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Could you kindly suggest me ways to find out closed form solution of the following equation type:

$$\tan(nx)=k\tan(x),$$

where $n$ and $k$ are some positive real numbers (that excludes zero). I can solve it using series expansion, but that gives me only an approximate solution for $x$.

Thanks in advance.

PS: I am in engineering stream, but I usually don't need to solve problems that seems very tough to me, mostly just couple of additions/subtractions etc.

Giovanni
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Amir
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  • except for $x=0,$ I see no reason to expect any closed form solution. Why would anyone want to do this? – Will Jagy Oct 27 '15 at 04:19
  • Why would you believe that there is a closed-form solution other than the trivial solution $x=0$? – Mark Viola Oct 27 '15 at 04:19
  • Thanks for your comment, but I am unsure if a closed-form expression of x could be obtained. please consider me not knowing anything about equations. – Amir Oct 27 '15 at 04:53
  • $\tan(x)=a\tan(bx)=0$ is equivalent to $\sin((1-b)x)+\frac{1-a}{1+a}\sin((1+b) x)=0$ which is addressed here – Тyma Gaidash Oct 23 '24 at 21:20

3 Answers3

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If you have a look to this page, you will notice that, for integer values of $n$, $\tan(nx)$ write as the ratio of two polynomials in $\tan(x)$ the degree of numerator being $(n-1)$ if $n$ is even and $n$ if $n$ is odd; the opposite happens for the degree of denominator.

This means that, for integer values of $n$, equation $$\tan(nx)=k\tan(x)$$ reduces to a polynomial of degree $(n+1)$ if $n$ is even and of degree $n$ if $n$ is odd. I suppose that you already see the problem (the periodity of $\tan(x)$ being the source of many other difficulties).

So, even for integer values of $n$, I do not think that there is any hope to have any closed form solution and only numerical methods would be able to solve the equation for given $n,k$ and a range.

For sure, if the solution you are looking for is small, expansions could provide quite good estimates for $$f(x)=\tan(nx)-k\tan(x)$$ Taylor expansion would give $$f(x)=(n-k)x+\frac{1}{3} \left(n^3-k\right)x^3+\frac{2}{15} \left(n^5-k\right)x^5+\frac{17}{315} \left(n^7-k\right)x^7 +O\left(x^8\right)$$ which is basically a cubic polynomial is $x^2$.

If you consider numerical methods, I would suggest that you remove the discontinuities induced by the tangent and consider instead the equation $$g(x)=\sin(nx)\cos(x)-k\cos(nx)\sin(x)$$ whic is much better conditioned.

Claude Leibovici
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  • Thanks a lot for your tips. After Dr. MV's comment i tried to find a solution for n=3 and k=2, that made me clear that closed-form expression may not be available at all. Many many thanks, all. – Amir Oct 27 '15 at 05:35
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I think i have a trivial solution . If $n=k=1$ and $x=45$ then we have $\tan(1\cdot45)=1\tan(45)$ so $1=1$.

Archis Welankar
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If $n$ and $k$ are positive integers $ \tan n x $ can be expressed as a fraction as indicated by Claude Leibovici. A common factor $ \tan x $ can be cancelled out, you have an $ n^{th} $ degree polynomial:

$$ \dfrac{\tan(nx)} {\tan(x)} - k = 0 $$

Else numerical solutions by Newton iteration etc. with correct seed value so that you do not land up in any of $ \pi$ intervals of argument.

Narasimham
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