I'm trying to find the limit value of this for large values of $x$, in terms of a closed form formula. However when I try to plot this using different representations I get different curves.

For $\cos(x)\cosh(x)-1$:

pic1

For $\cosh(x)-1/\cos(x)$:

pic2

For $\cos(x)-1/\cosh(x)$:

pic3

The answer was that the $\cos(x)-1/\cosh(x) $ gives the correct picture, and that $x=(n+1/2)\pi$ is the correct approximation. Why do I get these different graphs?

  • Well, picture 2 is about a function that has many asymptotes, I wouldn't use it. – egreg Dec 6 at 16:33

All functions are good, except that the second one creates problems for the drawing tool, because it has many asymptotes and they're in the proximity of the zeros.

As you see from the picture below, both graphs give the same zeros.

The curve $\cos x-\dfrac{1}{\cosh x}$ is best, because the function is bounded.

enter image description here

  • Ah so although the curves are different, they all have the same roots. Now its just a matter of which representation is best. Thank you! – aldo Dec 6 at 20:05

You may get numerical errors because cosh(x) grows very quickly.

Write the equation as

$\cos(x)=\frac{1}{\cosh{x}}$,

When $x$ is large, the solutions are going to be approximately

$\cos(x)=0$.

*** $\cos(x)\cosh(x)-1=0$ is the frequency equation of an Euler-Bernoulli beam under free-free or fixed-fixed boundary conditions.

As you noticed and as said in answers, the solutions are closed to $(2n+1)\frac \pi 2$.

As @egreg suggested, it is better to look for the zero of function $$f(x)=\cos (x)-\dfrac{1}{\cosh (x)}$$

Let us make one Newton iteration using $x_0=(2n+1)\frac \pi 2$; this will give as first iterate $$x_1=(2n+1)\frac \pi 2+\frac{\text{sech}\left((2n+1)\frac \pi 2\right)}{(-1)^{n+1}+\tanh \left((2n+1)\frac \pi 2\right)\, \text{sech}\left((2n+1)\frac \pi 2\right)}$$ Now, let us look at $$d_n=|x_1-x_0|_n=\left|\frac{\text{sech}\left((2n+1)\frac \pi 2\right)}{(-1)^{n+1}+\tanh \left((2n+1)\frac \pi 2\right)\, \text{sech}\left((2n+1)\frac \pi 2\right)} \right|$$ and better at $\log_{10}(d_n)$. You will get these numbers $$\left( \begin{array}{cc} n & \log_{10}(d_n) \\ 1 & -1.75330 \\ 2 & -3.10957 \\ 3 & -4.47430 \\ 4 & -5.83866 \\ 5 & -7.20304 \\ 6 & -8.56742 \\ 7 & -9.93179 \\ 8 & -11.2962 \\ 9 & -12.6605 \\ 10 & -14.0249 \end{array} \right)$$ Using a quick and dirty linear regression $\log_{10}(d_n)=a+b n$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -0.381456 & 0.000156 & \{-0.381766,-0.3811462\} \\ b & -1.364372 & 0.000003 & \{-1.364377,-1.3643667\} \\ \end{array}$$ We could even go further in the approximation since the term $\tanh(.)\text{sech}(.)$ is very small compared to $1$ and make $$d_n \sim \text{sech}\left((2n+1)\frac \pi 2\right)\approx 2 e^{-(2n+1)\frac{\pi}{2} }$$ which would give $\log_{10}(d_n)=-0.381158 -1.36438 \,n$ (quite close to the regression line) and then $$x_1 \sim (2n+1)\frac \pi 2+ 2(-1)^{n+1}e^{-(2n+1)\frac{\pi}{2} }$$

Let us try for $n=2$; this very last approxation will give $x_1=7.8532052$ while the "exact" solution would be $7.8532046$.

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