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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?

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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.2011-10-31

3 Answers 3

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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.

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    @Avatar: I used *Mathematica* to generate those cartoons.2012-05-09
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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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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.