Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In this topic i learned how to approximate a function with a high degree polynomial and how to derive the Maclaurin series:

$$ f (x) = P_n(x) = f(0)+{f'(0)\over 1!}x+{f''(0)\over 2!}x^2+{f'''(0)\over3!} x^3+\cdots+{f^{(n)}(0)\over n!}x^n $$

In the Maclaurin series we have $a=0$ which is not true for a Taylor series right? So taylor series equation goes like this:

$$ f (x) = P_n(x) = f(a)+{f'(a)\over 1!}(x-a)+{f''(a)\over 2!}(x-a)^2+{f'''(a)\over3!} (x-a)^3+\cdots+{f^{(n)}(a)\over n!}(x-a)^n $$

Q: To get the Taylor series out of Maclaurin we swaped $x \rightarrow(x-a)$ and $f(0) \rightarrow f(a)$. Could anyone show me (like drawing a picture) a geometrical impact of swapping these terms?

share|cite|improve this question
The MacLaurin expansion only consists in setting $a=0$ in the Taylor expansion. – Dominique Feb 13 '13 at 14:16
Be careful, if $f$ is equal to its MacLaurin series (so around $0$), it does not mean that it will equalt to its Taylor series centered at, say, $1$. Note the difference between the coefficients. Swapping is not ok. – 1015 Feb 13 '13 at 14:28
@ julien I don't understand this. – 71GA Feb 13 '13 at 15:50
@julien: if the Maclaurin series converges for $|x|<r$, then the Taylor series at $a$ will converge for $|x-a|\lt r-|a|$. – robjohn Feb 13 '13 at 20:21
@robjohn Yes, of course, thanks for the precision. So if $r=1/2$, you agree that we don't know if the Taylor series at $1$ converges. Also, I was just trying to tell the OP that you don't deduce the Taylor series from the MacLaurin series by simply swapping. Coeeficients are different. – 1015 Feb 14 '13 at 1:45

The number $a$ is called the center of the Taylor approximation. In general, a Taylor polynomial will only be a 'good' approximation of a function in an interval centered around $x=a$. The further we move from $a$, the greater the error in the approximation. Here are three pictures that should help.

Here is the fourth degree Taylor polynomial of $f(x)=\cos x$ centered at $a=0$. Notice that it is a good approximation of the actual function only for $x$ values near $0$.

enter image description here

Here is the fourth degree Taylor polynomial of $f(x)=\cos x$ centered at $a=\pi$. Now, the polynomial approximates $f(x)$ near $\pi$, but not very well near $0$.

enter image description here

Finally, here we're centered at $a=2\pi$.

enter image description here

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.