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I know that

$f_x=e^{t(x)}$

(where the notation $f_x=\frac{df}{dx}$)

(EDIT: $f=f(x)$ and $t$ parameterizes $x$, so $x=x(t) \Leftrightarrow t=t(x)$)

and that therefore

$\frac{d^n f_x}{dx^n}=\frac{e^t}{\dot{x}^n}$

(where $\dot{x}=\frac{dx}{dt}$)

I need to find out what the function $f(x)$ is from this information. Help appreciated.

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    @joriki: Nevermind, I don't know what I was thinking.2012-08-15

1 Answers 1

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My idea, excecution heavily dependend on $t(x)$:

$\ \ \ \ \frac{\text df(x)}{\text dx}=\text{e}^{t(x)}=\sum_{n=0}^\infty \frac{1}{n!}t(x)^n$

$\ \ \ \Longrightarrow\ \ \ f(x)=\sum_{n=0}^\infty \frac{1}{n!}\int t(x)^n \text d x,$

$\ \ \ \ \int t(x)^n \text d x\ \ \ \ ?$

Ansatz $t(x)^ng(x)$:

$\ \ \ \ \frac{\text d}{\text dx}(t^n g) =n t^{n-1}t'g+t^ng' =t^n(ngt'/t+g') =t^n(g\ \text{log}(t^n)'+g') \overset{!}{=}t^n$

To solve

$\ \ \ \ g'=1-\text{log}(t^n)'\ g,$

which will probably look something like $h_1(x)+h_2(x) \text e^{\int\ h_3(x) g(x)\ \text d x}$.