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## Hot answers tagged analysis

17

Suppose for some $x_0$ you have $f(x_0)\neq0$. Then you can solve the differential equation $$\frac{f'(x)}{f^2(x)}=1$$ with the initial condition $f(x_0)$, which gives $$\frac{-1}{f(x)}+\frac{1}{f(x_0)}=x-x_0\iff f(x)=\frac{1}{c-x}$$ where $c$ is a constant, and this holds for every $x$ that is in the same component of $\mathbb R\backslash\{c\}$ with $x_0$, ...

13

$\log(e^x)=\int_1^{e^x}\frac{1}{t}dt=\int_0^x\frac{1}{e^u}e^u du=x$ and since it's easy to prove that $e^x$ is bijective then $\log$ is its inverse.

11

No, it is not possible. Suppose otherwise. Then, there is some $N\in\mathbb N$ such that$$n\geqslant N\implies\lvert a_n-\varepsilon\rvert<\varepsilon\iff0<a_n<2\varepsilon.$$But this is impossible, since you are assuming that you always have $a_n\leqslant0$.

8

Well, if one puts $v=\frac{1}{u}$ then: $$I=\int_0^\infty\frac{v^2}{1+v^4} dv =\int_0^\infty\frac{1}{1+u^4} du$$ So summing up the two integrals from above gives: $$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$

8

The answer is yes. Let $S = \mathbb{N} \setminus \mathbb{N}^2$ be the set of numbers which are not perfect squares. Notice that $a\in S^m$ means that $a = b^m$ for some (unique) $b \in \mathbb{N}$ which is not a perfect square. Clearly $\mathbb{N} = \bigcup_{k \ge 0} S^{2^k}$ and this union is disjoint. Define $f : \mathbb{N} \to \mathbb{N}$ as $$f(a) = \... 7 This is not always true. For example if f is a constant map then the preimage of any point is either empty or \Bbb R^k. If you add the assumption that y\in f(\Bbb R^k) and that for any x\in f^{-1}(\{y\}), f is a submersion at x, then you have that f^{-1}(\{y\}) is a (k-l)-dimensional submanifold of \Bbb R^k (in this case we say that y ... 7 Mark a= 3x and b= 2x-1 then we get:$$a(\sqrt{a^2+3}+2)+b(\sqrt{b^2+3}+2)=0$$Now the function f(t) = t(\sqrt{t^2+3}+2) is odd and strictly increasing so it is also injective. Since f(a)+f(b)=0 we get$$f(a) = -f(b) = f(-b)\implies a=-b \implies x={1\over 5}$$7 This is a reasonable thing to be confused about since the terminology reveals an inconsistency between the way computer-scientists talk about functions, pure mathematicians talk about functions, and engineers talk about functions. First, as you say, there's no way the normal \sin function$$ \sin: \mathbb{R} \to \mathbb{R} $$whose graph is the wave ... 7 On the set where f(x) \neq 0 the derivative of -1/f is 1  so we get f(x) (x+c)=-1 for some constant c. It follows that the continuous function f(x) (x+c) takes only two values 0 and-1. Hence it is a constant. But f(0)=0 so f must vanish identically. [We get f(x)=0 for x \neq -c but f(-c) is also 0 by continuity]. Some additional ... 7 In the first part, I answer Question 2: I prove that the recursion$$\label{rec}\tag{1}a_{n+1}=\frac n{a_n}-a_n-a_{n-1}$$yields a positive sequence for the initial values a_0=0, a_1=2\Gamma(\frac34)/\Gamma(\frac14). (\ref{rec}) is actually a special discrete Painlevé 1 equation. One survey article mentioned this article by J.\ Shohat about ... 6 \lim_{x \to \infty} \frac x{1+x} =1. So if a_{n_k} \to \infty then \frac {a_{n_k}} {1+a_{n_k}} \to 1. Since \frac {a_n} {1+a_n} <1 for all n we can conclude that the limit superior is 1. 6 This probably isn't the slickest way of presenting the answer, but here goes. We're going to repeatedly use the change of variables theorem in what follows. Let B denote the unit ball in \Bbb{R}^3. Then, by the change of variables formula, if \alpha,\beta,\gamma are non-negative integers, then \int_B x^{\alpha}y^{\beta}z^{\gamma} \, dV ... 5 If you expand the squares, then you get$$\begin{split} \sum_{k=1}^{n-1} \Big(a+\frac{k}{n}\Big)^2&=\sum_{k=1}^{n-1} \Big(a^2+2a\frac{k}{n}+\frac{k^2}{n^2}\Big) \\ &=(n-1)a^2+2a\cdot \frac{1}{n}\cdot\frac{n(n-1)}{2}+\frac{1}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}. \\ &=(n-1)a^2+a(n-1)+\frac{(n-1)(2n-1)}{6n}, \end{split}$$and thus$$\begin{split} \...

5

Telescopic Solution Notice that $$\frac{2^k}{5^{2^k}-1}-\frac{2^k}{5^{2^k}+1}=\frac{2^{k+1}}{5^{2^{k+1}}-1}$$ Therefore, we get a telescoping series: \begin{align} \sum_{k=0}^\infty\frac{2^k}{5^{2^k}+1} &=\sum_{k=0}^\infty\left(\frac{2^k}{5^{2^k}-1}-\frac{2^{k+1}}{5^{2^{k+1}}-1}\right)\\[6pt] &=\frac14 \end{align} Analysis of the Solution ...

5

The value of the integral doesn't change if you change the integrand at a single point, so you can assume the integrand is $$f(t)=\cases{\sin t/t ,&t\neq0\\1,&t=0}$$ and apply the fundamental theorem of calculus.

5

Let $g(x)=e^x-f(x)$, it is obvious that $0\leq g^{(n)}(x)\leq 2e^x$. By Taylor's formula, we have $$g(x)=g(0)+g'(0)x+\cdots+\frac{g^{(n)}(0)}{n!}x^n+\frac{g^{(n+1)}(\eta_n)}{(n+1)!}x^{n+1},$$ where $\eta_n$ between $0$ and $x$. Since $$\left|\frac{g^{(n+1)}(\eta_n)}{(n+1)!}x^{n+1}\right|\leq\frac{2e^{\eta_n}}{(n+1)!}|x|^{n+1}\leq\frac{2e^{\max{\{0,x\}}}}{(n+... 5 Let$$P_s=\prod_{n=2}^\infty \frac{1}{1-n^{-s}}$$Using a CAS, there are some nice expressions such as$$\left( \begin{array}{cc} s & P_s \\ 2 & 2 \\ 3 & 3 \pi \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right) \\ 4 & 4 \pi \text{csch}(\pi ) \\ 6 & 6 \pi ^2 \text{sech}^2\left(\frac{\sqrt{3} \pi }{2}\right) \end{array} \right)$$P_5... 4 Partial fractions give$$f(x) = \frac{-2}{x-1} + \frac{4}{x-2}.$$Then$$f''(x) = \frac{-4}{(x-1)^3} + \frac{8}{(x-2)^3}.$$The numerator of that, after you add the fractions is x^3-6x+6, which doesn't have pretty roots. The real root is -(\sqrt[3]{4}+\sqrt[3]{2}). 4 Between 0 and 1, \exp(2\pi x) increases from 1 to e^{2\pi} while (1+x)/(1-x) increases from 1 to \infty. Do the graphs meet? For small positive x, \exp(2\pi x)\sim 1+2\pi x while (1+x)/(1-x)\sim 1+2x, so the graph of the former is above that of the latter near 0. The opposite is true near 1, so we can apply IVT. 4 Yes, f_n \to f in L^{1} implies that there is a subsequence which converges to f almost everywhere and the subsequence also converges to g almost everywhere. Hence f=g almost everywhere. 4 Hint 1:$$a_{n+1}-a_n=\left( \frac{1}{n+1}+ \frac{1}{n+1} +\dots + \frac{1}{2n+2} \right)-\left(\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}\right)\\ = \frac{1}{2n+1}+\frac{1}{2n+2}- \frac{1}{n}<\frac{1}{2n}+\frac{1}{2n}- \frac{1}{n}=0$$Hint 2:$$a_n =\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \geq \frac{1}{2n}+ \frac{1}{2n} +\dots + \frac{1}...

4

The notion you seem to want is Kolmogorov complexity, which informally speaking measures the size of the shortest program or expression that produces a given string. It is a fundamental result that not all strings have short descriptions. Some (most?) strings need descriptions at least as long as themselves, and so cannot be compressed. Moreover, ...

4

Note that $\Bbb R^n$ has a countable base: all finite products of open intervals with rational endpoints, or all open ball around rational points with rational radii. Enumerate them as $\{B_n: n \in \Bbb N\}$, say. Now for any (not just closed) subset $X$ of $\Bbb R^n$ pick $x_n \in B_n \cap X$ whenever this intersection is non-empty, and this gives us a ...

4

A function $f:X\to Y$ has an inverse if and only if it is bijective. If a function is $f:X\to Y$ is injective and not necessarily surjective then we "create" the function $g:X\to f(X)$ prescribed by $x\mapsto f(x)$. This function $g$ (closely related to $f$ and carrying the same prescription) is bijective so it has an inverse $g^{-1}:f(X)\to X$.

4

Fits into one sentence: $$\int_k^{k+1}f(t)\,dt-f(k)=\int_k^{k+1}(f(t)-f(k))\,dt=\int_k^{k+1}\int_k^tf'(s)\,dsdt,$$so \begin{align}\left|\int_k^{k+1}f(t)\,dt-f(k)\right| &\le\int_k^{k+1}\int_k^t|f'(s)|\,dsdt \\&\le\int_k^{k+1}\int_k^{k+1}|f'(s)|\,dsdt \\&=\int_k^{k+1}|f'(s)|\,ds,\end{align}hence $$\sum_{k=1}^\infty\left|\int_k^{k+1}f(t)\,dt-f(... 4 Your equation can be written as$$\left(\frac{d^2}{dx^2}+u(x)\right)\Psi=\lambda\Psi.If we define A:=\frac{d^2}{dx^2}+u, then it will suffice to show that A is formally self-adjoint. If we let f,g\in C_0^\infty(\mathbb{R}), then we can calculate \begin{align*}(Af,g)_{L^2}&=\int\limits_\mathbb{R} \left(\frac{d^2}{dx^2}+u(x)\right)f(x)g(x)\, dx\\ &... 4 Conformal and angle-preserving are the same thing, and refer to maps that preserve the angles between intersecting curves and the shapes of infinitesimally small objects. This can be applied to any mapping between manifolds, and is not restricted to \mathbb C or \mathbb R. Analytic functions can be represented in terms of a locally convergent power ... 4 Not true. f(a,0) = \frac{1}{\sqrt{a}} \to \infty as a \to 0+. 4 f(x)=2\sum_{n \ge 1}\frac{\cos nx}{n}=-2\ln{(2\sin{\frac{x}{2}})}, 0<x<2\pi (this is a classic Fourier series and the result is obtained by integration by parts for n \ge 1 and the usual result for the integral of \ln{(2\sin{\frac{x}{2}})} for n=0 once you know it; another way is to use Fourier Stieltjes series for \sum{\sin{nx}} and ... 3 I wonder if the question is well posed. If you want fit the functionf(t)=\sin(kt^2)$$with a given data$$\{(t_1,f(t_1)),\cdots(t_n,f(t_n))\}$$simply change the data with \quad\begin{cases} x_1=t_1^2 , \cdots, x_n=t_n^2 \\ y_1=f(t_1) , \cdots, y_n=f(t_n) \end{cases} and fit the function$$y(x)=\sin(kx)$$to the new data$$\{(x_1,y_1),\cdots(x_n,...

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