3
$\begingroup$

I know and can prove that any non-constant polynomials are unbounded, and then I was wondering whether this is true for power series (like infinite degree polynomials). For expamle this is true for the exponential function (when expressed as a power series). After some research, I know that this is true by Liouville's theorem via the Cauchy's integral formula. But is there any simpler way of proving this result?

  • 0
    What are you calling Cauchy's integral formula ? Is it the one connecting the coefficients of the series to the function ? – 2017-02-06
  • 0
    @Max https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula – 2017-02-06
  • 0
    Then I think you should rather try with this formula (which is essentially the same but maybe you'd like it better) : if $f(x) = \displaystyle\sum_{n=0}^{\infty} a_n x^n$ has radius $R$, and if $r$n\in \mathbb{N}$, $a_n = \frac{1}{2\pi r^n} \displaystyle\int_0^{2\pi} f(r e^{it} ) e^{-int} dt$ if I remember correctly, or something like that – 2017-02-06
  • 0
    Having redone the calculations, I confirm that this formula is correct, quite easy to prove, and that it's enough to prove Liouville's theorem – 2017-02-06
  • 0
    Why do you need the integral formula? If the power series converges everywhere then it's an entire non-constant function and you only need to invoke Liouville – 2017-02-07
  • 0
    @Max Thanks, so one does need some form of integral formula to prove this result – 2017-02-07
  • 0
    @benji but the proof of Liouville seems to rely on the integral formula – 2017-02-07
  • 0
    Lamkingming : I don't know if one *needs* an integral formula, but I've never seen any proof of Liouville without it (although I have to admit I haven't seen many proofs of Liouville) – 2017-02-07

0 Answers 0