parts? I'm especially interested in the following. Question Let [imath](\Omega, \mathcal{F}, \mu)[/imath] be a non-atomic measure space and [imath]f \in L^1(\Omega), f \ge 0[/imath]. Are there measurable and disjoint [imath]A, B \subset \Omega[/imath] such that [imath]A \cup B = \Omega[/imath] and [imath]\int_A f(x)\, d\mu= \int_B f(x)\, d\mu=\frac{1}{2}\int_\Omega f(x)\, d\mu?[/imath] |
871218 | Hard integral, low hints...
[imath]\int_{ - \pi /2}^{\pi /2} \frac1{2007^{x} + 1}\cdot \frac {\sin^{2008}x}{\sin^{2008}x + \cos^{2008}x} \, dx .[/imath] This integral stuns me for a while, I just can't solve it! I tried integration by recurrence since here it seems that we have something that looks like a recurrence, but nothing comes up. I also tried all the integration techniques I'm aware of but they just don't work. So I fear like this integral uses something like Fourier, or some Laplace transform thing or something of that sort but I don't know them. That's why I will be very thankful if they could explain me how to solve it. Thank you. | 741580 | Integral [imath] \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx [/imath]
I am trying to solve this integral [imath] \int_{-\pi/2}^{\pi/2} \frac{1}{2007^x+1}\cdot \frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx [/imath] A closed form does exist despite the looks of the integrand. This problem is from some old high school IMO training courses. I am not sure how to solve it. The only information that may be of help is [imath] \sin x=\sum_{n=1}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x=\sum_{n=1}^\infty \frac{ x^{2n}}{(2n)!} \quad \forall \ x [/imath] Thanks, I am looking for a complete solution, not a description as of what to do. |
871930 | [imath]\int_0^{2\pi}e^{\cos x}\cos(\sin x)dx[/imath]
[imath]\int_0^{2\pi}e^{\cos x}\cos(\sin x)dx[/imath] I tried Integration by parts but failed. Wolfram alpha gives answer in decimal points which are same as of [imath]2\pi[/imath]. Any hints or suggestions will be helpful. | 358548 | integrate [imath]\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta[/imath]
How to integrate [imath] 1)\displaystyle \int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta[/imath] [imath] 2)\displaystyle \int_0^{2\pi} e^{\cos \theta} \sin ( \sin \theta) d\theta[/imath] |
871740 | Clarification of Contour Integration
I apologise if this seems like an elementary and silly question, but I am confused about the integral [imath]I=\int^{\infty}_{-\infty}\frac{\cos{x}}{1+x^2}dx=\frac{\pi}{e}[/imath] If I consider a semicircular contour with infinite radius in the upper half on the plane, I can write [imath]I=\Re\oint_C \frac{e^{iz}}{1+z^2}dz=\Re\left[2\pi i\lim_{z \to i}\frac{e^{iz}}{z+i}\right]=\frac{\pi}{e}[/imath] since the integral over the arc vanishes as the radius approaches infinity. However, I get an incorrect result if I instead use the integral [imath]I=\oint_C\frac{\cos{z}}{1+z^2}dz=\pi i\lim_{z \to i}\frac{e^{iz}+e^{-iz}}{z+i}=\frac{\pi}{2}(e+e^{-1})\neq \frac{\pi}{e}[/imath] May I know where I have made a mistake? Why are we not allowed to use (or are we?) [imath]\cos{z}[/imath] directly in the complex integrand but have to use [imath]e^{iz}[/imath] instead and extract the real part? And why is it that the 2 answers turn out to be different? Help offered to clarify my doubts will be appreciated. Thank you. | 738139 | How does the integral [imath]\int_{D_C} e^{ia z}P(z)/Q(z)\,\mathrm{d}z[/imath] blow up.
In my book I have a theorem that goes something like the following Let [imath]P(x)[/imath] be [imath]Q(x)[/imath] polynomials such that [imath]\deg(Q) \geq \deg(P) + 2[/imath]. Then \begin{align*} \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} e^{iax} \,\mathrm{d}x = 2 \pi i \frac{a}{|a|} \sum_{k=1}^{m} \mathrm{Res}\left[ \frac{P(x)}{Q(x)} , z_k \right] \end{align*} where [imath]a[/imath] is a real constant and [imath]z_1,\,\ldots\,,z_m[/imath] are the singularities to [imath]P(x)/Q(x)[/imath] in the upper halfplane if [imath]a>0[/imath] and the lower halfplane if [imath]a<0[/imath]. I am having a bit of problems understanding this. I know the reason why we have to switch contour is that [imath]e^{iz}[/imath] blows up. But I really can not see why or how it blows up. Take the canonical example for why this theorem is useful [imath] J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1} [/imath] Now the clue here is to show that the integral along the curve tends to zero, and then use the residue theorem. To show that the integral tends to zero I did this \begin{align*} \left| \int_{C_1} \frac{e^{iz}}{1+z^2} \,\mathrm{d}z\right| \leq \int_{C_1} \left| \frac{e^{iz}}{1+z^2} \right| \,\mathrm{d}z \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{\left|1+z^2\right|} \right) \int_{C_1} |e^{iz}| \,\mathrm{d}z \end{align*} where the [imath]ML[/imath]-inequality was used in the last inequality On the circle with radius [imath]R[/imath] we have [imath]|e^{iz}|=R[/imath], and we can use the inequality [imath]|a+b|\leq|a|-|b|[/imath] to simplify further \begin{align*} \left|\int_{D_R} \frac{e^{iz}}{1+z^2}\,\mathrm{d}z \right| \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{|z|^2-1}\right) \int_{D_R} R \,\mathrm{d}z \leq \pi \frac{R}{R^2-1} \end{align*} which tends to zero as [imath]R \to \infty[/imath] as wanted. but if one instead had [imath]e^{-iz}[/imath] then the theorem states that one has to use the lower half plane. But I do not see where my calculations err if one persists in using the contour in the upper half plane? What goes wrong, and why does it go wrong? [imath]|e^{-iz}|[/imath] should still be [imath]R[/imath] on the semi-circle. How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane? |
20670 | Theorem of Arzelà-Ascoli
The more general version of this theorem in Munkres' 'Topology' (p. 290 - 2nd edition) states that Given a locally compact Hausdorff space [imath]X[/imath] and a metric space [imath](Y,d)[/imath]; a family [imath]\mathcal F[/imath] of continuous functions has compact closure in [imath]\mathcal C (X,Y)[/imath] (topology of compact convergence) if and only if it is equicontinuous under [imath]d[/imath] and the sets [imath] \mathcal F _a = \{f(a) | f \in \mathcal F\} \qquad a \in X[/imath] have compact closure in [imath]Y[/imath]. Now I do not see why the Hausdorff condition on [imath]X[/imath] should be necessary? Why include it then? Am I maybe even missing something here (and there are counterexamples)? btw if you are looking up the proof: Hausdorffness is needed for the evaluation map [imath]e: X \times \mathcal C(X,Y) \to Y, \, e(x,f) = f(x)[/imath] to be continuous. But the only thing really used in the proof is the continuity of [imath]e_a: \mathcal C(X,Y) \to Y, \, e_a(f) = f(a)[/imath] for fixed [imath]a \in X[/imath]. Cheers, S.L. | 561131 | Is Hausdorffness necessary for the classical ascoli theorem?
Munkres - topology p.278 I exactly followed the argument in the text, and I cannot find where I used hausdorffness. Where in the argument used Hausdorffness? The reason why I am asking is that the article in wikipedia requires the Hausdorffness: Wikipedia - Let [imath]X[/imath] be a compact Hausdorff space. Then a subset [imath]F[/imath] of [imath]C(X)[/imath] is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and pointwise bounded. |
871953 | Using l'Hôpital rule to find [imath]\lim_{x\to-\infty} xe^x[/imath]
I'm trying to solve this limit: [imath]\lim_{x\to-\infty} xe^x[/imath] I'm trying to solve this limit using the l'Hôpital rule. My question is: can I use this rule in the last limit below? [imath]\lim_{x\to-\infty} xe^x=\lim_{x\to-\infty} \frac{x}{e^{-x}}[/imath] Note the numerator tends to [imath]-\infty[/imath], while the denominator tends to [imath]\infty[/imath], can I use the l'Hôpital rule in this case? Thanks | 767781 | Why does [imath]\lim\limits_{t\to-\infty}te^t=0[/imath]?
Intuitively, [imath]\lim\limits_{t\to -\infty}te^t=0[/imath], since [imath]e^t\to 0[/imath] much faster than [imath]t\to -\infty[/imath]. Is there a way to more rigorously compute this? Writing it as [imath]e^t/(1/t)[/imath] seems resistant to l'Hopital's rule. |
872276 | How to show that [imath](a+b)^p\le 2^p (a^p+b^p)[/imath]
If I may ask, how can we derive that [imath](a+b)^p\le 2^p (a^p+b^p)[/imath] where [imath]a,b,p\ge 0[/imath] is an integer? | 793348 | Bounding [imath](x+y)^n[/imath]
Let [imath]n[/imath] be a natural number. Is it possible to write [imath](x+y)^n \leq C(x^n + y^n)[/imath] for some constant [imath]C[/imath]?? It is obvious for [imath]n=2[/imath] (using Young's inequality) but not obvious to me for other [imath]n[/imath]. Let [imath]x[/imath] and [imath]y[/imath] be positive reals. |
872817 | Prove that [imath]|Im\{f(z)\}|≤ [/imath][imath]\frac{2}{\pi} \log \frac{1+|z|}{1 - |z|},[/imath] [imath]z \in \mathbb{D}[/imath].
Let [imath]f(z)[/imath] be an analytic function defined on the unit disk [imath]\mathbb{D} = \{z : |z| < 1\}[/imath] so that [imath]f(0)=0[/imath] and [imath]−1<Re\{f(z)\}<1[/imath] for all [imath]z \in \mathbb{D}[/imath]. Prove that [imath]|Im\{f(z)\}|≤ [/imath][imath]\frac{2}{\pi} \log \frac{1+|z|}{1 - |z|},[/imath] [imath]z \in \mathbb{D}[/imath]. I want to use the fact that [imath]Imf[/imath] and [imath]Ref[/imath] satisfy maximum modulus principle (strong version. But not sure if that is useful or not. Would love some help. Thanks | 624888 | The estimate of real and imaginary part of a holomorphic function
Let [imath]f[/imath] be holomorphic in the unit disk [imath]D(0,1)[/imath], and [imath]f(0)=0[/imath]. Suppose [imath]|\Re f(z)|\leq 1[/imath] for all [imath]|z|<1[/imath]. Show that [imath]|\Re f(z)|\leq\frac{4}{\pi}\arctan|z|;[/imath] [imath]|\Im f(z)|\leq \frac{2}{\pi}\ln\frac{1+|z|}{1-|z|}.[/imath] What I could do is just to observe that [imath]\frac{e^{i\frac{\pi}{2}f(z)}-1}{e^{i\frac{\pi}{2}f(z)}+1}:D(0,1)\to D(0,1)[/imath] with [imath]0[/imath] being fixed. And Schwarz lemma applies. However, I could not get any information on the real and imaginary part of [imath]f[/imath]. |
873353 | Linear Algebra subspaces
Determine whether the set [imath]W= \{(2a-2,3b,2a-3b)\}[/imath] is a subspace in [imath]\mathbb R^3[/imath] Describe the set. I know that in order for a set to be a subspace, it must be closed under multiplication and addition.. How would I go about showing this? Thanks | 873164 | Determine whether the set [imath]W[/imath] is a subspace in [imath]\mathbb{R}^{3}[/imath]
Determine whether the set [imath]W=\{(2a-2,3b,2a-3b)\}[/imath] is a subspace in [imath]\mathbb{R}^{3}[/imath]. Describe the set. I have tried putting the set into matrix form but don't know which is correct. I also know that it needs to be closed under addition and multiplication but can't figure out how to show that. |
873442 | Square of [imath]7+\sum_{k=1}^n6\times10^k[/imath]
If we build a number as follow: [imath]N=7+\sum_{k=1}^n6\times10^k[/imath] we find: [imath]N^2=9+\sum_{k=1}^n8\times10^k+\sum_{j=n+1}^{2n+1 }4\times10^{j}[/imath] that means for example: [imath]67^2=4489[/imath], [imath]667^2=444889[/imath], [imath]6667^2=44448889[/imath] and so on. How can we prove that given an arbitrary [imath]n[/imath] and a number [imath]N[/imath] built as above, the square of [imath]N[/imath] can be obtained by th equation for [imath]N^2[/imath]? Thanks | 529597 | Prove that in a sequence of numbers [imath]49 , 4489 , 444889 , 44448889\ldots[/imath]
Prove that in a sequence of numbers [imath]49 , 4489 , 444889 , 44448889\ldots[/imath] in which every number is made by inserting [imath]48[/imath] in the middle of previous as indicated, each number is the square of an integer. |
173730 | Elements of [imath]\mathbb{F}_p[/imath] having cube roots in [imath]\mathbb{F}_p[/imath]
Let [imath]p[/imath] be a prime number, and let [imath]\mathbb{F}_p[/imath] be the field with [imath]p[/imath] elements. How many elements of [imath]\mathbb{F}_p[/imath] have cube roots in [imath]\mathbb{F}_p[/imath]? I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated. | 886203 | Modulo equation with prime
Let [imath]p[/imath] be prime with [imath]p\equiv 2 \bmod 3[/imath]. Show that for any integer [imath]a,p\nmid a[/imath] there is an integer [imath]x:x^3 \equiv a\bmod p[/imath] |
873654 | Prove that [imath]g[/imath] is continuous.
Let [imath]f:[a,b]\to \mathbb{R}[/imath] be a continuous function and let [imath]g:[a,b]\to \mathbb{R}[/imath] be a function such that [imath]g(a)=f(a)[/imath] and [imath]g(x)=\sup_{t\in [a,x]}f(t)[/imath]. Prove that [imath]g[/imath] is continuous on [imath][a,b][/imath]. I noticed that [imath]g[/imath] is increasing so if somehow I show that it satisfies Intermediate value property then it must be continuous. | 748639 | Prove functions defined by sup and inf are continuous
Suppose [imath]f[/imath] is continuous on [imath][a,b][/imath]. Show that the functions defined by [imath]m(x)=\inf\{f(y):y\in[a,x]\}[/imath] and [imath]M(x)=\sup\{f(y):y\in[a,x]\}[/imath] are well defined and are also continuous on [imath][a,b][/imath] I have already managed to prove that they are well defined, since [imath]f[/imath] is continuous on [imath][a,b][/imath] so it is bounded. Therefore for every [imath]x\in[a,b][/imath], the set [imath]\{f(y):y\in[a,x]\}[/imath] has a supremum and an infimum. To prove that they are continuous on [imath][a,b][/imath], I've taken an arbitrary sequence [imath]\{x_n\}[/imath] contained in [imath][a,b][/imath] converging to some number [imath]c\in[a,b][/imath] and am attempting to show that [imath]m(x_n)\rightarrow m(c)[/imath] and [imath]M(x_n)\rightarrow M(c)[/imath], but I'm not quite sure how. Any help would be appreciated, thanks! |
873644 | proving closure of a subset
Let B be a set, and let * be a binary operation in B. Suppose * satisfies the associative law. Let [imath]P=\{b \in B : b * w = w * b \quad\forall\, w \in B\}[/imath] Prove that P is closed under *. | 496498 | Is associative binary operator closed on this subset?
Here is the problem: Suppose that [imath]*[/imath] is an associative binary operation on a set [imath]S[/imath]. Let [imath]H:= \{a \in S\mid a * x = x * a \mbox{ for all }x\in S\}.[/imath] In other words, [imath]H[/imath] is consisting of all the elements of [imath]S[/imath] that commute with every element in [imath]S[/imath]. Show that [imath]H[/imath] is closed under [imath]*[/imath]. |
874026 | Set and cardinality injection and surjection proof
Let X be a set. Prove there is an injection from [imath]X \rightarrow 2^X[/imath]. Prove that there is not a surjection from [imath]X \rightarrow 2^X[/imath]. My try- Assume to the contrary that [imath]f: X \rightarrow 2^X[/imath]. is a surjection and consider the set [imath]M=\{x\in X | x \not\in f(x)\} [/imath]. Then show to show that M doesn't have a pre-image. So, there is no surjection. | 354556 | Cardinality of a set A is strictly less than the cardinality of the power set of A
I am trying to prove the following statement but have trouble comprehending/going forward with some parts! Here is the statement: If [imath]A[/imath] is any set, then [imath]|A|[/imath] [imath]<[/imath] [imath]|P(A)|[/imath] Here is what I have so far: We need to show that there is an injection from [imath]A[/imath] to [imath]P(A)[/imath] but not a surjection. A natural choice for an injection is the function [imath] f(x)[/imath] [imath]=[/imath] [imath]\{x \}[/imath], which in plain English, takes any element [imath]x[/imath] (that is in [imath]A[/imath]) and sends it to the one-element set [imath]\{x \}[/imath]. Thus [imath]f(x)[/imath] is injective! To show that there is no surjection, for the sake of contradiction, assume there is a surjection. Here is where I start to have trouble. Surjectivity means that every element of the co-domain is mapped to an element of the domain, correct? Consequently, in this particular case, we are "matching" sets (from [imath]P(A)[/imath]) to elements (from [imath]A[/imath]) right? If the above is correct, my problem arises here. I am not sure how to prove that [imath]f[/imath] is not surjective. Unfortunately, I am easily confused by notation so please explain in English. Thank you in advance!! :) |
874430 | Show [imath](a+b, a-b) = 1[/imath] or [imath]2[/imath] if [imath](a,b)=1[/imath]
Here was my take on the proof. We already know that since [imath](a,b)=1[/imath], there exist integers [imath]x,y[/imath] such that [imath]ax+by=1[/imath]. Let [imath]d=(a+b,a-b)[/imath]. Then [imath]d|(a+b)[/imath] and [imath]d|(a-b)[/imath]. In particular, there exist integers [imath]k_1, k_2[/imath] such that [imath]a+b=dk_1[/imath] [imath]a-b=dk_2[/imath] Thus depending whether we add or subtract the two equation we get both [imath]2a=d(k_1+k_2)[/imath] [imath]2b=d(k_1-k_2)[/imath] Thus, [imath]d|2a[/imath] and [imath]d|2b[/imath]. But then also [imath]d|(2ax+2by)\Rightarrow d|2(ax+by)[/imath] Thus, [imath]d|2[/imath] and thus [imath]d=1[/imath] or [imath]2[/imath]. I don't know why, but I feel like there is a problem with my argument. Perhaps something I haven't said...I feel like it's almost there, but missing some sort of justification, but it's late and I can't figure it out.... | 457296 | Prove that if [imath]a[/imath] and [imath]b[/imath] are relatively prime, then [imath]\gcd(a+b, a-b) = 1[/imath] or [imath]2[/imath]
Prove that if [imath]a[/imath] and [imath]b[/imath] are relatively prime, then [imath]\gcd(a+b, a-b) = 1[/imath] or [imath]2[/imath]. I started off by putting [imath]\gcd(a+b, a-b) = d[/imath]. This implies that there are two relatively prime integers [imath]x_1, x_2[/imath], such that [imath]dx_1 = a+b[/imath] [imath]dx_2 = a -b[/imath] Adding the first equation to the second gives us: [imath]d(x_1 + x_2) = 2a[/imath], and subtracting the second from the first gives us [imath]d(x_1 - x_2) = 2b[/imath]. This implies that [imath]d\mid2a, d\mid2b \implies \gcd(2a, 2b) = d \implies d = 2\cdot \gcd(a,b) \implies d = 2[/imath]. This is based on the fact that [imath]\gcd(x_1 + x_2, x_1 - x_2) = 1[/imath], which is in fact the statement of the problem. How do I prove this? Can I somehow use the Euclidean Algorithm for this. |
115714 | Calculate [imath]\lim_{t\to\infty}\frac 1t\log\int_0^1 \cosh(tf(x))\mathrm d x[/imath]
Given that [imath]f\in C([0,1])[/imath], evaluate [imath]\lim_{t\to\infty}\frac 1t\log\int_0^1 \cosh(tf(x))\mathrm d x.[/imath] I have thought about it for an entire day, but unfortunately I wasn't able to solve it. Now I'm asking you, more than a solution, which is in either case welcomed, an explanation on how to approach such a kind of problems. Thank you very much. | 2414621 | Compute [imath] \lim\limits_{t\to +\infty}\log \frac{1}{t}\int_0^1 \cosh(tf(x))dx. [/imath]
Please give some hints and not the whole solution. Problem Let [imath]f ∈ C([0, 1]).[/imath] Compute [imath] \lim\limits_{t\to +\infty}\log \frac{1}{t}\int_0^1 \cosh(tf(x))dx. [/imath] |
875266 | Proof by induction [imath]\sum_{k=1}^{n}[/imath] [imath]k \binom{n}{k}[/imath] [imath]= n\cdot 2^{n-1}[/imath] for each natural number [imath]n[/imath]
Prove by induction that [imath]\sum_{k=1}^{n}[/imath] [imath]k \binom{n}{k}[/imath] [imath]= n\cdot 2^{n-1}[/imath] for each natural number [imath]n[/imath]. | 683733 | Sum of [imath]k {n \choose k}[/imath] is [imath]n2^{n-1}[/imath]
Proof that [imath]\suṃ̣_{k=1}^{n}k {n \choose k}[/imath] for [imath]n \in \mathbb N[/imath] is equal to [imath]n2^{n-1}[/imath]. As a hint I got that [imath]k {n \choose k} = n {n-1\choose k-1} [/imath]. I tried solving this by induction but, in the inductive step I'm not arriving to the correct result. |
672936 | Prove: The product of any three consecutive integers is divisible by [imath]6[/imath].
I'm new to number theory and was wondering if someone could help me with this proof. Prove: The product of any three consecutive integers is divisible by [imath]6[/imath]. So far I have [imath]\cfrac{x(x+1)(x+2)}{6}[/imath]; How would I go about proving this? Should I replace [imath]x[/imath] with [imath]k[/imath] and then [imath]k[/imath] with [imath]k+1[/imath] and see if the statement is true? | 527300 | Prove that [imath]6[/imath] divides [imath]n(n + 1)(n + 2)[/imath]
I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows: Let [imath]n[/imath] be an integer such that [imath]n ≥ 1[/imath]. Prove that [imath]6[/imath] divides [imath]n(n + 1)(n + 2)[/imath]. Note: An integer [imath]a[/imath] divides an integer [imath]b[/imath], written [imath]a|b[/imath], if there exists [imath]q ∈ Z[/imath] such that [imath]b = qa[/imath]. Alternatively, [imath]a|b[/imath] if dividing [imath]b[/imath] by [imath]a[/imath], [imath]b ÷ a[/imath], results in an integer. Should I do a proof by induction? All help/input is appreciated! |
873615 | Matrix Semi-Definite Inequality
Does the following inequality hold? If matrix [imath]A[/imath] is a [imath]n \times n [/imath] positive semi-definite, [imath]A \succeq 0[/imath], and [imath]U[/imath] is one [imath]n \times k[/imath] unit column-orthogonal matrix ([imath]k \leq n[/imath]), [imath]U^{T}U=I[/imath], do we have [imath]A \succeq UU^{T} A U U^{T}[/imath] ? | 667064 | Matrix semi-positive definite
If [imath]n \times n[/imath] matrix [imath]A \succeq 0[/imath], and one [imath] n \times q[/imath] column orthogonal matrix [imath]U[/imath], does this inequality hold? [imath]A- UU^{T} A UU^{T} \succeq 0[/imath] |
874044 | Proof that the continuous image of a compact set is compact
Let [imath]X\subset \mathbb R^{n}[/imath] be a compact set, and [imath]f :\mathbb R^{n}\to \mathbb R [/imath] a continuous function. Then, [imath]$F(X)$[/imath] is a compact set. I know that this question may be a duplicate, but the problem is that I have to prove this using real analysis instead of topology. I'm struggling with proving that [imath]$F(X)$[/imath] is bounded. I know that the image of a continuous function is bounded, but I'm having trouble when it comes to prove this for vectorial functions. If somebody could help me with a step-to-step proof, that would be great. | 26514 | Proving continuous image of compact sets are compact
How to prove: Continuous function maps compact set to compact set using real analysis? i.e. if [imath]f: [a,b] \rightarrow \mathbb{R}[/imath] is continuous, then [imath]f([a,b])[/imath] is closed and bounded. I have proved the bounded part. So now I need some insight on how to prove [imath]f([a,b])[/imath] is closed, i.e. [imath]f([a,b])=[c,d][/imath]. From Extreme Value Theorem, we know that [imath]c[/imath] and [imath]d[/imath] can be achieved, but how to prove that if [imath]c < x < d[/imath], then [imath]x \in f([a,b])[/imath] ? Thanks! |
875014 | Evaluation of [imath] \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath]
Evaluation of [imath]\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx[/imath] [imath]\bf{My\; Try::}[/imath] Given [imath]\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx[/imath] Now Let [imath]\displaystyle \tan x = t^2\;,[/imath] Then [imath]\sec^2 xdx = 2tdt[/imath] or [imath]\displaystyle dx = \frac{2t}{1+t^4}dt[/imath] So Integral is [imath]\displaystyle \int\frac{t}{1+t}\cdot \frac{2t}{1+t^4}dt = 2\int\frac{t^2}{(1+t)\cdot (1+t^4)}dt[/imath] Now How can I solve after that Help me Thanks | 530779 | Integral of [imath]\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}[/imath]dx
Question: [imath]\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.[/imath] What we did: we tried using [imath]t=\tan (\frac x2)[/imath] and also dividing both numerator and denominator by [imath]\sqrt {\cos x}[/imath], eventually using the second method we got to this: [imath]\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1} [/imath], for which we know how to solve the first and second integral but not the third... Thanks |
734856 | Prove a real-valued function is monotonic if it is continuous on an open interval and has no local extremes
My approach is to assume that [imath]f[/imath] is not monotonic and then construct a local extreme by using the fact that [imath]f[/imath] attains its extreme values on a compact set. I can assume that [imath]\exists c\lt\delta[/imath] with [imath]f(c)\gt f(\delta)[/imath] (or equal). I can do something similar with the fact that [imath]f[/imath] is not decreasing. Still, aside some a vague vision of a torturous proof by cases, I'm stuck. I have a feeling that this proof can be done in a simple way that I am missing. Could someone give me a hint? Assure me that there is a simple proof of this? | 875976 | A continuous function that attains neither its minimum nor its maximum at any open interval is monotone
Let [imath]f: \mathbb R\to \mathbb R[/imath] be a continuous function such that [imath]f[/imath] attains neither its minimum nor its maximum at any open interval [imath]I \subseteq \mathbb R[/imath] , then how to prove that [imath]f[/imath] is monotone ? |
876287 | Prove that if the sum of each row of A equals s, then s is an eigenvalue of A.
Consider an [imath]n \times n[/imath] matrix [imath]A[/imath] with the property that the row sums all equal the same number [imath]s[/imath]. Show that [imath]s[/imath] is an eigenvalue of [imath]A[/imath]. [Hint: Find an eigenvector] My attempt: By definition: [imath]Ax = sx[/imath] which implies that [imath](A - sI)x = 0[/imath] [imath]s[/imath] is an eigenvalue for [imath]A[/imath] iff [imath]\det(A - sI) = 0[/imath] When you do [imath]A - sI[/imath] the sum of each row is now [imath]0[/imath]. I think that's important but I don't know what it means. So this is where I'm stuck | 347408 | Prove that if the sum of each row of [imath]A[/imath] equals [imath]s[/imath], then [imath]s[/imath] is an eigenvalue of [imath]A[/imath].
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix. [imath]i)[/imath]Prove that if the sum of each row of [imath]A[/imath] equals [imath]s[/imath], then [imath]s[/imath] is an eigenvalue of [imath]A[/imath]. [imath]ii)[/imath]Prove that if the sum of each column of [imath]A[/imath] equals [imath]s[/imath], then [imath]s[/imath] is an eigenvalue of [imath]A[/imath]. I think that being an eigenvalue of [imath]A[/imath] implies that [imath]sv=Av[/imath] for some vector [imath]v[/imath]. Furthermore, I know that [imath][a_i] = s[/imath] if we let [imath]a_i[/imath] denote the i-th row of [imath]A[/imath]. However, I do not seem to be able to find a link between these two facts. Could anyone please help me out? |
876341 | Question about Maximum Modulus Principle
Let [imath]f_1, f_2, \ldots ,f_n[/imath] be holomorphic functions on a region [imath]\Omega[/imath]. Show that if [imath]\phi (z)=|f_1(z)|+|f_2(z)|+\ldots +|f_n(z)|[/imath] attains a maximum value on [imath]\Omega[/imath], then [imath]f_i[/imath] is constant for each [imath]i=1,\ldots ,n[/imath]. | 289114 | Show that holomorphic [imath]f_1, . . . , f_n [/imath] are constant if [imath]\sum_{k=1}^n \left| f_k(z) \right|[/imath] is constant.
While studying for an exam in complex analysis, I came across this problem. Unfortunately I was not able to solve it. Any help would be greatly appreciated. Let [imath]U ⊂ \mathbb{C}[/imath] be a domain and [imath]f_1, . . . , f_n : U \rightarrow \mathbb{C}[/imath] holomorphic functions, such that [imath]\sum_{k=1}^n \left| f_k(z) \right|[/imath] is constant on [imath]U[/imath]. Show that [imath]f_1, . . . , f_n[/imath] are constant. |
876482 | A Set That Is "Precisely" Measure-Dense
This question asks for a set of real numbers that is measure-dense, whose complement is also measure dense. In terms of [imath][0,1][/imath], the question asks for an [imath]S[/imath] such that for every open interval [imath]I[/imath] we have [imath]0 < \mu(I \cap S) < 1[/imath]. This question can be made a little more precise: is there a set of real numbers [imath]S[/imath] such that for every open interval [imath]I[/imath], we have [imath]\mu(I \cap S) = \frac{1}{2} \mu(I)[/imath]? The Cantor-set construction referenced above doesn't really lend itself to finding the volume of [imath]I[/imath] contained in [imath]S[/imath], just guaranteeing that it isn't zero. You can modify that construction to get a set that has measure [imath]1/2[/imath], but it's not clear at all that it satisfies the property mentioned here. | 614503 | For set of positive measure [imath]E[/imath], [imath]\alpha \in (0, 1)[/imath], there is interval [imath]I[/imath] such that [imath]m(E \cap I) > \alpha \, m(I)[/imath]
I am a graduate student at Iowa State University attempting to return after a five-year hiatus and take the Real/Complex Analysis qualifier on January 8 for potential reinstatement. Since the professors here are on hiatus, I hope you guys don't mind if I bombard the board with a few past questions from our quals here, like this one from Spring 2013: Let [imath]m[/imath] denote Lebesgue measure on [imath]\mathbb R[/imath] and [imath]M[/imath] the [imath]\sigma[/imath]-algebra of Lebesgue measurable subsets. Given [imath]E \in M[/imath] with [imath]m(E) > 0[/imath] and [imath]\alpha \in (0, 1)[/imath], prove that there is an interval [imath]I[/imath] such that [imath]m(E \cap I) > \alpha \, m(I)[/imath]. I've attempted to force the issue using the definition of outer measure but it doesn't seem to work. I've also attempted using the definition of Lebesgue measurability but that doesn't seem to work either - so I'm at a loss! Thanks in advance! -Darrin Rasberry |
17914 | Cardinality of the set of all real functions of real variable
How does one compute the cardinality of the set of functions [imath]f:\mathbb{R} \to \mathbb{R}[/imath] (not necessarily continuous)? | 1336784 | Which is the cardinality of the set of all functions [imath]\mathbb{R} \to \mathbb{R}[/imath]?
Which is the cardinality of the set of all functions [imath]\mathbb{R} \to \mathbb{R}[/imath] ? For the relations in [imath]\mathbb{R}[/imath] I think the cardinality is the cardinality of [imath]\mathcal{P(\mathbb{R^2})}[/imath] , but I'm not sure. |
20016 | Bijection between [imath]2^{\mathbb R}[/imath] and [imath]\mathbb{ R ^ R}[/imath]
I'm well aware of the standard proof based on cardinality arithmetic to show that these two sets have the same cardinality and the question of defining a bijection between the two sets came up. I worked on it a little while and was wondering if there were any gaps, or a simpler method of coming up with such a bijection. The main idea I had was to take a family of sequences indexed by the real numbers. So that each one of these sequences corresponds to a unique real number. Then I can map each one of these sequences to another number, by applying the binary function from [imath]\mathbb R[/imath] to [imath]\{0,1\}[/imath] and mapping the infinite binary sequence to a real number, use one of the standard bijections. First let [imath]E[/imath] be a family of sequences indexed by the real numbers. Such that [imath]E_{xi}\neq E_{yj}[/imath] for all [imath]x,y \in \mathbb R[/imath] and [imath]i,j \in \mathbb N[/imath] also with the property that [imath]\mathbb R= \bigcup_{x \in \mathbb R, i \in \mathbb N} E_{xi}[/imath]. What sort of theorems would I need to prove such a family existed, is it enough to note that the reals can be represented as a uncountable union of countably infinite sets? Also let [imath]P: 2^{\mathbb N} \rightarrow \mathbb R[/imath] be you favorite bijection. Now consider an element of [imath]2^{\mathbb R}[/imath] as a function [imath]f: \mathbb R \rightarrow \{0,1\}[/imath] and [imath]g \in \mathbb{ R^R}[/imath] in a similar manner. Then our bijection between the two sets will be the function [imath]\Phi[/imath] which maps [imath]f[/imath] to a function [imath]g[/imath], such that [imath]g(x)=P(F(E_x))[/imath] where [imath]F[/imath] applies [imath]f[/imath] to each element of the sequence [imath]E_x[/imath]. I'm fairly sure this is an onto and one-to-one function. The one-to-one property should follow because the sequences are disjoint, cover [imath]\mathbb R[/imath] and [imath]P[/imath] is a bijection. The onto argument is a little hazier, but it should be possible to recreate the function [imath]f[/imath], by first inverting each element with [imath]P[/imath] and then defining [imath]f[/imath] based on these sequences. | 1353383 | Explicit bijection [imath]\Bbb R^{\Bbb R} \to P(\Bbb R) [/imath].
Is there any simple way to construct a bijective function: [imath]\Bbb R^{\Bbb R} \to P(\Bbb R) [/imath] to see that [imath]\Bbb R^{\Bbb R}[/imath] is isomorphic to [imath]P(\Bbb R)[/imath]? |
428295 | For [imath]G[/imath] group and [imath]H[/imath] subgroup of finite index, prove that [imath]N \subset H[/imath] normal subgroup of [imath]G[/imath] of finite index exists
Let [imath]G[/imath] be a group and [imath]H[/imath] be a subgroup of [imath]G[/imath] with finite index. I want to show that there exists a normal subgroup [imath]N[/imath] of [imath]G[/imath] with finite index and [imath]N \subset H[/imath]. The hint for this exercise is to find a homomorphism [imath]G \to S_n[/imath] for [imath]n := [G:H][/imath] with kernel contained in [imath]H[/imath]. The standard solution suggests to choose [imath]\varphi[/imath] as the homomorphism induced by left-multiplication [imath]\varphi: G \to S(G/H) \cong S_n[/imath]. I'm not 100% sure if I understand this correctly. What exactly does [imath]\varphi[/imath] do? We take [imath]g \in G[/imath] and send it to a bijection [imath]\varphi_g: G/H \to G/H, xH \mapsto gxH[/imath]? If so, how can I see that its kernel is contained in [imath]H[/imath]? Also, the standard solution claims its image is isomorphic to [imath]G/N[/imath] and thus [imath]N[/imath] has a finite index in [imath]G[/imath], how can I see that the image is isomorphic to [imath]G/N[/imath]? Thanks in advance for any help. | 2296956 | If [imath]H[/imath] is a subgroup of [imath]G[/imath] having finite index, then [imath]H[/imath] contains a normal subgroup of [imath]G[/imath] of finite index
I'm trying to show that any subgroup [imath]H[/imath] of a group [imath]G[/imath] having finite index must contain a normal subgroup of [imath]G[/imath] of finite index. I tried to define a homomorphism [imath]\psi:G/H \to G/H[/imath] given by [imath]\psi(xH) =gxH[/imath] and prove that [imath]\ker(\psi)\subset H[/imath]. Is there another way to solve this problem? |
877330 | Does differentiability imply absolute continuity?
Suppose [imath]f:[a,b] \rightarrow \mathbb{R}[/imath] is a function which is (i) differentiable at all [imath]x \in (a,b)[/imath] (ii) the right-derivative at [imath]x=a[/imath] exists and the left-derivative at [imath]x=b[/imath] exists. Does it follow that [imath]f[/imath] is absolutely continuous? | 185303 | Does the everywhere differentiability of [imath]f[/imath] imply it is absolutely continuous on a compact interval?
Suppose [imath]f[/imath] is differentiable everywhere on [imath][0,1][/imath]. Must [imath]f[/imath] be absolutely continuous on [imath][0,1][/imath]? I know this is true if [imath]f'[/imath] is integrable but I'm not sure in this more general case. |
877305 | Limit of factorial function: [imath]\lim\limits_{n\to\infty}\frac{n^n}{n!}.[/imath]
I am studying for a test and I am given this problem: [imath]\lim_{n\to\infty}\frac{n^n}{n!}.[/imath] How do I go about solving this limit? Intuitively I see how the numerator is growing much faster, but how do I express this precisely? Thanks! | 535226 | Limit of the sequence [imath]\{n^n/n!\}[/imath], is this sequence bounded, convergent and eventually monotonic?
I am trying to check whether or not the sequence [imath]a_{n} =\left\{\frac{n^n}{n!}\right\}_{n=1}^{\infty}[/imath] is bounded, convergent and ultimately monotonic (there exists an [imath]N[/imath] such that for all [imath]n\geq N[/imath] the sequence is monotonically increasing or decreasing). However, I'm having a lot of trouble finding a solution that sufficiently satisfies me. My best argument so far is as follows, [imath]a_{n} = \frac{n\cdot n\cdot n\cdot \ldots\cdot n}{n(n-1)(n-2)(n-3)\dots(2)(1)} = \frac{n}{n}\cdot \frac{n}{n-1}\cdot \ldots \cdot \frac{n}2\cdot n[/imath] so [imath]\lim a_{n}\rightarrow \infty[/imath] since [imath]n<a_{n}[/imath] for all [imath]n>1[/imath]. Since the sequence is divergent, it follows that the function must be ultimately monotonic. This feels a little dubious to me, I feel like I can form a much better argument than that, or at the very least a more elegant one. I've tried to assume [imath]\{a_{n}\}[/imath] approaches some limit [imath]L[/imath] so there exists some [imath]N[/imath] such that [imath]|a_{n} - L| < \epsilon[/imath] whenever [imath]n>N[/imath] and derive a contradiction, but this approach got me nowhere. Finally, I've also tried to use the fact that [imath]\frac{a_{n+1}}{a_n}\rightarrow e[/imath] to help me, but I couldn't find an argument where that fact would be useful. |
157397 | proof that translation of a function converges to function in [imath]L^1[/imath]
Let [imath]f \in L^1(\mathbb{R})[/imath], for [imath]a\in \mathbb{R}[/imath] let [imath]f_a(x)=f(x-a)[/imath], prove that: [imath]\lim_{a\rightarrow 0}\|f_a -f \|_1=0[/imath] I know that there exists [imath]g\in C(\mathbb{R})[/imath] s.t [imath]\|f-g\|_1 \leq \epsilon[/imath], this is also true for [imath]f_a[/imath] and [imath]g_a[/imath]. Now I have the next estimation: [imath]\begin{align*} \|f_a-f\|_1 &= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\ &= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\ &\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)| \end{align*}[/imath] my question is: I can argue that [imath]\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T[/imath], now I think that I can change the order of the limits here, but I am not sure why? P.s [imath]x_0[/imath] is some point in [imath][-T,T][/imath], and the above is valid from the intermediate integral theorem for continuous functions, right? Thanks. | 1180558 | Point wise convergent
Suppose that [imath]f\in L^1(\mathbb{R})[/imath] and that [imath]f(x)=0[/imath] if [imath]x\notin[0,1][/imath]. For [imath]n=1,2,\ldots,[/imath] let [imath]f_n(x)=f\left(x+\frac{1}{n}\right)[/imath]. Prove that [imath]\|f_n-f\|_1\to 0[/imath] as [imath]n\to\infty[/imath]. |
790451 | Prove that [imath]u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw\rightarrow 0[/imath] if [imath]x\rightarrow \infty[/imath]
I have the following problem: Be the equation: [imath]u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw[/imath] Show that [imath]u\rightarrow 0[/imath] as [imath]x\rightarrow \infty[/imath], even when [imath]e^{-iwx}[/imath] does not falter if [imath]x\rightarrow \infty[/imath]. The problem gives the hint to use integration by parts. I was hoping you explain this problem or help me solve it, for your attention and any help: thank you very much. | 791679 | Prove that [imath]u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw\rightarrow 0[/imath] if [imath]x\rightarrow \infty[/imath]
I have the following problem: Be the equation: [imath]u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw[/imath] Show that [imath]u\rightarrow 0[/imath] as [imath]x\rightarrow \infty[/imath], even when [imath]e^{-iwx}[/imath] does not falter if [imath]x\rightarrow \infty[/imath]. The problem gives the hint to use integration by parts. I was hoping you explain this problem or help me solve it. I asked this same question in Prove that [imath]u(x,t)=\int_{-\infty}^{\infty}c(w)e^{-iwx}e^{-kw^2t}dw\rightarrow 0[/imath] if [imath]x\rightarrow \infty[/imath], however, I found no answer. But I remembered that this equation is a solution of many physical systems, so I decided to put it here hoping for a little more luck. |
788634 | Where does the second '1' appear?
Where does the second '1' appear in a value of following series? [imath]\frac{1}{9}+\frac{1}{99}+\frac{1}{999}+\dots[/imath]=[imath]\sum_{n=1}^\infty {\frac{1}{10^n-1}}[/imath] I already have a value of Value of [imath]x=\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}} [/imath] and location of second digit [imath]1[/imath] of [imath]x[/imath] (link) But I want to know where the second '1' appears without this result. There is easy way to determine poistion of the second '1'? | 788573 | Value of [imath]x=\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}} [/imath] and location of second digit [imath]1[/imath] of [imath]x[/imath]
What is the value of the series [imath]\sum\limits_{n=1}^\infty {\frac{1}{10^n-1}}[/imath] and can one find closed forms for the sums [imath]\sum\limits_{n=1}^N {\frac{1}{10^n-1}}[/imath]? I found that [imath]\sum_{n=1}^N {\frac{1}{10^n-1}} [/imath] = [imath]\sum_{n=1}^N (\lim_{m\to\infty }{\frac{10^{-n}-10^{-nm}}{1-10^{-n}}})[/imath] = [imath]0.111\ldots+0.0101\ldots+\ldots[/imath] but I can't go farther. The original question is: "Where does the second '1' appear?" |
877549 | Image of a category under a functor need not be a category?
I've been trying to understand the following counterexample posted here: https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics?page=2&tab=votes#tab-top "You only need two objects 0, 1 and a single non-identity arrow 0->1. Consider the functor from this to Set which takes 0 and 1 to the natural numbers and the arrow to the successor function. " I don't get why the image of [imath]0\to 1[/imath] in [imath]Set[/imath], which is [imath]0\to 1[/imath] in [imath]\mathbb N[/imath] is not a category...what am I missing? All the axioms seem to check out. | 413138 | Can it happen that the image of a functor is not a category?
On Hilton and Stammbach's homological algebra book, end of chap. 2, they wrote in general [imath]F(\mathfrak{C})[/imath] is not a category at all in general. But I don't quite get it. I checked the axioms of a category for the image, and I think they are all satisfied. Am I missing something? Thanks. |
878041 | Question about writing a proof with continuous functions
How would I write a proof for this example? We know that all polynomial functions on the reals are continuous by using the sequential definition of continuity. In particular, we know that the function [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] given by [imath]f(x) = x^2[/imath] is continuous. Prove this using the [imath]\epsilon[/imath]-[imath]\delta[/imath] definition. | 264081 | prove [imath]x \mapsto x^2[/imath] is continuous
I am to show the continuity of this function with the help of [imath]\epsilon[/imath]-[imath]\delta[/imath] argument. The function is: [imath]g: \Bbb{R} \rightarrow \Bbb{R}[/imath], [imath]x \mapsto x^2[/imath]. Given the [imath]\epsilon[/imath]-[imath]\delta[/imath] definition of limit, I tried as follows: We must have: [imath]|f(x)-f(x_0)|<\epsilon[/imath], so then [imath]|x^2-x_0^2|=|(x-x_0)(x+x_0)|=|(x-x_0)||(x+x_0)|[/imath], so [imath]|x-x_0|<\frac{\epsilon}{|x+x_0|}[/imath]. So now I will choose [imath]\delta=\frac{\epsilon}{|x+x_0|}[/imath]. Is it enough? I am writing this sentences like a machine, but I am not understanding intuitively. |
426932 | Why are Vandermonde matrices invertible?
A Vandermonde-matrix is a matrix of this form: [imath]\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}[/imath]. condition ☀ : [imath]\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j[/imath] Why are Vandermonde-matrices with ☀ always invertible? I have tried to find a short argument for that. I know some ways to show that in principle: rank is equal to dimension all lines / rows are linear independence determinant is not zero find inverse According to proofwiki, the determinant is [imath]\displaystyle V_n = \prod_{1 \le i < j \le n} \left({x_j - x_i}\right)[/imath] There are two proofs for this determinant, but I've wondered if there is a simpler way to show that such matrices are invertible. | 1754669 | For what values of [imath]x_1, x_2, x_3, x_4[/imath] is the matrix [imath]A[/imath] is invertible?
For what values of [imath]x_1, x_2, x_3, x_4[/imath] is the matrix [imath]A=\begin{pmatrix} 1 & 1 & 1 &1 \\ x_1 & x_2 & x_3 &x_4 \\ x_1^2& x_2^2 & x_3^2 & x_4^2\\ x_1^3& x_2^3 & x_3^3 &x_4^3 \end{pmatrix}[/imath] is invertible Do I need to calculate det 2x2, 3x3 and so on or? |
878409 | How many continuous functions are differentiable?
Consider the set of continuous functions [imath]\mathbb{R} \to \mathbb{R}[/imath]. I assume that the subset that are not everywhere differentiable accounts for almost all of them. Is this true? What is the precise formulation of this idea, and how do you prove it? | 756681 | Are "most" continuous functions also differentiable?
Let [imath]A[/imath] be a nonempty open subset of [imath]\mathbb{R}[/imath]. Consider a function [imath]f : A \rightarrow \mathbb{R}[/imath]. Given that [imath]f[/imath] is continuous, what is the probability that it is differentiable? I suspect it is [imath]0[/imath]. |
663304 | Find the equation of a plane
Find the equation of a plane that passes through point [imath]P(1,5,1)[/imath], and is perpendicular to the planes [imath]2x+y-2z=2[/imath] and [imath]x+3z=4[/imath] My only guess so far is that we can obtain the plane's normal vector using [imath]2x+y-2z=2[/imath] but I'm clueless on how to involve [imath]x+3z=4[/imath] | 663619 | Find a plane that passes through a point and is perpendicular to 2 planes
Find an equation of a plane that passes through [imath]p(1,5,1)[/imath] and is perpendicular to planes [imath]2x+y-2z = 2[/imath] and [imath]x+3z=4[/imath]. I basically need the 2 other points to make the vector and perform the cross product. Since [imath]ax+by+cz+d=0[/imath] is the form of a plane. Can I obtain the points as [imath](a,b,c)[/imath] of the 2 planes given? Also, how can I set them up to obtain the points? |
878897 | Why is a raised to the power of Zero is 1?
Why is [imath]a^0=1[/imath] [imath]\forall a \in Z, a\neq0[/imath]. I understand [imath]2^4=2\cdot2\cdot2\cdot2[/imath] How can I express [imath]a^0[/imath]. I am serious about the practical proof of this | 238300 | How to understand why [imath]x^0 = 1[/imath], where [imath]x[/imath] is any real number?
Alright, so the idea of an exponent, [imath]x[/imath], is that you are multiplying its base by itself [imath]x[/imath] number of times. With base [imath]5[/imath] and [imath]x=3[/imath], we have that [imath]5^3[/imath] = [imath]5 \cdot 5 \cdot 5[/imath] I understand that the logarithm with base [imath]a[/imath] of [imath]x = c[/imath], tells us that [imath]a^c = x[/imath] and for [imath]c =[/imath] positive; values for [imath]x[/imath] are greater than [imath]1[/imath], and for [imath]c =[/imath] negative; values for [imath]x[/imath] are less than [imath]1[/imath], and for [imath]c = 0[/imath], values for [imath]x[/imath] are...[imath]1[/imath]. So in short, I understand how, by means of observation of the graph of [imath]f(x) = \log x[/imath], we can see that [imath]f(1) = 0[/imath], BUT, I see no other way to understand why [imath]x^0 = 1[/imath], apart from the graph and everything around that very point. I honestly cannot get my head around the idea, "[imath]5[/imath] times itself [imath]0[/imath] times... is one". Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand [imath]x^0 = 1[/imath] on its own? |
879059 | Exactly why coefficient of [imath]x^ky^{n-k}[/imath] is [imath]C(k,n)[/imath]
in combination when we have a binomial lattices like [imath](x+y)^n[/imath] the coefficient of [imath]x^ky^{n-k}[/imath] is equal with [imath]C(k,n)[/imath] ... for example we have [imath](x+y)^4[/imath] so we have this [imath]4[/imath] factor [imath](x+y)(x+y)(x+y)(x+y)[/imath] and we want to find the coefficient of [imath]x^2y^2[/imath] ? I want a complete answer thanks !! | 119480 | Derivation of binomial coefficient in binomial theorem.
How was the binomial coefficient of the binomial theorem derived? [imath]\frac{n!}{k!(n-k)!}[/imath] |
879148 | an example of a continuous bijection which is not a homeomorphism
I need an example of a continuous bijection [imath]f:X \to Y[/imath], where [imath]X[/imath] is NOT compact and [imath]Y[/imath] is Hausdorff, such that [imath]f[/imath] is not a homeomorphism. (It is easy to show that if [imath]X[/imath] is compact, then [imath]f[/imath] is necessarily a homeomorphism) Any help is appreciated, Thanks ! | 378717 | Finding counterexamples: bijective continuous functions that are not homeomorphisms
Let [imath]f: X \to Y[/imath] be a bijective continuous function. If [imath]X[/imath] is compact, and [imath]Y[/imath] is Hausdorff, then [imath]f:X \to Y[/imath] is a homeomorphism. My goal is to demonstrate the necessity of both the compact and Hausdorff property of [imath]X[/imath] and [imath]Y[/imath] respectively. I want to know if my counter examples are correct: Let [imath]f: X \to Y[/imath] be the identity function. (1) Let [imath]X=[0,1][/imath] with the standard topology and let [imath]Y=[0,1][/imath] with the indiscrete topology. Then [imath]f: X \to Y[/imath] is continuous and bijective but not a homeomorphism. Since [imath]U=(1/3,1/2)[/imath] is open in [imath]X[/imath], but [imath](f^{-1})^{-1}(U)=f(U)[/imath] is not open in [imath]Y[/imath]. (2) Let [imath]X=Y=\mathbb{R}[/imath]. Let [imath]Y[/imath] have the standard topology and [imath]X[/imath] have the discrete topology. Let [imath]U=\{x\}[/imath] since open sets are singletons. [imath]f(^{-1})^{-1}(x)=f(x)[/imath] is closed in [imath]Y[/imath] since [imath]Y[/imath] has the standard topology. Hence [imath]f[/imath] is not a homeomorphism. This is the old question My goal is to demonstrate the necessity of both the compact and Hausdorff property of [imath]X[/imath] and [imath]Y[/imath] respectively. I want to know if my counter examples are correct: Let [imath]f[/imath] be the identity function. Then [imath]f: X \to Y[/imath] be a bijective continuous function. If [imath]X=[0,1][/imath] with the standard topology and [imath]Y=[0,1][/imath] with the indiscrete topology we have that [imath]f:X \to Y[/imath] is not a homeomorphism. Let [imath]f: X \to Y[/imath] be a bijective continuous function, [imath]X=(0,1)[/imath], and [imath]Y=[0,1][/imath] both with the standard topology. It's not a homeomorphism because [imath]f^{-1}([0,1])[/imath] is not connected, but [imath][0,1][/imath] is connected. The first shows why Hausdoff is necessary for [imath]Y[/imath] and the second shows compactness is necessary for [imath]X[/imath]. Are both of the examples correct? I have no confidence in my first example because I'm having problems showing [imath]f^{-1}[/imath] is not continuous. However, I am confident in my second example. |
879397 | Intermediate Value Theorem, an application problem
The question is: if [imath]\frac{a_0}{1}+\frac{a_1}{2}+\dots+\frac{a_n}{n+1}=0[/imath] then, [imath]a_0+a_1x+\dots+a_nx^n=0[/imath] for some [imath]x[/imath] in the interval [imath][0,1][/imath]. My approach is to let [imath]f(x)=a_0+a_1x+\dots+a_nx^n[/imath] and the find [imath]x_0[/imath] and [imath]x_1[/imath] such that, for example, [imath]f(x_0)\leq0[/imath] and [imath]f(x_1)\geq0[/imath] and then use IVT to make the conclusion. However, I do not know how to use the given condition, especially I am not sure about if [imath]n[/imath] is odd or even and if the leading coefficient, [imath]a_n[/imath] is positive or negative. Any help will be appreciated. Thanks so much. | 363052 | Does the equation have a real solution...please help!
Suppose that [imath]c_0, c_1, \cdots , c_n[/imath] are real numbers and that [imath]c_0 + \frac {c_1}{2} + \frac {c_2}{3} + \cdots + \frac {c_{n-1}}{n} + \frac {c_n}{n+1} = 0[/imath]. Prove that the equation: [imath]c_0 + (c_1)x + \cdots + (c_{n-1})x^{n-1} + (c_n)x^n = 0[/imath] has at least one real solution between 0 and 1. |
880007 | What is the proof to the fact that all prime numbers are 1 above or below a 6 multiple?
I was just having an argument with my friend and I dunno how we got here. But he suddenly said all primes are 1 above or below a multiple of 6. At first I tried a lot of primes but couldn't disprove this. I tried googling but the stuff is too complicated for me. Is there a simple to understand proof for this statement? [imath]p \equiv \pm 1 \pmod{6}[/imath], where [imath]p[/imath] is prime. As pointed out by the answers. I forgot to mention that p > 3. I never checked 2 and 3 when talking to my friend. Somehow thought of them as corner cases. | 64414 | Show that every prime [imath]p>3[/imath] is either of the form [imath]6n+1[/imath] or of the form [imath]6n+5[/imath]
Show that every prime [imath]p>3[/imath] is either of the form [imath]6n+1[/imath] or of the form [imath]6n+5[/imath], where [imath]n=0,1,2, \dots[/imath] |
839504 | the indicator function of a increasing poset
Can someone help me understand this If [imath]X[/imath] is a poset, [imath]X'[/imath] is a subset of [imath]X[/imath], and [imath]X\cap[x,\infty)[/imath] is a subset of [imath]X'[/imath], then [imath]X'[/imath] is an increasing set. Equivalently, a subset [imath]X'[/imath] of a poset [imath]X[/imath] is an increasing set if the indicator function of [imath]X\cap[x,\infty)[/imath] is an increasing function on [imath]X[/imath] for each [imath]x[/imath] in [imath]X'[/imath] First, i do not really understand 1. and second I do not understand how the indicator function can be increasing. | 839818 | increasing subset of a partial order and characteristic function
Can someone help me understand this? Suppose that [imath]\preceq[/imath] is a partial order on a set [imath]S[/imath] and that [imath]A\subseteq S[/imath]. If [imath]\mathbf{1}_A[/imath] is the indicator function then [imath]A[/imath] is increasing if and only if [imath]\mathbf{1}_A[/imath] is increasing. [imath]A[/imath] is decreasing if and only if [imath]\mathbf{1}_A[/imath] is decreasing. |
880079 | Trouble understanding Hoffman / Kunze exercise
I am finishing up a number theory class and will be studying graduate Linear Algebra in the fall so I thought I'd start early getting familiar with the text and authors by doing some of the early text exercises. I'm already stumped with the second problem, but because I think it seems difficult, but I've never seen it written this way before. Here is the problem Let [imath]\mathbb{F}[/imath] be the field of complex numbers. Are the following two systems of linear equations equivalent? If so, express each equation in each system as a linear combination of the equations in the other system. [imath] \begin{cases} x_1-x_2=0\\ 2x_1+x_2=0 \\ \end{cases}[/imath][imath] \begin{cases} 3x_1+x_2=0\\ x_1+x_2=0 \\ \end{cases}[/imath] Now, they are equivalent if they produce the same answers for [imath]x_1[/imath] and [imath]x_2[/imath], which they are because [imath]x_1=x_2=0[/imath] in both cases. I'm really struggling to understand how to write linear combinations of equations of equations. I'm sure that it is really simple but with no example, I'm feeling extremely dumb right now. How I interpret it is this. Let's start writing the second system in terms of the first [imath] \begin{cases} a(x_1-x_2)+b(2x_1+x_2)=3x_1+x_2 \Rightarrow a=\frac1{3}, b=\frac4{3}\\ c(x_1-x_2)+d(2x_1+x_2)=x_1+x_2 \Rightarrow a=\frac{-1}{3}, b=\frac2{3} \\ \end{cases}[/imath] [imath] \begin{cases} \frac1{3}(x_1-x_2)+\frac4{3}(2x_1+x_2)=3x_1+x_2 \\ \frac{-1}{3}(x_1-x_2)+\frac2{3}(2x_1+x_2)=x_1+x_2 \\ \end{cases}[/imath] I should do the same thing in the second case for the in terms of the first system. Is this what they are asking of the reader? | 708187 | an exercise problem in the book "linear algebra" by Kenneth Hoffman and Ray Kunze, pg #5, Q.2
I am quite confused with this problem. "Let F be the field of complex numbers. Are the following two systems of linear equations equivalent ? If so, express each equation in each systems as a linear combination of the equations in the other system." [imath]x_1-x_2=0, \qquad 3x_1+x_2=0;[/imath] on one hand, and on the other: [imath]2x_1+x_2=0, \qquad x_1+x_2=0.[/imath] |
855299 | Greatest value of the binomial coefficient.
How should I prove the greatest value of the binomial coefficient [imath]C(n,r)[/imath] occurs for [imath]r=\left[\cfrac{(n+1)}{2}\right][/imath] ? | 722952 | How do you prove [imath]{n \choose k}[/imath] is maximum when [imath]k[/imath] is [imath] \lceil \tfrac n2 \rceil[/imath] or [imath] \lfloor \tfrac n2\rfloor [/imath]?
How do you prove [imath]n \choose k[/imath] is maximum when [imath]k[/imath] is [imath]\lceil n/2 \rceil[/imath] or [imath]\lfloor n/2 \rfloor[/imath]? This link provides a proof of sorts but it is not satisfying. From what I understand, it focuses on product pairings present in [imath]k! (n-k)![/imath] term which are of the form [imath]i \times (i-1)[/imath]. Since these are minimized when [imath]i=n/2[/imath], we get the result. But what about the reasoning for the rest of the terms? |
880901 | Entire function with [imath]L^2[/imath] modulus is identically zero
I want to show that if [imath]f[/imath] is entire and [imath]\int_{\mathbb{R}^2}\left| \:f\: \right|^2 < \infty[/imath], then [imath]f \equiv 0[/imath]. I was thinking of assuming [imath]f[/imath] is not identically zero; then, since a bounded entire function is constant, we know [imath]|\:f\:|[/imath] must get arbitrarily large. Intuitively, I thought that for both this and [imath]|| \: |f| \: ||_2 < \infty[/imath] to hold, the "spikes" in the modulus would have to be concentrated in small areas, and then [imath]|f|[/imath] would have a local maximum, contradicting the maximum modulus principle. But this is very hand-wavy, and on second thought, it may not be true: what if the graph of the modulus has an infinite, slender "ridge" which gets thinner and taller as it runs away from the origin? Then both the unboundedness AND the [imath]L^2[/imath] condition could be satisfied, without contradicting the maximum modulus principle! Any ideas? | 6340 | How do you show that an [imath]L^p[/imath] entire (holomorphic on the complex plane) function is [imath]0[/imath]?
Just to clarify, I want to show that: If [imath]f[/imath] is entire and [imath]\int_{\mathbb{C}} |f|^p dxdy <\infty[/imath], then [imath]f=0[/imath]. I think I can show that this is the case for [imath]p=2[/imath], but I'm not sure about other values of [imath]p[/imath]... |
837144 | On the number of countable models of complete theories of models of ZFC
Fix the language of set theory [imath]\mathcal{L}=\{\in\}[/imath]. Let [imath]\langle M,\in\rangle[/imath] be a set or proper class model of ZFC (e.g. [imath]M[/imath] could be [imath]L[/imath], [imath]HOD[/imath], [imath]V_{\kappa}[/imath] for some inaccessible cardinal [imath]\kappa[/imath], etc.) consider the complete theory of this model in the language [imath]\mathcal{L}[/imath] that is [imath]T_{M}:=Th(\langle M,\in\rangle)=\{\sigma\in\mathcal{L}~|~\langle M,\in\rangle\models \sigma\}[/imath]. My questions are about the possible number of countable models of this complete theory up to isomorphism [imath]I(T_{M},\aleph_0)[/imath] when [imath]M[/imath] varies over different models of [imath]ZFC[/imath]. Question 1: What are [imath]I(T_{L},\aleph_0)[/imath] and [imath]I(T_{HOD},\aleph_0)[/imath], [imath]I(T_{V_{\kappa}},\aleph_0)[/imath] (for [imath]\kappa[/imath] inaccessible)? Question 2: Is there a set or proper class model [imath]M[/imath] of ZFC such that [imath]I(T_{M},\aleph_0)=\lambda[/imath] for each cardinal [imath]\lambda\in\{1,3,4,\cdots,\aleph_{0},\aleph_1,2^{\aleph_{0}}\}[/imath]? | 638201 | Number of Non-isomorphic models of Set Theory
Assume that the meta theory allows for model theoretic techniques and handling infinite sets etc (The meta theory itself is, informally, "strong as ZFC"). Also assume that I'm studying ZFC inside this meta-theory and that I'm working with the assumption that ZFC is consistent. Let [imath]ZFC\subset{T}[/imath], where [imath]T[/imath] is a complete theory. I would like to know the number of non-isomorphic models of [imath]T[/imath] for some given [imath]\kappa\geq{\aleph_{0}}[/imath]. I feel as if this should be the maximum number possible, i.e. [imath]2^{\kappa}[/imath]. My idea for showing this was to use Shelah's theorem that says that unstable theories have [imath]2^{\kappa}[/imath] many models for any [imath]\kappa>{\aleph_{0}}[/imath] (which establishes the result for all uncountable [imath]\kappa[/imath]). But in order to do this I first need to show that [imath]T[/imath] is unstable. However I'm not sure about how to do this (or even if I'm on the right track). Any help would be much appreciated. |
881184 | Show that T2-space is preserved by continuous map.
Let (X,[imath]\tau[/imath]) and (Y,[imath]\tau_1[/imath]) be topological spaces and f : (X,[imath]\tau[/imath])[imath]\rightarrow[/imath](Y,[imath]\tau_1[/imath]) a continuous map. If f is one-to-one, prove that (Y,[imath]\tau_1[/imath]) is Hausdorff implies that (X,[imath]\tau[/imath]) is Hausdorff. I already proved that T1 is preserved by continuous map, but I don't know how to do with T2. Please give me some idea. | 696082 | If a continuous function[imath]f: (X, \mathscr T_X) \to (Y, \mathscr T_Y)[/imath] is injective (Given [imath]Y[/imath] is Hausdorff), show that X is hausdorff
[imath](X, \mathscr T_X)[/imath] and [imath](Y, \mathscr T_Y)[/imath] be 2 topological spaces. [imath]Y[/imath] be Hausdorff. [imath]f[/imath] be a continuous function, [imath]f: X \to Y[/imath]. To show that if [imath]f[/imath] is injective [imath]\implies[/imath] [imath]X[/imath] is Hausdorff. Here's how I tried: Let [imath]f(x_1)[/imath] and [imath]f(x_2)[/imath] be two elements of Y. Since [imath]Y[/imath] is Hausdorff we can find 2 disjoint neighborhoods around [imath]f(x_1)[/imath] and [imath]f(x_2)[/imath]. Let it be [imath]U, V[/imath] respectively. Since [imath]f[/imath] is continuous there exists an open neighborhood around [imath]x_1[/imath] and [imath]x_2[/imath] in X (let it be [imath]E(x_1)[/imath] and [imath] F(x_2)[/imath]) such that [imath]f(E) \subset U[/imath] and [imath]f(F) \subset V[/imath]. After this how to use injectivity property to prove that E is disjoint with F?? |
236083 | Determine all primes [imath]p[/imath] for which [imath]5[/imath] is a quadratic residue modulo [imath]p[/imath]
I need to determine all primes [imath]p[/imath] for which [imath]5[/imath] is a quadratic residue modulo p. I think I'll need to use quadratic recprocity laws to do this, i.e., I need to need to find numbers [imath]p[/imath] where [imath]x^2[/imath] is congruent to [imath]5 \bmod p[/imath]. I'm ok doing this for single values of [imath]p[/imath]. But how do I find all primes for which this holds? Thanks. | 1832429 | Prove that there are infinitely many primes [imath]p[/imath] such that [imath]\left(\dfrac{p}{5} \right) = 1[/imath]
Let [imath]\left(\dfrac{a}{p}\right)[/imath] denote the Legendre symbol. Prove that there are infinitely many primes [imath]p[/imath] such that [imath]\left(\dfrac{p}{5} \right) = 1[/imath]. Since there are infinitely many primes there must be infinitely many primes [imath]p[/imath] such that [imath]\left(\dfrac{p}{5} \right) = -1,0,[/imath] or [imath]1[/imath]. How do we prove that [imath]1[/imath] is achieved infinitely many times? |
881001 | Can we express the following in a closed form?
I want to evaluate the integral: [imath]I=\int_{0}^{\pi/2}\ln \left ( \frac{(1+\sin x)^{1+\cos x}}{1+\cos x} \right )\,dx[/imath] Well, the sub [imath]u=\pi/2-x[/imath] does not give me any result. In fact it makes the integral more complicated that it actually is, unless I do not see something. The method above is the only one I used since I do not see something else in this point. Any help would be grateful. | 626763 | Evaluate: [imath]I=\int\limits_{0}^{\frac{\pi}{2}}\ln\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}dx[/imath]
Evaluate: [imath]I=\int\limits_{0}^{\frac\pi2}\ln\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}dx[/imath] |
352950 | Prove that in a ring with [imath]x^3 = x[/imath] we have [imath]x+x+x+x+x+x=0[/imath].
This was an exercise on a course on abstract algebra at the University of Groningen. I have been working on this for ages, but I can't seem to figure it out. Problem Let [imath]R[/imath] be a ring with [imath]\forall x \in R: x^3 = x[/imath]. Prove that [imath]x+x+x+x+x+x=0[/imath]. Tried If [imath]x=0[/imath], the statement is of course trivial. If [imath]x \neq 0[/imath], we have [imath]x(x^2-1)=0=(x^2-1)x[/imath], so either [imath]x^2 = 1[/imath], so [imath]x[/imath] is a unit, or [imath]x^2-1[/imath] and [imath]x[/imath] are zero-divisors. And this is as far as I get... Any help would be appreciated! :) Edit: guide to answer The following hint was provided by Abel: [imath]\begin{align} x \in R &\implies x+x \in R\\ &\implies x+x=(x+x)^3 \end{align}[/imath] If you then work out all the terms, the answer will follow quickly. | 257190 | If [imath]x^3 =x[/imath] then [imath]6x=0[/imath] in a ring
Let [imath]R[/imath] be a ring with unity where [imath]x^3=x,\;\;\; \forall x \in R[/imath] How do I prove that [imath]x+x+x+x+x+x=0[/imath] |
881241 | the continuity of total variation function of a continuous function of bounded variation
Let f be a continuous function of bounded total variation (refer to http://en.wikipedia.org/wiki/Total_variation for the definition) on [imath][0,1][/imath], i.e., [imath]\text{Var}_{[0,1]}f<\infty[/imath]. Then the total variation of [imath]f[/imath] on [imath][0,x][/imath], denoted by [imath]\text{Var}_{ [0,x]} (f):=V(x)[/imath], is continuous on [imath][0,1][/imath]. How to prove it? Or any counterexample? | 144162 | Is the total variation function uniform continuous or continuous?
I have been doing some excercises on total variation when the following questions came up to my mind: (1) Let [imath]f[/imath] be continuous on the interval [imath][0,1][/imath] and be of bounded variation. Is it true that its total vatiation function [imath]TV(f_{[0,x]})[/imath] is uniformly continuous? i.e. Is it true that for [imath]\forall\space\epsilon>0[/imath], [imath]\exists\space\delta>0[/imath], such that for arbitrary interval [imath][a,b][/imath] with [imath]|b-a|<\delta[/imath], we have [imath]TV(f_{[a,b]})<\epsilon[/imath]? (2) Alternatively, if it is not uniformly continuous, can I say that for [imath]\forall\space{}x\in[0,1]\text{ and } \forall\space\epsilon>0[/imath] , [imath]\exists\space\text{ nondegenerate interval }I \text{ such that }x\in{}I\text{ and } TV(f_{I})<\epsilon[/imath]? Thank you! |
881464 | Evaluate [imath]\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt[/imath]
How can we compute [imath]I=\int_0^{+\infty } \frac{\log(t)}{1+t^2} \, dt[/imath] Mathematica gives [imath]I=0[/imath]. | 808678 | Evaluate [imath]\int_0^\infty\frac{\ln x}{1+x^2}dx[/imath]
Evaluate [imath]\int_0^\infty\frac{\ln x}{1+x^2}\ dx[/imath] I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated. Thanks. |
881494 | [imath]F[/imath] is a closed set in [imath]\mathbb{R}[/imath], Then [imath]\frac{d(x+y, F)}{|y|}\rightarrow 0, |y|\rightarrow 0[/imath] for each [imath]x\in F[/imath].
[imath]F[/imath] is a closed set in [imath]\mathbb{R}[/imath] and denote [imath]d(y,F)=\inf\{|y-z|: z\in F\}.[/imath] Then [imath]\frac{d(x+y, F)}{|y|}\rightarrow 0, |y|\rightarrow 0[/imath] for a.e [imath]x\in F[/imath]. I have no idea even though the hint says that the Lebesgue identity is useful. | 850601 | Prove that [imath]\lim_{x\to y} \frac{d(x, F )}{|x−y|} = 0[/imath] for a.e. [imath]y \in F[/imath].
Let [imath]F \subset \mathbb{R}[/imath] be a closed set and define the distance from [imath]x \in \mathbb{R}[/imath] to [imath]F[/imath] by [imath]d(x,F)= \inf_{y \in F} |x−y|.[/imath] Prove that [imath]\lim_{x\to y} \frac{d(x, F )}{|x−y|} = 0[/imath] for a.e. [imath]y \in F[/imath]. Hint: Consider Lebesgue points of F. I am not sure where to begin on this question. Any suggestions? Thanks. |
881557 | integrating square root of tanx
[imath]\int \sqrt{\tan (x)}dx [/imath] Let [imath]\tan(x)=t^{2}[/imath] then [imath]dx[/imath] will become [imath]\frac{2t}{1+t^{4}}[/imath] Hence [imath]\int \sqrt{\tan (x)}dx =\int\frac{2t}{1+t^4} dt [/imath] But I cannot proceed from this step. | 828640 | Evaluating the indefinite integral [imath] \int \sqrt{\tan x} ~ \mathrm{d}{x}. [/imath]
I have been having extreme difficulties with this integral. I would appreciate any and all help. [imath] \int \sqrt{\tan x} ~ \mathrm{d}{x}. [/imath] |
881128 | All topologies on [imath]X=\{ a,b \}[/imath]
I am trying to find the possible topologies on [imath]X=\{ a,b \}[/imath]. [imath]\varnothing ,\{ a,b \}[/imath] [imath]\varnothing ,\{ a \},\{ a,b \}[/imath] [imath]\varnothing ,\{ b \},\{ a,b \}[/imath] [imath]\varnothing ,\{ a \},\{ b \},\{ a,b \}[/imath] First off, a topology [imath]\tau [/imath] on a set [imath]X[/imath] consists of subsets of [imath]X[/imath] following the properties: The empty set [imath]\varnothing[/imath] and the space [imath]X[/imath] are both sets in the topology. The union of any collection of sets in [imath]\tau[/imath] is contained in [imath]\tau[/imath]. The intersection of any finitely many sets in [imath]\tau[/imath] is also contained in [imath]\tau[/imath] All four I think are actual topologies. Am I correct? | 740564 | describing all possible topologies on a set
Assuming the following: let [imath]X[/imath] be a set with two elements, [imath]X[/imath] = {[imath]a,b[/imath]}. what are all the possible topologies on [imath]X[/imath] ? The answer I've come up with is: [imath]\tau_{T}=\{ \emptyset, X\} [/imath] [imath] \tau_{a}=\{\emptyset,\{a\},X\} [/imath] [imath]\tau_{b}=\{\emptyset,\{b\},X \} [/imath] [imath] \tau_{D}=\{\emptyset,\{a\},\{b\},X\}[/imath] My question is, are these correct? and are there any more? thanks in advance |
471768 | Why do repunit primes have only a prime number of consecutive [imath]1[/imath]s?
Repunit primes are primes of the form [imath]\frac{10^n - 1}{9} = 1111\dots11 \space (n-1 \space ones)[/imath]. Each repunit prime is denoted by [imath]R_i[/imath], where [imath]i[/imath] is the number of consecutive [imath]1[/imath]s it has. So far, very few of these have been found: [imath]R_2, R_{19}, R_{23}, R_{317}, R_{1031}, R_{49081}, R_{86453}, R_{109297}, R_{270343}[/imath]. One thing to observe in all of these is that all of these have a prime subscript. Is this a prerequisite for all repunit primes. Does each repunit prime written as [imath]R_p[/imath] have to have [imath]p[/imath] as prime? I tried my hand at this: A repunit prime [imath]R_i[/imath], can be written as [imath]10^{i-1} + 10^{i-2} \dots + 10^{2} + 10 + 1[/imath]. We are required to prove that if [imath]i[/imath] is composite then [imath]R_i[/imath] is composite. Clearly, if [imath]i[/imath] is even, then [imath]R_i[/imath] is divisible by [imath]11[/imath]. [imath]3|i[/imath], then [imath]3|R_i[/imath]. So, [imath]i[/imath] has to be of the form [imath]6n \pm 1[/imath]. I could not put any more constraints of this. Is there an elementary proof for this? | 966747 | Show that if a positive integer [imath] n [/imath] is composite then [imath] R(n) = \frac{10^{n}-1}{9}= 111...11 (n times) [/imath] is composite
Show that if a positive integer [imath] n [/imath] is composite then [imath] R(n) = \frac{10^{n}-1}{9}= \underset{n\text{ times}}{\underbrace{111...11}} [/imath] is composite I attempted a both a normal proof and proof by contradiction by trying: [imath] n = ak [/imath] show if [imath] ak\equiv 0 \mod a [/imath] then [imath] (10^{ak}-1) \equiv 0\mod (10^a-1) [/imath] I reached a point where I began to go around in cirlces. Maybe I'm going down the wrong track. Any ideas ? |
802770 | Matsumura, Exercise 18.8: Cohen-Macaulay and (not) Gorenstein
I need an answer to the exercise 18.8 of Matsumura's book:" Commutative Ring Theory", and generate an algorithm if possible. Let [imath]k[/imath] be a field and [imath]t[/imath] an indeterminate. Consider the subring [imath]A = k[[t^3, t^5,t^7]][/imath] of [imath]k[[t]][/imath] and show that [imath]A[/imath] is a one-dimensional Cohen-Macaulay ring which is not Gorenstein. How about [imath]k[[t^3,t^4,t^5]][/imath] and [imath]k[[t^4,t^5,t^6]][/imath]? ADDED: Is there an algorithm for the general case: [imath]A=k[[t^a,t^b,t^c]][/imath], where [imath]a,b,c[/imath] are natural numbers? My attempt: [imath]A[/imath] is Cohen-Macaulay, because it's a domain of [imath]\dim\, 1[/imath]. | 165246 | Are the rings [imath]k[[t^3,t^4,t^5]][/imath] and [imath]k[[t^4,t^5,t^6]][/imath] Gorenstein? (Matsumura, Exercise 18.8)
Here is question 18.8 of Matsumura's Commutative Ring Theory. It asks whether the rings [imath]k[[t^3,t^4,t^5]][/imath], [imath]k[[t^4,t^5,t^6]][/imath] are Gorenstein. I got that 1) is not Gorenstein, but 2) is Gorenstein (by computing the socle). Just wanted to check if I am correct. I don't need the answer necessarily, a yes or a no will suffice. Thanks. |
882042 | Show that the series [imath]\sum\limits_n P(A \cap A_n)[/imath] converges for [imath]A=\bigcap\limits_{n=M}^{\infty} A_n^{c}[/imath].
Show that the series [imath]\sum\limits_n P(A \cap A_n)[/imath] converges for [imath]A=\bigcap\limits_{n=M}^{\infty} A_n^{c}[/imath]. From here https://math.stackexchange.com/a/878635/140308 (proof attempt is there too) Sorry for the confusion. How does that show that the series converges? What about n=1,...,m-1? Does our choice of A mean that the index now starts at M? Why/why not? I guess I got the idea right. Just don't know how to say it exactly | 877818 | [imath]P(\limsup A_n)=1 [/imath] if [imath]\forall A \in \mathfrak{F}[/imath] s.t. [imath]\sum_{n=1}^{\infty} P(A \cap A_n) = \infty[/imath]
Let [imath]\mathfrak{F} = (A_n)_{n \in \mathbb{N}}[/imath]. Prove that [imath]P(\limsup A_n)=1[/imath] if [imath]\forall A \in \mathfrak{F}[/imath] s.t. [imath]P(A) > 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty[/imath]. (Side question 1 Is second Borel-Cantelli out because we don't know if the [imath]A_n[/imath]'s are independent?) Suppose [imath]\forall A \in \mathfrak{F}[/imath] s.t. [imath]P(A) > 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty[/imath], but [imath]P(\limsup A_n)<1[/imath]. My prof gave us this convenient hint: Show [imath]\exists M > 0[/imath] s.t. [imath]P(\bigcap_{j=M}^{\infty} A_j^{c}) > 0[/imath]. [imath]1 > P(\limsup A_n)[/imath] [imath]1 - P(\limsup A_n) > 0[/imath] [imath]P(\liminf A_n^{c}) > 0[/imath] By definition of [imath]\liminf A_n^{c}[/imath], [imath]\exists M > 0 [/imath] s.t. [imath]P(\bigcap_{j=M}^{\infty} A_j^{c}) > 0[/imath]. (Side questions 2 and 3 Isn't this =1? ...Wait, is that what we're trying to prove? Haha) [imath]\to 1 - P(\bigcup_{j=M}^{\infty} A_j) > 0[/imath] [imath]\to 1 > P(\bigcup_{j=M}^{\infty} A_j)[/imath] Main question: Now I guess we have to come up with some A s.t. [imath]\to 1 > P(\bigcup_{j=M}^{\infty} A_j) \geq \sum_{n=1}^{\infty} P(A \cap A_n)[/imath], or is there something else to do? |
882344 | How to prove {[imath]a_n[/imath] } is increasing where [imath]a_1 = \sqrt{2}[/imath] and [imath]a_{n+1} = \sqrt{ 2+ a_n}[/imath]
I already found out that this sequence is bounded above and [imath]a_n <2 \forall n \in \mathbb Z_+ [/imath] I think I'm missing a point as I can't think of a way to prove that the sequence is increasing. | 449592 | Evaluating the limit of a sequence given by recurrence relation [imath]a_1=\sqrt2[/imath], [imath]a_{n+1}=\sqrt{2+a_n}[/imath]. Is my solution correct?
Problem The sequence [imath](a_n)_{n=1}^\infty[/imath] is given by recurrence relation: [imath]a_1=\sqrt2[/imath], [imath]a_{n+1}=\sqrt{2+a_n}[/imath]. Evaluate the limit [imath]\lim_{n\to\infty} a_n[/imath]. Solution Show that the sequence [imath](a_n)_{n=1}^\infty[/imath] is monotonic. The statement [imath]V(n): a_n < a_{n+1}[/imath] holds for [imath]n = 1[/imath], that is [imath]\sqrt2 < \sqrt{2+\sqrt2}[/imath]. Let us assume the statement holds for [imath]n[/imath] and show that [imath]V(n) \implies V(n+1)[/imath]. We have that [imath]a_n < a_{n+1}.[/imath] Adding 2 to both sides and taking square roots, we have that [imath]\sqrt{2+a_n} < \sqrt{2+a_{n+1}},[/imath] that is [imath]a_{n+1} < a_{n+2}[/imath] by definition. Find bounds for [imath]a_n[/imath]. The statement [imath]W(n): 0 < a_n < 2[/imath] holds for [imath]n=1[/imath], that is [imath]0 < \sqrt2 < 2[/imath]. Let us assume the statement holds for [imath]n[/imath] and show that [imath]W(n) \implies W(n+1)[/imath]. We have that [imath]0 < a_n < 2.[/imath] Adding two and taking square roots, we have that [imath]0 < \sqrt2 < \sqrt{2+a_n} < \sqrt4 = 2.[/imath] The limit [imath]\lim_{n\to\infty} a_n[/imath] exists, because [imath](a_n)_{n=1}^\infty[/imath] is a bounded monotonic sequence. Let [imath]A = \lim_{n\to\infty} a_n[/imath]. Therefore the limit [imath]\lim_{n \to\infty} a_{n+1}[/imath] exists as well and [imath]\lim_{n \to\infty} a_{n+1} = A[/imath]. (For [imath](n_k)_{k=1}^\infty = (2,3,4, \dots)[/imath], we have that [imath](a_{n_k})_{k=1}^\infty[/imath] is a subsequence of [imath](a_n)_{n=1}^\infty[/imath], from which the statement follows.) We have that [imath]a_{n+1} = f(a_n)[/imath]. That means that [imath]A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}[/imath]. Solving the equation [imath]A = \sqrt{2 + A}[/imath], we get [imath]A = -1 \lor A = 2[/imath]. Putting it all together, we get that [imath]A = 2[/imath], because the terms of the sequence are increasing and [imath]a_1 > 0[/imath]. Is my solution correct? |
846444 | continuous function on compact set, can be approximated by polynomial
show that if [imath]f[/imath] is homeomorphism of [imath][0,1][/imath] onto itself, then there is a sequence of polynomials [imath]P_n(x)[/imath], [imath]n=1,2,\ldots[/imath], such that [imath]P_n[/imath] converge to [imath]f[/imath] uniformly on [imath][0,1][/imath] and each [imath]P_n[/imath] is homeomorphism of [imath][0,1][/imath] onto itself. | 881296 | Show that there is sequence of homeomorphism polynomials on [0,1] that converge uniformly to homeomorphism
Let [imath]f:[0,1]\rightarrow [0,1][/imath] be a homeomorphism. Show that , there exists a sequence of polynomials [imath](P_n(x))_n[/imath] such that [imath]P_n(x)[/imath] converge uniformly to [imath]f[/imath] on [imath][0,1][/imath] and every [imath]P_n(x)[/imath] is a homeomorphism from [imath][0,1][/imath] to itself. I think that if we put additional condition that f be [imath]C^1[/imath] then [imath]f'[/imath] will be non-negative and continuous. Moreover, Weierstrass theorem says that there will be exist non-negative sequence of polynomials [imath](Q_n(x))_n[/imath] such that it converges uniformly to [imath]f'[/imath] on [imath][0.1][/imath]. Now it seems that [imath]P_n(x)=\int_{\mathbb{0}}^{x}Q_n(t)\:dt[/imath] works . Is this true ? Is there other ways? Thanks |
882904 | Groups of order [imath]n^2[/imath] with no subgroup of order [imath]n[/imath]
Is it possible to classify those groups whose order is [imath]n^2[/imath] for some natural number [imath]n[/imath] but which do not have any subgroups of order [imath]n[/imath]? To be a bit more specific (in case a full classification is not known): Will a solvable group of order [imath]n^2[/imath] always have a subgroup of order [imath]n[/imath]? In fact, I have not been able to find any examples at all of such groups (obviously, the order would have to be fairly large). My motivation for asking is that in my question Sudokus as composition tables of finite groups it turned out that having a subgroup of order [imath]n[/imath] is sufficient for being able to arrange the composition table to form a sudoku, but it seems unknown if this condition is necessary, and a natural starting point for determining if it is would be to look at examples. | 882859 | Groups of order [imath]n^2[/imath] that have no subgroup of order [imath]n[/imath]
For which [imath]n[/imath] is there a group of order [imath]n^2[/imath] without a subgroup of order [imath]n[/imath]. Such groups can not be nilpotent. This question is related to Sudokus as composition tables of finite groups. |
883165 | Why generalize the derivative for multivariable functions?
Sorry if this is a dupe (did a search, couldn't find anything). In single variable calculus, if the following limit exists: [imath]\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h},[/imath] then this expression itself is the derivative of [imath]f[/imath] at [imath]x[/imath]. This is nicely motivated geometrically and otherwise. This definition gives of [imath]f'(x)[/imath] which gives the best linear approximation [imath]\tilde{f}[/imath] to [imath]f[/imath] at [imath]x[/imath] by [imath]\tilde{f}(t) = f'(x)t + f(x).[/imath] For functions of several variables, most of the time I see the derivative defined in terms of the best linear approximation to the function explicitly. Specifically, the derivative of a multivariable function [imath]g[/imath] at [imath]\mathbf{y}[/imath] is some linear operator [imath]L(\mathbf{y})[/imath]. Concretely, if the following expression is satisfied: [imath]\lim_{\mathbf{h} \rightarrow 0}\frac{\|g(\mathbf{y} + \mathbf{h}) - g(\mathbf{y}) + L(\mathbf{y})\|}{\|\mathbf{h}\|} = 0,[/imath] then [imath]L(\mathbf{y})[/imath] is the derivative of [imath]g[/imath] at [imath]\mathbf{y}[/imath]. We can show that [imath]L(\mathbf{y})[/imath] is unique in this case. My question is why is this generalization necessary? What is the problem with simply defining is analogously to the single variable case as [imath]L(\mathbf{y}) = \lim_{\mathbf{h} \rightarrow 0}\frac{g(\mathbf{y} + \mathbf{h}) - g(\mathbf{y})}{\|\mathbf{h}\|}?[/imath] | 618937 | Derivative of function with 2 variables
I've leart in Calculus 1 that the derivetive of [imath]f(x)[/imath] is: [imath]\lim_{h\to0} \frac{f(x+h) - f(x)}{h}[/imath]. suppose [imath]f(x,y)[/imath] is a function with 2 variables, does [imath]f'(x,y) = \lim_{h\to0} \frac{f(x+h, y+h) - f(x,y)}{h}[/imath] ? If not , what is the derivetive of function with 2 variables? |
883251 | Proof of multiplicative inverse for polar complex numbers
Use the polar form of complex numbers to show that every complex number [imath]z\neq0[/imath] has multiplicative inverse [imath]z^{-1}[/imath]. If [imath]z=a+bi[/imath], then the polar form is [imath]z=r(\cos(\alpha)+i\sin(\alpha))[/imath]. I can do it, not using polar coordinates: If [imath]z=a+ib[/imath]. Since [imath]1+0i[/imath] is the multiplicative identity, if [imath]x+iy[/imath] is the multiplicative inverse of [imath]z=a+ib[/imath], then [imath](a+ib)(x+iy)=1+i0[/imath] [imath]\Rightarrow[/imath] [imath](ax-by)+i(ay+bx)=1+i0[/imath] [imath]\Rightarrow[/imath] [imath]ax-by=1bx+ay=0[/imath] [imath]\Rightarrow[/imath] [imath]x=\frac{a}{a^2+b^2}[/imath], [imath]y=\frac{-b}{a^2+b^2}[/imath], if [imath]a^2+b^2\neq0[/imath]. Thus, the multiplicative inverse of [imath]a+ib[/imath] is [imath]\frac{a}{a^2+b^2} +\frac{-b}{a^2+b^2}[/imath]. Now, how can I use this information to find it using [imath]z=r(\cos(\alpha)+i\sin(\alpha))[/imath]?? I'm stuck... | 882991 | Use polar complex numbers to find multiplicative inverse
Use the polar form of complex numbers to show that every complex number [imath]z\neq0[/imath] has multiplicative inverse [imath]z^{-1}[/imath]. If [imath]z=a+bi[/imath], then the polar form is [imath]z=r(cos(\alpha))+i(sin(\alpha))[/imath]. I can do it, not using polar coordinates: Let [imath]z=a+ib[/imath]. Since [imath]1+0i[/imath] is the multiplicative identity, if [imath]x+iy[/imath] is the multiplicative inverse of [imath]z=a+ib[/imath], then [imath](a+ib)(x+iy)=1+i0[/imath] [imath]\Rightarrow[/imath] [imath](ax-by)+i(ay+bx)=1+i0[/imath] [imath]\Rightarrow[/imath] [imath]ax-by=1bx+ay=0[/imath] [imath]\Rightarrow[/imath] [imath]x=\frac{a}{a^2+b^2}[/imath], [imath]y=\frac{-b}{a^2+b^2}[/imath], if [imath]a^2+b^2\neq0[/imath]. The multiplicative inverse of [imath]a+ib[/imath] is therefore [imath]\frac{a}{a^2+b^2} +i\frac{-b}{a^2+b^2}[/imath]. Now, should I use this same process and just replace [imath]z=a+ib[/imath] with [imath]z=r(cos(\alpha))+i(sin(\alpha))[/imath]?? Or is there another way to go about this process? |
203083 | Infinite set and countable subsets
Prove that a set [imath]A[/imath] is infinite if and only if [imath]A[/imath] contains a countable subset [imath]C[/imath]. I know I have to build a sequence and then I'll get a countable subset, but I don't know how to build that sequence from a infinite set. | 918148 | Prove that for any infinite set [imath]A[/imath], [imath]|\mathbb{N}|\le |A|[/imath]
How can you show that for any infinite set [imath]A[/imath], [imath]|\mathbb{N}|\le |A|[/imath]? thanks |
884381 | Proof:"Infinite subset of [imath]\mathbb N[/imath] is countable
I've read a proof of the statement:"An infinite subset of [imath]\mathbb N[/imath] is countable; that is, if [imath]A \subset \mathbb N[/imath] and if [imath]A[/imath] is infinite, then [imath]A[/imath] is equivalent to [imath]\mathbb N[/imath]." in Carothers' textbook and there is one part of the proof I don't understand. Proof Recall that [imath]\mathbb N[/imath] is well ordered. That is, each nonempty subset of [imath]\mathbb N[/imath] has a smallest element. Thus, since [imath]A \ne \emptyset[/imath], there is a smallest element [imath]x_1 \in A[/imath]. Then [imath]A \setminus \{x_1\} \ne \emptyset[/imath], and there must be a smallest [imath]x_2 \in A \setminus \{x_1\}[/imath]. But now [imath]A \setminus \{x_1,x_2\} \ne \emptyset[/imath], and so we continue, setting [imath]x_3=\min(A \setminus \{x_1,x_2\})[/imath]. By induction, we can find [imath]x_1,x_2,x_3,...,x_n,... \in A[/imath], where [imath]x_n=\min(A \setminus \{x_1,...,x_{n-1}\})[/imath]. How do we know that this process exhausts [imath]A[/imath]? Well, suppose that [imath]x \in A \setminus \{x_1,x_2,...\} \ne \emptyset[/imath]. Then the set [imath]\{k : x_k>x\}[/imath] must be nonempty (otherwise we would have [imath]x \in A[/imath] and [imath]x<x_1=\min A[/imath]), and hence it has a least element. That is, there is some [imath]n[/imath] with [imath]x_1<...<x_{n-1}<x<x_n[/imath]. But this contradicts the choice of [imath]x_n[/imath] as the first element in [imath]A \setminus \{x_1,...,x_{n-1}\}[/imath]. Consequently, [imath]A[/imath] is countable. My questions It is affirmed that the set [imath]\{k : x_k>x\}[/imath] can't be empty (I don't get the reason Carothers gives for this being impossible). As I see it, if [imath]\{k : x_k>x\}[/imath] is empty, then [imath]x_k \leq x[/imath] for all [imath]k[/imath]. So the set [imath]\{x_1,x_2,...\}[/imath] obtained by the method of extracting the smallest element from each remaining nonempty subset of [imath]\mathbb N[/imath] is finite, but we've seen that the set constructed from choosing [imath]x_n=\min(A \setminus \{x_1,...,x_{n-1}\})[/imath] is infinite. So, wouldn't this be the reason why the set [imath]\{k : x_k>x\}[/imath] can't be empty rather than "otherwise we would have [imath]x \in A[/imath] and [imath]x<x_1=\min A[/imath]"?. I couldn't even understand why would [imath]x[/imath] be in [imath]A[/imath] instead of [imath]x \in A \setminus \{x_1,x_2,...\}[/imath] or why [imath]x<x_1[/imath]. I know it is a very small part of the proof but I would like to understand all the steps of it so I would appreciate if someone could clear up my doubt. | 856228 | Help with a proof. Countable sets.
This is a Lemma from N.L. Carothers Real Analysis. Lemma. An infinite subset [imath]A[/imath] of [imath]\mathbb{N}[/imath] is countable. Proof. Since [imath]A\ne\emptyset[/imath], there is a smallest element [imath]x_1\in A[/imath]. Then [imath]A\backslash\{x_1\}\ne\emptyset[/imath], and there must be a smallest [imath]x_2\in A\backslash\{x_1\}[/imath]. By induction we can find [imath]x_1,x_2,\dotsc,x_n,\dotsc\in A[/imath], where [imath]x_n=\min(A\backslash\{x_1,\dotsc,x_{n-1}\})[/imath]. Now, suppose that [imath]x\in A\backslash\{x_1,x_2,\dotsc\}\ne\emptyset[/imath]. Then the set [imath]\{k:x_k>x\}[/imath] must be nonempty, and hence it has a least element. That is, there is some [imath]n[/imath] with [imath]x_1<x_2<\cdots<x_{n-1}<x<x_n[/imath]. But this contradicts the choice of [imath]x_n[/imath]. The part of the proof that I don't understand is that the set [imath]\{k:x_k>x\}[/imath] must be nonempty. Carothers claims that otherwise we would have [imath]x\in A[/imath] and [imath]x<x_1=\min A[/imath]. I can't prove that if [imath]\{k:x_k>x\}=\emptyset[/imath] then [imath]x<x_1[/imath]. Help appreciated please. |
884464 | Trigonometric identity on [imath]\cos \pi/7[/imath]
I found this in a book I used to train myself for the Math Olympics a bunch of years ago: Prove that [imath]\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2} [/imath] I couldn't solve it then and I can't solve it now. As far as I know, polynomials might be involved somehow since it was in the polynomials section of the book, and it's not suposed to use complex numbers since it's in the section previous to that. Using the double and triple angle formulaes I got the LHS to: [imath]4\cos^3\frac{\pi}{7} -2\cos^2\frac{\pi}{7} - 2\cos\frac{\pi}{7}+1 = \frac{1}{2}[/imath] Using the sum-to-product formulaes I got to the same result. And I'm pretty much stuck at this point. | 347112 | How to prove [imath]\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)[/imath]
Is there an easy way to prove the identity? [imath]\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )[/imath] While solving one question, I am stuck, which looks obvious but without any feasible way to approach. Few observations, not sure if it would help [imath] \begin{align} \dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\ \dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7} \end{align} [/imath] |
884590 | Almost Dedekind (not Noetherian)
A domain [imath]R[/imath] is a Dedekind domain iff it is integrally closed, Noetherian with Krull dimension at most one. I think they put Noetherian in the definition not only to make proofs easier but also because it's important. To see that it's really necessary to have Noetherian in the definition do you have an example for me of a domain [imath]R[/imath] which is integrally closed, not Noetherian with Krull dimension one? All my examples of domains which are not Noetherian fail to have Krull dimension one. | 64838 | Integral domain that is integrally closed, one-dimensional and not noetherian
I've tried to construct examples of rings that match all except one of the properties in the definition of a Dedekind domain. (This is an old number theory qual question from Berkeleys MGSA website). The only starting point that I can think of would be [imath]R:=K[X_1,X_2,\ldots][/imath] for [imath]K[/imath] a field. This is clearly integrally closed as any polynomial relation is contained in [imath]K(X_1,\ldots,X_n)[/imath] for some large enough [imath]n[/imath] and [imath]K[X_1,\ldots,X_n][/imath] is integrally closed being a UFD. The problem is then to get rid of enough ideals from this ring [imath]R[/imath], so that its dimension would be 1, but I can't seem to figure out what to do. Anyone have any other ideas for rings that could work as a starting point? The polynomial ring with an infinite number of variables is pretty much the only example that I know of for how to construct a non-noetherian ring that's still a domain. |
884697 | Show that [imath]\int^{\infty}_0 \frac{\sin^4 x}{x^4}=\frac{\pi} 3[/imath].
Show that [imath]\int^{\infty}_0 \left( \frac{\sin x}{x}\right)^4=\frac{\pi} 3[/imath] Although I know the integral with the index is [imath]1[/imath] and [imath]2[/imath], I have no idea on this one. Please help. | 318037 | Prove [imath]\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}[/imath]
I need to show that [imath] \int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3} [/imath] I have already derived the result [imath]\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}[/imath] using complex analysis, a result which I am supposed to start from. Using a change of variable [imath] x \mapsto 2x [/imath] : [imath] \int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi [/imath] Now using the identity [imath]\sin^2(2x) = 4\sin^2x - 4\sin^4x [/imath], we obtain [imath] \int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4} [/imath] [imath] \frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} [/imath] [imath] \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} [/imath] But I am now at a loss as to how to make [imath]x^4[/imath] appear at the denominator. Any ideas appreciated. Important: I must start from [imath] \int_0^\infty \frac{\sin^2x}{x^2}dx [/imath], and use the change of variable and identity mentioned above |
600916 | functions [imath]\sin(x)/x[/imath] is between [imath]1[/imath] and [imath]2/\pi [/imath]
We have to prove that for [imath]0 < x < \pi/2[/imath] ; [imath]1 > \frac{\sin(x)}{x} > \frac{2}{\pi}[/imath] . This is a simple problem . I just want to know what I am doing is correct or not . At [imath]0[/imath] function is very close to [imath]1[/imath] and [imath]f(\pi/2) = 2/\pi[/imath] . If I prove that [imath]f(x)[/imath] is monotonically decreasing in this domain , then that will be enough , isn't it ? | 596634 | Mean Value Theorem: [imath]\frac{2}{\pi}<\frac{\sin x}{x}<1[/imath]
I need to show that [imath]\dfrac{2}{\pi}<\dfrac{\sin(x)}{x}<1[/imath] for [imath]0<x<\dfrac{\pi}{2}[/imath]. I know I need to use the mean value theorem, would I just say that since [imath]f[/imath] is continuous in the interval (call it I). We know [imath]\exists c\in I[/imath] such that [imath]f'(c) = \dfrac{f\left(\dfrac{\pi}{2}\right)-f(0)}{\dfrac{\pi}{2}}[/imath]. I don't know where this is going though. Also, would I need to show [imath]\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1[/imath]? Is there a way to do this without L'H Rule? |
885104 | Prove [imath]x^2=t[/imath] for any [imath]t>0[/imath]
Prove for any positive number [imath]t[/imath], there is a solution for [imath]x^2=t[/imath]. So we want to show that [imath]x^2=t[/imath] for [imath]t\geq0[/imath]. We can break this into two cases: Case 1: Assume [imath]t=0[/imath], then we have [imath]x^2=0[/imath] [imath]\Rightarrow[/imath] [imath]x=0[/imath], and we are done. Case 2: Assume [imath]t>0[/imath], then we have [imath]x^2=t[/imath] [imath]\Rightarrow[/imath] [imath]x=\sqrt{t}[/imath] And here is where I'm stuck, not sure what I should do next...? | 46832 | Existence of square root
I am self-studying real analysis for fun and came across the following question (from A First Course in Real Analysis by Berberian). (i) Prove the following: If [imath]r[/imath] is a real number such that [imath]0 \leq r \leq 1[/imath], then there exists a unique real number [imath]s[/imath] such that [imath]0 \leq s \leq 1[/imath] and [imath]s^2 = r[/imath]. (ii) Generalize (i) to the following: If [imath]a \in \mathbb{R}[/imath], [imath]a \geq 0[/imath], then there exists a unique [imath]b \in \mathbb{R}[/imath], [imath]b \geq 0[/imath] such that [imath]b^2 = a[/imath]. Put [imath]y = 1-r[/imath] and [imath]x = 1-s[/imath]. Then we want to find a real number [imath]x \in [0,1][/imath] such that [imath](1-x)^2 = 1-y[/imath]. I am trying to find an increasing sequence [imath](x_n)[/imath] such that [imath]0 \leq x_n \leq 1[/imath] (defined recursively). For (ii) how could I use (i) to get the result? |
885357 | Decide convergence divergence of [imath]\sum \dfrac{1}{(\ln n)^{\ln n}}[/imath]
Does the series [imath]\sum \dfrac{1}{(\ln n)^{\ln n}}[/imath] converges? I can intuitively say that it converges, because [imath](\ln n)^{\ln n} [/imath] is going to [imath]\infty[/imath] on a hayabusa | 840924 | Prove that [imath]\sum_{n=2}^\infty (\ln n)^{- \ln n}[/imath] converges
As the title suggests, I'd like to prove that the sum [imath] \sum_{n=2}^\infty (\ln n)^{- \ln n} [/imath] is finite. The root and ratio test both fail here, but WA suggests that there is a comparison that can be used to show convergence. The only thought I have is that it may help to write the terms as [imath]e^{-\ln(n)\ln(\ln(n))}[/imath], but this has not led me to any particular insight. Any ideas are appreciated. |
633824 | Prove [imath]\int_{0}^{\pi}\frac{x^2}{\sqrt{5}-2\cos{x}}dx=\frac{\pi^3}{15}+2\pi\ln^2{\left(\frac{1+\sqrt{5}}{2}\right)}[/imath] without contour integration
Show that [imath]\int_{0}^{\pi}\dfrac{x^2}{\sqrt{5}-2\cos{x}}dx=\dfrac{\pi^3}{15}+2\pi\ln^2{\left(\dfrac{1+\sqrt{5}}{2}\right)}[/imath] This thread demonstrates how contour integration can be used to solve the above integral, but I'm interested in finding alternative methods. I believe one might utilize the Polylogarithm in some way. My attempt: I tried to rely on the identity [imath]2\cos{x}=e^{ix}+e^{-ix}.[/imath] Can someone think of a solution to this integral which doesn't involve the residue theorem? Thank you | 502007 | How do I evaluate this integral [imath]\int_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x[/imath]?
Show that [imath]\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right).[/imath] I don't have any idea how to start, but maybe I could use the Polylogarithm. |
885675 | Is the [imath] L^1[/imath] norm continuous in the following sense?
Suppose I have [imath]f \in L^1\cap L^\infty[/imath], and I want to take the following limit [imath]\lim_{h\to0} \int|f(x+h)-f(x)|dx[/imath] Does this follow from the dominated convergence theorem? Since [imath]|f(x+h)-f(x)| \leq |f(x+h)|+|f(x)|[/imath], and [imath]\|f(x+h)\|_{L^1(\mathbb{R})}<\infty[/imath] and [imath]\|f(x)\|_{L^1(\mathbb{R})}<\infty[/imath], we then have [imath]\lim_{h\to0} \int|f(x+h)-f(x)|dx = \int\lim_{h\to0}|f(x+h)-f(x)|dx = 0[/imath]? I'm not convinced, and I think a better argument would be to involve the density of continuous function in [imath]L^1[/imath]. If the approach in this post does not work, can anyone explain why? | 458230 | Continuity of [imath]L^1[/imath] functions with respect to translation
Let [imath]f\in L^1[/imath], consider the map [imath]t\mapsto f_t=f(x-t)[/imath], then how can one show that [imath]t\mapsto f_t[/imath] is continuous? More explicitly one wants to show that [imath]\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=0[/imath]. I tried to use approximation by [imath]C_0(\mathbb{R})[/imath] functions [imath]g^n[/imath] to approximate [imath]f[/imath] in [imath]L^1[/imath] norm. Then one has [imath]\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0[/imath], but then I came across the problem: how can one show that the two limits can exchange so that one has [imath]\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=\lim_{h\to 0}\lim_{n\to\infty}|g^n_{t+h}-g^n_t|_{L^1}=\lim_{n\to\infty}\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0.[/imath] Can someone help me with some conditions on which two limits can be exchanged, or do you have a better way of proving the continuity? Thank you! |
885753 | Does every inner product fail to turn [imath]C[0,1][/imath] into a Hilbert space?
I have the following question: With the usual inner products that come in mind, [imath]C[0,1][/imath] is not a Hilbert space. But is it true that every inner product fails to turn [imath]C[0,1][/imath] into a Hilbert space? If so, then I would be curious, how to prove this. Thank you very much for the help! Yes, I would be interested in the case that the inner product on [imath]C[0,1][/imath] gives an equivalent topology as the sup norm on [imath]C[0,1][/imath], but I forgot to clarify this, when typing the question. | 540302 | The sup norm on [imath]C[0,1][/imath] is not equivalent to another one, induced by some inner product
Let [imath]\mathrm{C}[0,1][/imath] be the space of continuous functions [imath][0,1]\rightarrow \mathbb{R}[/imath] endowed with the norm [imath]||x||_{\infty}=\mathrm{max}_{t\in [0,1]}|x(t)|[/imath]. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for [imath]x(t)=t[/imath] and [imath]y(t)=1[/imath]). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties... |
886204 | Evaluate [imath]\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx[/imath]
[imath]\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx[/imath] The square root really got me confused here. I've tried using the standard trick with [imath]x = x + \sqrt{x^2+4}[/imath] but failed. That might still be the best approach. What's the best way to solve this and how to do it? | 54467 | The integral [imath]\int_0^8 \sqrt{x^4+4x^2}\,dx[/imath]
[imath]\displaystyle \int\nolimits_0^8 \sqrt{x^4+4x^2}\,dx[/imath]. Alright, so I thought I had this figured out. Here's what I did: I factor out an [imath]x^2[/imath] to get [imath]\sqrt{x^2(x^2+4)}[/imath]. I let [imath]x = 2\tan(\theta)[/imath], therefore the integrand is [imath]\sqrt{4\tan^2(\theta) (4\tan^2(\theta) + 4)}[/imath]. Factor out a 4 and it becomes [imath]\sqrt{(16\tan^2(\theta) (\tan^2(\theta) + 1))}[/imath] Which equals [imath]\sqrt{16\tan^2(\theta) \sec^2(\theta)}[/imath] This is easy to take the sqrt of. The integrand becomes [imath]4\tan(\theta)\sec(\theta)[/imath]. Now, the integral of this is [imath]4\sec(\theta)[/imath] And it's evaluated from [imath]0[/imath] to [imath]\arctan(4)[/imath] right? Because as [imath]x[/imath] goes to [imath]0[/imath], so does [imath]\theta[/imath], and as [imath]x[/imath] goes to [imath]8[/imath], [imath]\theta[/imath] goes to [imath]\arctan(4)[/imath]... But the end result [imath](4 (\sec(\arctan(4)) - 1) )[/imath] isn't the correct answer I put it into WolframAlpha and I get [imath](8/3) (17\sqrt{17} - 1)[/imath], which is the right answer. How did they get that? (there's no "show steps" option) Any help is greatly appreciated! PS, what's the syntax for doing sqrts and exponentials? |
886579 | How to calculate the integral [imath]\int_{-\infty}^{\infty} \frac{dx}{(\exp(x)- x)^2 + \pi^2} = \frac{1}{1 + W(1)}[/imath]
On Mathworld one finds without proof the integral [imath]\int_{-\infty}^{\infty} \frac{dx}{(\exp(x) - x)^2 + \pi^2} = \frac{1}{1 + W(1)}[/imath] where [imath]W[/imath] denotes the Lambert W function. How can one show this? The link given on Mathworld is broken. | 45745 | Interesting integral related to the Omega Constant/Lambert W Function
I ran across an interesting integral and I am wondering if anyone knows where I may find its derivation or proof. I looked through the site. If it is here and I overlooked it, I am sorry. [imath]\displaystyle\frac{1}{\int_{-\infty}^{\infty}\frac{1}{(e^{x}-x)^{2}+{\pi}^{2}}dx}-1=W(1)=\Omega[/imath] [imath]W(1)=\Omega[/imath] is often referred to as the Omega Constant. Which is the solution to [imath]xe^{x}=1[/imath]. Which is [imath]x\approx .567[/imath] Thanks much. EDIT: Sorry, I had the integral written incorrectly. Thanks for the catch. I had also seen this: [imath]\displaystyle\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+W(1)}=\frac{1}{1+\Omega}\approx .638[/imath] EDIT: I do not what is wrong, but I am trying to respond, but can not. All the buttons are unresponsive but this one. I have been trying to leave a greenie and add a comment, but neither will respond. I just wanted you to know this before you thought I was an ingrate. Thank you. That is an interesting site. |
887025 | Is Sobolev space [imath]H^{s}(\mathbb R),[/imath] for [imath]s>\frac{1}{2},[/imath] closed under point wise multiplication?
We note that, [imath]L^{2}(\mathbb R)[/imath] is not closed under point wise multiplication. Let [imath]s>\frac{1}{2};[/imath] and we define Sobolev space, as follows: [imath]H^{s}(\mathbb R)=\{f\in L^{2}(\mathbb R):[\int_{\mathbb R} |\hat{f}(\xi)|^{2}(1+|\xi|^{2})^{s}d\xi]^{1/2}<\infty \}.[/imath] My Question is: Is Sobolev space [imath]H^{s}(\mathbb R),[/imath] for [imath]s>\frac{1}{2},[/imath] closed under point wise multiplication, that is, if [imath]f,g \in H^{s}(\mathbb R),[/imath] can we expect [imath]fg\in H^{s}(\mathbb R)[/imath] ? Thanks, | 314820 | Sobolev space [imath]H^s(\mathbb{R}^n)[/imath] is an algebra with [imath]2s>n[/imath]
How do you prove that the Sobolev space [imath]H^s(\mathbb{R}^n)[/imath] is an algebra if [imath]s>\frac{n}{2}[/imath], i.e. if [imath]u,v[/imath] are in [imath]H^s(\mathbb{R}^n)[/imath], then so is [imath]uv[/imath]? Actually I think we should also have [imath]\lVert uv\rVert_s \leq C \lVert u\rVert_s \lVert v\rVert_s[/imath]. Recall that [imath]\lVert f\rVert_s=\lVert(1+|\eta|^2)^{s/2}\,\hat{f}(\eta)\rVert[/imath], the norm on [imath]H^s(\mathbb{R}^n)[/imath]. This is an exercise from Taylor's book, Partial differential equations I. |
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