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Math Help - how to prove this theorem

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by szpengchao View Post
    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.
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  3. #3
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    Quote Originally Posted by szpengchao View Post
    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.
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    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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