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.

  • $\begingroup$ 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. $\endgroup$ Feb 24, 2018 at 21:03
  • $\begingroup$ I agree, there is little point in approximating a polynomial with a polynomial! $\endgroup$ Feb 24, 2018 at 21:05
  • $\begingroup$ @JoséCarlosSantos Did you have a chance to take a look at the linked question? $\endgroup$ Feb 24, 2018 at 21:10
  • $\begingroup$ I think the Taylor series is a red herring in the other post: see my answer to it. $\endgroup$
    – Rob Arthan
    Feb 24, 2018 at 21:12
  • 1
    $\begingroup$ @gimusi Noted. Thank you for your answer. $\endgroup$ Feb 26, 2018 at 0:14

1 Answer 1


Taylor series can be applied to polynomial function as a way of shifting toward a point where the higher degree terms converges to 0 faster than lower degree terms, so that the approximation can get by with fewer terms.

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 $f(x)\approx 4$

  • $\begingroup$ Very enlightening. Can you include an example? $\endgroup$ Feb 24, 2018 at 21:57
  • $\begingroup$ You are confusing change of variables with Taylor series expansion. The Taylor series for a polynomial is the same as the polynomial. $\endgroup$
    – Rob Arthan
    Feb 24, 2018 at 23:06
  • $\begingroup$ 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. $\endgroup$ Feb 25, 2018 at 2:59
  • $\begingroup$ And, incidentally, it is true, like Rob mentions, that $4+(x-3)+(x-3)^2=x^2-6x+9+4x-12=x^2-5x+10.$ $\endgroup$ Feb 25, 2018 at 3:12
  • $\begingroup$ What do you mean by the center point $x_0$ of the polynomial? $\endgroup$ Feb 25, 2018 at 3:14

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