# Why do we assume Taylor series to be polynomials?

Consider the Taylor series of a function $f(x)$ we usually assume $f(x)$ to be a polynomial: $$f(x)=a_0+a_1x+a_2x^2+a_3x^3...$$

But why not assume $$f(x)=a_0+a_{0.5}x^{0.5}+a_1x+a_{1.5}x^{1.5}+...$$

or

$$f(x)=a_0+a_{0.1}x^{0.1}+a_{0.2}x^{0.2}+a_{0.3}x^{0.3}+...$$

what are the advantages of polynomials?

• Not gonna lie: looking forward to the answers. Nov 28, 2017 at 21:59
• Polinomials are simple, manageable and you have Weierstrass theorem... This should be more than motivation to use polinomials. Nov 28, 2017 at 22:00
• Some of the answers here may be relevant and interesting, as well as this Nov 28, 2017 at 22:15
• How would you find the coefficients? Nov 29, 2017 at 0:35

According to the Stone-Weierstrass theorem, any collection of functions that separates points is dense in the continuous functions. Hence there are many ways to approximate any continuous function (not just the analytic ones) by any collection you want, including the $x^{1/2},x^{3/2},...$ example you give.

So what's so special about the polynomial approximation? The statement that a function is differentiable means that the function can be approximated at small distances as linear. If the function is twice differentiable, that means its derivative is approximately linear, so the function is approximately quadratic. By induction, an $n$ times differentiable function is approximated by a polynomial of degree $n$.

So why is it polynomials? To sum up, because those are the things you get by taking the antiderivatives of linear functions. It is the natural approximation scheme that is given by virtue of the function being differentiable.

• To piggy-back on this, is this another reason not to use roots: you would lose differentiability at the center. Nov 28, 2017 at 22:04

Before people had calculators and such, polynomials were the only continuous functions which could be calculated by hand. So finding a Taylor series for a function allowed you to estimate its value using polynomials. Actually, your calculator uses a Taylor series to calculate exponential and trig functions.

You need to use polynomials to actually calculate $x^{0.5}, x^{1.5}$ etc.

Take a look at the fourier series, a series made of trig functions. The Bessel series is a complicated example of a useful infinite series. In fact, any series of orthogonal functions can be made into an infinite series, each better at different things.

Sometimes you don't want to use polynomials for a series, it depends what you are trying to accomplish. Polynomial series are very simple and good at approximations, but all series can be useful as a tool in math.

Simple answer, we don't assume polynomials. We use polynomials for many things, but not everything.

You are taking the limit of polynomials to get a power series.

In the second case you can put $y=x^{0.5}$ to recover the same kind of expansion. But this highlights that you have $y^2=x$ and $x$ cannot be negative. Likewise your second example reduces to a power series in $y=x^{0.1}$ - but there is the same issue with taking a tenth root - $x$ cannot be a negative real number.

On the real line, if you take an odd root, then you can dodge this problem. But there are issues which arise when you start to try to work in the complex plane, because the cube root for example has three possible values at each point and you need to cut the plane or make a Riemann Surface to avoid discontinuities.

Taking integer powers of $x$ avoids some of these issues.

For a polynomial, you only use 2 operations: multiplication and addition. For the taylor series you also include taking a limit. That's it. This means you only use the underlying ring structure (and topology for the limits) which is theoretically interesting, but also practical as pointed out in other answers.