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Consider

s[t_] := Exp[-t^2] Sin[5 t]
sf[w_, n_] = FourierTransform[Exp[-t^2] Sin[n t], t, w];

Then

Table[Plot[Abs[sf[w, n]], {w, 1, 15}, WorkingPrecision -> 100], {n, {1, 5, 10}}]

but when I

Table[LogLogPlot[Abs[sf[w, n]], {w, 1, 15}, WorkingPrecision -> 100], {n, {1, 5, 10}}]

enter image description here the last plot remains empty.

(note that WorkingPrecision -> 100 is near the minimum amount for which sf gets plotted correctly with Plot; increasing it won't solve the problem with the last plot)

This has happened to me also in other situations, with different functions.

I wonder

  1. What is happening in the last and other similar plots
  2. How can I handle this situation, not on this particular case but in general

Version: 11.3.0 for Mac OS X x86 (64-bit) (March 7, 2018)

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  • 1
    $\begingroup$ Are you sure? Mathematica 10.4 returns this imgur.com/a/0BElEyM $\endgroup$ – J42161217 Oct 11 '18 at 22:23
  • $\begingroup$ Yes, the plots are correct. I think it is something related to the newer version $\endgroup$ – pp.ch.te Oct 11 '18 at 22:32
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Use SetDelayed (:=) when defining functions in Mathematica. It's better practice:

s[t_] := Exp[-t^2] Sin[5 t]
sf[w_, n_] := FourierTransform[Exp[-t^2] Sin[n t], t, w];

Table[Plot[Abs[sf[w, n]], {w, 1, 15}, 
  WorkingPrecision -> 100], {n, {1, 5, 10}}]

Table[LogLogPlot[Abs[sf[w, n]], {w, 1, 15}, 
  WorkingPrecision -> 100], {n, {1, 5, 10}}]

enter image description here

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  • $\begingroup$ What is the rationale behind this behaviour? I mean, how can the results be different? I understand the difference between = and := but I didn't think it could make such a difference $\endgroup$ – pp.ch.te Oct 13 '18 at 8:45
  • $\begingroup$ Also, this was to me the typical case in which using = would be preferred over := since the symbolic calculation of the transform is a bit slow (as in mathprogramming-intro.org/book/node381.html) $\endgroup$ – pp.ch.te Oct 13 '18 at 9:00

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