# Proof Taylor Series

1. Oct 26, 2007

### rock.freak667

How does one prove taylor series? Is it proven the same way as Maclaurin's Series(Which i know is a special case of taylor series)

$$f(x)=A_0+A_1x+A_2x^2+A_3x^3+...$$
$$f(\alpha)=A_0+A_1\alpha+A_2(\alpha)^2+A_3(\alpha)^3+...$$

this kinda doesnt seem like a good way to prove it...as that is how I know how to prove maclaurin series

2. Oct 26, 2007

### mathman

Can you state precisely what is "it" that you are trying to prove?

3. Oct 26, 2007

### ZioX

If I understand your problem, f(g(x))=f(x-a) is just as differentiable as f is. Although, the logical approach taught in analysis courses is to prove Taylor's theorem and then the result about Maclaurin series holds at a=0.

I'm not even sure if you prove the result for Maclaurin series without implicitly proving Taylor series.

Last edited: Oct 26, 2007
4. Oct 26, 2007

### rock.freak667

Well I am trying to show that if a function is differentiable at x=a
then f(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2/2!+...
I never learned Taylor series...but I was yet shown Maclaurin's series and how to prove it without know that maclaurin's series is just a special case of Taylor series

5. Oct 26, 2007

### Kummer

The Taylor series is not so easy as it looks. The general result is the following.

First, a class $$C^n$$ function on $$(a,b)$$ (with $$a<0<b$$) is continously differentiable $$n$$ times. Then the polynomial $$T_n(x) = \sum_{j=0}^{n} \frac{f^{(j)}(0)}{j!}x^j$$ has the property that the error, i.e. $$R_n(x) = f(x) - T_n(x)$$ satisfies $$\lim_{x\to 0}\frac{R_n(x)}{x^n} = 0$$. This is a result which becomes of theoretic important.

6. Oct 26, 2007

### Gib Z

Wikipedia's article on Taylor Series has several different proofs, I'm sure you can read them there. The simplest one by memory was the integration by parts one.

7. Oct 26, 2007

### rock.freak667

I didnt see any proofs when i checked it

EDIT:I see it ...thanks

8. Nov 5, 2007

### yasiru89

Well Taylor series is a simple consequence of Cauchy's Integral formula. That is if you allow a complex analysis based proof(real ones aren't all that elegant).
For any analytic function f(z), represent with the integral formula for any suitable disc centred at say a and expand the denominator in a geometric series such that the intgration variable isn't caught up in it. Seperate the integral inside the sum and there's your Taylor coefficient. The fact that this is a higher derivative of the original function is again a simple consequence of the Cauchy Integral Formula(which isn't terribly difficult to prove using Cauchy's theorem once you understand the basic notions of holomorphism)