# Taylor Series Help!

1. Apr 10, 2007

### IMGOOD

I don't get how these two forms of the taylor series are equivalent:

$$f(x+h)= \sum_{k=0}^{\infty} \frac{f^k(x)}{k!} h^k$$

$$f(x) = \sum_{k=0}^{\infty} \frac{f^k(0)}{k!}x^k$$

The second one makes sense but I just can't derive the first form using the second. I know its something very simple but I keep confusing myself! :grumpy:

Last edited: Apr 10, 2007
2. Apr 11, 2007

### christianjb

Replace x by x' in the second eqn.
Then the second eqn. is gotten from the first by

x=0
x'=h

3. Apr 11, 2007

### quasar987

Or, let g(h)=x+h. Then, f(x+h)=(f o g)(h).

The first equation is the Taylor (MacLaurin) series of (f o g)(h).

$$(f\circ g)^{(k)}(0) = (f^{(k)}\circ g)(0)\cdot 1=f^{(k)}(x+0)=f^{(k)}(x)$$

$$\Rightarrow f(x+h)=\sum_{k=0}^{\infty} \frac{f^{(k)}(x)}{k!}h^k$$

4. Apr 11, 2007

### IMGOOD

Your approach makes sense but could you explain the above line in a little more detail. Specifically, I don't get how you got $$f^{(k)}(x+0)$$ in the above equation.

Last edited: Apr 11, 2007
5. Apr 11, 2007

### AiRAVATA

$$(f^{(k)}\circ g)(0)\cdot 1=f^{(k)}(g(0))=f^{(k)}(x+0)=f^{(k)}(x)$$.

Sorry for steping in, I was bored.

6. Apr 11, 2007

### IMGOOD

Thanks!....

7. Apr 12, 2007

### IMGOOD

Actually, I am still kinda confused. I know now how you got $$f^{(k)}(x)$$ but how did you get $$h^k$$?

8. Apr 12, 2007

### quasar987

Because we'Re computing the Taylor series of a function of h. Recall, I set g(h)=x+h, a function of h. x is considered constant. And this h dependence is passed on to f(x+h): f(x+h)=(f o g)(h).

9. Apr 12, 2007

### ObsessiveMathsFreak

Taylor series are something that is never written consistently, with some authors choosing to expand the series in x about zero, and others choosing to expand the series in h(or a) about x. Still more choose to evaluate the function at x, with the series expanded around a point a, and the powers being of (x-a). So you can see lots of things like:

$$f(x) = \sum_{k=0}^{\infty} \frac{f^k(0)}{k!}x^k$$

$$f(x+h)= \sum_{k=0}^{\infty} \frac{f^k(x)}{k!} h^k$$

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

Now personally, I prefer to expand in x about the point a, as there is no guarantee that you will be able to expand about zero. The function may not even be defined there, or may be singular. Secondly, it's nice to be able to just write f(x), and not have to worry too much about the "arbitrary but fixed" point a (detest this phrase). You can keep the regular f(x) notation and then just change a at will.

By the way, anyone interested in Taylor series in higher dimensions should look into the rather nice multi-index notation for multi-variable analysis. It enables you to wite things like.

$$f(\mathbf{x})=\sum_{|\grave{\alpha}| \ge 0} \frac{\mathbf{D}^{\grave{\alpha}}f(\mathbf{x})}{\grave{\alpha}!} (\mathbf{x}-\mathbf{a})^{\grave{\alpha}}$$

Here $$\mathbf{x}$$ is vector of n variables $$(x_1,x_2,.....x_n)$$, and $$\grave{\alpha}$$ is... complicated. It's called multi-index notation and is very useful for compacting the oft times awkward Taylor expansions in n dimensions. It takes a bit of getting used to but is worth it.

My personal favourite is it's compression of the inherantly forgettable, but undeniably useful multinomial expansion.
$$(x_1 + x_2 + \cdots + x_n )^k = k! \sum_{|\grave{\alpha}|=k} \frac{\mathbf{a}^{\grave{\alpha}}}{\grave{\alpha}!}$$

Which is a good deal more memorable than the usual expansion.

Last edited: Apr 13, 2007
10. Apr 12, 2007

### christianjb

For taylor series in multiple dimensions, the easiest form to use is the operator

$e^{\mathbf{r}.\hat{\nabla}$

which gives

$e^{\mathbf{r}.\hat{\nabla}} f(0)=f(\mathbf{r})$