# Why and how to use Taylor expansions for polynomial equations?

This is likely incredibly naïve and basic, but ignorance is worse... so here it goes...

I understand that Taylor series are great to approximate a difficult function (trigonometric, for example) with a polynomial built up via derivatives of the original function at a nearby point we know well in the original function. The derivatives become the coefficients of the newly-created polynomial (modified by the factorials in the denominators).

But I don't see what is the point or how to use Taylor expansions when we already have a polynomial, and we are talking about the roots or zeros.

The context of this question is this other post. I am trying to build up my background knowledge to actually understand the answer on the prior post.

• I surely would love to help here, since I enjoy working with Taylor polynomials and Taylor series. But I also don't see the point of using them for polynomials. Feb 24, 2018 at 21:03
• I agree, there is little point in approximating a polynomial with a polynomial! Feb 24, 2018 at 21:05
• @JoséCarlosSantos Did you have a chance to take a look at the linked question? Feb 24, 2018 at 21:10
• I think the Taylor series is a red herring in the other post: see my answer to it. Feb 24, 2018 at 21:12
• @gimusi Noted. Thank you for your answer. Feb 26, 2018 at 0:14

In fully expanded form like $f(x)=\sum_i c_i x^i$, if the center point $x_0$ of the polynomial is not $0$, then higher degree terms may actually increase faster than lower degree terms near $x_0$, which is not what we want. So we want to shift the center toward $x_0$ so that the function becomes $f(x)=\sum_i c'_i (x-x_0)^i$, in which case the higher degree terms will tend toward $0$ faster as $x$ approaches $x_0$, so we can get rid of the higher degree terms without too much error.
Example: We want to approximate near $x_0=3$, and we have $f(x)=x^2-5x+10$. If we put $x=3$, $x^2$ is still required for good result since it adds some value in the result. If we apply Taylor series on $x_0=3$, we get $f(x)=(x-3)^2+(x-3)+4$. If we put $x$ near $3$, we can see $(x-3)^2$ converges to 0 faster than other terms, so we can approximate the function as $f(x)\approx (x-3)+4$ or even $f(x)\approx 4$
• I made a careless mistake in my prior follow-up question (erased for good measure, and too uninteresting to mention). Yes, the Taylor series of $f(x)$ at $x=3$ is indeed $(3^2-5\cdot 3+10) + \frac{2\cdot 3-5}{1!}(x-3)+\frac{2}{2!}(x-3)^2=4+(x-3)+(x-3)^2.$ But I was the only one who didn't know... Hahaha. Feb 25, 2018 at 2:59
• And, incidentally, it is true, like Rob mentions, that $4+(x-3)+(x-3)^2=x^2-6x+9+4x-12=x^2-5x+10.$ Feb 25, 2018 at 3:12
• What do you mean by the center point $x_0$ of the polynomial? Feb 25, 2018 at 3:14