# Maclaurin vs Taylor and their geometrical difference

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?

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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 $0$.

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 $0$.

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

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