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Proving equality of a function

  1. Jul 11, 2012 #1
    I have the solution to the question, but I don't understand the first step so I am just gonna paste the first step here.I need people to explain to me.

    Question:
    For the function f(x) = (logx)/x, prove the following equality.

    fn (1) = (-1)(n-1) n![1+ 1/2 +1/3 +...+1/n]

    First step of the solution(The bolded ones are what I don't understand):
    Let y(x) = 1 - x and g(x) = f(y(x)) = log(1 - x) / (1 - x). Then:

    g(y(x)) = log(1 - (1 - x)) / (1 - (1 - x)) = log(x) / x = f(x)

    Thus:

    f'(x) = g'(y(x)) * y'(x) ... (chain rule)
    = g'(y(x)) * -1
    = -g'(y(x))

    f''(x) = -g''(y(x)) * y'(x)
    = -g''(y(x)) * -1
    = g''(y(x))

    So on, by induction, we can see that:

    f^(n)(x) = (-1)^n * g^(n)(y(x))
    ===> f^(n)(1) = (-1)^n * g^(n)(y(1)) = (-1)^n * g^(n)(0)

    Ok, first off, why and how do we know to put y(x) = 1 - x?
    Secondly, the chain rule part. Where does he get y(x) in that chain rule?
    I thought f'(x)=g'(y(x))
     
  2. jcsd
  3. Jul 11, 2012 #2

    arildno

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    Dearly Missed

    As to your first question: "How and why?", that is a misfocus!
    Instead, you should focus on understanding that introducing the auxiliary function y(x) is a LEGITIMATE mathematical operation.
    Do you agree that it is?
     
  4. Jul 11, 2012 #3
    Yes, but that's not my question. I want to know how y(x) is 1- x, not why y(x) is introduced.
     
  5. Jul 11, 2012 #4

    chiro

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    With regard to the chain rule, the derivative of y(z(x)) with respect to x is z'(x)y'(z(x)) and not y'(z(x)) which you stated in your last sentence which needs to be taken into account.
     
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