Number of Real Solutions to Inverse Trigonometric Infinite Series Equation

To find the number of real solutions to the given equation in the open interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$, we first simplify the infinite geometric series involved on both sides of the equation.

For $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$, the common ratios of all infinite series satisfy $|r| < 1$, ensuring convergence.

Step 1: Simplify the argument of the LHS

Let $A$ be the expression inside $\sin^{-1}$: $A = \sum_{i=1}^{\infty} x^{i+1} - x \sum_{i=1}^{\infty} \left(\frac{x}{2}\right)^i$

Using the sum formula for an infinite geometric progression $S = \frac{a}{1 - r}$: $\sum_{i=1}^{\infty} x^{i+1} = \frac{x^2}{1 - x}$ $\sum_{i=1}^{\infty} \left(\frac{x}{2}\right)^i = \frac{\frac{x}{2}}{1 - \frac{x}{2}} = \frac{x}{2 - x}$

Substituting these sums back into $A$: $A = \frac{x^2}{1 - x} - x \left(\frac{x}{2 - x}\right) = x^2 \left( \frac{1}{1 - x} - \frac{1}{2 - x} \right) = \frac{x^2}{(1 - x)(2 - x)}$

Step 2: Simplify the argument of the RHS

Let $B$ be the expression inside $\cos^{-1}$: $B = \sum_{i=1}^{\infty} \left(-\frac{x}{2}\right)^i - \sum_{i=1}^{\infty} (-x)^i$

Using the sum formula for an infinite geometric progression: $\sum_{i=1}^{\infty} \left(-\frac{x}{2}\right)^i = \frac{-\frac{x}{2}}{1 - \left(-\frac{x}{2}\right)} = -\frac{x}{2 + x}$ $\sum_{i=1}^{\infty} (-x)^i = \frac{-x}{1 - (-x)} = -\frac{x}{1 + x}$

Substituting these sums back into $B$: $B = -\frac{x}{2 + x} - \left(-\frac{x}{1 + x}\right) = \frac{x}{1 + x} - \frac{x}{2 + x} = \frac{x}{(1 + x)(2 + x)}$

Step 3: Simplify the Equation

The given equation is: $\sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(B)$

Using the identity $\frac{\pi}{2} - \cos^{-1}(B) = \sin^{-1}(B)$ for $B \in [-1, 1]$, the equation becomes: $\sin^{-1}(A) = \sin^{-1}(B)$

For $x \in \left(-\frac{1}{2}, \frac{1}{2}\right)$, we check the domains of $A$ and $B$:

• $A = \frac{x^2}{(1 - x)(2 - x)} \in \left[0, \frac{1}{3}\right) \subset [-1, 1]$
• $B = \frac{x}{(1 + x)(2 + x)} \in \left(-\frac{2}{3}, \frac{2}{3}\right) \subset [-1, 1]$

Since $A, B \in [-1, 1]$, $\sin^{-1}(A) = \sin^{-1}(B)$ reduces to: $A = B \implies \frac{x^2}{(1 - x)(2 - x)} = \frac{x}{(1 + x)(2 + x)}$

Step 4: Solve for $x$

Case 1: $x = 0$ Direct substitution yields $A = 0$ and $B = 0$, giving: $\sin^{-1}(0) = \frac{\pi}{2} - \cos^{-1}(0) \implies 0 = 0$ Thus, $x = 0$ is a valid real solution in $\left(-\frac{1}{2}, \frac{1}{2}\right)$.

Case 2: $x \neq 0$ Dividing both sides by $x$: $\frac{x}{(1 - x)(2 - x)} = \frac{1}{(1 + x)(2 + x)}$

Cross-multiplying and expanding: $x (1 + x)(2 + x) = (1 - x)(2 - x)$ $x(x^2 + 3x + 2) = x^2 - 3x + 2$ $x^3 + 3x^2 + 2x = x^2 - 3x + 2$ $x^3 + 2x^2 + 5x - 2 = 0$

Let $h(x) = x^3 + 2x^2 + 5x - 2$.

1. Monotonicity: $h'(x) = 3x^2 + 4x + 5$. The discriminant of $h'(x)$ is $\Delta = 16 - 60 < 0$, so $h'(x) > 0$ for all $x \in \mathbb{R}$. This means $h(x)$ is strictly increasing on $\mathbb{R}$ and has at most one real root.
2. Existence of root in $\left(-\frac{1}{2}, \frac{1}{2}\right)$:
   • $h(0) = -2 < 0$
   • $h\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 2 = \frac{1}{8} + \frac{1}{2} + \frac{5}{2} - 2 = \frac{9}{8} > 0$

By the Intermediate Value Theorem, $h(x) = 0$ has exactly one root in the interval $\left(0, \frac{1}{2}\right) \subset \left(-\frac{1}{2}, \frac{1}{2}\right)$.

Conclusion

Combining both cases, there are two real solutions in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$: $x = 0$ and the unique non-zero root in $\left(0, \frac{1}{2}\right)$.

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