Calculus: solving an equation

• Apr 7th 2009, 01:48 PM
NonCommAlg
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! (Smile)
• May 21st 2009, 02:16 AM
chisigma
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 (Hi) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 21st 2009, 04:04 AM
NonCommAlg
Quote:

Originally Posted by chisigma
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 (Hi) ...

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! (Happy)

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! (Evilgrin) also we define $\displaystyle 0^0=1,$ so the formula for $\displaystyle b_n$ will be valid for $\displaystyle n \geq 0$)

therefore $\displaystyle f(x)=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty} \frac{b_n}{n!}x^n=\sum_{n=0}^{\infty}\frac{(1-n)^{n-1}}{n!}x^n.$ (Cool)
• May 21st 2009, 04:08 AM
chisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$
• May 21st 2009, 05:23 AM
chisigma
Quote:

Originally Posted by NonCommAlg

.... 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? (Nerd) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 21st 2009, 05:28 AM
NonCommAlg
Quote:

Originally Posted by chisigma
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? (Nerd) ...

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. $\displaystyle y(0)=1.$