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I want to approximate the solution to the following sum, in such a way that I can plot the function for the variable $\phi$, for a fixed value of $\mu$.

\begin{equation} f(r(\phi), \mu)=4 e^{-\mu } r^2\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{ \left(\frac{r^2}{r^2+1}\right)^k \binom{n}{k} \mu ^{k+n} \left(\frac{1}{r^2+1}\right)^{n-k} \left(n-\mu\frac{ \, _1F_1\left(-n;n+2;-r^2 \mu \right)}{\, _1F_1\left(-n;n+1;-r^2 \mu \right)}\right)^2}{\Gamma (k+n+1)} \end{equation}

with $r(\phi)=\tan \phi$

So now I want to represent that expression as a function of $\phi$ for fixed $\mu$. I define the Function to be summed

F[μ_, Φ_] := 4*((μ^(k + n)*Binomial[n, k]*(Cos[Φ]^2)^(-k + n)*(-n + ((1 + n)*μ*n!
*Hypergeometric1F1[-n, 2 + n, (-μ)*Tan[Φ]^2])/
((1 + n)!*Hypergeometric1F1[-n, 1 + n, (-μ)*Tan[Φ]^2]))^2*(Sin[Φ]^2)^k)/(E^μ*(k + n)!))*Tan[Φ]^2; 

Now the following way to perform the summation will take indefinite time to compute even for a given value of $\phi$ and $\mu$ so I cannot use it to plot:

 (NSum[NSum[N[F[μ,Φ]],{k,0,∞}],{n,0,∞}])/.{μ->1,Φ->Pi/4}

Sometime ago I would just truncate the series in a value that I think it is appropiate for example $n=10$ and $k=10$:

((NSum[NSum[N[F[μ, Φ]], {k, 0, 10}], {n, 0, 
      10}]) Tan[Φ]^2) /. {μ -> 1, Φ -> Pi/4}
(*2.23421*)

Note that if you sum all the terms up to $n=20$ and $k=20$ the result is the same. If you try to sum up to $n=30$ and $k=30$ the computation start to become so lengthy. It is an option to do the plot using a summation up to 10. But this arbitrary truncation doesn't look very scientific and for some other values of $\phi$ the number of terms contributing to the sum could change. Do you have an alternative way to make this plot?

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    $\begingroup$ In your $\LaTeX$ expression, the $\rho$ parameter is unused and $\bar{n}$ is not defined. $\endgroup$ – Roman Apr 29 at 9:55
  • $\begingroup$ thank you, mate $\endgroup$ – Popeye Apr 29 at 9:57
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You can do the sum over $k$ analytically up to infinity, which simplifies things greatly:

G[μ_, Φ_, n_] = Assuming[Element[Φ, Reals], 
  Sum[F[μ, Φ], {k, 0, Infinity}] // FullSimplify]
(* analytic answer *)

Then do the $n$-sum numerically. Also, substitute the numerical values of μ and Φ before doing the numerical sum, not afterwards.

Here's a very efficient way of doing the sum until the result no longer changes (i.e. until machine-precision convergence has been reached):

S[μ_?NumericQ, Φ_?NumericQ] := 
  Module[{n = 0}, FixedPoint[# + G[N@μ, N@Φ, n++] &, 0]]

Try it out:

S[1, π/4]

2.23421


update: if analytic summing is impossible

For other situations, where analytic summation is not possible, we can still rewrite the doubly infinite sum into a singly infinite one:

$$ \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}F(n,k) = \sum_{m=0}^{\infty}\left[\sum_{i=0}^m F(i,m-i)\right] $$

This singly-infinite sum can then be done with the same FixedPoint trick as above:

F[μ_, Φ_, n_, k_] = 4*((μ^(k + n)*Binomial[n, k]*(Cos[Φ]^2)^(-k + n)*(-n + ((1 + n)*μ*n!
*Hypergeometric1F1[-n, 2 + n, (-μ)*Tan[Φ]^2])/
((1 + n)!*Hypergeometric1F1[-n, 1 + n, (-μ)*Tan[Φ]^2]))^2*(Sin[Φ]^2)^k)/(E^μ*(k + n)!))*Tan[Φ]^2; 

S[μ_?NumericQ, Φ_?NumericQ] := Module[{m = -1}, 
  FixedPoint[(m++; # + Sum[F[N@μ, N@Φ, i, m - i], {i, 0, m}]) &, 0]]

Try it out:

S[1, π/4]

2.23421

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  • $\begingroup$ Hello, I'm gonna try this. I have redone the calculations and It turns out I have made a mistake, so the actual function I have to sum is different, but I think I may use most of the things you have written. Thanks $\endgroup$ – Popeye Apr 29 at 16:58

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