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Thread: Calculus: solving an equation

  1. #1
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    Calculus: solving an equation

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

    Note: Don't worry about details such as where $\displaystyle x$ and $\displaystyle f(x)$ 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 $\displaystyle f(x)=y$ we have...

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

    ... with the condition...

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

    Since is ...

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

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

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

    ... that can be written as...

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

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; May 21st 2009 at 04:44 AM. Reason: error in (5)
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Setting $\displaystyle f(x)=y$ we have...

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

    ... with the condition...

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

    Since is ...

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

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

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

    ... that can be written as...

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

    $\displaystyle \chi$ $\displaystyle \sigma$
    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 $\displaystyle (x+y)y'=y.$ i'll call this (1). from $\displaystyle y \ln y = x$ it's clear that $\displaystyle y(0)=1$ and so from (1) we have $\displaystyle y'(0)=1.$

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

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

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

    therefore $\displaystyle f(x)=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty} \frac{b_n}{n!}x^n=\sum_{n=0}^{\infty}\frac{(1-n)^{n-1}}{n!}x^n.$
    Last edited by NonCommAlg; May 21st 2009 at 04:36 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...

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

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

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

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

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

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; May 21st 2009 at 04:58 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 $\displaystyle (x+y)y'=y$... i'll call this (1)....from $\displaystyle y \ln y = x$ it's clear that $\displaystyle y(0)=1$ and so from (1) we have $\displaystyle y'(0)=1$
    Could you explain better this point, please!...

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

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

    ... is ...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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...

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

    ... is ...

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

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

    Kind regards

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