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Math Help - Calculus: solving an equation

  1. #1
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    Calculus: solving an equation

    Suppose f(x)=\sum_{n=0}^{\infty}a_nx^n satisfies the equation f(x)\ln(f(x))=x. Find a closed-form expression for a_n, \ n \geq 0.

    Note: Don't worry about details such as where x and f(x) are defined, etc. You may assume whatever is needed!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting f(x)=y we have...

     y= \sum_{n=1}^{\infty} a_{n}\cdot x^{n} (1)

    ... with the condition...

     y\cdot \ln y = x \rightarrow \ln y = \frac{x}{y} (2)

    Since is ...

     \frac{d}{dx} \ln \{f(x)\} = \frac {f^{'}(x)}{f(x)} (3)

    ... the condition (2) becomes...

     \frac{d}{dx} \ln y= \frac{y^{'}}{y} = \frac {y - x\cdot y^{'}}{y^{2}} (4)

    ... that can be written as...

    y^{'}\cdot (1+\frac{x}{y}) = 1 (5)

    The (5) is a first order ODE and we have to find a solution that for (1) has to be analythic... end of the first step ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 21st 2009 at 05:44 AM. Reason: error in (5)
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Setting f(x)=y we have...

     y= \sum_{n=1}^{\infty} a_{n}\cdot x^{n} (1)

    ... with the condition...

     y\cdot \ln y = x \rightarrow \ln y = \frac{x}{y} (2)

    Since is ...

     \frac{d}{dx} \ln \{f(x)\} = \frac {f^{'}(x)}{f(x)} (3)

    ... the condition (2) becomes...

     \frac{d}{dx} \ln y= \frac{y^{'}}{y} = \frac {y - x\cdot y^{'}}{y^{2}} (4)

    ... that can be written as...

    y^{'}\cdot (1+\frac{x}{y}) = \frac{1}{y} (5)

    The (5) is a first order ODE and we have to find a solution that for (1) has to be analythic... end of the first step ...

    Kind regards

    \chi \sigma
    this is probably the most interesting problem i've ever posted in MHF and i was kind of disappointed to see that it was ignored! so, thanks for replying!

    ok, your last result, (5), is not correct. it should be (x+y)y'=y. i'll call this (1). from y \ln y = x it's clear that y(0)=1 and so from (1) we have y'(0)=1.

    now using Leibniz rule (n-th derivative of product of two functions) in (1) we'll get: (x+y)y^{(n+1)} + n(1+y')y^{(n)} + \sum_{k=2}^n \binom{n}{k}y^{(n-k+1)}y^{(k)}=y^{(n)}. call this (2).

    let y^{(n)}(0)=b_n. we already showed that b_0=b_1=1. so if, in (2), we put x=0, we'll get: b_{n+1} + (2n-1)b_n + \sum_{k=2}^n \binom{n}{k}b_{n-k+1}b_k = 0. we will call this (3).

    finally use induction in (3) to show that b_n=(1-n)^{n-1}, (this won't be an easy induction! also we define 0^0=1, so the formula for b_n will be valid for n \geq 0)

    therefore f(x)=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}  \frac{b_n}{n!}x^n=\sum_{n=0}^{\infty}\frac{(1-n)^{n-1}}{n!}x^n.
    Last edited by NonCommAlg; May 21st 2009 at 05:36 AM. Reason: nothing important!
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  4. #4
    MHF Contributor chisigma's Avatar
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    The ODE we arrived in previous post [and that I have corrected...] can be written simply as...

    y^{'}= \frac{y}{x+y} (1)

    ... and now we have to search a solution of it of the form...

    y(x)= \sum_{n=1}^{\infty} a_{n}\cdot x^{n} (2)

    The 'brute force' approach to (1) is difficult because it is a non linear ODE, som that we swap the role of variables x and y 'transforming' the (1) in ...

     x^{'} = 1 + \frac{x}{y} (3)

    The (3) is a linear ODE and its general solution is easy to find...

    x= y\cdot (\ln y + c ) (4)

    But for (2) must be x(0)=0 so that is c=1 and the inverse of the function we are searching is...

    x=f^{-1} (y) = y\cdot \ln y (5)

    ... and we don't achieve any progress because (5) is not a surprise!... never mind!...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 21st 2009 at 05:58 AM. Reason: mistakes corrected
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by NonCommAlg View Post

    .... it should be (x+y)y'=y... i'll call this (1)....from y \ln y = x it's clear that y(0)=1 and so from (1) we have y'(0)=1
    Could you explain better this point, please!...

    ... the fact is that if we define x as function of y in this way...

     x(y) = y\cdot \ln y (1)

    ... is ...

     \lim_{y \rightarrow 0} y\cdot \ln y=0 (2)

    ... so that x(0)=0 and that means that, by reflessive property, is also y(0)=0... am I whrong? ...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by chisigma View Post
    Could you explain better this point, please!...

    ... the fact is that if we define x as function of y in this way...

     x(y) = y\cdot \ln y (1)

    ... is ...

     \lim_{y \rightarrow 0} y\cdot \ln y=0 (2)

    ... so that x(0)=0 and that means that, by reflessive property, is also y(0)=0... am I whrong? ...

    Kind regards

    \chi \sigma
    we have y(0) \ln(y(0)) = 0. but y(0) \neq 0 because otherwise \ln(y(0)) would be undefined. so we must have \ln(y(0)) = 0, i.e. y(0)=1.
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