I met a interesting equation: $$\sin\left [\cos\left (x \right ) \right ]=\cos\left [\sin\left (x \right ) \right ]$$

(And of course, the equation has no roots).

So, Let $f\left ( x \right )$ and $g\left ( x \right )$ be 2 continuous function with variable $x$.

I wonder whether the general equation $$\color{red}{f\left [g\left (x \right ) \right ]=g\left [ f\left ( x \right ) \right ]}$$ has roots, and when if it has.

You can solve it by the way you want.

The last, it's just a inquiry of mine. I'm only a high school student, please don't ask me anything too subliminal.

  • 1
    $\begingroup$ Well one thing I see which doesn't give solutions is when $f(x)=C$ and $g(C)\neq C$. $\endgroup$ – Tucker Aug 10 '15 at 5:32
  • $\begingroup$ @Tucker Thanks, but it's only true when $f\left ( x \right )$ and $g\left ( x \right )$ are 2 constants. $\endgroup$ – mja Aug 10 '15 at 5:36
  • 2
    $\begingroup$ for my example $g(x)$ does not have to be a constant. For, instance $f(x) = 1$, $g(x)=cos(x)$ your equation then reads $1=cos(1)$, which is not true. I do see your point that this discussion is a bit degenerate. $\endgroup$ – Tucker Aug 10 '15 at 5:41
  • $\begingroup$ This works when your functions are linear transformations ($f(x) = ax$ and $g(x) = bx$) since $1 \times 1$ matrices over $\mathbb{R}$ commute. In that case, every $x$ is a root :P $\endgroup$ – Mohamad Ali Baydoun Aug 10 '15 at 5:44
  • 1
    $\begingroup$ I would be surprised if anything can be said in general. I'd be surprised to learn anything even if both functions are arbitrary polynomials, let alone for arbitrary continuous functions. $\endgroup$ – pjs36 Aug 10 '15 at 5:48

The equation $\sin(\cos(x)) = \cos(\sin(x))$ does have roots, just not real roots. For example, $ \pm 0.7853981634 \pm 0.4663385348 i$ (approximately) are roots.

In some cases when $f$ and $g$ are polynomials with real coefficients, real roots exist. Suppose the leading terms of $f(x)$ and $g(x)$ are $\alpha x^m$ and $\beta x^n$. Then the leading terms of $f(g(x))$ and $g(f(x))$ are $\alpha \beta^m x^{mn}$ and $\beta \alpha^n x^{mn}$. If $m$ and $n$ are both odd and $\alpha \beta^m \ne \beta \alpha^n$, then $f(g(x)) - g(f(x))$ is a polynomial of odd degree, so it has at least one real root.

On the other hand, here are some polynomial examples where there are no real roots. Try $f(x) = x^2$, $g(x) = a x^2 + b$ with $a, b > 1$. Then $f(g(x)) - g(f(x)) = (a^2 - a) x^4 + 2 a b x^2 + b^2 - b > 0$ for all real $x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.