# Question #08c52

Nov 2, 2017

The answer will be long. plz wait

#### Explanation:

There are some steps:

[Step1] Prove that ${e}^{x} \cos x = 1$ has no negative root.

If $x < 0$, the following must be satisfied:
$0 < {e}^{x} < 1$, $- 1 \le \cos x \le 1$
$\to \left\mid {e}^{x} \cos x \right\mid < 1$
So, the roots ${e}^{x} \cos x = 1$ must be 0 or greater.

[Step2] Find where the roots of ${e}^{x} \cos x = 1$ lie.

Let $f \left(x\right) = {e}^{x} \cos x - 1$.
$f \left(0\right) = {e}^{0} \cdot \cos 0 - 1 = 1 \cdot 1 - 1 = 0$ and $x = 0$ is the smallest root.

Differentiate $f \left(x\right)$:
$f ' \left(x\right) = {e}^{x} \left(\cos x - \sin x\right)$
$= - \sqrt{2} {e}^{x} \sin \left(x - \frac{1}{4} \pi\right)$

$x = \left(n + \frac{1}{4}\right) \pi$ satisfies $f ' \left(x\right) = 0$ if $n \ge 0$ is an integer.
Indeed, $f \left(\left(2 n + \frac{1}{4}\right) \pi\right)$ is a local maxima and $f \left(\left(2 n + \frac{5}{4}\right) \pi\right)$ is a local minima.

Here is a graph for $y = {e}^{x} \cos x - 1$. Let $x = \alpha$ is a positive root for $f \left(x\right) = 0$.
$\alpha$ must satisfy $\left(2 n + \frac{1}{4}\right) \pi < \alpha < \left(2 n + \frac{1}{2}\right) \pi$ or $\left(2 n + \frac{3}{2}\right) \pi < \alpha < \left(2 n + 2\right) \pi$ since:
$f \left(\left(2 n + \frac{1}{4}\right) \pi\right) = {e}^{\left(2 n + \frac{1}{4}\right) \pi} / \sqrt{2} - 1 > 0$ (Note that ${e}^{\frac{\pi}{4}} = 2.193 \ldots$ is larger than $\sqrt{2}$)
$f \left(\left(2 n + \frac{1}{2}\right) \pi\right) = - 1 < 0$

$f \left(\left(2 n + \frac{3}{2}\right) \pi\right) = - 1 < 0$
$f \left(\left(2 n + 2\right) \pi\right) = {e}^{\left(2 n + 2\right) \pi} - 1 > 0$

There are no roots between $2 n \pi < x \le \left(2 n + \frac{1}{4}\right) \pi$ because f(x) is monotonically increasing in this range($f ' \left(x\right) > 0$).
Similarly, there are no roots between $\left(2 n + \frac{1}{2}\right) \pi \le x \le \left(2 n + \frac{3}{2}\right) \pi$. f(x) is always negative here.

So,
$\textcolor{red}{\text{it is sufficient to prove that at least one root for " e^xsinx-1=0 " exists in the range } 2 n \pi < x < \left(2 n + \frac{1}{4}\right) \pi \mathmr{and} \left(2 n + \frac{1}{2}\right) \pi < x < \left(2 n + \frac{3}{2}\right) \pi .}$

[Step3] Prove the statement above(in red characters)
Let $g \left(x\right) = {e}^{x} \sin x - 1$.
$g \left(2 n \pi\right) = - 1 < 0$
$g \left(\left(2 n + \frac{1}{4}\right) \pi\right) = {e}^{\left(2 n + \frac{1}{4}\right) \pi} / \sqrt{2} - 1 > 0$
$g \left(\left(2 n + \frac{1}{2}\right) \pi\right) = {e}^{\left(2 n + \frac{1}{2}\right) \pi} - 1 > 0$
$g \left(\left(2 n + 1\right) \pi\right) = - 1 < 0$

The proof is completed via the intermediate value theorem.

Visualization: Roots for ${e}^{x} \cos x - 1 = 0$ is in the red area and roots for ${e}^{x} \sin x - 1 = 0$ is in the green area.