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Approaching transcendental equations is in general new to me. My experience with numerical methods is limited, and this equation seems to require such a method. But there's a catch - it contains an undetermined constant $a$ used to describe a family of such equations.

$$ \pm20 = at \sin(at)$$

My goal is to find a solution for $t$, where $t>0$.

Graphing the original equation with several values of $a$ was useful. It led me to believe that the solution should occur around the seventh extrema, or around three and a fourth periods: $t \approx 3.25\dfrac{2\pi}{a} $.

Any ideas on how to increase my accuracy?

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    $\begingroup$ You could solve $x\sin(x)=\pm20$ numerically. Then $x=at$, so $t=\frac{x}{a}$, giving you an expression for $t$ in terms of $a$. $\endgroup$ – Peter Huxford Oct 30 '14 at 10:49
  • $\begingroup$ Well, that makes too much sense. Looks like I made a mountain out of a molehill! $\endgroup$ – zahbaz Oct 30 '14 at 17:24
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Consider $a\times t$ as a global variable, say $y$. So, you have to solve $$f(y)=y \sin(y)-20=0$$ As you showed, the solution is close to $y=\frac{13 \pi }{2}$ (what you wrote $\approx 3.25 \frac{2}{\pi}$). So, let us write a Newton iteration which, starting from a "reasonable" guess $y_0$ will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ So, we use $$y_0=\frac{13 \pi}{2}$$ and the first iterates are $$y_1=20$$ $$y_2=\frac{20+400 \cos (20)}{\sin (20)+20 \cos (20)} \approx 20.1919$$ Continuing the iterative process will lead to iterates $20.2519$, $20.2598$, $20.2600$ which is the solution for six significant figures.

Another thing you could do is to develop as a Taylor series up to second order; this would give $$f(y)=\left(\frac{13 \pi }{2}-20\right)+\left(y-\frac{13 \pi }{2}\right)-\frac{13}{4} \pi \left(y-\frac{13 \pi }{2}\right)^2+O\left(\left(y-\frac{13 \pi }{2}\right)^3\right)$$ and solve the quadratic the two roots of which being $$\frac{16+676 \pi ^2-\sqrt{256-66560 \pi +21632 \pi ^2}}{104 \pi } \approx 20.2606$$ and $$\frac{16+676 \pi ^2+\sqrt{256-66560 \pi +21632 \pi ^2}}{104 \pi } \approx 20.6781$$ while the exact solutions are $20.2600$ and $20.6769$.

Edit

There is another solution to this problem : building a $[2,2]$ Pade approximant of $y\sin(y)$ at $y=\frac{13 \pi }{2}$ leads to $$y \sin(y)\approx\frac{-\frac{65 \pi }{24}z^2+\frac{\left(48+169 \pi ^2\right) }{6 \left(8+169 \pi ^2\right)}z+\frac{13 \pi }{2}}{\frac{\left(48+169 \pi ^2\right) }{12 \left(8+169 \pi ^2\right)}z^2-\frac{65 \pi }{3 \left(8+169 \pi ^2\right)}z+1}$$ and we are left with a quadratic in $z=y-\frac{13 \pi }{2}$ and solving, the solutions are, for $y$, $\approx 20.2600$ and $20.6769$ which are the exact solutions for six significant figures.

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  • $\begingroup$ Thank you Claude! Wow... It seems so straightforward after treating $a \times t$ as a global variable. I appreciate the nod towards both the Taylor series and Newton's method. $\endgroup$ – zahbaz Oct 30 '14 at 17:21
  • $\begingroup$ You are very welcome ! Funny, isn't ? Cheers :-) $\endgroup$ – Claude Leibovici Oct 30 '14 at 18:32

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