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

Doing some exercises i found this function expressed by power series, someone recognize a friend? $$F_{n,m}(x)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\frac{x^{2(k+n+1)+1}}{2(k+n+1)+1}m^{2k+1}$$ It's possible to get out a different rappresentation for $m=0,1,2,3$ at least?

share|cite|improve this question
up vote 6 down vote accepted

Note that

$$\frac{d}{dx} F_{n,m}(x) = x^{2 n+2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} (m x)^{2 k+1} = x^{2 n+2} \sin{(m x)} $$

We may then integrate to find $F_{n,m}(x)$; when we do, we find that it satisfies a recurrence:

$$ F_{n,m}(x) = -\frac{1}{m} x^{2 n+2} \cos{m x} + \frac{2 n+2}{m^2} x^{2 n+1} \sin{m x} - \frac{(2 n+2) (2 n+1)}{m^2} F_{n-1,m}(x)$$

$$F_{0,m}(x) = \frac{\left(2-m^2 x^2\right) \cos (m x)+2 m x \sin (m x)-2}{m^3}$$

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.