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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?

J. M. ain't a mathematician
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deem
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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. – hmakholm left over Monica 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
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    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. – hmakholm left over Monica Oct 31 '11 at 22:23
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    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
  • OK. I did something like this on wolframalpha http://www.wolframalpha.com/input/?i=series%5Bsinx%2C%7Bx%2C2%2C4%7D%5D why are there these 3 curves? – deem Oct 31 '11 at 22:31
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    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

3 Answers3

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...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:

$\exp\,x$:

Taylor polynomials for e^x

$\dfrac1{1-x}$:

Taylor polynomials for 1/(1-x)

$\ln(1+x)$:

Taylor polynomials for ln(1+x)

$\arctan\,x$:

Taylor polynomials for arctan(x)

$\sin\,x$:

Taylor polynomials for 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.

J. M. ain't a mathematician
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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.

Ross Millikan
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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.

Valerio Capraro
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