Spivak, Ch. 24, Problem 6: Is there a relatively easy way to find $f^{(k)}(0)$ for $f(x)=\sin{x}/x, f(0)=1$ using power series? 


*If $f(x)=\frac{\sin{x}}{x}$ for $x\neq 0$ and $f(0)=1$, find $f^{(k)}(0)$. Hint: Find the power series for $f$.


Here is what the solution manual says

Since
$$\sin{x}=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
we have
$$f(x)=\sum\limits_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$$
(notice that the right side is $1$ for $x=0$). So
$$f^{(k)}(0)=\begin{cases} \frac{(-1)^n}{2n+1}, k=2n \\ 0, k \text{ odd } \end{cases}\tag{1}$$

As is customary with the solution manual, it skips all the intermediate steps. The comment in parentheses indicates that every time the exponent on $x$ is even then there is some $n$ for which it is zero and thus the entire term is a constant. I understand the solution intuitively. All you have to do is write out the series to notice that for even derivatives the first term is a constant, but for odd derivatives all terms contain an $x$.
I can't seem to see this analytically though.
Here is what I had
$$f(x)=\sum\limits_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n+1)!}$$
$$=\sum\limits_{n=0}^\infty \frac{(-1)^n}{2n+1}\cdot\frac{x^{2n}}{(2n)!}$$
$$f'(x)=\sum\limits_{n=0}^\infty \frac{(-1)^n}{2n+1}\cdot\frac{x^{2n-1}}{(2n-1)!}$$
$$f^{(k)}(x)=\sum\limits_{n=n_0}^\infty \frac{(-1)^n}{2n+1}\cdot\frac{x^{2n-k}}{(2n-k)!}$$
Now, if $k$ is even, then $n_0=k/2$, and the first term is $\frac{(-1)^{n_0}}{2n_0+1}$.
If $k$ is odd, then $n_0=\frac{k-1}{2}$ (this is kind of a hunch, I didn't actually prove this, and being comfortable believing it made me waste a lot of time) and the first term is $\frac{(-1)^{n_0}}{2n_0+1}x$.
Therefore,
$$f^{(k)}(0)=\begin{cases} \frac{(-1)^n}{2n+1}, k \text{ even } \\ 0, k \text{
odd } \end{cases}\tag{1}$$
It seems I have found the same result as the solution manual. The reason I am asking this question is that I spent an absurd amount of time to figure this out, and even now it seems like a sketchy process I used.
Is there an easier way to manipulate the sum so as to reach (1) with less pain and suffering?
 A: The main issue with my solution (and that makes reasoning about it painful) is the fact that I manipulated the summation terms in a way that wasn't ideal.
We have
$$f(x)=\sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n+1)!}\tag{1}$$
$$f'(x)=\sum\limits_{n=0}^\infty (-1)^n \cdot 2n \cdot \frac{x^{2n-1}}{(2n+1)!}\tag{1a}$$
and
$$f^{(k)}(x)=\sum\limits_{n=0}^\infty (-1)^n \cdot 2n\cdot (2n-1)\cdot (2n-2)...\cdot (2n-(k-1)) \cdot \frac{x^{2n-k}}{(2n+1)!}\tag{1b}$$
From (1b) we can see that if $k$ is even the terms for $n=1,2,...\left ( \frac{k-2}{2}\right )$ will all be $0$ because some one term in
$$2n\cdot (2n-1)\cdot (2n-2)...\cdot (2n-(k-1))$$
is $0$.
For $n=k/2$, however, we have the term
$$(-1)^n\cdot 2n\cdot (2n-1)\cdot (2n-2)...\cdot (2n-(k-1))\frac{1}{(2n+1)!}$$
$$=k!\frac{(-1)^{k/2}}{(k+1)!}$$
$$=\frac{(-1)^{k/2}}{k+1}\tag{2}$$
and all subsequent terms have an $x$ factor. Therefore $f^{(k)}(0)$ is just (2).
On the other hand, if $k$ is odd, then the terms for $n=1,2,...,\left (\frac{k-1}{2}\right )$ are all $0$, and all subsequent terms have an $x$ factor. Therefore $f^{(k)}(0)=0$.
A: I think the thing that you are not seeing is that the series
$$\frac{\sin x}{x}=\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots$$
Is missing some terms. Really it should read,
$$\frac{\sin x}{x}=1+0x-\frac{x^2}{3!}-0x^3+\frac{x^4}{5!}+\cdots$$
So you should write
$$\frac{\sin x}{x}=\sum_{n=0}^\infty a_n x^n \\ a_n=\begin{cases}\frac{1}{(n+1)!} & \operatorname{mod}(n,4)=0 \\ \frac{-1}{(n+1)!} & \operatorname{mod}(n,4)=2 \\ 0 & \text{otherwise}\end{cases}$$
Or better yet as
$$\frac{\sin x}{x}=\operatorname{sinc}x=\sum_{n=0}^\infty \frac{(-1)^{n+1}\sin(\pi n/2)}{(n+1)!}x^n \\ =\sum_{n=0}^\infty \frac{1}{n!}\frac{(-1)^{n+1}\sin(\pi n/2)}{n+1}x^n$$
Now that we have included all the terms, even the odd ones which are zero, this now precisely the Taylor series ($f(x)=\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(0)~x^n$), and so we can say
$$\operatorname{sinc}^{(m)}(0)=\frac{(-1)^{m+1}\sin(\pi m/2)}{m+1}$$
Now tell me, what happens when $m=2k$ is even or $m=2k+1$ is odd?
A: Start with
$\sin' = \cos
$,
$\cos' = -\sin
$,
$\sin(0) = 0$,
$\cos(0) = 1$,
and
$\sin^2+\cos^2 = 1$.
For small $t$,
$1 \ge \cos(t)
\ge 0
$
so
$\sin(x)
=\int_0^x \cos(t)dt
\le x
$.
Therefore
$1-\cos(x)
=\int_0^x \sin(t) dt
\le \int_0^x t dt 
= \frac{t^2}{2}
$
so
$\cos(t)
\ge 1-\frac{t^2}{2}
$.
Therefore
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$.
So we already have
$x-\frac{x^3}{6}
\le \sin(x)
\le x
$.
Doing this again,
$1-\cos(x)
=\int_0^x \sin(t) dt
\ge \int_0^x (t-\frac{t^3}{6}) dt 
= \frac{t^2}{2}-\frac{x^4}{24}
$
so
$\cos(t)
\le 1-\frac{x^2}{2}+\frac{x^4}{24}
$.
Doing this one more time,
$\sin(x)
=\int_0^x \cos(t)dt
\le\int_0^x (1-\frac{t^2}{2}+\frac{t^4}{24})dt
=x-\frac{x^3}{6}+\frac{x^5}{120}
$.
By induction,
we can get the power series
for $\sin$ and $\cos$.
Note:
This is not original.
I first saw this in
"100 Great Problems of Elementary Mathematics"
by Heinrich Dorrie
(less that $15 from Dover).
Get it.
