While working on a homework problem, I am hitting a point where I can kind of see where I am headed, but I'm not sure how to resolve a couple of items.

I am currently sitting at the following expression:

$$\frac{1}{2\sqrt{2\pi y}} e^{\frac{-\mu^2}{2}}e^{\frac{-y^2}{2}}(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$$

and need to get to:

$$\sum_{k = 0}^\infty \frac{(e^{\frac{-\mu^2}{2}})(\frac{\mu^2}{2})^k}{k!} * \frac{1}{2^{\frac{2k + 1}{2}}\gamma(\frac{2k + 1}{2})}y^{\frac{2k + 1}{2}-1}e^{\frac{-y}{2}}$$ The second part is a Gamma distribution $(\frac{2k+1}{2}, 2)$

I can see that the first two listed exponentials are likely to move to the gamma distribution, but is there some identify that will allow me to take $(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$ and change it in order to cut out the even $k$ values of the infinite sum?

  • 2
    $\begingroup$ Equation? I see no equation...where is the "$=$"? And yes, I'm being intentionally obtuse. Learning the basic terminology is fundamental to seeing what the problem is. I think you mean "expression". $\endgroup$
    – MPW
    Sep 2, 2016 at 2:24
  • $\begingroup$ @MPW Thank you, I've corrected that mistake. $\endgroup$
    – Aggie Kidd
    Sep 2, 2016 at 2:38
  • $\begingroup$ How does $y$ become $x$? And $u=\mu$? Or not? $\endgroup$
    – Did
    Sep 2, 2016 at 6:48

1 Answer 1


For the $(e^{\mu\sqrt{y}} + e^{-\mu\sqrt{y}})$ question, note that this is essentially the $\cosh$ function defined by $\cosh(x) =\frac12(e^x + e^{-x}) $. This cancels out the odd terms, because $\dfrac{x^{2k+1}}{(2k+1)!} +\dfrac{(-x)^{2k+1}}{(2k+1)!} =0 $.

Similarly, to cancel out the even terms, use the $\sinh$ function defined by $\sinh(x) =\frac12(e^x - e^{-x}) $. This cancels out the even terms, because $\dfrac{x^{2k}}{(2k)!} -\dfrac{(-x)^{2k}}{(2k)!} =0 $.


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