Question

The number of real solutions of the equation
$sin^{-1} \left(\sum\limits^{\infty}_{i = 1} x^{i+1}-x \sum\limits^{\infty}_{i = 1} \left(\frac{x}{2}\right)^{i}\right) = \frac{\pi}{2}-cos^{-1} \left(\sum\limits^{\infty}_{i = 1} \left(-\frac{x}{2}\right)^{i} - \sum\limits^{\infty}_{i = 1} \left(-x\right)^{i}\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is_____.
(Here, the inverse trigonometric functions $sin^{-1}$ x and $cos^{-1}$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left[0, \,\pi\right]$, respectively.)

Answer 1

$\displaystyle\sum_{i=1}^{\infty} X^{i+1}=\frac{X^{2}}{1-X}$
$\displaystyle\sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^{i}=\frac{x}{2-x} \displaystyle\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^{i}$
$=\frac{-x}{2+x} \displaystyle\sum_{i=1}^{\infty}(-x)^{i}=\frac{-x}{1+x}$
To have real solutions
$\displaystyle\sum_{i=1}^{\infty} X^{i+1}-X \sum_{i=1}^{\infty}\left(\frac{X}{2}\right)^{i}=\displaystyle\sum_{i=1}^{\infty}\left(\frac{-X}{2}\right)^{i}-\displaystyle\sum_{i=1}^{\infty}(X)^{i}$
$\frac{X^{2}}{1-X}-\frac{X^{2}}{2-X}=\frac{-X}{2+X}+\frac{X}{1+X}$
$X\left(X^{3}+2 X^{2}+5 X-2\right)=0 $
$\therefore x=0 $ and let $ f(x)=X^{3}+2 X^{2}+5 X-2$
$f\left(\frac{1}{2}\right) \cdot f\left(-\frac{1}{2}\right) < 0$
Hence two solutions exist