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I ran into this problem when I try to solve a partial differential equation. Here is my code:

dd = NDSolve[{D[tes[t, x], t] ==D[tes[t, x], x, x] + Exp[-1/(tes[t, x])],
   tes[t, 0] == 1, tes[t, -1] == 1, tes[0, x] == 1}, {tes[t, x]}, {t, 0, 5}, {x, -1, 0}]

f[t_, x_] = tes[t, x] /. dd
kkk = FunctionInterpolation[Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}], {t, 0, 0.05}]
kkg[t_] = Integrate[Exp[-1.1/( Evaluate[f[t, x]])], {x, -1, 0}]
Plot[Evaluate[kkk[t]] - Evaluate[kkg[t]], {t, 0, 0.05}]
N[kkg[0.01] - kkk[0.01], 1]

It's strange that the deviation showed in the graph reaches up to more than 5*10^-7 around t=0.01, while it's only -3.88578*10^-16 when calculated by N[kkg[0.01] - kkk[0.01], 1], I wonder how this error comes out.

By the way, I feel it strange that the output of N[kkg[0.01] - kkk[0.01], 1] has so many decimal places, I've set the precision as 1, right?

xzczd
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1 Answers1

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Using Mathematica 7 the plot I get does not show a peak at 0.01:

Plot[kkk[t] - kkg[t], {t, 0, 0.05}, GridLines -> Automatic]

Mathematica graphics

There is a peak at about 0.00754:

kkk[0.00754] - kkg[0.00754] // N

{6.50604*10^-7}

Regarding N, it does not change the precision of machine precision numbers as it does for exact or arbitrary precision ones:

N[{1.23456789, Pi, 1.23456789`50}, 2]

Precision /@ %

{1.23457, 3.1, 1.2}

{MachinePrecision, 2., 2.}

Look at SetPrecision if you want to force (fake) a precision, and NumberForm if you want to print a number in a specific format.

Mr.Wizard
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  • Thank you for your answer about precision! Then, well, after seeing your graph I rechecked mine and sadly found that I misread the coordinate on the graph…… – xzczd Jul 26 '12 at 12:00
  • @user1551513 hopefully `GridLines` will help with that in the future. Also, you can right-click on the plot and choose **Get Coordinates** to help pin down a value. – Mr.Wizard Jul 26 '12 at 12:03