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I tried to validate this function from 0 to 50 but it takes very long time, is there a faster way to validate this function for t from 0 to 50 and add them in a list?


    Plot[InverseLaplaceTransform[
      s^(5/4)/(0.0048093124845766414 + 0.02440766170480293 s^(1/4) + 
         0.012905621492995754 E^(-35. s) s^(1/4) + 
         0.0032916706480934036 E^(0.7079967811634376 s) s^(1/4) Gamma[0, 
           35.70799678116344 s] - 
         0.0053059348427644286 Gamma[1.25, 35. s]), s, t], {t, 0, 50}]

I can get a chart from top code but the bottem one dosent work. I dont know how its possible. Maby I dont know the logic of the plot and table.

Thank you very much for the time you have taken. I am not a professional programmer in math and still trying to learn it.


    Table[InverseLaplaceTransform[
      s^(5/4)/(0.0048093124845766414 + 0.02440766170480293 s^(1/4) + 
         0.012905621492995754 E^(-35. s) s^(1/4) + 
         0.0032916706480934036 E^(0.7079967811634376 s) s^(1/4) Gamma[0, 
           35.70799678116344 s] - 
         0.0053059348427644286 Gamma[1.25, 35. s]), s, t], {t, 0, 50}]

enter image description here

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  • $\begingroup$ I will add from 0-5 because 0-50 really takes a lot of time. Just a sec. $\endgroup$ Aug 2 at 11:20
  • 1
    $\begingroup$ I tried up to 50 with increments of 1.0, but it slows the higher t gets. It might be just due to your function complexity the numerical inverse Laplace is having hard time. It is fast up to 15 or so, then slows down. But you might not need very high t value to verify this. Not sure. $\endgroup$
    – Nasser
    Aug 2 at 13:11

1 Answer 1

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but the bottem one dosent work.

Change your Table command to use real numbers, and do not start it from 0.

Table[InverseLaplaceTransform[
  s^(5/4)/(0.0048093124845766414 + 0.02440766170480293 s^(1/4) + 
     0.012905621492995754 E^(-35. s) s^(1/4) + 
     0.0032916706480934036 E^(0.7079967811634376 s) s^(1/4) Gamma[0, 
       35.70799678116344 s] - 
     0.0053059348427644286 Gamma[1.25, 35. s]), s, t], {t, 0.01,1, .01}]

Gives

Mathematica graphics

Mathematica graphics

Just make sure the starting value is not 0.0 else it will not work. It looks like the Plot command was smart enough to detect this and it also did not start at t=0

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