I'm trying to teach myself partial differential equations from Strauss' book. I have run into a very bizarre problem - I cannot figure out what is the Fourier series of $e^x$! And not even Google has helped.

The book's general formula is this with this.

The book's answer for the $e^x$ fourier series is this

But I derived this by hand (and with Mathematica) independently: this

So what gives? Did Mathematica and I both fail to do a simple straightforward computation? Or did the book just pull something out of a hat?

In addition, I cannot figure out how to transfer from a complex Fourier series to a real one. I assume one cannot just take the real part?

  • $\begingroup$ The latter question has a simple, elegant answer: the transform of a real-valued function at a negative frequency is the complex conjugate of the corresponding positive frequency. There is probably another answer for that, or you might ask a separate question. $\endgroup$ – Potatoswatter Mar 3 '14 at 10:25
  • $\begingroup$ Are the two formulas equivalent? Can you substitute in some values and see if they give the same result? $\endgroup$ – AnonSubmitter85 Mar 3 '14 at 10:32
  • $\begingroup$ @AnonSubmitter85 I'm not seeing the equivalency; I tried infinite-summing both of them but Mathematica claims they don't converge for some reason; anyway, what is the advantage of writing it the way the book does? And how would I get from one formula to the other - any hints? (Specifically, I don't see how I would conjure up a sinh(l) when none exists in my formula) $\endgroup$ – user2258552 Mar 3 '14 at 10:48
  • $\begingroup$ @Potatoswatter I'm not sure what you mean. I also want the real Fourier series of e^x (e.g. in terms of sines and cosines) as opposed to the e^ix one. $\endgroup$ – user2258552 Mar 3 '14 at 10:49
  • $\begingroup$ mathisfunforum.com/viewtopic.php?id=18347 $\endgroup$ – Claude Leibovici Mar 3 '14 at 10:54

$$\sinh(l-i n \pi)=\tfrac12(e^{l-i n \pi}-e^{-l+i n \pi})=(-1)^n\sinh(l)$$

since $e^{i\pi}=-1$, and

$$\frac1{l-i n \pi}=\frac{l+i n \pi}{(l-i n \pi)(l+i n \pi)}=\frac{l+i n \pi}{l^2+n^2 \pi^2}$$

so all your quoted expressions are actually equal.

  • $\begingroup$ Perfect! Thank you, I kept forgetting it's a l and not an 1 in the l - inπ, so my attempted derivation went awry (which is why I never use l as a variable unless it's already being used...) $\endgroup$ – user2258552 Mar 3 '14 at 18:52

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