# Limit of the solutions of a trigonometric equation

Consider the equation

$$2\cos(\sqrt{\lambda} \pi) \sin(\sqrt{\lambda} \frac{\pi}{n}) + \sin(\sqrt{\lambda} \pi) \cos(\sqrt{\lambda} \frac{\pi}{n})=0$$

for $n \in \mathbb{N}$. This came up when solving an eigenvalue equation. I want to show the existence of a unique solution $\lambda (n) \in (0,1)$ which satisfies $\lambda (n) \rightarrow 1$ as $n \rightarrow \infty$.

Intuitively the solution, if existent, should have this property since the first part of the sum approaches $\sin(\sqrt{\lambda} \pi)$ and the second part approaches $0$ as $n \rightarrow \infty$. However, I don't see how to prove this.

Bonus question: What if 2 is replaced by an arbitrary $k \in \mathbb{N}$. Do the solutions, if existent, still converge to $1$?

Edit: I'm sorry to break the symmetry, but I forgot a factor in the first part of the sum.

• Use the formula $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$. – Gregory Grant Jun 12 '16 at 11:26
• @NiU, it's considered a bit rude to edit the question to invalidate the answers already provided. Although no one has complained this time, but allow me to suggest that next time you should accept the answer and post your edits as a new question. – Wiley Jun 17 '16 at 16:08
• I'm sorry. I will definitely do that next time. – NiU Jun 17 '16 at 16:30
• For your new question, given $k>0$ and sufficiently large $n$, let's consider $$k\tan\left(\frac{x\pi}{n}\right)=-\tan(x\pi)=\tan((1-x)\pi)$$ for $x\in(1/2,1)$, where both sides are monotonic. RHS decreases from $+\infty$ to $0$, while LHS increases from $k\tan(\pi/(2n))$ to $k\tan(\pi/n)$. Using $\tan (x)\sim x$, you can easily see the unique solution within this range should satisfy $\sqrt{\lambda}=x=1-O(1/n)$, so $\lambda\to1$ when $n\to\infty$ as desired. – Wiley Jun 17 '16 at 16:36
• generally updated – Yuri Negometyanov Jun 18 '16 at 22:32

Solution

Let $$c=\cos(\sqrt\lambda\dfrac\pi{n}), \quad s=\sin(\sqrt\lambda\dfrac\pi{n}), \quad C=\cos(\sqrt\lambda\pi), \quad S=\cos(\sqrt\lambda\pi),$$ then $$c^2+s^2=1,\quad C^2+S^2 = 1, \quad 2Cs+Sc=0,\quad C+jS=(c+js)^n,\quad j=\sqrt{-1}$$ Trivial solutions are $$s=S=0,\quad c=\pm1,\quad C=\pm1;$$ $$c=C=0, \quad s=\pm1, \quad S=\pm1,$$ where the signs $\pm$ depends of $n$.

For $c\not=0,$ $$S=-\dfrac{2s}{c}C,\quad C^2+\dfrac{4s^2}{c^2}C^2=1,\quad C^2=\dfrac{c^2}{c^2+4s^2},$$ $$C=\pm\dfrac{c}{\sqrt{c^2+4s^2}}, \quad C+jS=\pm\dfrac{c-2js}{\sqrt{c^2+4s^2}},$$ $$(c+js)^{n} = \pm\dfrac{c-2js}{\sqrt{c^2+4s^2}}.$$ $$\Re(c+js)^{n}=\Re(c\pm j\sqrt{1-c^2})^{n}=\mathrm T_n(c),$$ $$\mathrm T_n(c) = \pm\dfrac{c}{\sqrt{4-3c^2}},\quad \mathrm T_n^2(c)=\dfrac{c^2}{4-3c^2},$$ where $\mathrm T_n(x)$ is Chebyshev polynomials of the first kind. $$2T_n^2(x)-1=T_{2n}(x),$$ so $$\mathrm T_{2n}(c) = \dfrac{5c^2-4}{4-3c^2},\quad \sqrt\lambda = \dfrac n\pi \arccos c,\qquad(1)$$ $$\boxed{(3x-4)\mathrm T_{2n}(\sqrt x) +5x-4 = 0,\quad \lambda = \left(\dfrac n\pi \arccos\sqrt x\right)^2.}\qquad(2)$$

Order of the first equation $(2)$ is $n+1,$ so maximal quantity of solutions is $n+1$.

Examples

Now we can consider solutions of $(2)$ for various $n.$

$\boxed{n=1}$

$$\mathrm T_2(\sqrt x) = 2x-1,$$ $$(3x-4)(2x-1) + 5x-4 = 0,\quad 6x(x-1) = 0,$$ $$(x,\lambda)\in\left\{ \left(0,\dfrac14\right), \left(1,0\right) \right\}.$$

$\boxed{n=2}$

$$\mathrm T_4(\sqrt x) = 8x^2 -8x +1,$$ $$(3x-4)(8x^2-8x+1) + 5x-4 = 0,\quad 8(x-1)^2(3x-1) = 0.$$ $x=1$ is root of multiplicity 2. $$(x,\lambda)\in\left\{ \left(\dfrac13, \left(\dfrac2\pi\arccos{\sqrt{\dfrac13}}\right)^2\right), \left(1,0\right) \right\},$$ $$(x,\lambda)\in\left\{ \left(0.333333, 0.369875\right), \left(1, 0\right) \right\}.$$

$\boxed{n=3}$

$$\mathrm T_6(\sqrt x) = 32x^3 -48x^2 +18x -1,$$ $$(3x-4)(32x^3-48x^2 +18x-1) + 5x-4 = 0,\quad 2x(12x-7)(x-1)(4x-5) = 0,$$ $$(x,\lambda)\in\left\{ \left(0,\dfrac94\right), \left(\dfrac7{12},\left(\dfrac3\pi\arccos{\sqrt{\dfrac7{12}}}\right)^2\right), \left(1,0\right), \left(\dfrac54, \left(\dfrac3\pi\arccos{\sqrt{\frac54}}\right)^2\right) \right\},$$ $$(x,\lambda)\in\left\{ \left(0, 2.25\right), \left(0.583333, 0.448966\right), \left(1,0\right), \left(1.25, -0.211162\right) \right\},$$

$\boxed{n=4}$

$$\mathrm T_8(\sqrt x) = 128x^4 -256x^3 +160x^2 -32x +1,$$ $$(3x-4)(128x^4 -256x^3 +160x^2 -32x +1) + 5x-4 = 0, \quad 8 (x-1)(4x^2-6x+1)(12x^2-10x+1) = 0,$$ $$(x,\lambda)\in\left\{ \left(\dfrac{5-\sqrt{13}}{12}, \left(\dfrac4\pi\arccos{\sqrt{\dfrac{5-\sqrt{13}}{12}}}\right)^2\right), \left(\dfrac{3-\sqrt5}4, \left(\dfrac4\pi\arccos{\sqrt{\dfrac{3-\sqrt5}4}}\right)^2\right), \left(\dfrac{5+\sqrt{13}}{12}, \left(\dfrac4\pi\arccos{\sqrt{\frac{5+\sqrt{13}}{12}}}\right)^2\right) \left(1,0\right), \left(\dfrac{3+\sqrt5}4, \left(\dfrac4\pi\arccos{\sqrt{\dfrac{3+\sqrt5}4}}\right)^2\right), \right\},$$ $$(x,\lambda)\in\left\{ \left(0.116204, 2.424530\right), \left(0.190983, 2.028178\right), \left(0.717129, 0.509826\right), \left(1,0\right), \left(1.309017, -0.456474\right) \right\},$$

Note that the roots $\dfrac{3\pm\sqrt5}4$ don't satisfy to issue equation.

Roots analysis

Determination of Chebyshev polynomials is given $$\mathrm T_i(y)=\cos(i\arccos y).$$ That translates $(1)$ as $$\cos(2n\arccos c) = \dfrac{5c^2-4}{4-3c^2}.$$ Considering it as equation $\cos(t)=a,$ we get soliution in indexed form $$2n\arccos c = \dfrac\pi2(2m+1)+(-1)^m\left(\arccos\dfrac{5c^2-4}{4-3c^2}\right), \quad m=0..2n-1,$$ ($\arccos c \in[0,\pi]$).

If to build function $$h(c,m)=\dfrac1{2\arccos(c)}\left(\dfrac\pi2(2m+1)+(-1)^m\left(\arccos\dfrac{5c^2-4}{4-3c^2}\right)\right),\qquad(3)$$ the roots of the given equation are the points of intersection of this function with horizontal lines $h(x,m) = n.$ The plot in Mathcad gives an indication of the number of nontrivial roots of the equation $(2)$ on the interval $(0,1)$.

Note that the zero branch of the graph passes through the point $c = 1$ and terminates at $c = \sqrt{4/3}$. However, the remaining branches are at $c = 1$ vertical asymptote. This fact is confirmed by the graphs of functions and their derivatives, as well as an analysis of the expression for the derivatives.

Thus, the hypothesis of the asymptotic behavior of the roots receives strong support.

Case of k > 2

Using the same way, we have the formulas $$\mathrm T_{2n}(c) = \dfrac{(k^2+1)c^2-k^2}{k^2-(k^2-1)c^2},\quad \sqrt\lambda = \dfrac n\pi \arccos c,\qquad(1')$$ $$((k^2-1)x-k^2)\mathrm T_{2n}(\sqrt x) + (k^2+1)x - k^2 = 0,\quad \lambda = \left(\dfrac n\pi \arccos\sqrt x\right)^2,\qquad(2'),$$ $$h(c,m,k)=\dfrac1{2\arccos(c)}\left(\dfrac\pi2(2m+1)+(-1)^m\left(\arccos\dfrac{(k^2+1)c^2-k^2}{k^2-(k^2-1)c^2}\right)\right), \qquad(3').$$ We have the next graphs of $h(c,m,3)$ and $h'_c(c,m,3)$ with the next formula for derivatives.

Сonclusions

$1.$ Initial trigonometric equation with parameter $n$ is reduced to a polynomial algebraic equation of order $n+1$ with known constant coefficients.

$2.$ Initial trigonomic equation determines no more than $n+1$ eigenvalues. We can get them all, but the part of them don't satisfy to issue equation for different reasons.

$3.$ The eigenvalue $\lambda = 0$ exists for all $n>2$.

$4.$ We obtain a resolutions of the equation for each branch of ambiguity.

$5.$ The hypothesis of the asymptotic behavior of the roots receives strong support

$6.$ If instead of $2$ to take an arbitrary value $k\in\mathbb N$, then both the methodology and the findings will change slightly.

The equation is equivalent to:

$$\sin\left( \sqrt{\lambda} \left( \frac{n+1}n \right) \pi \right) = 0$$

Which has the solutions:

$$\sqrt{\lambda}(n+1) = nk$$

From here, we see that $k \ge 0$.

We have the solution, for $k = 1$,

$$\lambda = \left( \frac{n}{n+1} \right)^2 = \lambda(n)$$

which is in $(0,1)$ and converges to $1$.

Any other solution must be for $k = 0$ (which gives $\lambda = 0 \notin (0,1)$) or $k \ge 2$, in which case:

$$\sqrt{\lambda} = \frac{kn}{n+1} \ge \frac{2n}{n+1} \ge \frac{n+1}{n+1} = 1$$

which gives $\lambda \ge 1$. Therefore, the above is the unique solution in $(0,1)$

• I'm sorry - I forgot a factor in the first sum. Using $\sin(a+b)=\cos(a) \sin(b) + \cos(b) \sin(a)$ doesn't seem to simplify the expression in this case. – NiU Jun 12 '16 at 15:08

HINT:

$$\implies\sin\sqrt\lambda\pi\dfrac{(n+1)}n=0$$

$$\implies\sqrt\lambda\pi\dfrac{(n+1)}n=m\pi$$ where $m$ is any integer

$$\implies\sqrt\lambda=\dfrac{mn}{n+1}$$

• I'm sorry - I forgot a factor in the first sum. Using $\sin(a+b)=\cos(a) \sin(b) + \cos(b) \sin(a)$ doesn't seem to simplify the expression in this case. – NiU Jun 12 '16 at 15:07