# Taylor series for different points… how do they look?

I can't understand what it means to do the Taylor series at the point $a$.

The best way would be showing me how it looks for different $a$ on a graph. Do I find those graphs on the Internet?

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Have you seen Taylor series at Wikipedia? Its graphs all have $a=0$, but that's just a matter of mentally translating everything to the left or right. –  Henning Makholm Oct 31 '11 at 22:12
I know it, but I don't know when it should be done around 0, and when around any another point. And why. –  deem Oct 31 '11 at 22:19
It's not a matter of when it "should" be done around which $a$. You can (except in particular pathological cases) choose for yourself which point you want to develop the function around. It can always be done; the difference is just in when the results will be helpful to you. And that depends entirely on what you're trying to do with the series afterwards. –  Henning Makholm Oct 31 '11 at 22:23
One reason to do it around a point is you know the value of a function at that point. For example, if you want $\log_{10} 997$ it is natural to make a Taylor series around $1000$ because you know that $\log_{10}1000=3$, so $\log_{10}(1000+x)\approx 3+(x-1000)\frac{d}{dx}\log_{10}x|_{1000}$ will be quite close –  Ross Millikan Oct 31 '11 at 22:29
The three blue curves represent different orders of the expansion. The highest and straight one only considers the constant and linear terms-that is why it is straight and has only one dot. The curve with two dots considers the second order term: $-\frac{1}{2}(x-2)^2\sin(2)$, so is more accurate. The one with three dots includes the third order term as well and is more accurate yet. –  Ross Millikan Oct 31 '11 at 23:11

...The best way would be showing me how it looks for different $a$ on a graph.

The others have done (most of) the math; I'll do the cartoons:

$\exp\,x$:

$\dfrac1{1-x}$:

$\ln(1+x)$:

$\arctan\,x$:

$\sin\,x$:

Note that the polynomials (except for the horizontal constant function) are "tangent" to the original function at the expansion point (shown in red above). That's sort of the idea: these polynomials are the unique $p$-th degree polynomials that have $p+1$-fold contact with the function being approximated.

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Awesome graphics. How were they created? –  AvatarOfChronos May 9 '12 at 4:25
@Avatar: I used Mathematica to generate those cartoons. –  Guess who it is. May 9 '12 at 4:29

I guess that you can find a lot of stuff just googling "Taylor series". Anyway, here is a very brief explanation.

Let $f:(a,b)\rightarrow\mathbb R$ differentiable infinitely many times. You can consider the series

$$\sum_{n=0}^\infty\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}$$

which under suitable hypotheses gives you $f(x)$ back in a neighborhood of $x_0$. So, the fact that you have fixed a base point $x_0$, that now I see you called $a$, explains why the expansion is around a point.

Now, exercise: write down the Taylor series for $f(x)=e^x$ about $x_0=0$ and then $x_0=1$ and see that they are different.

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If you do a Taylor series around $0$ (also called a MacLaurin series) it looks like $f(x)=b_0+b_1x+b_2x^2+\ldots$. If you do it around $a$ it looks like $f(x)=b_0+b_1(x-a)+b_2(x-a)^2+\ldots$. The expansion is generally more accurate the closer $x$ is to the expansion point.

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