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Math Help - Proof of the Lagrange formula

  1. #1
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    Proof of the Lagrange formula

    I'm not too sure how to do this question?

    Suppose that f is (n+1)-times differentiable on some open interval I containing 0 and that f^{(n+1)} is continuous on I. Suppose that x\in I and recall that the remainder term in Taylor's theorem is given by:

    R_{n+1}(x)=\frac{1}{n!}\int_0^x f^{n+1}(t)(x-t)^n dt

    Use the mean value theorem for integrals to deduce that:

    R_{n+1}(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}

    where c is some real number between 0 and x
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  2. #2
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    Quote Originally Posted by acevipa View Post
    I'm not too sure how to do this question?

    Suppose that f is (n+1)-times differentiable on some open interval I containing 0 and that f^{(n+1)} is continuous on I. Suppose that x\in I and recall that the remainder term in Taylor's theorem is given by:

    R_{n+1}(x)=\frac{1}{n!}\int_0^x f^{n+1}(t)(x-t)^n dt

    Use the mean value theorem for integrals to deduce that:

    R_{n+1}(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}

    where c is some real number between 0 and x
    There are many different versions of the MVT for integrals, one of them states that If
    f(x),g(x) are continuous on [a,b] then

    there exists a c \in (a,b) such that

    \displaystyle \int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx

    So in you case we have the interval [0,x] and the functions

    f^{n+1}(t),\text{ and }\frac{(x-t)^n}{n!} this gives

    \displaystyle \int_{0}^{x}f^{n+1}(t)\frac{(x-t)^n}{n!}dt=f^{n+1}(c)\int_{0}^{x}\frac{(x-t)^n}{n!}dt=f^{n+1}(c)\frac{-(x-t)^{n+1}}{(n+1)!}\bigg|_{0}^{x}=... for some c \in (0,x)
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