Let $S(x)=\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$. Define $C(x):=S'(x)$ and show that $C'(x)=-S(x)$.

Question

so this problem I am trying to solve says

Let $S(x)=\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$. Define $C(x):=S'(x)$ and show that $C'(x)=-S(x)$.

Well, I get that $S'(x)=\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}$, which is correct. But then I get that $C'(x)=\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k-1}}{(2k-1)!}$, but I don't get $$C'(x)=-\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!}$$ One thing I have noticed though is tht we cannot have $(2k-1)!$ for $k=0$, but I am not sure how to use that to prove the result required. My intuition tells me that

$$C'(x)=\sum\limits_{k=0}^\infty \frac{(-1)^{k+2} x^{2k+1}}{(2k+1)!}=-\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!},$$

but would that be correct? It would be great if someone could explain to me the rules for this kind of manipulation... Thanks! :)

Answer 1

Note that $S'(x)=\displaystyle\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}$. The first term here is $1$, so when you differentiate, it dies. Then $$S''(x)=\sum_{k\geqslant 1}\frac{(-1)^k x^{2k-1}}{(2k-1)!}$$

i.e. we start the sum at $k=1$. Can you continue? All you need to do now is shift an index. Note that we repeatedly used $(x^n)'=nx^{n-1}$ and $n!=(n-1)!\,n$.

Answer 2

You have to make an index shift:

$S'(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!} $

as you said but now, in

$S''(x) = \sum_{k=0}^{\infty} 2k \frac{(-1)^k x^{2k-1}}{(2k)!} $

the first term equals zero so you can forget about it which yields to

$S''(x) = \sum_{k=1}^{\infty} \frac{(-1)^k x^{2k-1}}{(2k-1)!} $

Now index shift from $k \rightarrow k+1$ leads to the desired result

$S''(x) = \sum_{k=0}^{\infty} \frac{(-1)^{(k+1)} x^{2(k+1)-1}}{(2(k+1)-1)!}= -\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!} $

Answer 3

If we look in the function $\sin(x)$ and $\cos(x)$. The function Taylor series we will get: $\sin(x)=S(x)$ and $\cos(x)=C(x)$. Now you prove what you need.