Counting Roots

While $\sin x$ is real for all real $x$ , $\ln(1+x)$ is only real for $x \in (-1, \infty)$ . By Maclaurin series , we have:

$\begin{cases} \begin{aligned} \sin x & = x - \dfrac {x^3}{3!} + \dfrac {x^5}{5!} - \dfrac {x^7}{7!} + \cdots & - \infty < x < \infty \\ \ln (1+x) & = x - \dfrac {x^2}2 + \dfrac {x^3}3 - \dfrac {x^4}4 + \cdots & -1 < x \le 1 \end{aligned} \end{cases}$

And we note that:

$\begin{cases} \sin x > \ln (1+x) & \text{for }-1 < x < 0 \\ \sin x = \ln (1+x) & \text{for }x = 0 \\ \sin x > \ln (1+x) & \text{for } 0 < x \le 1 \end{cases}$

This means that $\sin x \ge \ln(1+x)$ for $-1 < x \le 1$ and $\sin x = \ln(1+x)$ at $x=0$ .

Since $\dfrac d{dx} \ln (1+x) = \dfrac 1{1+x} > 0$ for $x \in (-1, \infty)$ , $\ln(1+x)$ is an increasing function. We note that $\sin x \le 1$ , and $\ln (1+x) = 1$ when $x = e-1 \approx 1,718$ . As $\sin x = 1$ , at $x = \dfrac \pi 2 \approx 1.571$ and decreasing to $0$ at $x = \pi \approx 3.142$ , there must be a $x_1 \in (1.571, 1.718)$ such that $\sin x = \ln (1+x)$ . And since $\ln (1+x)$ is increasing $\sin x < \ln (1+x)$ for $x > x_1$ . Therefore there are only $\boxed 2$ roots to the equation.

A hint to this problem. ## This proof involves only the mean value theorem.The following statements can be easily proved by that only particular theorem.

        try to prove each step!!

F(x)=Sin(x)-ln(1+x)

Clearly. (1+x)>0

x=0 is a root

F(x) is decreasing with F(x)>0 in (-1,0)

F(x)>0 in (0,π/2]

F(x) has {{only}} one root in (π/2,2)

F(x)<0 when x≥2

To show that F(x) has at least one root in (π/2,2) is easy,To prove that no other root is present ,one may use the fact that F'(x)≠0

Beyond the algebraic proof graphing is also a good tool to visualize

They meet when X=0 and when X~=1.6968
So 2 roots