Analytical solution(s) to $x^2 -1 = \sin(x)$

Question

Is there an analytical solution of this function?

$x^2 - 1 = \sin(x)$

I could only find numerical solutions for this equation.

Answer 1

There are analytic results for this equation involving summations over all the solutions of this equation in the complex plane. We can write the equation as $f(z) = 0$ with $f(z) = \sin(z) + 1 - z^2$. We can then consider the contour integral:

$$I(R) = \oint_{C(R)} \frac{1}{z^n}\frac{f'(z)}{f(z)}dz$$

where $C$ is a contour with the origin as center and radius R. Now, in this case for $n>1$, $I(R)$ tends to zero of we send $R$ to infinity such that the contour avoids the zeros of $f(z)$ . The contour integral can also be computed by summing all the residues at the poles of the integrand, which yields the summation of the reciprocals of the $n$th power of the roots plus the residue at $z = 0$. The summation over all the roots is thus minus the residue at $z = 0$.

We can compute the residue at $z=0$ by computing the series expansion of $\log\left[f(z)\right]$:

$$\log\left[f(z)\right] = z - \frac{3}{2}z^2 + \frac{7}{6}z^3 - \frac{19}{12}z^4 + \frac{15}{8}z^5 + \mathcal{O}(z^6)$$

The residue of the integrand at $z = 0$ is the coefficient of $z^{-1}$ in the expansion of the integrand, which is $n$ times the coefficient of $z^n$ of $\log\left[f(z)\right]$. Denoting the roots of $f(z)$ by $\alpha_j$, we thus have:

$$ \begin{split} \sum_{j}\frac{1}{\alpha_j^2} &= &3\\ \sum_{j}\frac{1}{\alpha_j^3} &= -&\frac{7}{2}\\ \sum_{j}\frac{1}{\alpha_j^4} &= &\frac{19}{3}\\ \sum_{j}\frac{1}{\alpha_j^5} &= -&\frac{75}{8} \end{split} $$

Answer 2

If $f(x) = x^2 - 1 - \sin(x) $, since $0 \le 1+\sin(x) \le 2$, then $f(x) > 0$ for $x > \sqrt{2}$ or $x < -\sqrt{2}$.

Numerically, according to Wolfy, there are only two real root s at $x≈-0.636733$ and $x≈1.40962$.

Answer 3

As alreasy said in comments and answer, there is no analytical solutions.

However, the function can be approximated using Pade approximants. To keep the problem simple, let us use degree $2$ for numerator and degree $n$ for denominator.

For $n=1$, the approximant would be $$\frac{\frac{7 x^2}{6}-\frac{5 x}{6}-1}{1-\frac{x}{6}}\implies x_{1,2}=\frac{1}{14} \left(5\pm\sqrt{193}\right)$$ For $n=2$ $$\frac{\frac{47 x^2}{42}-\frac{6 x}{7}-1}{\frac{x^2}{42}-\frac{x}{7}+1}\implies x_{1,2}=\frac{1}{47} \left(18\pm\sqrt{2298}\right)$$ For $n=3$ $$\frac{\frac{1043 x^2}{940}-\frac{203 x}{235}-1}{\frac{11 x^3}{2820}+\frac{5 x^2}{188}-\frac{32 x}{235}+1}\implies x_{1,2}=\frac{2 }{1043}\left(203\pm\sqrt{286314}\right)$$ I give below a table od the decimal representation of the solutions as functions of $n$ $$\left( \begin{array}{ccc} n & x_1 & x_2 \\ 0 & -0.61803 & 1.61803 \\ 1 & -0.63518 & 1.34946 \\ 2 & -0.63697 & 1.40292 \\ 3 & -0.63679 & 1.41531 \\ 4 & -0.63674 & 1.40812 \\ 5 & -0.63673 & 1.40966 \end{array} \right)$$ to be compared to the solutions given by marty cohen in his answer.