So I read about the Taylor series and it said you can choose to expand the series around a given point ($x=a$). Does it matter which point you choose in calculating the value of the series?

For example, if I wanted to calculate "$e^x$" at $x=1$ then would it matter if I'd expand the series around $a=1$ or $a=0$? Thanks in advance :)


3 Answers 3


Sticking to your example of $e^x$, if you can expand it around $a=1$, then you already know the value of $e^x$ at $x=1$. In other words, you would not use the Taylor expansion to approximate a function about a point you already can compute the value at.

Choosing the point for the expansion is largely a question of computational ease and what's available. It's a lot easier to compute the Taylor expansion of, say, $e^x$, $\sin(x)$, or $\cos (x)$ about the point $x=0$ then it would about the point $x=0.12345563$ or $x=\pi + 6.7$ for the simple reason that it's so easy to compute the value the derivatives attain at $x=0$, but less easy (and a lot more messy) at other points. Issues of suitably approximating the error are of importance here, as well as making a choice that will increase the speed of convergence could be relevant.

Also, when one tries to extrapolate a function from given empirical values you simply have to work with what you have. If you have more numerical information about a function and its derivatives at and about a point $a$ than you have at or about a point $b$, then use $x=a$ as the point for the Taylor expansion.

  • $\begingroup$ Thank you. That cleared things up! :) $\endgroup$
    – matanc1
    Apr 30, 2013 at 10:38
  • $\begingroup$ you're welcome. $\endgroup$ Apr 30, 2013 at 10:42
  • 1
    $\begingroup$ "Choosing the point for the expansion is largely a question of computational ease and what's available." - this is the most important point of this answer. +1. $\endgroup$ Apr 30, 2013 at 11:18

Yes, it would matter, because the idea of the Taylor expansion is that you want to evaluate f(x), when x is very close to a. This means that the closer x is to a, you'll need to sum up less terms to get to a certain accuracy.

For example, if I wanted to calculate $ e^2 $, then it would be smart to expand around a=1, because 2 is closer to 1 than to 0, but if I wanted to calculate $e^{0.234325}$, then it would be smarter to expand around a=0.

  • $\begingroup$ notice that to expand about $x=1$ you'll need to first find a good approximation of $e^1$ before you can start computing approximations for $e^2$. So, in fact it makes more sense to use $a=0$ even if you want to compute $e^2$ (and even if you want to computer $e^10$, the Taylor polynomials give really good approximations in this case). $\endgroup$ Apr 30, 2013 at 10:42

The point is that if you know the value of a function and all its derivatives at a single point then you can use this data to determine your function in some neighborhood of that point, namely an open ball around the point with radius equal to the radius of convergence of the series expanded about that point. The nice thing about functions like the exponential and trigonometric functions (or functions with infinite radius of convergence) is that knowing the countable data consisting of the value of the function along with all its derivatives at a single point is enough to determine the uncountable data consisting of its value at every point.


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