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Math Help - Help with some questions...

  1. #1
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    Help with some questions...

    The first one - I have no idea...

    the second - I guess we can use Cauchy's mean value theorem, but how??

    the third - Taylor's theorem?
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  2. #2
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    I do not really like the notation,
    y=y(x) and x=x(y).
    It makes me want to vomit.

    Let us assume,
    y=f(x) has an inverse on some open interval
    y^{-1}=f^{-1}(x) is the inverse function.
    Further, assume y is twice differenciable on the interval.
    Then, y^{-1} is twice differenciable on the interval.

    Thus,
    f(f^{-1}(x))=x throughout the interval.
    Take derivative (chain rule),
    (f^{-1}(x))'f'(f^{-1}(x))=1
    Thus,
    f'(f^{-1}(x))=\frac{1}{(f^{-1}(x))'}*)
    Take derivative again, use chain on left use quotient on right.
    (f^{-1}(x))'f''(f^{-1}(x))=-\frac{(f^{-1}(x))''}{f^{-1}(x))')^2}
    Thus, (I assume, but am lazy to check that).
    (f^{-1}(x))'f''(f^{-1}(x))=f''(x)
    Thus,
    f''(x)=-\frac{(f^{-1}(x))''}{(f^{-1}(x))')^2}
    In your notation,
    \frac{d^2 y}{dx^2}=-\frac{ \frac{d^2 x}{dy^2} }{\left( \frac{ dx }{dy}\right)^2}


    *)Note (f^{-1}(x))'\not = 0 at some point. There are two possibilities. Either zero throught the interval in that case, f^{-1}(x)=C but then f(x) cannot exists, because it is one-to-one map. And it cannot happen that it is zero at some point but not all, that will lead to non-differenciability. Thus, we can divide.
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  3. #3
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    Yes! Use the Extended Mean Value theorem.

    Work on the interval,
    [0,x].
    Both are differenciable on,
    I=(0,x)
    Both are continous on,
    [0,x].
    Now,
    f(x)=\ln (1+x) thus, f'(x)=\frac{1}{1+x}
    And,
    g(x)=\sin^{-1} (x) thus, g'(x)=\frac{1}{ \sqrt{1-x^2} }
    Where, g'(x)\not = 0 \forall x\in I.
    Thus, \exists c\in I
    \frac{\sqrt{1-c^2}}{1+c}=\frac{\ln (1+x)}{\sin^{-1} x}
    But, 0<c<x.
    Thus,
    \sqrt{\frac{1-x}{1+x}}<\sqrt{\frac{1-c}{1+c}}=\frac{\ln (1+x)}{\sin^{-1} x}
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  4. #4
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    ?

    thank you very much!

    i didn't understand one thing... (in the image)
    can you explain?
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  5. #5
    TD!
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    Quote Originally Posted by ThePerfectHacker View Post
    Thus, (I assume, but am lazy to check that).
    (f^{-1}(x))'f''(f^{-1}(x))=f''(x)
    This is wrong, probably because you trusted the problem statement to be correct.

    Suppose y is a function of x and you want to switch the roles of y and x, i.e. make y the independent variable of the function x. It is safer to introduce new variables, let me show it for Phoebe83 in particular. Let:

    <br />
\left\{ \begin{array}{l}<br />
 x = u \\ <br />
 y = t \\ <br />
 \end{array} \right.<br />

    Then:

    <br />
y' = \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dy}}}} = \frac{1}{{x'}}<br />

    Of course, y' means that y as a function of x is being differentiated with respect to x, while x' means that x as a function of y is being differentiated with respect to y. Taking the second derivative, which I'll take with respect to t again, using the chain rule:

    <br />
y'' = \frac{{dy'}}{{dx}} = \frac{{\frac{{dy'}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{1}{{\frac{{dx}}{{dt}}}}\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{1}{{\frac{{dx}}{{dt}}}}\frac{d}{{dt}}\left( {\frac{1}{{\frac{{dx}}{{dt}}}}} \right)\frac{1}{{\frac{{dx}}{{dt}}}}\left( {\frac{{0 \cdot \frac{{dx}}{{dt}} - 1 \cdot \frac{{d^2 x}}{{dt^2 }}}}{{\left( {\frac{{dx}}{{dt}}} \right)^2 }}} \right)<br />

    In the last step, I explicitly wrote the quotient rule. This simplifies to:

    <br />
y'' =  - \frac{{\frac{{d^2 x}}{{dt^2 }}}}{{\left( {\frac{{dx}}{{dt}}} \right)^3 }} =  - \frac{{x''}}{{x'^3 }}<br />

    So there should be a cube instead of a square in the denominator.
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  6. #6
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    How does that happen then?
    If these expressions are not equal then the original problem is not true*)


    *)Unless (f^{-1}(x))'f''(f^{-1}(x))-f''(x)=C
    They differ by some constant throught the interval.
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  7. #7
    TD!
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    That's why I said:

    "This is wrong, probably because you trusted the problem statement to be correct."

    The intial statement isn't correct, at least I think it's incorrect.
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