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I have the following summation I would like to simplify: $f(y)=\frac{1}{\left(2^{N}-1-N\right) \sqrt{2 \pi}} \sum_{k=2}^{N} \frac{\left(\begin{array}{l} N \\ k \end{array}\right)}{\sqrt{k \sigma^{2}+1}} e^{-0.5 y^{2} /\left(k \sigma^{2}+1\right)}$ $N>>1$ is a positive integer.

The Mathematica code I tried is below:

f=Sum[Binomial[N,k]Exp[(-0.5y^2)/(k *(sigma^2)+1)]/(Sqrt[k *(sigma^2)+1]), {k, 2, N}]
g=FullSimplify[f]

The output just displays what I have already written. It is not doing any simplification. Can someone tell me if this can be simplified using Mathematica? If yes, what should be the appropriate command?

NB: I am totally new to mathematica and hence asking.

wanderer
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    Please at least look up how to spell Exp and Sqrt, and make your code actually match the formula (using parentheses). Also, N is a reserved symbol in Mathematica, don't use it as a variable. – Roman Jun 13 '21 at 15:34
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    Post code, not pictures, please.

    https://mathematica.stackexchange.com/help/formatting

     – rasher Jun 13 '21 at 07:55
  • Is there a relationship between $x$ and $y$, or are they independent variables? – Roman Jun 13 '21 at 07:55
  • @Roman, $x$ is not related to $y$. – wanderer Jun 13 '21 at 08:01
  • So why exactly did you write $k=x/\sigma^2$? Where does $x$ come into play? – Roman Jun 13 '21 at 08:45
  • @Roman, Actually that's just an extra info. Not needed for computation. I will edit the question. – wanderer Jun 13 '21 at 10:12
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  • Please add in Mathematica format the code you have tried, Showing that you have made some effort will encourage readers to help. – bbgodfrey Jun 13 '21 at 14:02
  • @bbgodfrey I have made the edits by adding the Mathematica code I tried. – wanderer Jun 13 '21 at 15:12
  • @Roman, Thanks for pointing out. I will edit. – wanderer Jun 13 '21 at 15:38
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    I suspect that there is no simplification and you might be better off looking for an approximation. Note that for a fixed $y$ and $\sigma$ a plot of $\log f(y, N, \sigma)$ vs $N$ (with $N$ appropriately represented by n in Mathematica) looks very linear with a slope dependent only on $\sigma$. – JimB Jun 13 '21 at 16:13
  • @JimB, Thank you! One follow-up question: I am interested in an approximation to the summation which is in the form of a Gaussian function $e^{(y-a)^{2}/b}$ form. Is there a way to do that in Mathematica? – wanderer Jun 13 '21 at 16:31
  • That would mean that the log of the summation would have a form of $(y-a)^2/b$, so either LinearModelFit or NonlinearModelFit should be helpful. But you might want to write that up as a separate question (along with any attempts you've made with Mathematica). – JimB Jun 13 '21 at 16:35
  • In general, one thing you can do is add assumptions to FullSimplify, which amount to e.g. telling Mathematica that n is a positive integer, e.g. g = FullSimplify[f, n \[Element] Integers && n >2 && y \[Element] Reals && sigma \[Element] Reals && 1 + k sigma^2 > 0]. Unfortunately, that doesn't seem to help here, but just so you know the strategy! – thorimur Jun 14 '21 at 19:01
  • Also note that machine numbers like 0.5 will be treated differently than "exact" numbers like 1/2. When not doing numerics, it's important to avoid decimals! – thorimur Jun 14 '21 at 19:03

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