Assuming two functions are invertible, is it true that the inverse of the sum of the two functions is the sum of the inverses (assuming all functions are well behaved)?


Let $f(x) = x, g(x) = -x$, both obviously invertible. Then $(f+g)(x) == 0$, which is not invertible.


As gnometorule pointed in general the sum of invertible is not necessarily invertible.. Anyhow, if two functions are strictly increasing over $\mathbb R$ then their sum is invertible.. Still there is no connection in general between the inverse of the sum and the inverses.

Consider the following simple example:

$$f(X)=X^{5}, g(x)=X \,.$$

Then, both functions are invertible, and so is $f+g$. Anyhow, using Galois Theory, it can be shown that $(f+g)^{-1}(-1)$ is not a number expressible via radicals, let alone in terms of $(-1)$ and $\sqrt[5]{(-1)}=-1$.

  • $\begingroup$ Hi ! Though I have not taken galois theory,but still I want to know are you referring to differential galois theory? $\endgroup$ May 15 '19 at 21:16
  • 1
    $\begingroup$ @NewBornMATH No. Just standard algebraic Galois theory. The "simplest" equation which cannot be solved via radicals is $X^5+X+1=0$. $\endgroup$
    – N. S.
    May 16 '19 at 1:08

No, this one is wrong. The best invertible function is $f(x)=x$. Then $f^{-1}=x$ and $f(x)+f(x)=2f(x)$, but $$(2f(x))^{-1}=\frac{x}{2}$$


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