## Taylor Series

For the discussion of math. Duh.

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Generic Goon
Posts: 18
Joined: Sun Jan 27, 2008 5:20 pm UTC

### Taylor Series

So, we started learning about Taylor series in school today.

We learned that the general form for f(x) centered at c is:

f(c)+f'(C)(X-C)/1!+f''(C)(X-C)^2/2!...

The problem I have is applying that to the example we got in class: f(x)=e^(x^2) centered at 0.

f'(x) = 2xe^(x^2)...f'(0)=0
f '' (x) = 2e^(x^2)+4x^2e^(x^2)...f''(0)=2

1 + 0 + 2X/2 ...

But in class the answer given was:

1 + x^2 + (x^4)/2 ...

So, what am I missing?

Blatm
Posts: 638
Joined: Mon Jun 04, 2007 1:43 am UTC

### Re: Taylor Series

Generic Goon wrote:So, we started learning about Taylor series in school today.

We learned that the general form for f(x) centered at c is:

f(c)+f'(C)(X-C)/1!+f''(C)(X-C)^2/2!...

The problem I have is applying that to the example we got in class: f(x)=e^(x^2) centered at 0.

f'(x) = 2xe^(x^2)...f'(0)=0
f '' (x) = 2e^(x^2)+4x^2e^(x^2)...f''(0)=2

1 + 0 + 2X/2 ...

But in class the answer given was:

1 + x^2 + (x^4)/2 ...

So, what am I missing?

The bolded part should be 2x^2/2 = x^2. I'm assuming this was just a slip up, but if you don't understand why it's an x^2 term, look over the general formula carefully.

Stanford
Posts: 38
Joined: Fri Nov 02, 2007 6:52 pm UTC

### Re: Taylor Series

Generic Goon wrote:f'(x) = 2xe^(x^2)...f'(0)=0
f '' (x) = 2e^(x^2)+4x^2e^(x^2)...f''(0)=2

1 + 0 + 2X/2 ...

But in class the answer given was:

1 + x^2 + (x^4)/2 ...

I'm going to assume this is a homework question, so I won't spell it out completely. Your error is pretty simple, you missed an exponent somewhere; just look over the steps you took and you should find it fairly quickly.

hocl
Posts: 82
Joined: Thu Apr 16, 2009 3:18 am UTC

### Re: Taylor Series

When you're doing Taylor series, one strategy to help avoid this sort of error is to write down each power of x. So you would have 1x^0+0x^1+1x^2...

Of course, the lazy way to do this would be to define u to be equal to x^2, and then find that f(x) is then equal to e^u. Hopefully, you know what the Taylor series for e^u is. You can then substitute x^2 back in for u.

the tree
Posts: 801
Joined: Mon Apr 02, 2007 6:23 pm UTC
Location: Behind you

### Re: Taylor Series

For the sake of double checking and finding alternative routes, it's sometimes good to look for [imath]f^{(n)}(x)[/imath] as well.

Oh, and if you're going to hang around the forums for a while then take a look at the LaTeX plugin, because well, it's pretty and it makes what your trying to say easier to read. $f(x) = \sum_{i=0}^{n} \frac{ f^{(i)}(c) (x-c)^{i}}{i!}$See? Pretty.