Compute $f^{(22)}(0)$ where $f(x)= \sin(x)/x$ if $x\neq0$ and $1$ if $x=0.$ Let
$$f(x)=
\begin{cases}
\frac{\sin(x)}x &\text{ if x}\neq0\\
1 &\text{ if x}=0.
\end{cases}
$$
What is $f^{(22)}(0)$?
First I found that
$$
f'(0)=\lim_{h \to 0}\frac{\sin(h)/h-1}{h}
=\lim_{h \to 0}\frac{\sin(h)-h}{h^{2}}
=\lim_{h \to 0}\frac{\cos(h)-1}{2h}
=0. 
$$
Similarly I found that $f''(0)=-1/3$
This makes me think that 
$$f^{(n)}(0)=
\begin{cases}
(-1)^{n/2}/(n+1) &\text{for n even}\\
0 &\text{for n odd}
\end{cases}
$$
So I tried using induction.  Letting $g(x)=sin(x)/x$,
I found that 
$$
g^{(N)}(x)=\sum \limits_{n={\lceil N/2 \rceil}}^\infty
\frac{(-1)^{n}(2n)(2n-1)...(2n-N+1)x^{2n-N}}
{(2n+1)!}
$$
(which may be incorrect).  I cannot prove the inductive step, so either my entire approach is wrong, my hypothesis is wrong, or I am making a mistake.  Please help!
 A: *

*Your computation appears correct, and you can use induction to do this whole thing, but it's not, as you found, pretty. 

*Here's a different approach, which uses some theorems about Taylor series and convergence and why/when it's OK to differentiate a Taylor series, etc., all of which I've mentioned only in passing:
For every $x$, the series
$$
x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \ldots
$$
converges to $\sin x$, by Taylor's theorem. 
That means that away from $0$, we have
$$
\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} \ldots\\
= \sum_{n = 0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}.
$$
The right hand side happens to also agree with your function (known as "sinc") at $x = 0$. So we have
$$
sinc(x) = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}.
$$
Since we can differentiate both sides (and wave the convergence wand to prove that the results are in fact equal) as often as needed, we can see that the $22$nd derivative of the left side is the 22nd derivative of the right; at $x = 0$, this derivative will be the constant term of the 22nd derivative on the right, i.e., the result of differentiating
$$
(-1)^{22} \frac{x^{22}}{23!} =  \frac{1}{23}\frac{x^{22}}{22!}
$$
twenty two times. The 22 derivatives will put a $22!$ in the numerator, and we'll be left with $\frac{1}{23}$. 
A: If you know some Fourier analysis, this is a typical problem that is solved nicely by the use of the Fourier transform: 
Let us use the convention $\hat{f}(\xi)=\int_{-\infty}^{+\infty} e^{-i\xi x}f(x)\,dx$ with inversion formula $f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{i\xi x}\hat{f}(\xi)\,d\xi$.
It is a standard result that, with your $f$,
$$
\hat{f}(\xi) = 
\begin{cases} 
\pi & |x|<1\\
0 & |x|>1
\end{cases}
$$
Moreover, it follows by integration by parts that the fouriertransform of $\frac{d^nf}{dx^n}$ at $\xi$ equals $(i\xi)^n\hat{f}(\xi)$.
With the inversion formula (applied to $(i\xi)^{22}\hat{f}(\xi)$) and at $x=0$, we find that
$$
\frac{d^{22}f}{dx^{22}}(0)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}(i\xi)^{22}\hat{f}(\xi)\,d\xi=-\frac{1}{2}\int_{-1}^{1}\xi^{22}\,d\xi=-\frac{1}{23}.
$$
A: Note that
$$f(x):={\sin x\over x}=\int_0^1\cos(t\,x)\>dt\qquad(x\in{\mathbb R})\ .$$
It follows that
$$f^{(22)}(x)=-\int_0^1 t^{22}\cos(t\,x)\>dt\ ,$$
which implies
$$f^{(22)}(0)=-{1\over23}\ .$$
