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Can any continuous function be represented as an infinite polynomial?

Motivation: the antiderivative

$ \int^\ e^{-x^2}dx\ $

can be expressed as an infinite polynomial(write Taylor series for integrand function and integrate) but this antiderivative has no closed/elementary form expression according to Liouville's theorem but is clearly continuous. So are the rest of the non-elementary functions expressible as infinite polynomials? Fascinating.Any insights on how to proceed????

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    $\begingroup$ Look at the Stone-Weierstrass Theorem $\endgroup$ – Tyler Nov 5 '13 at 16:24
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    $\begingroup$ what is the definition of infinite polynomial? can constant and identity funtion x represented by infinite polynomials? $\endgroup$ – GA316 Nov 5 '13 at 16:24
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    $\begingroup$ @GA316 polynomials are equal to their Maclaurin series. $\endgroup$ – Tyler Nov 5 '13 at 16:26
  • $\begingroup$ @Tyler So any polynomial is an infinite polynomial? $\endgroup$ – GA316 Nov 5 '13 at 16:28
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    $\begingroup$ I didn't say that. The OP might be confused but it is obvious he means Taylor series when he says infinite polynomials. Look at his aside. If you want let him know the difference, be my guest, but I don't feel there is any need to beat around the bush about it. $\endgroup$ – Tyler Nov 5 '13 at 16:32
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No!

The functions that are given by a convergent power series are really quite rare in the whole scheme of things. They are called analytic functions.

There are a whole class of functions that are called flat functions. These have all of their derivatives zero at a given point and so, as far as Taylor series can tell, are identically zero. The classical example of a flat function is $x \mapsto \operatorname{e}^{-1/x^2}$. Where $0 \mapsto 0$. In this case all of the derivatives are zero at zero (you have to take limits) and so, as far as Taylor series are concerned, this is the zero function.

In addition, some Taylor series only hold in cetain regions. For example, the Taylor series of $(1-x)^{-1}$ is given by $1+x+x^2+x^3+\cdots+x^k+\cdots$. This is fine for all $-1 < x < 1$, but when $|x|>1$ we have serious trouble.

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The following function is not only continuous, but has continuous derivatives of all orders. However, it is not equal to any Taylor series. $$f(x)=\begin{cases} e^{-1/x^2} & x>0\\ 0 & x\le 0\end{cases}$$

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    $\begingroup$ To be fair, it is equal to its Taylor series locally away from $0$. $\endgroup$ – Chris Janjigian Nov 5 '13 at 16:33
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And there is something more weird such as the trajectory of one-dimensional Brownian motion, which is a continuous function but nowhere differentiable. Since power series are differentiable on interval of covergence(except for the endpoints), these nowhere differentiable continuous functions can not be represented as power series

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Not all continuous functions have a useful Taylor expansion (TE), such as $e^\frac{1}{x}$. (Although, you could say that its TE simply has a radius of convergence equal to zero.)

To answer your other point, the TE could be useful for integration, but not really. The problem is you still end up with an infinite sum, which is not elementary, and difficult for computers to calculate. I'll use your example with $e^{-x^2}$.

The TE for $e^{-x^2}$ is $\sum_{n=0}^{\infty}\frac{1}{n!}(-x^2)^n=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{2n}$. Integrating, we get $\int\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^{2n}dx=\sum_{n=0}^{\infty}\int\frac{(-1)^n}{n!}x^{2n}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!(2n+1)}x^{2n+1}$.

Great! Now we have a definite way of finding values of erf, the integral of $e^{-x^2}$. But, as you said there is no way to express this as not an infinite series.

Finally, I am pretty sure that as long as you choose a point on the curve where the limit exists, you can write a Maclaurin series (general Taylor series) that represents that function to a certain radius of convergence. The problem is, that series stops being effective outside that radius. So, you could write a Maclaurin series for the Gamma function, but it doesn't tell you a whole lot about the function.

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  • $\begingroup$ $\cos(x)$ is an infinite series too $\endgroup$ – reuns Jun 24 '17 at 21:46
  • $\begingroup$ Actually, cos(x) can be written as an infinite sum and an infinite product. Still waiting on a closed polynomial form though... $\endgroup$ – Marcus Luebke Jul 19 '17 at 21:05

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