Sine Cosine Sequence?

Question

Two real sequences $\{x\}$ and $\{y\}$ satisfy $$x_{n+2}=x_nx_{n+1}-y_ny_{n+1},$$ $$y_{n+2}=x_ny_{n+1}+y_nx_{n+1}.$$ Given $x_1=y_1=1/\sqrt 2$ and $x_2=y_2=1$, find closed forms of $x_n$ and $y_n$.

Since the equations look like sine cosine addition formulas (and the given values are common trig values), I guessed $x_n=\cos f(n)$ and $y_n=\sin g(n)$ for functions $f,g$. To obey the equations, it must be that $f(n)=g(n)=F(n)$ where $F$ satisfies the Fibonacci recurrence (with some chosen base case). These clearly satisfy the equations, but the base cases don't match because $x_2=y_2=1\implies \cos t=\sin t=1$ for some $t$ which is impossible.

Can they have any non-trig expression? What am I missing?

Answer 1

let us try $$z_n = x_n + iy_n, \quad z_nz_{n+1} =x_nx_{n+1}-y_ny_{n+1}+i(x_ny_{n+1} + x_{n+1}y_n) = z_{n+2} .$$ so we have the recurrence relation $$z_{n+2} = z_nz_{n+1}, z_0= e^{i\pi/4}, z_1 = \sqrt 2e^{i\pi/4}$$ the sequence of $z_n$ is $$e^{i\pi/4}, \sqrt2e^{i\pi/4}, \sqrt 2e^{i\pi/2}, 2e^{3i\pi/4}, 2\sqrt 2e^{5i\pi/4}, 4\sqrt 2, 16e^{5i\pi/4}, 64\sqrt 2e^{5i\pi/4}, \cdots $$

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$\bf p. s.$ on the suggestion of michael galuza, you get an easier fibonacci recurrence relation for the sequence $u_n = \ln(z_n).$ we have $$u_{n+2} = u_n+u_{n+1}, u_0 = i\pi/4, u_1 = \ln(2)/2+i\pi/2. $$ therefore the sequence $u_n$ is $$ i\pi/4, \ln(2)/2+i\pi/2, 3i\pi/4+\ln(2)/2, \ln(2) + 5i\pi/4, 3\ln(2)/2, \cdots$$

Answer 2

see that $(x_{n+2}^2+y_{n+2}^2) = (x_{n+1}^2+y_{n+1}^2)(x_n^2+y_n^2)$. Defining $F(n)=(x_n^2+y_n^2)$, we have this equation $F(n+2)=F(n+1)*F(n)$. Observe that for all natural $n$, $F(n)$ is major or equal to zero. we can see that $F(n)=0$ or $F(n)=1$ are solutions (in particular, when $F(n)=1$ we have the famous trigonometric equation $sin(x)^2+cos(x)^2=1$). See that the initial problem become find all solutions of the functional equation $F(n+2)=F(n+1)*F(n)$, and make $y_n=(F(n)-x_n^2)^{1/2}$. the general solutions of $F(n+2)=F(n+1)*F(n)$: take the logarithm, and define $Ln(f(n))=g(n)$, we have the Fibonnaci functional equation $g(n+2)=g(n+1)+g(n)$, just solve it with the initial parameters $F(1)=1$ $(g(1)=0)$ and $F(2)=2$ $(g(2)=ln(2))$