All roots to one variable transcendental equation

Question

How to find all possible roots for one variable transcendental equation? The equation is as follows: $\cosh(x)\cos(x)+1=0$

How should I solve the above equation if it has two variables? $\cosh(x)\cos(y)+1=0$

Answer 1

There is a real root between $n \pi$ and $(n+1) \pi$ for every integer $n$. For large $n$, writing $x = (n+1/2) \pi + t$, the equation becomes $$\sin(t) = \frac{(-1)^n}{\cosh(t + (n+1/2) \pi)}$$ No closed form, but we can approximate it. Let $\tau = e^{-(n+1/2) \pi}$, so we can expand $$ \sin(t) \approx 2 (-1)^n \left( \tau e^{-t} - \tau^3 e^{-3t} + \tau^5 e^{-5t} + \ldots \right)$$ and expand the solution in an asymptotic series:

$$t = (-1)^n \left( 2 \tau - 4 \tau^2 + \frac{34}{3} \tau^3 - \frac{113}{3} \tau^4 + \ldots\right) $$

Answer 2

In this question, there are two problems :

• the equation is transcendental which means that, more than likely, numerical methods would be required.
• because of the cosine, there will be an infinite number of roots.

Moreover, if $x$ is a root, $-x$ will be another one.

PLotting the function over a given range will give an idea about the root locations and using a "reasonable" guess $x_0$ for a specific root, Newton method will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In this specific case $$f(x)=\cos (x) \cosh (x)+1$$ $$f'(x)=\cos (x) \sinh (x)-\sin (x) \cosh (x)$$

For illustration purposes, let us consider the first roots $$\left( \begin{array}{cc} n & x_n \\ 0 & 2 \\ 1 & 1.88527 \\ 2 & 1.87518 \\ 3 & 1.87510 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 5 \\ 1 & 4.76087 \\ 2 & 4.69825 \\ 3 & 4.69411 \\ 4 & 4.69409 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 8 \\ 1 & 7.87238 \\ 2 & 7.85506 \\ 3 & 7.85476 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & 11 \\ 1 & 10.9956 \\ 2 & 10.9955 \end{array} \right)$$

As you can notice, more or less, the next root is more or less equal to the previous root plus $\pi$.

Naming $a_n$ the $n^{th}$ root,we can easily approximate the next one using a single step of Newton method writing $$0=f(a_{n+1})\approx f(a_n+\pi)+(a_{n+1}-a_n-\pi)f'(a_n+\pi)\implies a_{n+1}\approx$$