Equation including tangent function

Question

I've been studying one problem and I need to consider the following problem. Let $n\in\mathbb{N}$. Is $\tan\left(nx\right)=n \tan(x)$ solvable on $\left(0,\pi\right)$? If it is, what are solutions?

Answer 1

If you have a look to
http://www.trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesTangent.htm
you will notice that Tan[n x] can be expressed as the ratio of terms A and B such that

A = n Tan[x] + Sum[a[i] Tan[x]^(i+2),{i,1,n-2}]

B = 1 + Sum[b[i] Tan[x]^(2 i),{i,1,Floor[n /2]}]

So, multiplying lhs and rhs of your equation by B leads, before any simplification, to a polynomial equation of y = Tan[x] of degree (n+1) (I limit now the discussion to odd values of "n" in order to stay in phase with the nice case provided by user2943324). What is amazing is that, for n=9 as used by user2943324, the final equation reduces to

4 y^3 (60 - 252 y^2 + 45 y^4 - 20 y^6) = 0

where y = Tan[x].

So, what it seems is that your equation can be reformulated as a rather simple polynomial. Just by curiosity, I worked the case n=10; this leads to

2 y^3 (165 - 924 y^2 + 990 y^4 - 220 y^6 + 5 y^8) = 0

Notice that, inside the brackets, only even powers of "y" appear which means that for n=9, we have a cubic (as indicated by user2943324) and for n=10 a quartic.

Solving the polynomial will be definitely much easier than solving the trigonometric equation (in particular because of the removal of all vertical asymptotes).

There is another point of interest : if you develop the tangent as infinite Taylor series, the equation write

n x^3 [(n^2 - 1) / 3 + 2 (n^4 - 1) x^2 / 15 + 17 (n^6 - 1) x^4 / 315 + ...]

What is very interesting is what user2943324 added for the polynomials.

Answer 2

I guess that this cannot be done in an analytical manner (if we exclude the cases $x=0$ and $x=\pi$). Numerical methods look to me the only way assuming we know which root to look at since the function contains many vertical asymptotes on $(0,\pi)$.

Answer 3

In general $$\tan(nx) = \dfrac{1}{i} \dfrac{e^{inx} - e^{-inx}}{e^{inx} + e^{-inx}} = \dfrac{1}{i} \dfrac{(\cos(x) + i \sin(x))^n - (\cos(x) - i \sin(x))^n}{(\cos(x) + i \sin(x))^n + (\cos(x) - i \sin(x))^n} = \dfrac{1}{i} \dfrac{(1+i \tan(x))^n - (1-i\tan(x))^n}{(1+i\tan(x))^n + (1-i\tan(x))^n} = \dfrac{P_n(\tan(x))}{Q_n(\tan(x))}$$

$$\eqalign{P_n(t) &= \sum_{{k=1}\atop {k \text{ odd}}} {n \choose k} (-1)^{(k-1)/2} t^k\cr Q_n(t) &= \sum_{{k=0}\atop {k \text{ even}}} {n \choose k} (-1)^{k/2} t^k}$$

$$\dfrac{P_n(t)}{Q_n(t)} - n t = \dfrac{P_n(t) - n t Q_n(t)}{Q_n(t)}$$

The numerator's terms are all of odd degree, and the $t^1$ term is ${n \choose 1} t - n {n \choose 0}t = 0$, so the numerator is divisible by $t^3$. The other factor is then an even polynomial of degree $n-3$ if $n \ge 3$ is odd, $n-2$ if $n \ge 2$ is even (so of degree $(n-3)/2$ or $(n-2)/2$ if you substitute $t^2 = s$).