Roots of $ x^{13} - e^{-x} + x - sin(x) = 0$

Question

What can be said about roots of equation like

$$ x^{13} - e^{-x} + x - sin(x) = 0$$

If we assume f (x) to be above equation then f'(x) is positive, but I am not sure doing it this way. Is there some other general method?

Answer 1

A function $f(x)$ has a zero when its graph crosses the horizontal axis, so if one can decide where $f$ increases and decreases, and where it is positive and negative, then one can determine bounds on the number of its zeros, and some idea of where they can be found. Analyzing where a differentiable function increases and decreases amounts to finding the sign of its derivative, so looking at the derivative is basically inescapable. In your example the derivative of $f$ is everwhere positive, so $f$ is strictly increasing. A strictly increasing function crosses the horizontal axis at most once (as the function $g(x) = e^{x}$ shows, it need not cross the axis at all). Since as $x \to \infty$ your $f(x)$ tends to $+\infty$, it suffices to observe that $f(0) < 0$ to conclude that $f(x)$ has a zero, and that this zero is positive. Since $f(2) > 0$, one can in fact say that the zero is in $(0, 2)$ (the choice o $x = 2$ is quite arbitrary and meant simply to illustrate). To actually find the zero, one needs to use some approximation, such as the Newton method.

Answer 2

Like @Dan Fox: was saying:

$$f'(x) = e^{-x} + (1-\cos x) + 13 x^{12} $$

clearly $>0$ for all $x$.

The unique solution of the equation $f(x)=0$ is approximately $0.9076764..$ .