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Taylor expansion allows us to convert non-polynomial form of any function into polynomial form (though the function itself may not be polynomial, as the polynomail can contain infinite terms.).

The question, is this Taylor expansion the only way to do this?

Edit: I am changing the question into the following:

Suppose that there is a function $f(x)$. Is there a way to create a power series that equals to $f(x)$ without using Taylor expansion?

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Taylor expansion either gives you something which is not a polynomial but rather a power series (as it has an infinite number of terms), or it does not in general give you the function you started with (if you restrict to a finite number of terms), unless the function is very nice. So Taylor expansion does not do what you want either. –  Tobias Kildetoft Oct 10 '12 at 11:56
@Tobias So a function got to be infinitely differentiable in order to apply Taylor expansion? –  Subtomic Oct 10 '12 at 12:01
There are functions $f(x)$ for which all derivatives exist for any real $x$, $and$ all derivatives of $f$ are 0 at $x=0$, but which are $not$ the zero function $f(x)=0$. For these functions the taylor expansion around $x=0$ gives all coefficients 0, so that the taylor expansion is not the original function. This means it's not enough to assume infinitely differentiable in order to apply taylor expansion. –  coffeemath Oct 10 '12 at 12:12

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Since this is a very generally posed question, I'll keep it brief and commend further reading. I think of the Bernstein polynomials as a non-Taylorian way to approximate functions by polynomials:


However, the Bernstein approximation uses values of the function at different positions. Your question does not tell whether or not this is crucial to you. Maybe Stone-Weierstrass' Theorem will also interest you...

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