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I know the form of $f(x)$ and that of $g(x)$ and I would like to find an expression for a function $H$ such that $H[g(x)] = f(x)$. $f$ is a polynomial and $g$ is more complex and involves some $\exp$ and $\cos$. Is there any procedure to find $H$?

More details:

$f(x) = Ax^6+Bx^{12}$

$g(x) = e^x\left(\cos{x}+1\right)$

I know (numerically) $g(x)$ and $f(x)$ and, I would like to find a function such that $H[g(x)] = f(x)$ without having to evaluate it from $x$.

Thank you!

Alma Arjuna
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Begbi
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  • There is no general procedure for this sort of thing. You will have to be much more specific and tell us what is the domain and codomain of $H$, and how $f$ is determined from $g$. – Ben W Jan 02 '19 at 17:57
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    For example, if $f(x)=x$ you are asking for a universal expression which gives an inverse to $g(x)$. In general, though, inverse functions can be quite hard to construct. – lulu Jan 02 '19 at 18:09

1 Answers1

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Suppose there is such function $H$.

Because $\cos((2n+1)\pi) = -1$, we have $g((2n+1)\pi) = 0$ for $n\in\mathbb N$.

Therefore $f((2n+1)\pi) = H(g((2n+1)\pi)) = H(0)$ for $n\in\mathbb N$.

Thus, $f(x)-H(0)$ is a polynomial with infinite roots. This can only happen if $f(x)\equiv H(0)$.

By this point I think it's clear that what this question desires is not achievable, but, just for the sake of completion, we might go on for a bit more...

Since $f(x) = Ax^6+Bx^{12}$ and we conclude that $f$ must be constant, $A$ and $B$ shall be $0$.

Then, the desired relation is just $H(g(x))\equiv 0$. This can be accomplished by taking $H(x)\equiv 0$.

Alma Arjuna
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