# Math Help - how to prove this theorem

1. ## how to prove this theorem

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

2. 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.

3. 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.

4. 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.