2
$\begingroup$

I have an equation for $\xi$: $$\xi\gamma=\cos\xi,$$

where $\gamma\gg1$. I've tried solving it assuming that $\xi\approx0$ and approximating $\cos$ by Taylor's second order formula:

$$\xi\gamma\approx1-\frac{\xi^2}2,\tag1$$

then I get $$\xi\approx2\left(\sqrt{\gamma^2+1}-\gamma\right).\tag2$$

The book I read gives the approximate solution of it as $$\xi\approx\frac1\gamma\left(1-\frac1{2\gamma^2}\right).\tag3$$

When I tried to understand how they got this, my first thought was like "they made another simplification and assumed $\xi\approx\frac1\gamma$ in $(1)$". I thus thought that this approximation would be worse. But when plotting these solutions as functions of $\gamma$, I found that $(3)$ in fact converges much faster to numerical solution than $(2)$!

So I now wonder: how did they get $(3)$?

$\endgroup$
  • $\begingroup$ You can get it by Newton's method, starting with the approximation 0. Next approximation is $\frac{1}{\gamma}$, and then (3). $\endgroup$ – almagest Sep 8 '14 at 8:07
  • $\begingroup$ More precisely, the next approximation is $$\frac1{\gamma}+\frac{\cos\frac1{\gamma}-1}{\sin\frac1\gamma+\gamma},$$which is further approximated. $\endgroup$ – Yves Daoust Sep 8 '14 at 8:18
  • 1
    $\begingroup$ I think you made an error when applying the quadratic formula $\endgroup$ – Hurkyl Sep 8 '14 at 9:57
  • $\begingroup$ @Hurkyl oops, indeed. Funnily, when I correct it to say $\xi\approx\sqrt{\gamma^2+2}-\gamma$, this approximation appears better than $(3)$, so the question appears based on a false premise. $\endgroup$ – Ruslan Sep 8 '14 at 10:28
2
$\begingroup$

$f(x)=\cos x-\gamma x$ is a concave decreasing function over $(0,\pi/2)$, hence Newton's method gives that the first positive root of $f(x)$ is less than: $$ 0-\frac{f(0)}{f'(0)} = \frac{1}{\gamma} $$ as well as it is less than: $$\frac{1}{\gamma}-\frac{f(1/\gamma)}{f'(1/\gamma)}=\frac{1}{\gamma}-\frac{1}{2\gamma^3}+O\left(\frac{1}{\gamma^5}\right).$$ Further iterations do not change the appearance of the last asymptotics.

$\endgroup$
  • 1
    $\begingroup$ Indeed, looks like it. I suspect the authors in fact used so called "single tangent method" (see Russian wiki page) variation of Newton's method, where the derivative is always taken on the first point and function is evaluated at $x_n$. $\endgroup$ – Ruslan Sep 8 '14 at 10:44
1
$\begingroup$

Since you already received the good answers in comments and answers, let me show you another approximation based on the fact that, for $-\frac{\pi}{2}\leq \xi \leq \frac{\pi}{2}$, $$\cos(\xi) \simeq \frac{5 \pi ^2}{\xi ^2+\pi ^2}-4$$ So, consider the solution of $$ \xi \gamma=\frac{5 \pi ^2}{\xi ^2+\pi ^2}-4$$ that is to say $$\gamma \xi ^3+\pi ^2 \gamma \xi -4 \xi ^2+\pi ^2=0$$ Using Cardano,it seesm that there is only one real root as soon as $$\gamma \gt \frac{\sqrt{\frac{13 \sqrt{65}}{8}-\frac{83}{8}}}{\pi }$$ this root $$\xi= \frac{A \left(2^{2/3} A+8\right)+2 \sqrt[3]{2} \left(16-3 \pi ^2 \gamma ^2\right)}{6 A \gamma }$$ $$A=\sqrt[3]{128-63 \pi ^2 \gamma ^2+3 \sqrt{3} \pi \sqrt{4 \pi ^4 \gamma ^6+83 \pi ^2 \gamma ^4-256 \gamma ^2}}$$ can now be expanded for large values of $\gamma$ and the solution is approximated by $$\xi=\frac{1}{\gamma }-\frac{5}{\pi ^2 \gamma ^3}+\frac{55}{\pi ^4 \gamma ^5}+O\left(\left(\frac{1}{\gamma }\right)^6\right)$$ Limiting to the first terms you then have $$\xi\simeq\frac{1}{\gamma }-\frac{5}{\pi ^2 \gamma ^3}=\frac{1}{\gamma }\Big(1-\frac{5}{\pi ^2 \gamma ^2}\Big)$$ and see how close is $\frac{5}{\pi ^2}$ to $\frac{1}{2}$.

Let me consider an example with a small value $\gamma=2$; the approximation of the book is then $0.437500$, the one I proposed is $0.436674$, the solution of the cubic equation $0.449785$ and the exact solution is $0.449785$.

$\endgroup$
0
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align} \xi = {\cos\pars{\xi} \over \gamma} \end{align}

It leads to the iteration: $$ \xi_{n + 1} ={\cos\pars{\xi_{n}} \over \gamma}\,,\qquad \xi_{0} = 0 $$

$$ \xi_{1} = {1 \over \gamma}\,,\qquad\xi_{2}={1 \over \gamma}\,\cos\pars{1 \over \gamma}\approx\color{#66f}{\Large{1 \over \gamma}\pars{1 - {1 \over 2\gamma^{2}}}} $$ since $\ds{\cos\pars{x} \approx 1 - \half\,x^{2}}$ when $\ds{x \approx 0}$.

$\endgroup$
  • $\begingroup$ How does it lead to this iteration? Why does it have to converge? $\endgroup$ – Ruslan Sep 9 '14 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.