Solving for $\lambda$ in $\frac{k\lambda}{\alpha} \,\tan(\lambda R) -1 = 0$

Question

How to analytically solve for $\lambda$ in the following equation? $$ \frac{k\lambda}{\alpha}\,\tan(\lambda R) -1 = 0. $$ where $k$, $\alpha$, $R$ are positive constants, and $\lambda$ is a variable.

Help me, please.

Answer 1

Let us make a few changes in the notations : let $x=\lambda R$ and $c=\frac{R\alpha}k$ to make the equation $$x \tan(x)-c=0$$ which shows an infinite number of roots and symmetry.

For the time being, we shall consider the case of the first root only which is somewhere between $0$ and $\frac \pi 2$.

We can obtain a quite reasonable approximation of the solution using the $[4,4]$ Padé approximant built around $x=0$ $$x \tan(x) \approx \frac{x^2-\frac{2 }{21}x^4}{1-\frac{3 }{7}x^2+\frac{1}{105}x^4}\tag 1$$ which reduces the problem to a quadratic equation in $X=x^2$, namely $$\frac{c+10}{105} X^2-\frac{3 c+7}{7} X+c=0$$ giving as an estimate of the solution $$x_0=\sqrt{\frac{45 c+105-\sqrt{15} \sqrt{107 c^2+350 c+735}}{2(c+10)}}\tag 2$$ which must not be used for $c >3600$ since $$\lim_{c\to \infty } \, x_0 =\sqrt{\frac{1}{2} \left(45-\sqrt{1605}\right)}$$ which is slightly $> \frac \pi 2$.

In order to check the quality of the approximation, let us give a value of $x$, compute $c=x \tan(x)$ and compute $x_0$ as given by $(2)$. Below are the results $$\left( \begin{array}{ccc} x & c & x_0 \\ 0.00 & 0.00000 & 0.00000 \\ 0.05 & 0.00250 & 0.05000 \\ 0.10 & 0.01003 & 0.10000 \\ 0.15 & 0.02267 & 0.15000 \\ 0.20 & 0.04054 & 0.20000 \\ 0.25 & 0.06384 & 0.25000 \\ 0.30 & 0.09280 & 0.30000 \\ 0.35 & 0.12776 & 0.35000 \\ 0.40 & 0.16912 & 0.40000 \\ 0.45 & 0.21738 & 0.45000 \\ 0.50 & 0.27315 & 0.50000 \\ 0.55 & 0.33721 & 0.55000 \\ 0.60 & 0.41048 & 0.60000 \\ 0.65 & 0.49413 & 0.65000 \\ 0.70 & 0.58960 & 0.70000 \\ 0.75 & 0.69870 & 0.75000 \\ 0.80 & 0.82371 & 0.80000 \\ 0.85 & 0.96758 & 0.85000 \\ 0.90 & 1.13414 & 0.90000 \\ 0.95 & 1.32846 & 0.95000 \\ 1.00 & 1.55741 & 1.00001 \\ 1.05 & 1.83048 & 1.05001 \\ 1.10 & 2.16124 & 1.10002 \\ 1.15 & 2.56967 & 1.15002 \\ 1.20 & 3.08658 & 1.20003 \\ 1.25 & 3.76196 & 1.25005 \\ 1.30 & 4.68273 & 1.30007 \\ 1.35 & 6.01455 & 1.35010 \\ 1.40 & 8.11704 & 1.40015 \\ 1.45 & 11.9452 & 1.45021 \\ 1.50 & 21.1521 & 1.50028 \\ 1.55 & 74.5216 & 1.55039 \end{array} \right)$$ which seems to be quite good.

Now, if you need more accuracy, start Newton method which will update this estimate according to $$x_{n+1}=\frac{2x_n^2}{2x_n+\sin(2x_n)}$$ Let us try for $c=123.456$. The iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.5585808078448329063 \\ 1 & 1.5559605862010816830 \\ 2 & 1.5552780426163443014 \\ 3 & 1.5552451989102813111 \\ 4 & 1.5552451292564785464 \\ 5 & 1.5552451292561665886 \end{array} \right)$$ which is the solution for twenty significant figures.

All of that could easily be done using Excel.

Edit

For very large values of $c$ (say, for example, $c>1000$), I would prefer to use the expansion $$x \tan(x)=-\frac{\pi }{2 \left(x-\frac{\pi }{2}\right)}-1+\frac{1}{6} \pi \left(x-\frac{\pi }{2}\right)+O\left(\left(x-\frac{\pi }{2}\right)^2\right)$$ Ignoring the higher order terms, let $Y=x-\frac{\pi }{2}$ to get the equation $$\frac{\pi }{6}Y^2-(c+1) Y-\frac{\pi }{2}=0 \implies x_0=\frac{\pi }{2}-\frac{ \sqrt{9 (c+1)^2+3 \pi ^2}-3(c+1)}{\pi }$$ For example, using $c=123456.789$, this would give $x_0=1.5707836034475935978$ while the exact solution would be $1.57078360344759359783$.

In fact, for infinitely lage values of $c$, if you plan to polish the root using Newton method, a very good estimate would be $$x_0=\frac \pi 2 \left(1-\frac 1 c +\frac 1 {c^2}\right)$$ Applied to the first worked example, this would give $x_0=1.558176$.

Update

All of the above was done working around $x=0$. We can do better noticing that $$\lim_{x\to \frac{\pi }{2}} \, \left(1-\frac{4 x^2}{\pi ^2}\right)x\tan (x)=2$$ and that this product varies more or less as $x^2$.

So, minimizing with respect to $a$ $$\int_0^{\frac\pi 2} \left(\left(1-\frac{4 x^2}{\pi ^2}\right)x\tan (x) -a x^2\right)^2$$ we get $$a=\frac{40 \left(7 \pi ^2 \,\zeta (3)-60 \,\zeta (5)\right)}{\pi ^6}$$ and then, for the whole range, the approximation $$x \tan(x)=\frac{40 x^2 \left(7 \pi ^2 \zeta (3)-60 \zeta (5)\right)}{\pi ^4(\pi^2-4 x^2)}$$ from which an estimate $$x_0=\frac{\pi ^3}{2} \sqrt{\frac{c}{\pi ^4 c-600 \zeta (5)+70 \pi ^2 \zeta (3)}}\approx 15.5031 \sqrt{\frac{c}{97.4091 c+208.311}}$$ For the worked examples, this would give $x_0=1.55737$ and $x_0=1.57078$. One or two iterations of Newton method will probably more than sufficient for a quite high accuracy.