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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?

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Wouldn't that imply that 0 + -1 = 1? –  Rotwang Dec 9 '10 at 21:37
    
Could you tell us a little about your motivation? –  Mike Spivey Dec 9 '10 at 21:38
    
There was a typo, sorry. in the last equation there should be 1 –  Anixx Dec 9 '10 at 21:40
    
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. –  Rotwang Dec 9 '10 at 21:44
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Is that your final question? –  Aryabhata Dec 9 '10 at 21:51
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2 Answers 2

You are looking for a function $\displaystyle f(z) = \sum_{k \ge 0} \frac{c(k)}{k!} z^k$ satisfying

$$f(1) = 1$$ $$f(-1) = -1$$ $$f(i) = -i.$$

Infinitely many functions have this property. There is a unique quadratic polynomial $p(z)$ with this property (for example by Lagrange interpolation), and for any entire function $q(z)$ the function $p(z) + (x - 1)(x + 1)(x - i) q(z)$ has this property. In fact these are all entire functions with this property.

More generally I think the theory of interpolation by entire functions is fairly well-understood, but I don't know of a good reference. If the set of $z$ at which you fix the value of $f$ has a limit point, then $f$ is unique by the identity theorem. If the set of $z$ at which you fix the value of $f$ is countable and does not have a limit point, then $f$ is non-unique by Weierstrass factorization.

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Following Rotwang, define $b(k)=c(k)/k!$, then note that both the second and fifth can be satisfied as long as $\sum{b(2k)}=0$. Then the first and third are redundant and define $d(k)=b(2k+1)$. Now we have $\sum{d(k)}=1, \sum{(-1)^kd(k)}=-1$. Adding and subtracting, $\sum{d(2k)}=0, \sum{d(2k+1)}=1.$ So the final is that we must have $$\begin{align} \sum\frac{c(2k)}{k!}&=0\\ \sum\frac{c(4k+1)}{(4k+1)!}&=0\\ \sum\frac{c(4k+3)}{(4k+3)!}&=1 \end{align}$$

and any $c(k)$ that satisfies this will work.

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@Mike Spivey, Rotwang: Fixed. Sorry for that, Rotwang. –  Ross Millikan Dec 9 '10 at 22:20
    
Thank you. Seems the system is too loose. –  Anixx Dec 9 '10 at 22:22
    
@MathFacts: Yes. You can think of it as 3 equations and an infinite number of variables. –  Ross Millikan Dec 9 '10 at 23:08
    
@MathFacts: Making the last more explicit, there is nothing in the problem statement relating the c(k) to each other for different k. –  Ross Millikan Dec 9 '10 at 23:21
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