Why doesn’t induction extend to infinity? (re: Fourier series)

While reading some things about analytic functions earlier tonight it came to my attention that Fourier series are not necessarily analytic. I used to think one could prove that they are analytic using induction

  1. Let P(n) be some statement parametrized by the natural number n (in this case: the nth partial sum of the Fourier series is analytic)
  2. Show that P(0) is true
  3. Show that P(n1)P(n)
  4. (Invalid) conclusion: P(n) continues to be true as we take the limit n*

Why exactly is the conclusion not valid here? It seems very strange that even though P(n) is true for any finite n, it ceases to be valid when I remove the explicit upper bound on n. Are there circumstances under which I can make an argument of this form?


Example of invalid proof: Define the truncated Fourier series Fn(x) as the partial sum

Fn(x)=nk=0Aksin(kxT)+Bkcos(kxT)

where Ak and Bk are the Fourier coefficients for some arbitrary function f. Using the facts that sin(t) and cos(t) are analytic, and that any linear combination of analytic functions is analytic:

  1. P(n) is the statement “Fn(x) is analytic”
  2. F0(x) is clearly analytic because it is a linear combination of sine and cosine functions
  3. Fn(x) can be written as the linear combination

    Fn(x)=Fn1(x)+Ansin(nxT)+Bncos(nxT)

    So if Fn1(x) is analytic, Fn(x) is analytic.

  4. F(x)limnFn(x) is analytic. But F(x) is the Fourier series for f; therefore, the Fourier series for f is analytic.

*I’m assuming that P(n) is a statement about some sequence which is parametrized by n and for which taking the limit as n is meaningful

Answer

A trivial case where P(n) is true for all nN but P() is false is the statement “n is finite”.

Attribution
Source : Link , Question Author : David Z , Answer Author : Marc van Leeuwen

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