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## Hot answers tagged power-series

3

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then $\frac{\omega^n+\omega^{2n}+1}{3}$ equals $1$ iff $3\mid n$, hence: $$\sum_{n\geq 1}\frac{x^{3n}}{(3n)!} = \color{red}{-1+\frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3}}. \tag{1}$$

3

Define $$y(x)=\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}.$$ We observe that $$\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}+\sum_{k\geq0}\frac{x^{3k+1}}{\left(3k+1\right)!}+\sum_{k\geq0}\frac{x^{3k+2}}{\left(3k+2\right)!}=e^{x}$$ so we have the second order ODE $$y''\left(x\right)+y'\left(x\right)+y\left(x\right)=e^{x}.\tag{1}$$ Let start to solve the ...

2

To expound my comment, Consider: $$S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4+it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4-it} \right)$$ Where $a(t)=arg(\Gamma(1/4+it))$. Then: $$S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 ... 2 If 0<R<\infty, then the series \sum_{n=0}^{\infty}R^{-n}x^n has radius of convergence R, because for fixed x it is just a geometric series. 2 If you have power series \sum_{n\geq0}a_nq^n and \sum_{n\geq0}b_nq^n, and you know that they are equal, then you know that a_n=b_n for all n\geq0. That is «equating the like powers of q» in the two series. 1 Since f has only finitely many singularities z_1,……,z_m which are simple poles, then f can be extended to some B(0,R)(R>1) to become a meromorphic functions. Without lost of generality we set the non-zero residue Res(f,z_j)=c_j, then$$g(z):=f(z)-\sum_{j=1}^m \frac{c_j}{z-z_j} \in H(B(0,R))$$and set$$f(z)=\sum_{n=0}^{\infty}a_n z^n,z \in ...

1

Sure there is: If $$\sum_{n=0}^{n=\infty}c_n(z-a)^n$$ is a power series with radius of convergence R, then for any $\lambda > 0$ the series $$\sum_{n=0}^{n=\infty}c_n\left(\frac{z-a}{\lambda}\right)^n$$ converges iff $\left\lvert\frac{z-a}{\lambda}\right\rvert < R$, i.e., iff $|z-a| < \lambda R$. Hence, the power series ...

1

The following assumes that you already know $e^x = \sum_{n=0}^\infty x^n/n!$, ie the Taylor series of $e^x$ at $0$ (and that it converges to $e^x$). Given this, simply note $$\sum e (x-1)^n /n! = e \cdot \sum (x-1)^n/n! = e \cdot e^{x-1} = e^x.$$

1

We need to show that $\displaystyle\lim_{n\to\infty} R_n(x)=0$ for all x, where $\displaystyle R_n(x)=\frac{f^{n+1}(c)(x-1)^{n+1}}{(n+1)!}=\frac{e^c(x-1)^{n+1}}{(n+1)!}$ and $c$ is a number between 1 and $x$. 1) If $x< 1$, then $\displaystyle \lim_{n\to\infty}R_n(x)=0$ since $\displaystyle |R_n(x)|< \frac{e|x-1|^{n+1}}{(n+1)!}$ (since $e^c< e$) ...

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