Skip to main content

plotting - Plot Indefinite integral


I'm having a hard time to find a solution to a numerical limitation on mathematica:


PwD = 0.005*Integrate[(Exp[-(0.25/\[Tau])]/\[Tau])*
  Integrate[Exp[-((Tan[45*Degree]^2 - (z + 30)^2)/(4*\[Tau]))]*
         (1 + 
      2*Sum[Exp[-((n^2*Pi^2*\[Tau])/10000)]*Cos[0.8*n*Pi]*
         Cos[(n*Pi)/2 - (n*Pi*z)/100], {n, 1, 4}]), {z, -50, 
    50}], {\[Tau], 0, t}]


Mathematica graphics


After solving the integral in space {z,-50,50} I need to evaluate the indefinite integral in time {tau,0,t}, so I eventually generate a plot of PwD as function of Log[t]. But it seems that it does not converge.


Can someone provide an alternative way?



Hi everybody thanks for the comments... They sound about right! In the past 2 months I dedicated my time to derive these solutions again - once I had no help from the author. Well, I finally get it! The published expression has a type mistake... The right equation should be:


PwD = (1/200)*Integrate[
    (1/(E^(0.25/\[Tau])*\[Tau]))*Integrate[
      (1 + 2*Sum[(Cos[0.8*n*Pi]*Cos[(n*Pi)/2 - 
              (n*Pi*z)/100])/E^((n^2*Pi^2*\[Tau])/
             10000), {n, 1, 100}])/

       E^((Tan[\[Psi]*Degree]^2*(z + 30)^2)/(4*\[Tau])), 
      {z, -50, 50}], {\[Tau], 0, t}]

Now I'm able to generate the graphics. For different angles \[Psi] this is what I did:


Pw1 = (Exp[-0.25/t]/t)*
   Integrate[
    Exp[-(((Tan[0 Degree]^2)*(z + 30)^2)/(4*t))]*(1 + 
       2*Sum[Exp[-((n^2)*(Pi^2)*t)/(100^2)]*Cos[0.8*n*Pi]*
          Cos[(n*Pi/2) - (n*Pi*z/100)], {n, 1, 100}]), {z, -50, 50}];
Pw2 = (Exp[-0.25/t]/t)*

   Integrate[
    Exp[-(((Tan[30 Degree]^2)*(z + 30)^2)/(4*t))]*(1 + 
       2*Sum[Exp[-((n^2)*(Pi^2)*t)/(100^2)]*Cos[0.8*n*Pi]*
          Cos[(n*Pi/2) - (n*Pi*z/100)], {n, 1, 100}]), {z, -50, 50}];
Pw3 = (Exp[-0.25/t]/t)*
   Integrate[
    Exp[-(((Tan[60 Degree]^2)*(z + 30)^2)/(4*t))]*(1 + 
       2*Sum[Exp[-((n^2)*(Pi^2)*t)/(100^2)]*Cos[0.8*n*Pi]*
          Cos[(n*Pi/2) - (n*Pi*z/100)], {n, 1, 100}]), {z, -50, 50}];


PD1[y_] := 0.005*NIntegrate[Pw1, {t, 0, y}, MaxRecursion -> 20];
PD2[y_] := 0.005*NIntegrate[Pw2, {t, 0, y}, MaxRecursion -> 20];
PD3[y_] := 0.005*NIntegrate[Pw3, {t, 0, y}, MaxRecursion -> 20];

T1 = Table[{y, 
    PD1[y]}, {y, {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 2, 
     2.2, 2.5, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 
     18, 19, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 
     90, 100, 110, 120, 130, 140, 150, 200, 250, 300, 350, 400, 450, 
     500, 550, 600, 650, 700, 750, 800, 850, 900, 950, 1000, 1100, 

     1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000, 2100, 2200,
      2300, 2400, 2500, 2600, 2700, 2800, 2900, 3000, 3500, 4000, 
     4500, 5000, 5500, 6000, 6500, 7000, 7500, 8000, 8500, 9000, 9500,
      10000, 11000, 13000, 15000, 17000, 19000, 20000, 22000, 24000, 
     26000, 28000, 30000, 35000, 40000, 45000, 50000, 55000, 60000, 
     65000, 70000, 75000, 80000, 90000, 95000, 100000, 150000, 200000,
      250000, 300000, 350000, 400000, 450000, 500000, 550000, 600000, 
     700000, 800000, 900000, 1000000}}];
T2 = Table[{y, 
    PD2[y]}, {y, {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 2, 

     2.2, 2.5, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 
     18, 19, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 
     90, 100, 110, 120, 130, 140, 150, 200, 250, 300, 350, 400, 450, 
     500, 550, 600, 650, 700, 750, 800, 850, 900, 950, 1000, 1100, 
     1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000, 2100, 2200,
      2300, 2400, 2500, 2600, 2700, 2800, 2900, 3000, 3500, 4000, 
     4500, 5000, 5500, 6000, 6500, 7000, 7500, 8000, 8500, 9000, 9500,
      10000, 11000, 13000, 15000, 17000, 19000, 20000, 22000, 24000, 
     26000, 28000, 30000, 35000, 40000, 45000, 50000, 55000, 60000, 
     65000, 70000, 75000, 80000, 90000, 95000, 100000, 150000, 200000,

      250000, 300000, 350000, 400000, 450000, 500000, 550000, 600000, 
     700000, 800000, 900000, 1000000}}];
T3 = Table[{y, 
    PD3[y]}, {y, {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1, 2, 
     2.2, 2.5, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 
     18, 19, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 
     90, 100, 110, 120, 130, 140, 150, 200, 250, 300, 350, 400, 450, 
     500, 550, 600, 650, 700, 750, 800, 850, 900, 950, 1000, 1100, 
     1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000, 2100, 2200,
      2300, 2400, 2500, 2600, 2700, 2800, 2900, 3000, 3500, 4000, 

     4500, 5000, 5500, 6000, 6500, 7000, 7500, 8000, 8500, 9000, 9500,
      10000, 11000, 13000, 15000, 17000, 19000, 20000, 22000, 24000, 
     26000, 28000, 30000, 35000, 40000, 45000, 50000, 55000, 60000, 
     65000, 70000, 75000, 80000, 90000, 95000, 100000, 150000, 200000,
      250000, 300000, 350000, 400000, 450000, 500000, 550000, 600000, 
     700000, 800000, 900000, 1000000}}];

PwD1 = Interpolation[T1];
PwD2 = Interpolation[T2];
PwD3 = Interpolation[T3];


P1 = LogLogPlot[{PwD1[y], y*PwD1'[y]}, {y, 0.1, 1000000}, 
   PlotRange -> {0.01, 10}, PlotStyle -> {{Black}, {Dashed, Black}}, 
   Frame -> True, FrameLabel -> {tD, "PD e tD*PD'"}, 
   BaseStyle -> {FontSize -> 12}];
P2 = LogLogPlot[{PwD2[y], y*PwD2'[y]}, {y, 0.1, 1000000}, 
   PlotRange -> {0.01, 10}, PlotStyle -> {{Brown}, {Dashed, Brown}}];
P3 = LogLogPlot[{PwD3[y], y*PwD3'[y]}, {y, 0.1, 1000000}, 
   PlotRange -> {0.01, 10}, PlotStyle -> {{Purple}, {Dashed, Purple}}];


Show[P1, P2, P3]

The problem is that it takes a very long time to compute these codes... Since I don't have good programming skills, could anyone propose an alternative way to computing this plot?


thanks



Answer



You could do the indefinite integral and put in the limits afterwards. E.g., on Linux I get a speed-up of about 14 for one integral (I have no time to do more right now ...)


Mathematica 8.0 for Linux x86 (64-bit)
Copyright 1988-2011 Wolfram Research, Inc.

In[1]:= !!i

Print["timing 1: ", 
First @ AbsoluteTiming[
  Pw3 = ExpandAll[(Exp[-0.25/t]/t)*
     Integrate[
      Exp[-((Tan[60*Degree]^2*
           (z + 30)^2)/(4*t))]*
       (1 + 2*Sum[Exp[-(n^2*Pi^2*t)/
            100^2]*Cos[0.8*n*Pi]*
           Cos[n*(Pi/2) - n*Pi*
            (z/100)], {n, 1, 100}]), 

      {z, -50, 50}]]; ]
];
Print["timing 2 : ", 
First @ AbsoluteTiming[
  nPw3 = Expand[(Exp[-0.25/t]/t)*
     ((#1 /. z -> 50) - (#1 /. 
         z -> -50) & )[
      ParallelMap[(ExpandAll[Integrate[#1, z]] & ),
       (TrigToExp[Expand[
          Exp[-((Tan[60*Degree]^2*

            (z + 30)^2)/(4*t))]*
           (1 + 2*Sum[Exp[-(n^2*Pi^2*
            t)/100^2]*Cos[0.8*n*Pi]*
            Cos[n*(Pi/2) - n*Pi*
            (z/100)], {n, 1, 
            100}])]])]] ]; ]
]
Print["check ", Chop[Pw3 - nPw3]]

In[1]:= <
timing 1: 75.121517
Launching kernels...       Mathematica 8.0 for Linux x86 (64-bit) Copyright 1988-2011 Wolfram Research, Inc.                                                                                                            
timing 2 : 5.255553                                                                                                                    
check 0

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more