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I'm trying to derive a function f(x) that has the following properties:

$$f_x-\frac{f}{x}=g_x$$

(You might call this the Lagrangian?)

where $$\begin{align*} f&=f(x)\\ f_x&=\frac{df}{dx}\\ g_x&=\frac{d}{dx}\left(\frac{f}{x}\right) \end{align*}$$

By rearranging I've gotten to the point where: $$\frac{f}{f_x}=\frac{x^2-x}{x+1}.$$ But can't figure out what function satisfies this.

Thanks

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1 Answer 1

up vote 2 down vote accepted

Your derivation is incorrect (you've got a wrong sign). I'll use $f'$ instead of $f_x$ (too many $x$'s around...)

We have: $$f' - \frac{f}{x} = \frac{d}{dx}\frac{f}{x}.$$ If $f=0$, we get a solution. So assume that $f$ is not always $0$. Then the above equation is equivalent to $$\begin{align*} f' - \frac{f}{x} &= \frac{xf' - f}{x^2}\\ x^2f' - xf &= xf' - f\\ x^2f' - xf' &= xf - f\\ (x^2-x)f' &= (x-1)f\\ \frac{f'}{f} &= \frac{x-1}{x^2-x} = \frac{x-1}{x(x-1)} = \frac{1}{x}. \end{align*}$$ (you can see that you had $x+1$ instead of $x-1$).

Now, integrating we have: $$\begin{align*} \ln|f| &= \ln|x|+C\\ |f| &= Ax &&A\gt 0\\ f(x) &= Bx &&B\neq 0. \end{align*}$$ Adding in $B=0$ we get that the solutions are $f(x)=Cx$, with $C$ a constant. Indeed, notice that if $f(x) = Cx$, then $$f'-\frac{f}{x} = C - C = 0,$$ and $$\frac{d}{dx}\frac{f}{x} = \frac{d}{dx}C = 0,$$ so they all satisfy your desired equation.

(If you actually had $\frac{f'}{f} = \frac{x+1}{x^2-x}$, then this can be solved by integration as well: $$\begin{align*} \ln|f| &= \int\frac{x+1}{x^2-x}\,dx\\ &= \int\left(\frac{-1}{x} + \frac{2}{x-1}\right)\,dx\\ &= -\ln|x| + 2\ln|x-1| + C\\ &= \ln\left|\frac{(x-1)^2}{x}\right| + C, \end{align*}$$ and from this you can likewise obtain a formula for $f$.)

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Thanks a lot. Actually, I've noticed that a lot of functions (but not all) satisfy: $f'-\frac{f}{x}=xg'$ which is the same constraint I opened the post with except now the right side of the equation is multiplied by $x$. I wonder what this constraint says about the group of functions that satisfy it? For example I wonder if it says anything about their convexity/concavity? –  ben Oct 30 '11 at 7:08
    
@ben: That equation reduces to triviality. It is satisfied by every function that is differentiable. Because $$x\left(\frac{f}{x}\right)' = x\left(\frac{xf' - f}{x^2}\right) = f'-\frac{f}{x}.$$What function do you believe does *not* satisfy the equation $$f' - \frac{f}{x} = f' - \frac{f}{x}\ ?$$ –  Arturo Magidin Oct 30 '11 at 18:27

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