We know that corresponding to every analytic real valued function of a real variable there is a power series representation. I was just curious if the converse is true or not. The $a_n$ involved in the summation $$\sum_{n=0}^\infty a_nx^n$$ can be any random function of $n$ which could supress all possibilities of a closed form representation. If the answer happens to be true, please provide for sufficient details. Thanks in advance.

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    $\begingroup$ Your sum is just $\sum_{i=0}^n a_nx^n = (n+1)a_n x^n$. Do you mean $\sum_{n=0}^\infty a_nx^n$? $\endgroup$ – gammatester Sep 14 '17 at 14:09
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    $\begingroup$ Are you familiar with the notion of "radius of convergence"? What happens if you take $a_n = n^n$? $\endgroup$ – Mees de Vries Sep 14 '17 at 14:13
  • $\begingroup$ @gammatester Edited. Thanks $\endgroup$ – Madhur Panwar Sep 14 '17 at 14:28
  • $\begingroup$ @MeesdeVries I think the radius of convergence would then be 0 but what are its implications? $\endgroup$ – Madhur Panwar Sep 14 '17 at 14:32
  • $\begingroup$ There's no associated function, then. The series would converge only for $x = 0$. (You can prove this directly.) $\endgroup$ – Mees de Vries Sep 14 '17 at 14:35

If the power series has a positive radius of convergence $R$ then in the interval $(-R,R)$ it defines a real analytic function. That function is unlikely to have a closed form representation if by "closed form" you mean some rational expression involving polynomials and exponential/trigonometric functions.

In fact the term by term integral of your power series will define an analytic function that is even more unlikely to have a closed form expression: see

How can you prove that a function has no closed form integral?



  • $\begingroup$ Small note: the guaranteed interval would be $(-R,R)$, whether the function can be extended to either bound depends on the function. $\endgroup$ – Mees de Vries Sep 14 '17 at 14:40
  • $\begingroup$ @MeesdeVries Indeed. Fixed, thanks. $\endgroup$ – Ethan Bolker Sep 14 '17 at 14:42
  • $\begingroup$ In what cases will the function defined by power series in the interval $(-R,R)$ have a 'closed form' representation? Is there a criteria? $\endgroup$ – Madhur Panwar Sep 14 '17 at 14:54
  • $\begingroup$ By the way, yes I exactly mean that when I say 'closed form' i.e. a combination of some known functions. $\endgroup$ – Madhur Panwar Sep 14 '17 at 14:55

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