Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $n$ be an arbitrary integer. Define: $$\begin{align} c_0 &= 1;\\ c_m &= \frac1m \sum_{k = 1}^m (k n - m + k) \frac{(-1)^k}{(2k)!} c_{m - k}. \end{align}$$

This recursion turns up in my quest of computing integrals of functions of Bessel functions.

Instead of solving this recursion, I'm also satisfied with the power series of $\cos^n \alpha$.

I have tried solving that recursion using generating functions and then find some way to see this as the product of two series, however the $m^{-1}$ messes that up!

Any suggestions?

share|cite|improve this question
up vote 4 down vote accepted

The cosine power series can be expanded using

$$\left(\sum_{\ell=0}^\infty \frac{(-1)^\ell}{(2\ell)!}a^{2\ell}\right)^m=\sum_{\lambda_1,\cdots,\lambda_m=0}^\infty \frac{(-1)^{\lambda_1+\cdots+\lambda_m}}{(2\lambda_1)!\cdots(2\lambda_m)!}a^{2\lambda_1+\cdots+2\lambda_m}$$

$$=\sum_{n=0}^\infty \left(\frac{(-1)^{n}}{(2n)!}\sum_{|\lambda|=n}\binom{2n}{2\lambda_1,\cdots,2\lambda_m}\right)a^{2n}.$$

Note we sum over all nonnegative $\lambda_1,\cdots,\lambda_m$ that sum to $n$ within the inner sum, and rewrite the negative-one power using this, as well as both multiply and divide by $(2n)!$ so we can use a multinomial (because I feel like using that would make things look nice, I guess). Ultimately this boils down to collecting all terms that are in front of an $a^{2n}$ power into one sum. This may or may not be in a useful form for you, I don't know.

Note that we can evaluate the inner sum using the multinomial theorem and symmetry:

$$\sum_{|\lambda|=n}\binom{2n}{2\lambda_1,\cdots,2\lambda_m}=\frac{1}{2^m}\sum_{x_i=\pm1} (x_1+\cdots+x_m)^{2n}$$

Here the sum is over $x_i$'s being plus or minus one, independently. We can collect all terms with $v$ negative-ones and rewrite this as

$$\frac{1}{2^m}\sum_{v=0}^m \binom{m}{v}(m-2v)^{2n}. $$

Multiply this by $(-1)^n/(2n)!$ and you have the coefficient of $a^{2n}$ in $\cos^ma$. 8-)

share|cite|improve this answer
Where is the $m$ in the final expression? A useful form is something I can do easily using the computer (I need to sum over $m$), if that $\lambda_n$ needs to be $\lambda_m$, I need to add a variable each iteration and that isn't so nice. – Jonas Teuwen Mar 10 '12 at 22:21
@Jonas: That was indeed a typo I just fixed. I don't think this form will be that useful for computation. – anon Mar 10 '12 at 22:23
I believe that you mean $(2n)!$ in the last line. – Jonas Teuwen Mar 11 '12 at 18:24
@Jonas: Yes, fixed. – anon Mar 12 '12 at 0:50

The power series of $\cos^n \alpha$ can be found in the following article, in the guise of the power series of $\cosh^n \alpha$. Since $\cosh \alpha = \cos i\alpha$, the two are the same up to sign.

I found this by consulting the OEIS.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.