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Lemma: in any ring $R$, any power series in $R[[x]]$ with constant term $1$ is invertible. Explanation. Every power series with constant term $1$ can be written as $1-xf(x)$ for some other power series $f(x)$. Then $(1-xf(x))(1+xf(x)+x^2f(x)^2+\cdots)=1$. The geometric series can be simplified to a power series in $x$ by expanding all of the powers $f(x)^n$ ...


5

Pick any $B>0$. Then, if $|x|\leq B$, then one has that $|f_k(x)|=|x|^k/k!\leq B^k/k!$ for each non-negative integer $k$. Since $$\sum_{k=0}^{\infty}\frac{B^k}{k!}=\exp(B)$$ is convergent, Weierstrass's $M$-test reveals that the series $\sum_{k=0}^{\infty} x^k/k!$ converges uniformly for $x\in[-B,B]$. Together with the continuity of the partial sums ...


3

The number of solutions of the equation equals the coefficient of $z^n$ in the expression $$ \frac{1}{(1-z)(1-z^2)(1-z^3)\ldots }=\sum_{n}p(n)z^n, $$ where $p(n)$ is the Euler partition function, so that the number of combinations is $p(n)$. (Christoph): There is a bijection between the partition function and the desired function, since from a partition ...


3

Almost certainly not: if there were a "general formula" for $\sum_{i=1}^r p_i^n$ for even a single $n$, you'd also have a "general formula" for the prime numbers by $$p_r=\sqrt[n]{\left(\text{“general formula” for }\sum_{i=1}^r p_i^n\right) - \left(\text{“general formula” for }\sum_{i=1}^{r-1}p_i^n\right)}$$


2

We have: $$\begin{align}&f(x)=\sum_{n=0}^\infty\binom\alpha n x^n\\{}\\ &f'(x)=\sum_{n=1}^\infty\binom\alpha n nx^{n-1}\end{align}$$ Perhaps the trick you're looking for is $$n\binom\alpha n=\alpha\binom{\alpha-1}{n-1}$$ Added on request: $$\alpha f(x)=\sum_{n=0}^\infty \alpha\binom\alpha ...


1

Taylor expanding $\cos z$ about $z=\frac{\pi}{2}$ one finds that: \begin{equation} \cos z=-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots \end{equation} Similarly, Taylor expanding $\frac{1}{z+\frac{\pi}{2}}$ about $z=\frac{\pi}{2}$ one finds that: \begin{equation} ...


1

A very simple proof of this actually involves complex analysis. The set of complex numbers extends to the reals. $e^{z}=\sum_{0}^ \infty \frac{z^{k}}{k!} $ This is a series for $e^{z}$ expanded at $\alpha=0$. In complex analysis in order for a number to be inside the disc of convergence it has to follow the following conditions: Let $\beta$ be the ...


1

We will use the fact that $$ \sum_{n=0}^\infty\frac{x^n}{n!} $$ converges absolutely for all $x\in\mathbb{R}$. This can easily be shown using the ratio test and means that for any $x$, there is an $N$ so that $$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\epsilon $$ Find an $N$ so that we have $$ ...


1

One may recall that, as $x \to 0$, $$ \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad x \in \mathbb{C}, \tag1\\\\ \frac{1}{1-x^2}&=\sum_{n=0}^{\infty}x^{2n}=\frac1{2}\sum_{n=0}^{\infty}(1+(-1)^n)x^{n}, \quad |x|<1, \tag2 \end{align} $$ then using the Cauchy product we get $$ ...


1

One method of evaluating $\sum_{n=0}^\infty(1+n)x^n$ can be like this, we take the generating function f = $\sum_{n=0}^\infty x^n $, then $$\sum_{n=0}^\infty (n+1)x^n = (xD + 1) f $$ $$ \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{1}{(1-x)^2}$$here D means differentiation w.r.t x



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