Consider the power series $$\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$$

From this, it follows that its sum defines an infinitely differentiable function $f$, given by $$f(x):=\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$$

  • Determine whether $f$ is even or odd.

To do this, do I simply show that $f(x)=f(-x)$ and hence the function is even or is there more to the justification than that?

  • Is there a simple relationship between $f''$ and $f$?

I found that the series can be expressed as the function $\cosh x$. The power series of the first derivative is or $\sinh x$ or $\sum_{n=1}^\infty{\frac{x^{2n+1}}{(2n+1)!}}$ and the second derivative is $\sum_{n=0}^\infty{\frac{x^{2n}}{(2n)!}}$, but I'm not sure what is meant by 'a simple relationship between $f''$ and $f$.

  • $\begingroup$ does there can be something simpler than $f = f''$? :D $\endgroup$ Commented Dec 10, 2012 at 10:34

1 Answer 1


$f(x)=f(-x)$ is the definition of even function, so that is enough.

$$ f''(x) = \sum_{n=2} \frac{x^{2n-2} 2n\cdot(2n-1)}{(2n)!} = \sum_{n=2} \frac{x^{2n-2} }{(2n-2)!} = \sum_{k=0} \frac{x^{2k} }{(2k)!} = f(x) $$

  • $\begingroup$ Great, thanks for clearing that up. Would it please be possible to explain how the second proof shows there's a simple relationship? I'm a little unclear as to what the question means. $\endgroup$
    – user51462
    Commented Dec 10, 2012 at 10:51
  • $\begingroup$ @user51462: $f\,''=f$ is about as simple a relationship as you can get! $\endgroup$ Commented Dec 10, 2012 at 10:53
  • $\begingroup$ simple means 'can be expressed'. many functions have relations between function and its derivative (see for instance properties of Bessel function). $\endgroup$
    – 0x2207
    Commented Dec 10, 2012 at 12:12

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