11

We can ignore the $\tanh$ component momentarily and consider instead the function $$f(x) = \frac{\sin x + \sinh x}{x}.$$ This has series expansion $$f(x) \sim 2 + \frac{x^4}{60} + \frac{x^8}{181440} + O(x^{12}).$$ Therefore, on the open interval $x \in (-1,1)$, $f(x)$ has exceptionally small deviation from $2$, thus on the real line, $$g(x) = \frac{\sin \...


2

Since, for each $z\in\Bbb C\setminus\{0\}$,$$e^{1/z}=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots,$$you have (again, for each $z\in\Bbb C\setminus\{0\}$),\begin{align}(1-z)e^{1/z}&=-z+(1-1)+\left(1-\frac1{2!}\right)\frac1z+\left(\frac1{2!}-\frac1{3!}\right)\frac1{z^2}+\left(\frac1{3!}-\frac1{4!}\right)\frac1{z^4}+\cdots\\&=-z+\sum_{n=0}^\infty\left(\...


2

If $(f_n)$ converges uniformly to $f$, then you have for every finite-length path $\gamma$, that $$\left|\int_{\gamma} f_n(z) dz - \int_{\gamma} f(z) dz \right| =\left|\int_{\gamma} f_n(z) -f(z) dz \right| \leq \int_{\gamma} \left| f_n(z) -f(z) \right| dz \leq ||f_n-f||_{\infty} L(\gamma)$$ where $L(\gamma)$ denotes the length of the path $\gamma$. Because $|...


2

A Taylor series of infinite radius of convergence corresponds to an entire function, i.e. a function analytic on all $\mathbb C$. A linear differential equation whose coefficients are analytic in a region of the complex plane, and whose leading coefficient is $1$, has solutions that are analytic in that region. In particular if the coefficients are entire ...


2

Along the lines of @Mark Viola 's hint, we can rewrite as $$f(x)=\frac{6x^2}{16^2}\cdot \frac{1}{\left(1-\frac{-x^3}{8}\right)^2}=\frac{3x^2}{128}\cdot g\left(-\left(\frac{x}{2}\right)^3\right)$$ Where of course $$g(t):=\frac{1}{(1-t)^2}$$ Notice that $$g(t)=\frac{\mathrm{d}}{\mathrm{d}t}\left\{\frac{1}{1-t}\right\}$$ Can you continue?


1

This formula only works for simple zeroes for which df/dx does not cancel. A good proof can be found in Kanwal's book "Generalized Functions" pp. 49-50


1

There is no need to expand in a series; indeed there is no need for $f$ to have more than one derivative at all in order to make sense of $\delta(f(x))$. What is really happening is that you locally invert $f$. That is, $f$, being a differentiable function that is assumed to never have zero derivative at any of its zeros, can be locally inverted around each ...


1

First consider uniform convergence on any interval $(0,\delta)$ where $\delta \leqslant1$. We have $$\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|= \sum_{k=n+1}^{\infty}\frac{(1-x)^k}k \geqslant\sum_{k=n+1}^{2n}\frac{(1-x)^k}k ,$$ and $$\sup_{x \in (0,\delta)}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|\geqslant \sup_{x \in (0,\...


1

A Maclaurin series is given by $$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}x^k$$ In our case we have: $$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{5}\frac{f^{(k)}\left(a\right)}{k!}x^k$$ I stop me for $n=5$. You can calculate the derivates and you will have $$f\left(x\right)\approx\frac{0}{0!}x^{0}+\frac{0}{1!...


1

$ \displaystyle\int_0^1\frac{\ln^2(x)\ln(1+x)}{1-x}dx=\displaystyle\int_0^1\displaystyle\int_0^1\frac{x\ln^2(x)}{(1+xy)(1-x)}dxdy$ $=\displaystyle\int_0^1\frac{1}{1+y}\left(\displaystyle\int_0^1\frac{\ln^2(x)}{1-x}-\frac{\ln^2(x)}{1+yx}dx\right)dy$ $=\displaystyle\int_0^1\frac{1}{1+y}\left(2\zeta(3)-\displaystyle\sum_{n=1}^{\infty}(-y)^{n-1} \displaystyle\...


1

Let us first briefly try to understand the differences between the Laurent and the Taylor series. You have correctly observed that the Taylor series is only defined in a region in which the function is analytic. To be a little bit more specific: If $f$ is analytic at $z_0$, its Taylor series converges uniformly in the biggest circle centered around $z_0$ in ...


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