0
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I try

Coefficient[E^(I a (t - b)), E^(I a t)]

and expect the output

E^(-I a b)

but in fact I get 0.

What should be corrected in my code? I cannot solve this problem for a week.

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8
  • 2
    $\begingroup$ Since the documentation on Coefficient[] speaks of polynomials only, why do you expect anything sensible? $\endgroup$
    – Igor Rivin
    Oct 6 '14 at 17:19
  • $\begingroup$ @IgorRivin So is it possible to do what I want? $\endgroup$
    – Vladimir
    Oct 6 '14 at 17:20
  • $\begingroup$ I am sure it is possible, but it may be a little painful, the question is: what is it you are really trying to do? $\endgroup$
    – Igor Rivin
    Oct 6 '14 at 17:22
  • $\begingroup$ I need to collect the coefficients of exponents with particular powers e.g. E^(I a t), E^(I b t) and so on. Now I see that it is not a Coefficient[] but any other method. Somehow I did not pay attention that Coefficient[] processes only polynomials. $\endgroup$
    – Vladimir
    Oct 6 '14 at 17:24
  • 2
    $\begingroup$ Possibly you can get the sort of behavior you want by using TrigExpand[ExpToTrig[expression]] and the trying to extract the coefficients of Cos[a*t] and Sin[a*t]. Another approach would be to recast as an explicit polynomial, by converting Exp[I*a*t] to a "variable" eiat and likewise for Exp[I*a*b]. $\endgroup$ Oct 6 '14 at 17:54
2
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Coefficient[E^(I a (t - b)) // ExpandAll, E^(I a t)]

(* Exp[-I a b] *)
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1
  • $\begingroup$ You can delete your comment, just click on the "X" sign next to the time. I'll delete mine, I can't delete yours. $\endgroup$
    – RunnyKine
    Oct 6 '14 at 19:48

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