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Math Help - Induction Problem

  1. #1
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    Induction Problem

    having problems with the following problem:

    Consider,

    f(x)=\frac{1}{\sqrt{1-4x}}


    Use the principle of induction to prove that for n>=1 the n^{th} derivative f^{n}(x) of f(x) is:

    f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}

    if anyone could show me how to solve this problem it would be much appreciated.
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  2. #2
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    Quote Originally Posted by Jonny-123 View Post
    having problems with the following problem:

    Consider,

    f(x)=\frac{1}{\sqrt{1-4x}}


    Use the principle of induction to prove that for n>=1 the n^{th} derivative f^{n}(x) of f(x) is:

    f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}


    if anyone could show me how to solve this problem it would be much appreciated.
    note that f'(x)=\frac{2}{(1-4x)^{3/2}}

    First verify the base case when n=1

    f^{1}(x)=\frac{2!}{1!}(1-4x)^{-(1+\frac{1}{2}}=\frac{2}{(1-4x)^{3/2}}

    So the base case checks

    Now we assume it is true for n and show n+1

    f^{n}(x)=\frac{(2n)!}{n!}(1-4x)^{-(n+\frac{1}{2})}

    Now take the derivative of the induction hypothsis to get

    f^{n+1}(x)=\frac{(2n)!}{n!}[-(n+\frac{1}{2})](1-4x)^{-(n+1+\frac{1}{2})}(-4)

    lets simplify a bit

    f^{n+1}(x)=\frac{(2n)!}{n!}(4n+2)(1-4x)^{-(n+1+\frac{1}{2})}

    and now factor out a 2 to get

    f^{n+1}(x)=2\frac{(2n)!}{n!}(2n+1)(1-4x)^{-(n+1+\frac{1}{2})}

    f^{n+1}(x)=2\frac{(2n+1)!}{n!}(1-4x)^{-(n+1+\frac{1}{2})}

    but this isn't quite what we want so lets build the fraction by multiplying by (n+1)

    f^{n+1}(x)=2\frac{(2n+1)!}{n!}\frac{n+1}{n+1}(1-4x)^{-(n+1+\frac{1}{2})}

    f^{n+1}(x)=2(n+1)\frac{(2n+1)!}{(n+1)!}(1-4x)^{-(n+1+\frac{1}{2})}

    f^{n+1}(x)=\frac{(2n+2)!}{(n+1)!}(1-4x)^{-(n+1+\frac{1}{2})}

    f^{n+1}(x)=\frac{[2(n+1)]!}{(n+1)!}(1-4x)^{-[(n+1)+\frac{1}{2}]}

    and we are done.
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