What did i do wrong with this derivation?

Question

$$ \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} $$ Therefore \begin{align} \frac{1}{\cos(x)} &= \frac{1}{1-(\frac{x^2}{2} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots)} \\ &= \sum_{n=0}^\infty (\frac{x^2}{2} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots)^n \\ &= \sum_{n=0}^\infty (1-\cos(x))^n \end{align}

This part i am okay with, but now i will attempt to find a power series of $\tan(x)$ in terms of $\cos(x)$ with the following reasoning:

I realised that by differentiating The secant function i would get $\sec(x) \tan(x)$ and by differentiating the series would also yield a $\sin(x)$, therefore i would end up with $\sec^2(x)$ (which is the derivative of $\tan(x)$) therefore by integration i would find myself a power series for $\tan(x)$ in terms of $\cos(x)$ I will do as i explained below:

$$ \frac{d}{dx}[\sec(x)] = \sum_{n=0}^\infty \frac{d}{dx}(1-\cos(x))^n $$

$$ \frac{\sin(x)}{\cos^2(x)} = \sin(x) \sum_{n=0}^\infty (n+1) (1-\cos(x))^n $$

$$ \frac{1}{\cos^2(x)} = \sum_{n=0}^\infty (n+1) (1-\cos(x))^n $$

Now to redefine $(1-\cos(x))^n$ by using the following identity:

$$ (1+x)^{\alpha} = \sum_{k=0}^\alpha {\alpha \choose k} x^k $$

Therefore:

$$ (1-\cos(x))^n = \sum_{k=0}^n {n \choose k}(-\cos(x))^k $$

Now this is where i run into some doubt:

$$ \sec^2(x) = \sum_{n=0}^\infty (n+1) \sum_{k=0}^n {n \choose k}(-\cos(x))^k $$ Since the following is true for ordinary power series: $$ fg \longleftrightarrow \left\{\sum_k a_k b_{n-k} \right\} $$

Therefore by setting $a_k = \frac{(-\cos(x))^k}{k!}$ and $b_{n-k} = \frac{1}{(n-k)!}$

$$ \sum_{n=0}^\infty (n+1) \sum_{k=0}^n {n \choose k}(-\cos(x))^k = \sum_{n=0}^\infty (n+1)! e^{1-\cos(x)} $$

This is clearly wrong obviously as the series does not converge... What was my error?

Answer 1

$\sum_{n=0}^\infty (n+1) \sum_{k=0}^n {n \choose k}(-\cos(x))^k = \sum_{n=0}^\infty (n+1)! e^{1-\cos(x)} $

The problem is that, in your series, only one of the terms has the $x^n$.

Here is a more detailed examination of what happens:

Let $u = \cos(x) $. We have

$\frac1{u} =\frac1{1-(1-u)} =\sum_{n=0}^{\infty} (1-u)^n $.

Differentiating term-by-term, and then using the binomial theorem,

$$\begin{align} \frac{-1}{u^2} &=\sum_{n=1}^{\infty} (-n)(1-u)^{n-1}\\ &=-\sum_{n=0}^{\infty} (n+1)(1-u)^{n}\\ &=-\sum_{n=0}^{\infty} (n+1)\sum_{k=0}^n\binom{n}{k}(-1)^ku^k \text{ ( you need }u^n \text{ as well as } u^k)\\ &=-\sum_{k=0}^{\infty} \sum_{n=k}^{\infty} (n+1)(\frac{n!}{k!(n-k)!})(-1)^ku^k \text{ (here is the problem)}\\ &=-\sum_{k=0}^{\infty} \frac{(-1)^ku^k}{k!}\sum_{n=k}^{\infty} \frac{(n+1)!}{(n-k)!}\text{ (the inner sum diverges violently)}\\ &=-\sum_{k=0}^{\infty} \frac{(-1)^ku^k}{k!}\sum_{n=0}^{\infty} \frac{(n+k+1)!}{n!}\\ \end{align} $$

You can only do the series multiplication if the same variable occurs in both series.