Finding All Solutions For $\sin(x) = x^2$ Hello everyone how can I find the count of the solution for $\sin(x) = x^2$?
I know there is a one solution in $x = 0$ and for the other solutions I tried to find the extreme point of the function: $y = x^2 - \sin(x)$ and $y'$ is:
$y' = 2x -\cos(x)$ but I don't know how to solve this equation.
 A: It's quite obvious that there are no solutions when $x<0$, so we will look for $x\ge0$. You have found that $x=0$ satisfies the equation. Let's analyze for $x>0$:
Take $f(x)=x^2$ and $g(x)=\sin(x)$.
For $x=\frac{\pi}{4}$, some calculations give $f(\frac{\pi}{4})\approx 0.625$ while $g(x) \approx 0.7$: $$f(\frac{\pi}{4}) < g(\frac{\pi}{4})$$
For $x=1$, $f(1)=1$ but $g(1)<1$ since $\sin(x)$ is increasing for $x\in[0,\pi/2]$ and $\sin(\pi/2)=1$, then $$f(1)>g(1)$$
which means $f(x)$ exceeds $g(x)$ between $(\pi/4,1)$ and intersect in this inteval. Now you just need to prove that they can't intersect more than once.
A: Just for the fun of it !
There is no explicit solution for the zero of function $$f(x)=2x -\cos(x)=0$$ If you need it, use Newton method which will converge quite fast as shown in the table
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.000000 \\
 1 & 0.500000 \\
 2 & 0.450627 \\
 3 & 0.450184
\end{array}
\right)$$ Another solution could be a series expansion
$$2x -\cos(x)=1-2 x-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ and use series reversion to get
$$x=t-\frac{t^2}{4}+\frac{t^3}{8}-\frac{11 t^4}{192}+\frac{3 t^5}{128}-\frac{121
   t^6}{23040}-\frac{19 t^7}{5120}+O\left(t^8\right)\quad \text{where}\quad t=\frac{1-f(x)}2$$ Making $f(x)=0$ that is to say $t=\frac 12$, you should get, as an approximation,
$$x =\frac{531037}{1179648}\approx 0.450166$$
Amazing would be to use the $\color{red}{1,400}$ years old approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$  which would lead to the cubic equation
$$2 x^3+4 x^2+2 \pi ^2 x-\pi ^2=0$$
$$x=-\frac{2}{3} \left(1-\sqrt{3 \pi ^2-4} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{63 \pi ^2-32}{4 \left(3 \pi
   ^2-4\right)^{3/2}}\right)\right)\right)\approx 0.449785$$
A: $$ \text {We know that } -1 \le sin(x) \le 1$$
$$ \text {So, x has to be within [-1,1]. }$$
$$ \text {For any value of x beyond this bound, } x^2 \text {will be more than 1.} $$
$$ \text {Also, x cannot be negative. For x} \lt 0, \text {sin(x) is negative whereas } x^2 \text { is positive.}$$
$$ \text {So, x is within [0,1].}$$
In fact there will be only one value of x beyond x = 0, where they will be equal.
You can use multiple methods like Taylor's Series etc. to get an approximate value.
$$ \text {If you try with } \frac {\pi} {6} \text {, } \frac {\pi} {4} \text { and } \frac {\pi} {3}, \text { you realize the x is somewhere between }\frac {\pi} {4} \text { and } \frac {\pi} {3}.$$
A: The parabola $y=x^2$ is concave up and the sine curve $y=\sin x$ is concave down on the interval $[0,1]$, so the second root (which must lie in $[0,1]$) is unique. We can get a reasonable approximation by truncating $\sin x=x-{1\over6}x^3+{1\over120}x^5-\cdots$ at the cubic term, giving $x^2\approx x-{1\over6}x^3$, for which the resulting quadratic, $x^2+6x-1\approx0$, tells us $x\approx-3+\sqrt{9+6}=\sqrt{15}-3\approx0.873$. The actual solution is closer to $0.877$.
