3
$\begingroup$

Let's consider the series: $$ F(t) = \sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{\infty} \frac{(-b)^k(-a)^n\binom{n+k}{k} t^{2n+k(2-\alpha)}}{\Gamma(2n+k(2-\alpha)+2)} $$

where $a,b$ are positive reals, $0 < \alpha < 1$, and $\Gamma(z)$ is the gamma function.

I would like to numerically compute the sum $F$. I replace the doubly infinite sum by a double finite sum in the following way:

$$ \sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{\infty}\left(\ldots\right) \approx \sum\limits_{i=0}^{N}\sum\limits_{{n+k = i}\atop{0\leq n,k\leq i}} \left(\ldots\right) \ \ \ (\star)$$

where bound $N$ is set in such a way, that $\left|\frac{S_{N+1}-S_{N}}{S_{N+1}}\right|<\varepsilon$ ($S_j$ - sum from right hand path of the ($\star$) for $N=j$; $\varepsilon$ - some small number, for example $\varepsilon = 10^{-6}$).

Unfortunately there is something wrong with my approach. For example for $a=1$, $b=1$, $\alpha=0.5$ and $0\leq t \leq 30$ I get following graph:

Graph of approximation of the F(t) for 0\leq t \leq 30

Definitely, there is something wrong for $t\geq 25$.

My question: How to compute approximation of $F(t)$ correctly also for $t\geq 25$?

Code (in this way I define function $F(t)$):

F[t_, α_] := Block[{convergence, N1, S, SN, tab, re, i},
  N1 = 0;
  S = 1;
  convergence = False;
  While[! convergence && N1 <= 1000,
   N1++;
   (* ponizej k = i, n = N1-i *)
   tab = Table[((-v0)^i (-w0^2)^(N1 - i) Binomial[N1, i] t^(
      2 (N1 - i) + i (2 - α)))/
     Gamma[2 (N1 - i) + i (2 - α) + 2], {i, 0, N1}];
   SN = S + Sum[tab[[i]], {i, 1, Length[tab]}];
   re = Abs[(S - SN)/S];
   If[re < 10^-6, convergence = True];
   S = SN;
   ];
  Return[{S, N1, re}];
  ]
$\endgroup$
2
  • $\begingroup$ Sorry for such strange formating but I could not add this post due to error "Your post appears to contain code that is not properly formatted as code. Please indent all code by 4 spaces using the code toolbar button or the CTRL+K keyboard shortcut. For more editing help, click the [?] toolbar icon." (it was some problem with LaTeX code) $\endgroup$
    – azonips
    Sep 24, 2018 at 10:54
  • $\begingroup$ You can probably do one summation symbolically, then your problem should be much easier. $\endgroup$ Sep 24, 2018 at 12:03

1 Answer 1

3
$\begingroup$

This is a very rough solution I just hacked out; I might try to refine it later.

I had previously talked about Cuyt's transformation of a double series as an aid in evaluating them numerically in this answer. Applied to this case, we have

Plot[With[{a = 1, b = 1, t = SetPrecision[tx, 25], α = 1/2}, 
          NumericalMath`NSequenceLimit[Accumulate[Table[Sum[
          Function[{n, k}, (-a)^n (-b)^k Binomial[n + k, k] t^(2 n + k (2 - α))/
          Gamma[2 n + k (2 - α) + 2]] @@ v,
          {v, FrobeniusSolve[{1, 1}, k]}], {k, 0, 29}]]]], {tx, 0, 30}, 
     MaxRecursion -> 0, PlotPoints -> 75, PlotRange -> All]

plot

where I use NumericalMath`NSequenceLimit[] for performing the Shanks transformation on the result of Cuyt's transformation.

The method is a bit finicky numerically, which is why I needed the use of SetPrecision[]. I also very arbitrarily chose the cutoff 29 in Cuyt's transformation; for other parameter choices, less or more terms might be necessary. All of this will need to be experimented with.

$\endgroup$
3
  • $\begingroup$ Thank you for very nice solution of my problem (unfortunately I use version <11.2 of Mathematica, so function NumericalMath`NSequenceLimit[] in unavailable for me) $\endgroup$
    – azonips
    Sep 24, 2018 at 20:23
  • 1
    $\begingroup$ Replace that with SequenceLimit[], then. $\endgroup$ Sep 24, 2018 at 20:29
  • $\begingroup$ It works, thank you :) $\endgroup$
    – azonips
    Sep 26, 2018 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.