how to prove this theorem

**szpengchao** · May 15th 2008, 08:58 AM

prove that if the function is infinitely differentiable on an interval (r,s) containing a, then for any x in (r,s) f(x) can be expand :

f(a) + (x-a) f'(a) + (x-a)^2*f''(a)/2! +....+ (x-a)^n * f(n')(a) + Rn(f,a,x)

Rn is the remainder

**ThePerfectHacker** · May 15th 2008, 10:27 AM

Originally Posted by szpengchao

prove that if the function is infinitely differentiable on an interval (r,s) containing a, then for any x in (r,s) f(x) can be expand :

f(a) + (x-a) f'(a) + (x-a)^2*f''(a)/2! +....+ (x-a)^n * f(n')(a) + Rn(f,a,x)

Of course a function can be expanded in this way. Just expand a function as a polynomial. Then add the remainder term to make it equal to the function. I assume you are not asking what you want to ask. You want to bound the remainder, and therefore say that the approximation is a good approximation.

**Opalg** · May 15th 2008, 11:05 AM

Originally Posted by szpengchao

prove that if the function is infinitely differentiable on an interval (r,s) containing a, then for any x in (r,s) f(x) can be expand :

f(a) + (x-a) f'(a) + (x-a)^2*f''(a)/2! +....+ (x-a)^n * f(n')(a) + Rn(f,a,x)

Rn is the remainder

There are three standard ways to express the remainder term R_n(f,a,x): the Lagrange form, the Cauchy form and the integral form. They are all described here, together with proofs.

**ThePerfectHacker** · May 15th 2008, 03:02 PM

Here is another proof of Lagrange's estimate using repeated application of Rolle's theorem.

Let $x\not =0$ and let $a$ be the solution to the equation:

$f(x)=\sum_{k=0}^{n}\frac{f^k(0)}{k!}x^k + \frac{ax^{n+1}}{(n+1)!}$ .

The argument will be complete if we can show $a=f^{(n+1)}(\mu)$ for some $\mu$ between $0$ and $x$ .
Now define the function,

$g(y) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}y^k + \frac{ay^{n+1}}{(n+1)!} - f(y)$ .

Notice that $g(0)=g'(0)=g''(0)=...=g^{(n)}(0)=0$ .

Since $g(x) = 0$ by definition of $a$ we know by Rolle's theorem there is $x_1$ between $0$ and $a$ so that $g'(x_1)=0$ . But since $g'(0)=g'(x_1)$ by Rolle's theorem again there is an $x_2$ between $0$ and $a$ so that $g''(x_2)=0$ . This continues until we get an $x_{n+1}$ between $0$ and $x_n$ so that $g^{(n+1)}(x_{n+1})=0$ . But then by definition of $g$ it follows that $a = f^{(n+1)}(x_{n+1})$ , choose $\mu = x_{n+1}$ and the proof is complete.

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