# Tag Info

7

Here's a funny one: Define the function $f$ as $$f:\mathbb{R}\longrightarrow\mathbb{R},\ a\longmapsto\int_{-\infty}^{+\infty}\mathrm{e}^{-x^2/2}\cos(a x)\,\mathrm{d}x.$$ It is well-known that the improper integral defining $f$ is convergent. We can apply the differentiation theorem to differentiate this improper integral (proof left to the reader) and we ...

3

Let $f(x)= A\sin(\alpha x)+B\cos(\alpha x)$. We will assume that $f=0$ and arrive at a contradiction. If $f=0$ for all $x$, then so is its derivative $f'$. Therefore, we have both $$A\sin(\alpha x)+B\cos(\alpha x)=0 \tag 1$$ and $$\alpha (A\cos (\alpha x)-B\sin(\alpha x))=0 \tag 2$$ for all $x$. If $\alpha \ne 0$, then $(1)$ and $(2)$ imply ...

2

That is because $\alpha$,$A$ and $B$ are constants, and when since $x$ can be arbitrary, the expression $Acos(\alpha x)+B sin(\alpha x)$ can not be zero for all $x$.

2

Another chance is given by the expansion of $\cos(\pi n x)$ as a Taylor series: $$\cos(\pi n x) = \sum_{m\geq 0}\frac{(-1)^m (\pi n)^{2m} x^{2m}}{(2m)!}\tag{1}$$ together with the fact that: $$\int_{\mathbb{R}} x^{2m} e^{-x^2/2}\,dx = \int_{0}^{+\infty} x^{m-1/2} e^{-x/2}\,dx = 2^{m+1/2}\cdot\Gamma\left(m-\frac{1}{2}\right).\tag{2}$$ These identities ...

2

You are solving the so-called wave equation. After separating variables we obtain for the space-dependent part $X(x) = C\sin\alpha x + D\cos\alpha x$. The first boundary condition yields $X(0) = 0$, and results in $D=0$. The second boundary condition becomes $X(\pi) = 0$ and leads to $C\sin\alpha\pi = 0$. If $B=0$, then the solution is trivial, i.e. $X ... 1 With your hypotheses, and for the function$f(x)=x$, the Fourier series of sines is equal to$f$everywhere, except at$x=\pi$, where it is$0$(as you can check explicitly, at this point, simply substituting$x=\pi$in the series that you already wrote). More generally, the Fourier series of sines of a function$f$with the hypotheses that you described ... 1 The way I view the use of the FT using periodic BCs is that the problem is really about expressing the solution in terms of Fourier series with the "transform" being the Fourier coefficients. The periodic BC's imply that $$u(x,t) = \sum_{n=-\infty}^{\infty} c_n(t) e^{i n \pi x}$$ where $$f(x) = \sum_{n=-\infty}^{\infty} c_n(0) e^{i n \pi x}$$ Thus, ... 1 $$\int_{-\infty}^{\infty} e^{-|x| + xi\xi} \ dx = \int_{-\infty}^0 e^{x(1+i\xi)} \ dx + \int_0^{\infty}e^{-x(1-i\xi)} \ dx$$ And then my friend pointed out the following:$e^{x+ix\xi}=e^xe^{ix\xi}=e^x (\cos(x\xi)+i\sin(x\xi))\leq e^x (1+i)$Now:$e^x (1+i)$- so we can ignore the$(1+i)$as it is bounded and we only care about$e^x$which goes to zero as ... 1 HINT: $$\int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}\left(\cos\left(\pi nx\right)\right)dx= \Re\left[ \int_{\mathbb{R}}e^{-\frac{x^{2}}{2}}e^{i\pi nx}dx\right]$$ Then observe that $$-\frac{x^2}2+i\pi nx=-\left(\frac x{\sqrt2}-\sqrt2i\pi n\right)^2+2\pi^2 n^2$$ and use the well known integral $$\int_{\Bbb R}e^{-s^2}\,ds=\sqrt{\pi}\;\;$$ with a suitable ... 1 Let$F_{n+1}$be the$(n+1)\times(n+1)$Fourier matrix. Let$C_n$be the column vector whose coefficients are constituted by the$n$th line of Pascal's triangle. The image of$C_n$by$F_{n+1}$is not proportional to$C_n$but to a vector whose entries' moduli are proportional to the$n$th power of a cosine function$A cos^n(...)$. Explicitly: the ... 1 Let$A \subseteq \Bbb{R}$be Borel measurable, and$T$a dense subset of$\Bbb{R}$. Suppose for every$t \in T$that $$m((A+t)\setminus A)=0,$$ where$m$is the Lebesgue measure. Then$m(A)=0$or$m(\Bbb{R}\setminus A) = 0$. Proof: We define$-A$as$-A=\{-x : x \in A\}$. Note that for all$t, x \in \Bbb{R}\$\chi_{A+t}(x) = ...

1

Here is Euler's Other Proof by Gerald Kimble \begin{align*} \frac{\pi^2}{6}&=\frac{4}{3}\frac{(\arcsin 1)^2}{2}\\ &=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}}\,dx\\ &=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx ...

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