I am trying to express this formula in terms of N:

$$ A=\frac{a^2N} {a\cdot \tan{\frac{180}{N}}} $$

I really don't know how to do this. I tried and got this:

$$ A \cdot a \cdot \tan{\frac{180}{N}} = a^2N $$ $$ \tan{\frac{180}{N}}=\frac{a^2N}{Aa} $$ $$ \frac{180}{N}=\tan^{-1}{\frac{a^2N}{Aa}} $$ $$ N=\frac{180}{\tan^{-1}{\frac{a^2N}{Aa}}} $$

But I just can't get the N on the right side. The $\tan$ is so annoying! How do I do this?

  • 1
    $\begingroup$ For largish $N$, we have $\tan(180^\circ/N)\approx \frac{\pi}{N}$. This yields a nice (approximate) equation for $N$, easily solved. Unfortunately not exact. But then you can use a numerical procedure, such as fixed point iteration. $\endgroup$ – André Nicolas Feb 6 '16 at 5:45

There is no way to to this. Many implicit equations, especially those involving trig functions, don't have any way to represent them explicitly.

You can use the approximation Andre points out to obtain $$N=\sqrt{\frac{A\pi}{a}}$$

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    $\begingroup$ So if the value of $a$ and $A$ is known, we still can't find $N$? $\endgroup$ – Sweeper Feb 6 '16 at 5:41
  • $\begingroup$ Correct. There are various approximations though, which I can explain if you'd like, but no algebraic solutions. $\endgroup$ – Stella Biderman Feb 6 '16 at 5:43
  • $\begingroup$ Please explain it. Thank you! $\endgroup$ – Sweeper Feb 6 '16 at 5:44
  • $\begingroup$ I was actually about to recommend what André commented a minute ago. It gives $A\pi/a=N^2$ $\endgroup$ – Stella Biderman Feb 6 '16 at 5:47

Equations which mix polynomial and trigonometric terms do not show analytical solutions and numerical methods should be involved.

Let us consider the equation $$A=\frac{a^2N} {a\cdot \tan({\frac{\pi}{N})}}=\frac{aN} {\tan({\frac{\pi}{N})}}$$ So, we can consider that we look for the zero of the function $$f(N)=\frac{aN} {\tan({\frac{\pi}{N})}}-A$$ $$g(N)={aN} -A\,{\tan({\frac{\pi}{N}})}$$ $$h(N)=a N \cos(\frac{\pi}{N})-A \sin(\frac{\pi}{N})$$ Probably, the last equation will be better to consider since discontinuties have been removed.

For illustration purposes, let us consider the case $a=2$, $A=1$. A plot of the function $h(N)$ shows a solution somewhere between $2$ and $3$. So, let us start Newton method using $N_0=2$; this will generates the following iterates $$N_1=2.31830988618379$$ $$N_2=2.31348787275960$$ $$N_3=2.31348827662319$$ which is the solution for fifteen significant figures.

If we consider the case where $N$ could be large (this would correspond to large values of ratio $\frac A a$, Taylor expansion would give $$h(N)=a N-\frac{\frac{\pi ^2 a}{2}+\pi A}{N}+O\left(\frac{1}{N^2}\right)$$ that is to say $$N\approx\sqrt{\frac{\pi }{2}} \sqrt{\frac{\pi a+2 A}{a}}$$ In the case considered above, this would lead to $N\approx \frac{1}{2} \sqrt{\pi (2+2 \pi )}\approx 2.55061$ which is not too bad.

Using $a=2$, $A=35$, the estimate would be $7.74033$ while the solution would be $7.63580$.


You could be amazed by the fact that, using a $[2,2]$ Pade approximant (do not worry : you will learn about them sooner or later), the solution can be approximated by $$N=\frac 12 \sqrt{\frac \pi {3a}}\sqrt{2 A \left(\frac{2 A}{\pi a+2 A}+5\right)+5 \pi a}$$ which, for the two worked cases gives $\approx 2.34191$ and $\approx 7.63777$.

Still more funny, using a $[2,4]$ Pade approximant, the solution can be approximated by $$N= \frac 15 \sqrt{\frac \pi {6a}}\sqrt{2 A \left(\frac{4 A (37 \pi a+84 A)}{5 \pi ^2 a^2+20 \pi a A+24 A^2}+61\right)+61 \pi a}$$ which, for the two worked cases gives $\approx 2.31681$ and $\approx 7.63584$.


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