# Taylor's theorem: What does the a represent?

dawn_pingpong

## Homework Statement

Taylor's theorem states:
$$f(x)=\displaystyle\sum_{k=0}^{infinity}{\frac{(f(^k(a)(x−a)^k)}{n!}}$$

When approximating a function, we always set a=0, and it gives an approximation for the whole function.

Why set the a=0? What does the a reperesent?
Thank you.

## The Attempt at a Solution

Last edited:

Muphrid
You don't always set $a=0$. You can set $a$ to whatever number you want. In all cases, the point $a$ is the point that you expand the series about--a point that, in principle, you should know several derivatives of the function at in order to construct an approximation. There is no reason this point must be the origin, though often times, it is.

For example, in physics one often Taylor expands around the bottom of a potential well. Even if the well is not centered on the origin, this is a valid thing to do. One takes $a$ to be where the minimum of the well lies, and immediately, one knows that $f'(a) = 0$ because the point is a minimum. This derivative being zero is equivalent to saying the force at an equilibrium point is zero, and knowing that the first non-vanishing term in the expansion is quadratic in $x-a$ tells us that, very close to the minimum, any potential can be approximated as a simple harmonic potential, which also goes as $(x-a)^2$.

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Dearly Missed

## Homework Statement

Taylor's theorem states:
$$f(x)=\displaystyle\sum_{k=0}^{infinity}{\frac{(f(^k(a)(x−a)^k)}{n!}}$$

When approximating a function, we always set a=0, and it gives an approximation for the whole function.

Why set the a=0? What does the a reperesent?
Thank you.

## The Attempt at a Solution

Your statement "When approximating a function, we always set a=0, and it gives an approximation for the whole function" is false: we do not always set a = 0, and even when we do we do not necessarily approximate the "whole function". For example,
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots,$$ but this is valid only if -1 < x < 1. So, if you want to expand f(x) = 1/(1-x) in a power series near x = 2, for example, you would need to expand in powers of (x-2), not x. In fact,
$$\frac{1}{1-x} = \frac{1}{-1 + 2-x} = \frac{-1}{1 + (x-2)} = -1 + (x-2) - (x-2)^2 + \cdots.$$ This is valid for |x-2| < 1, or 1 < x < 3.

RGV

Last edited:
dawn_pingpong
Okay, thank you! so basically, it is the estimation around the point x=a, and the further it deviates from a the more inaccurate? For example, evaluating the value of e, without using a calculator, we still set a=0, because then$$f^n(a)$$ will be 1, and it will be easy to calculate, though there will be an error term?

Okay, thank you! so basically, it is the estimation around the point x=a, and the further it deviates from a the more inaccurate? For example, evaluating the value of e, without using a calculator, we still set a=0, because then$$f^n(a)$$ will be 1, and it will be easy to calculate, though there will be an error term?