# Taylor series question

1. Jan 11, 2013

### cytochrome

What does it mean to calculate the Taylor series ABOUT a particular point?

I understand the formula for the Taylor series but how do you solve it about a particular point for a function? It's the about the particular point that confuses me.

Could someone please explain this and provide examples?

Thanks!

2. Jan 11, 2013

### jbunniii

A Taylor series of a function $f$, about the point $x = 0$ is a representation of $f$ as a sum of polynomials centered at $x = 0$, namely $x, x^2, x^3, x^4, \ldots$

Thus it will have the form

$$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \ldots$$

More generally, we can express $f$ as a sum of polynomials centered at some other point, say $x = c$. Then it will look like

$$f(x) = a_0 + a_1 (x - c) + a_2(x - c)^2 + a_3 (x - c)^3 + a_4 (x - c)^4 + \ldots$$

This is a more general form: if I set $c = 0$ then it reduces to the previous equation. Note that the values of the coefficients generally change if I change $c$. Here is the formula for the k'th coefficient:

$$a_k = \frac{f^{(k)}(c)}{k!}$$

where $f^{(0)}$ is taken to mean $f$, and $f^{(k)}$ means the k'th derivative of $f$. So to calculate $a_k$, you find the k'th derivative of $f$ and evaluate it at $x = c$.

3. Jan 11, 2013

### jbunniii

As a simple example, take $f(x) = e^x$. This is a nice example because all of the derivatives also equal $e^x$. Thus, if I center the Taylor series at $c$, the coefficients will be
$$a_k = \frac{f^{(k)}(c)}{k!} = \frac{e^c}{k!}$$
and the series will therefore be
$$e^x = \frac{e^c}{0!} + \frac{e^c}{1!}(x - c) + \frac{e^c}{2!}(x - c)^2 + \ldots$$
For the special case $c = 0$, we have $e^c = e^0 = 1$ so the coefficient simplifies to
$$a_k = \frac{1}{k!}$$
and the series will therefore be
$$e^x = \frac{1}{0!} + \frac{1}{1!}x + \frac{1}{2!}x^2 + \ldots$$

4. Jan 11, 2013

### Simon Bridge

The Taylor series is only defined to be about a particular point. The Taylor series expansion of y(x) about point x=a would be: $$y(x)_a = \sum_{n=0}^\infty \frac{f^{(n)}}{n!}(x-a)^n$$ It will be exact at that point - see:

... this is the Taylor series expansion (red line), taken about x=0, for the exponential function. See how after each term is added in, the red line has the same value as the blue at x=0? As an exercise, try doing the expansion for $y(x)=e^x$ about $x=2$, and compare.

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