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If we are given a function

$g(x) = x - 1/x$

And another one given in terms of composition

$f(g(x)) = x^3 - 1/x^3$

By which general method does one find $f(x)$ ?

Can it be found for arbitrary $g(x)$ and $f(g(x)$?

vallev
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    Hint: see what $,\big(g(x)\big)^3,$ is. – dxiv Jul 21 '23 at 03:30
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    Thanks. I can see that the solution in the given example can be found by playing with algebra.

    $(g(x))^3=(x-1/x)^3= x^3 - 1/x^3 - 3( x - 1/x)= f(g(x))-3g(x)$

    which gives

    $(g(x))^3+-3g(x) = f(g(x))$ and thus $x^3-3x = f(x)$

    I was wondering if there is any mathematical theory or method already existing for these kind of problems in the general case, since the problem feels unintuitive and such trick may not always exist.

     – vallev Jul 21 '23 at 03:41
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    If $g(x)$ is invertible then with $x \mapsto g^{-1}(x)$ you get $f(x) = \big(g^{-1}(x)\big)^3 - 1/\big(g^{-1}(x)\big)^3$. But this particular $g(x)$ is not invertible, though you can use a variation of the idea to get the final result, albeit with some more calculations. But that won't be the case in general, and even if $g(x)$ is bijective, thus invertible, it's not always possible to get an explicit closed form for the inverse. So the answer is basically no, there is no universal method to solve this kind of problems. – dxiv Jul 21 '23 at 04:01
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    Thank you! I think you answered the question. – vallev Jul 29 '23 at 17:49

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