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Infinite series works only to order k=20

MEK MUS

MEK MUS, Physicist, Oran University, Algeria

Posted 2 years ago

Hi Here is a function g defined by an infinite series which works only for k<= 20. The series is convergent as we check it by another mean Needs["PlotLegends`"]; g[j_] := (1/2)^0.5Sin[Pi0.5]/Pi* NSum[Gamma[-k + 0.5, 0, 10]/k!*(j/2)^(2 k), {k, 0, 30}]; Omega = Table[g[j], {j, 0.1, 20, 0.1}]; ListLinePlot[{Omega}, PlotStyle -> Thick, PlotRange -> {-2, 2}]

Hi Here is a function g defined by an infinite series which works only for k<= 20. The series is convergent as we check it by another mean

Needs["PlotLegends`"];
g[j_] :=  (1/2)^0.5*Sin[Pi*0.5]/Pi*     
   NSum[Gamma[-k + 0.5, 0, 10]/k!*(j/2)^(2 k), {k, 0, 30}];
Omega = Table[g[j], {j, 0.1, 20, 0.1}];
ListLinePlot[{Omega}, PlotStyle -> Thick, PlotRange -> {-2, 2}]

POSTED BY: MEK MUS

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Updating Name

Posted 2 years ago

Thank you M.I for your instantaneous help. I appreciate the the way you compute the second series. I however have some remarks about it. First the sequence you get L[k] from the first 10 terms may not be unique so one has to check the equality of the sequence with the original one -2j^(2k) Gamma[-2k]Sin[kPi]/Sqrt[Pi] Second the sequence you get start from zero Table[L[k],{k,0,10}]=0,Sqrt[[Pi]], -(1/2) j^2 Sqrt[[Pi]],...... I don't know why but that will not impact the final result you get . To prove the equality one has to make the shift k->k+1 first which result in the sequence (-1)^kj^(2k)Sqrt(Pi)/Gamma(1+2k) then apply the identity Gamma[a]Gamma[1-a]=Pi/Sin[Pi*a] to get the equality ( I do it analytically because I' not not good in mathematica)

POSTED BY: Updating Name

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 2 years ago

$\sum _{k=0}^{\infty } -\frac{2^{-2 k} j^{2 k} \Gamma \left(\frac{1}{2}-k,10\right)}{\Gamma (1+k)}+\sum _{k=0}^{\infty } -\frac{2 j^{2 k} \Gamma (-2 k) \sin (k \pi )}{\sqrt{\pi }}$ Sum[#, {k, 0, Infinity}] & /@ (Gamma[-k + 1/2, 0, 10]/k!(j/2)^(2 k) // FunctionExpand) (Sum[-((j^(2k)Gamma[1/2 - k, 10])/(2^(2k)Gamma[1 + k])), {k, 0, Infinity}] + Sum[(-2j^(2k)Gamma[-2k]Sin[kPi])/Sqrt[Pi], {k, 0, Infinity}]) First series we can compute from identity $\Gamma (a,z)=\int_z^{\infty } t^{a-1} \exp (-t) \, dt$ Integrate[ Sum[-((2^(-2 k) j^(2 k) (t^(a - 1) Exp[-t] /. a -> 1/2 - k))/ Gamma[1 + k]), {k, 0, Infinity}], {t, 10, Infinity}, Assumptions -> j >= 0] (-(1/2) E^(-I j) Sqrt[\[Pi]] (Erfc[(20 - I j)/(2 Sqrt[10])] + E^(2 I j) Erfc[(20 + I j)/(2 Sqrt[10])])) Second series we can compute like this: L = Table[ Limit[-((2 j^(2 k) Gamma[-2 k] Sin[k \[Pi]])/Sqrt[\[Pi]]), k -> m, Assumptions -> j >= 0], {m, 0, 10}] Sum[FindSequenceFunction[L, k] // FunctionExpand, {k, 0, Infinity}] (Sqrt[\[Pi]] Cos[j]*) Regards M.I.

$\sum _{k=0}^{\infty } -\frac{2^{-2 k} j^{2 k} \Gamma \left(\frac{1}{2}-k,10\right)}{\Gamma (1+k)}+\sum _{k=0}^{\infty } -\frac{2 j^{2 k} \Gamma (-2 k) \sin (k \pi )}{\sqrt{\pi }}$

Sum[#, {k, 0, Infinity}] & /@ (Gamma[-k + 1/2, 0, 10]/k!*(j/2)^(2 k) //
    FunctionExpand)
 (*Sum[-((j^(2*k)*Gamma[1/2 - k, 10])/(2^(2*k)*Gamma[1 + k])), {k, 0, Infinity}] + 
  Sum[(-2*j^(2*k)*Gamma[-2*k]*Sin[k*Pi])/Sqrt[Pi], {k, 0, Infinity}]*)

First series we can compute from identity $\Gamma (a,z)=\int_z^{\infty } t^{a-1} \exp (-t) \, dt$

Integrate[
 Sum[-((2^(-2 k) j^(2 k) (t^(a - 1) Exp[-t] /. a -> 1/2 - k))/
   Gamma[1 + k]), {k, 0, Infinity}], {t, 10, Infinity}, Assumptions -> j >= 0]

(*-(1/2) E^(-I j) Sqrt[\[Pi]] (Erfc[(20 - I j)/(2 Sqrt[10])] + 
   E^(2 I j) Erfc[(20 + I j)/(2 Sqrt[10])])*)

Second series we can compute like this:

L = Table[
  Limit[-((2 j^(2 k) Gamma[-2 k] Sin[k \[Pi]])/Sqrt[\[Pi]]), k -> m, 
   Assumptions -> j >= 0], {m, 0, 10}]
Sum[FindSequenceFunction[L, k] // FunctionExpand, {k, 0, Infinity}]

(*Sqrt[\[Pi]] Cos[j]*)

Regards M.I.

POSTED BY: Mariusz Iwaniuk

MEK MUS

MEK MUS, Physicist, Oran University, Algeria

Posted 2 years ago

It's about the sum you give at the beginning : Looks like ...... If I ask Mathematica to perform the sum naively that is if I write Sum[Gamma[-k - 1/2, 0, 10]/k!*(j/2)^(2 k), {k, 0, [Infinity]}] I get no answer . My question is how did you get the above result: -1/2 exp[-ij]......by using Mathematica software then how or have you copied it from some textbook

POSTED BY: MEK MUS

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 2 years ago

What do you mean by Will you please give some hints about how you proceeded ?

POSTED BY: Mariusz Iwaniuk

MEK MUS

MEK MUS, Physicist, Oran University, Algeria

Posted 2 years ago

Thanks for the answer I understand that the terms of the series k>=50 have insignificant contributions to the sum. On the other hand I didn't succeed in reproducing the series giving at the beginning for fixed j. Will you please give some hints about how you proceeded Thanks again

POSTED BY: MEK MUS

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 2 years ago

Looks like: $\sum _{k=0}^{\infty } \frac{\Gamma \left(-k+\frac{1}{2},0,10\right) \left(\frac{j}{2}\right)^{2 k}}{k!}=-\frac{1}{2} e^{-i j} \sqrt{\pi } \left(\text{erfc}\left(\frac{20-i j}{2 \sqrt{10}}\right)+e^{2 i j} \text{erfc}\left(\frac{20+i j}{2 \sqrt{10}}\right)\right)+\sqrt{\pi } \cos (j)$ g[j_] := (1/2)^(1/2)Sin[Pi(1/2)]/Pi* Sum[Gamma[-k + 1/2, 0, 10]/k!(j/2)^(2 k), {k, 0, 50}]; Omega = Table[g[j], {j, 1/10, 20, 2/10}]; L = ListLinePlot[{Omega}, PlotStyle -> Thick, PlotRange -> {-2, 2}] g1[j_] := (1/2)^(1/2) Sin[Pi(1/2)]/ Pi(-(1/2) E^(-I j) Sqrt[\[Pi]] (Erfc[(20 - I j)/(2 Sqrt[10])] + E^(2 I j) Erfc[(20 + I j)/(2 Sqrt[10])]) + Sqrt[\[Pi]]*Cos[j]); Omega1 = Table[g1[j], {j, 1/10, 20, 2/10}]; L2 = ListLinePlot[{Omega1}, PlotStyle -> {Red, Thick}, PlotRange -> {-2, 2}] L3 = ListLinePlot[{Omega - Omega1}, PlotStyle -> {Red, Thick}]

Looks like: $\sum _{k=0}^{\infty } \frac{\Gamma \left(-k+\frac{1}{2},0,10\right) \left(\frac{j}{2}\right)^{2 k}}{k!}=-\frac{1}{2} e^{-i j} \sqrt{\pi } \left(\text{erfc}\left(\frac{20-i j}{2 \sqrt{10}}\right)+e^{2 i j} \text{erfc}\left(\frac{20+i j}{2 \sqrt{10}}\right)\right)+\sqrt{\pi } \cos (j)$

g[j_] := (1/2)^(1/2)*Sin[Pi*(1/2)]/Pi*
   Sum[Gamma[-k + 1/2, 0, 10]/k!*(j/2)^(2 k), {k, 0, 50}];
Omega = Table[g[j], {j, 1/10, 20, 2/10}];
L = ListLinePlot[{Omega}, PlotStyle -> Thick, PlotRange -> {-2, 2}]

g1[j_] := (1/2)^(1/2)*
  Sin[Pi*(1/2)]/
   Pi*(-(1/2) E^(-I j)
      Sqrt[\[Pi]] (Erfc[(20 - I j)/(2 Sqrt[10])] + 
       E^(2 I j) Erfc[(20 + I j)/(2 Sqrt[10])]) + 
    Sqrt[\[Pi]]*Cos[j]); Omega1 = Table[g1[j], {j, 1/10, 20, 2/10}];
L2 = ListLinePlot[{Omega1}, PlotStyle -> {Red, Thick}, 
  PlotRange -> {-2, 2}]

L3 = ListLinePlot[{Omega - Omega1}, PlotStyle -> {Red, Thick}]

POSTED BY: Mariusz Iwaniuk

MEK MUS

MEK MUS, Physicist, Oran University, Algeria

Posted 2 years ago

Hi Please execute first Needs["PlotLegends`"]; g[j_] := (1/2)^0.5Sin[Pi0.5]/Pi* NSum[Gamma[-k + 0.5, 0, 10]/k!*(j/2)^(2 k), {k, 0, 30}]; Omega = Table[g[j], {j, 0.1, 20, 0.1}]; ListLinePlot[{Omega}, PlotStyle -> Thick, PlotRange -> {-2, 2}] to see what's going on. Note that you made a print mistake it's (j/2)^(2 k and not (k/2)^(2 k. Now I said convergence because I plotted the same function but written in terms of reduced Bessel function and found no errors. The aim is to make the plot with k=0 to infinity. Thanks.

POSTED BY: MEK MUS

Shenghui Yang

Shenghui Yang, WOLFRAM

Posted 2 years ago

Hi, Could you please share the proof of the convergence? I did quick asymptotic analysis on the last term of your sum and it appears the absolute value grows exponential. Both the asymptotic value and direct computation show the same trend.

POSTED BY: Shenghui Yang

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