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Plot the first terms of Taylor series of the below function?

Rafik zeraoulia

Rafik zeraoulia, University Batna2

Posted 4 years ago

I have tried to plot the first term of Taylor expansion of the below function but I didn't come up to the plot . Any help, Where is the problem in my code ? The Function is : $$I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}({(\sqrt{2\pi})t))}\operatorname {erf}({(\sqrt{2\pi})t)}) dt$$ The same Copy of that question is mentioned [here] , But the given code doesn't work at me in Wolfram cloud page, any help ? My Code : Clear[\[Lambda], Ze, Z, ZTaylor]; \[Lambda] = NIntegrate[Exp[-t^2* Erf[(Sqrt[2Pi])t]* Erfi[(Sqrt[2Pi])t]],{t,-Infinity,Infinity}]; Ze[a_] := NIntegrate[Exp[-t^2* Erf[(Sqrt[2Pi])t]* Erfi[(Sqrt[2Pi])t]],{t,-Infinity,a}]; Z[x_] := Integrate[Exp[-t^2* Erf[(Sqrt[2Pi])t]* Erfi[(Sqrt[2Pi])t]], {t, 0, x}]; ZTaylor[n_][x_] := Series[Z[x], {x, 0, n}]; TaylorSeries == ZTaylor[11][x] //TraditionalForm Plot[Evaluate[{Normal[ZTaylor[11][x]], Ze[x]}], With[{x = 1}, Plot[Z[x] , {x, (x-(8x^5/5)), ( -32/405 (-45+4Pi^2)x^9)}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400, ]]]

I have tried to plot the first term of Taylor expansion of the below function but I didn't come up to the plot . Any help, Where is the problem in my code ?
The Function is : $$I(x)=\int_{-\infty}^{x} \exp(-t^2 \operatorname {erfi}({(\sqrt{2\pi})t))}\operatorname {erf}({(\sqrt{2\pi})t)}) dt$$ The same Copy of that question is mentioned [here] , But the given code doesn't work at me in Wolfram cloud page, any help ? My Code :

Clear[\[Lambda], Ze, Z, ZTaylor];
        \[Lambda] = NIntegrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]],{t,-Infinity,Infinity}];
        Ze[a_] := NIntegrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]],{t,-Infinity,a}];
        Z[x_] := Integrate[Exp[-t^2* Erf[(Sqrt[2*Pi])*t]* Erfi[(Sqrt[2*Pi])*t]], {t, 0, x}];
        ZTaylor[n_][x_] := Series[Z[x], {x, 0, n}];
        TaylorSeries == ZTaylor[11][x] //TraditionalForm
        Plot[Evaluate[{Normal[ZTaylor[11][x]], Ze[x]}],
        With[{x = 1}, 
         Plot[Z[x] , {x, (x-(8x^5/5)), (
          -32/405 (-45+4Pi^2)x^9)}, {x, -1, 2}, 
          PlotRange -> {0, 1.5}, ImageSize -> 400, 


        ]]]

POSTED BY: Rafik zeraoulia

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Claudio Chaib

Claudio Chaib, Consultant at Wolfram Research Inc.

Posted 4 years ago

Hi Rafik, are you looking for something like this? Clear[\[Lambda], Ze, Z, ZTaylor]; \[Lambda] = NIntegrate[ Exp[-t^2Erf[(Sqrt[2Pi])t]Erfi[(Sqrt[2Pi])t]], {t, -Infinity, Infinity}]; Ze[a_] := NIntegrate[ Exp[-t^2Erf[(Sqrt[2Pi])t]Erfi[(Sqrt[2Pi])t]], {t, -Infinity, a}]; Z[x_] := Integrate[ Exp[-t^2Erf[(Sqrt[2Pi])t]Erfi[(Sqrt[2Pi])t]], {t, 0, x}]; ZTaylor[n_][x_] := Series[Z[x], {x, 0, n}]; TaylorSeries == ZTaylor[11][x] // TraditionalForm Ze[1] Plot[{Ze[1], Z[x], {x, (x - (8 x^5/5)), (-32/405 (-45 + 4 Pi^2) x^9)}}, {x, -1, 2}, PlotRange -> {0, 1.5}, ImageSize -> 400]

Hi Rafik, are you looking for something like this?

Clear[\[Lambda], Ze, Z, ZTaylor];
\[Lambda] = 
  NIntegrate[
   Exp[-t^2*Erf[(Sqrt[2*Pi])*t]*Erfi[(Sqrt[2*Pi])*t]], {t, -Infinity, 
    Infinity}];
Ze[a_] := 
  NIntegrate[
   Exp[-t^2*Erf[(Sqrt[2*Pi])*t]*Erfi[(Sqrt[2*Pi])*t]], {t, -Infinity, 
    a}];
Z[x_] := Integrate[
   Exp[-t^2*Erf[(Sqrt[2*Pi])*t]*Erfi[(Sqrt[2*Pi])*t]], {t, 0, x}];
ZTaylor[n_][x_] := Series[Z[x], {x, 0, n}];
TaylorSeries == ZTaylor[11][x] // TraditionalForm
Ze[1]
Plot[{Ze[1], 
  Z[x], {x, (x - (8 x^5/5)), (-32/405 (-45 + 4 Pi^2) x^9)}}, {x, -1, 
  2}, PlotRange -> {0, 1.5}, ImageSize -> 400]

im1

POSTED BY: Claudio Chaib

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 4 years ago

This gives an answer, which may or may not be what you have in mind: i[x_] = Integrate[Exp[-t^2Erf[(Sqrt[2Pi])t] Erfi[(Sqrt[2Pi])t]], {t, -Infinity, x}] Series[i[x], {x, 0, 1}] // Normal // N Plot[%, {x, -1, 1}]

POSTED BY: Gianluca Gorni

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