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Math Help - Lagrange & Taylor

  1. #1
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    Lagrange & Taylor

    Find the Lagrange form of the remainder Rn(x).

    f(x) = sqrt(x+1) ; n=3



    So I know the remainder in taylor's theorem can be written as...

    Rn(x) = [f^(n+1) (c)] / [(n+1)!] *[x^(n+1)]

    not sure to actually do the whole application for these kinds of problems.
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  2. #2
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    Quote Originally Posted by WartonMorton View Post
    Find the Lagrange form of the remainder Rn(x).

    f(x) = sqrt(x+1) ; n=3



    So I know the remainder in taylor's theorem can be written as...

    Rn(x) = [f^(n+1) (c)] / [(n+1)!] *[x^(n+1)]

    not sure to actually do the whole application for these kinds of problems.
    You have the formula- just plug in the data for this problem.

    The only "hard" part is finding the general derivative.

    If f(x)= \sqrt{x+ 1}= (x+ 1)^{1/2}, then f'(x)= (1/2)(x+ 1)^{-1/2}, f''(x)= -1/4(x+ 1)^{-3/2}, f'''(x)= 3/8(x+ 1)^{-5/2}, f^{(4)}= -15/16(x+1)^{-7/2}...

    Do you see the pattern? Obviously, there is an alternating + and -: (-1)^{n+1} since this is + for n odd, - for n even.

    The denominator of the fraction is a power of 2: 2^n.

    The numerator is a little harder- it is a product of odd integers from 2n-1 down: 1*3*5*7... We can write that more compactly by filling in the missing evens: \frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}. And we can write that denominator as (2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!.

    The fraction is \frac{(2n-1)!}{(2^n)(2^{n-2}(n-2)!} and the exponent on x+ 1 is \frac{-(2n-1)}{2}.

    That is, f^(n)(x)= \frac{(-1)^{n+1}(2n-1)!}{2^{2n-2}(n-2)!}(x+ 1)^{-\frac{2n-1}{2}}.

    Using that, find a maximum on f^{n+1} between 0 and x.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    You have the formula- just plug in the data for this problem.

    The only "hard" part is finding the general derivative.

    If f(x)= \sqrt{x+ 1}= (x+ 1)^{1/2}, then f'(x)= (1/2)(x+ 1)^{-1/2}, f''(x)= -1/4(x+ 1)^{-3/2}, f'''(x)= 3/8(x+ 1)^{-5/2}, f^{(4)}= -15/16(x+1)^{-7/2}...

    Do you see the pattern? Obviously, there is an alternating + and -: (-1)^{n+1} since this is + for n odd, - for n even.

    The denominator of the fraction is a power of 2: 2^n.

    The numerator is a little harder- it is a product of odd integers from 2n-1 down: 1*3*5*7... We can write that more compactly by filling in the missing evens: \frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}. And we can write that denominator as (2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!.

    The fraction is \frac{(2n-1)!}{(2^n)(2^{n-2}(n-2)!} and the exponent on x+ 1 is \frac{-(2n-1)}{2}.

    That is, f^(n)(x)= \frac{(-1)^{n+1}(2n-1)!}{2^{2n-2}(n-2)!}(x+ 1)^{-\frac{2n-1}{2}}.

    Using that, find a maximum on f^{n+1} between 0 and x.
    I know I have the forumula, but where am I getting the c from? Do I have to solve some sort of equation to get the max?
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