1
$\begingroup$

I need to find the function c(k), knowing that

$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$

$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$

$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$

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

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

Is it possible?

  • 0
    Wouldn't that imply that 0 + -1 = 1?2010-12-09
  • 0
    Could you tell us a little about your motivation?2010-12-09
  • 0
    There was a typo, sorry. in the last equation there should be 12010-12-09
  • 0
    The $k!$ seems to be a complete red herring - define $b(k) \equiv c(k)/k!$ and the equivalent question is more elegant IMO. Anyway, there are clearly an infinite number of functions that will work (e.g. $c(k_0) = k_{(0)}!$ for any odd $k_0$, with all other $c(k)$'s equal to $0$). e: obviously this no longer applies since your edit.2010-12-09
  • 0
    I have added some equations and it seems that the first equation is not necessary because it follows from the second and third.2010-12-09
  • 2
    Is that your final question?2010-12-09
  • 0
    Yes, if these equations are enough to find the function in question. I believe they are enough.2010-12-09
  • 0
    @Math: No they aren't enough. There are still infinitely many solutions.2010-12-09
  • 0
    No, I believe that there is still an infinite number of possible solutions. E.g. $c(2k_0 + 1) = (2k_0 + 1)!$ for some odd $k_0$, with all other $c(k)$ equal to zero.2010-12-09
  • 0
    @Rotwang: Unless inconsistent, given a finite number of equations like those, there will always be an infinitely many number of solutions, by setting the "tail" of the series to 0 and getting a set of linear equations which have more variables than equations.2010-12-09
  • 0
    @Moron: Indeed.2010-12-09
  • 0
    Well obviously this is not enough, this system is satisfied by c(1)=1, otherwise 0 function. I have to add more equations.2010-12-09

2 Answers 2