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I want to solve recurrence relation as following : $$ g_{l+1} = (1-2t)g_l+2\sqrt{t(1-t)}b_l \quad and \quad b_{l+1}=(1-2t)b_l-2\sqrt{t(1-t)}g_l \quad $$ $$ g_0=\sqrt{t} \quad and \quad b_0 = \sqrt{1-t}$$ where t is constant

I tried generating function method

Let $ G(x) = \sum_{m=0}^{\infty} g_mx^m $ and $ B(x)= \sum_{m=0}^{\infty} b_mx^m $

Then the recurrence relation becomes $$ g_{l+1}x^{l+1}=(1-2t)g_lx^{l+1}+2\sqrt{t(1-t)}b_lx^{l+1} \quad and \quad b_{l+1}x^{l+1}=(1-2t)b_lx^{l+1}-2\sqrt{t(1-t)}g_lx^{l+1}$$

Continuing calculation, $$ G(x) = (1-2t)xG(x) + 2\sqrt{t(1-t)}xB(x) \quad and \quad B(x)=(1-2t)xB(x)-2\sqrt{t(1-t)}xG(x)$$ $$ B(x) = -\frac{2\sqrt{t(1-t)}x}{1-(1-2t)x}G(x) $$

Substituting B(x) to LHS equation, $$G(x)=(1-2t)xG(x)-\frac{4t(1-t)x^2}{1-(1-2t)x}G(x) $$ $$[x^2-2(1-2t)x+1]G(x)=0$$

Then I get the result $ x=(1-2t) \pm 2\sqrt{t^2-t} $

But there is no information about $g_l$ and $ b_l $

How can I solve this recurrence formula?

Quantum John Do
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1 Answers1

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Let $x_l=\left (\begin{split} b_l\\g_l\end{split}\right)$ and $M=\left (\begin{array}{cc} 1-2t &&-2\sqrt{t(1-t)}\\ 2\sqrt{t(1-t)}&&1-2t\end{array}\right)$

Then $$x_{l+1}=Mx_l=M^{l+1}x_0$$ Note that $M$ is a rotation since $(1-2t)^2 + \left(2\sqrt{t(1-t)}\right)^2=1$. Thus if you define $\theta$ such that $$\left\{\begin{array}{lll} \cos \theta &= \sqrt{1-t}&=b_0\\ \sin\theta &= \sqrt{t}&=g_0\\ \end{array}\right.$$ then $$\left\{\begin{array}{lll} 1-2t &= 1-2\sin^2(\theta) &=\cos(2\theta)\\ 2\sqrt{t(1-t)} &= 2\sin(\theta)\cos(\theta)&=\sin(2\theta)\\ \end{array}\right.$$ and $$M^l = \left (\begin{array}{cc} \cos 2\theta &&-\sin2\theta\\ \sin 2\theta&&\cos 2\theta\end{array}\right)^l=\left (\begin{array}{cc} \cos 2l\theta &&-\sin 2l\theta\\ \sin 2l\theta&&\cos 2l\theta\end{array}\right)$$ You can see that $$ \left\{\begin{array}{ll} b_l &= \cos (2l\theta)\cos\theta - \sin(2l\theta)\sin\theta&=\cos((2l+1)\theta)\\ g_l &= \sin (2l\theta)\cos\theta+ \cos (2l\theta)\sin\theta&=\sin((2l+1)\theta)\\ \end{array}\right. $$ In conclusion $$ \left\{\begin{array}{ll} b_l &=\cos((2l+1)\arcsin\sqrt{t})\\ g_l &= \sin((2l+1)\arcsin\sqrt t)\\ \end{array}\right. $$

Stefan Lafon
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