TAOCP 1.2.8 Exercise 33
By the addition formulas for the sine function, \sin\left(\frac{\pi}{2}+w\right)=\cos w.
Section 1.2.8: Fibonacci Numbers
Exercise 33. [HM24] Given that $z=\pi/2 + i\ln\phi$, show that $\sin nz/\sin z = i^{,1-n}F_n$.
Verified: no
Solve time: -
Solution
By the addition formulas for the sine function,
$$ \sin\left(\frac{\pi}{2}+w\right)=\cos w. $$
Hence, with
$$ z=\frac{\pi}{2}+i\ln\phi, $$
we have
$$ \sin z=\cos(i\ln\phi). $$
Since
$$ \cos(ix)=\frac{e^{-x}+e^x}{2}, $$
it follows that
$$ \sin z =\frac{\phi+\phi^{-1}}{2}. $$
Now $\phi^2=\phi+1$, therefore
$$ \phi+\phi^{-1} =\phi+\phi-1 =2\phi-1 =\sqrt5, $$
because $\phi=\frac12(1+\sqrt5)$. Thus
$$ \sin z=\frac{\sqrt5}{2}. \tag{1} $$
Next,
$$ nz=\frac{n\pi}{2}+in\ln\phi. $$
Using
$$ \sin(a+b)=\sin a\cos b+\cos a\sin b, $$
together with
$$ \cos(ix)=\frac{e^x+e^{-x}}{2}, \qquad \sin(ix)=i\frac{e^x-e^{-x}}{2}, $$
we obtain
$$ \sin nz =\sin\frac{n\pi}{2}\cdot\frac{\phi^n+\phi^{-n}}{2} +i\cos\frac{n\pi}{2}\cdot\frac{\phi^n-\phi^{-n}}{2}. \tag{2} $$
The values of $\sin(n\pi/2)$ and $\cos(n\pi/2)$ depend on the parity of $n$.
If $n=2m$, then $\sin(n\pi/2)=0$ and $\cos(n\pi/2)=(-1)^m$. Equation (2) gives
$$ \sin nz =i(-1)^m\frac{\phi^{2m}-\phi^{-2m}}{2}. $$
Since
$$ \hat\phi=-\phi^{-1} $$
by equation (13), equation (14) yields
$$ F_{2m} =\frac{1}{\sqrt5}\left(\phi^{2m}-\phi^{-2m}\right). $$
Therefore
$$ \sin nz =i(-1)^m\frac{\sqrt5}{2}F_{2m}. $$
Dividing by (1),
$$ \frac{\sin nz}{\sin z} =i(-1)^mF_{2m}. $$
Now
$$ i^{,1-2m} =i(-1)^m, $$
hence
$$ \frac{\sin nz}{\sin z}=i^{,1-n}F_n $$
when $n$ is even.
If $n=2m+1$, then $\sin(n\pi/2)=(-1)^m$ and $\cos(n\pi/2)=0$. Equation (2) gives
$$ \sin nz =(-1)^m\frac{\phi^{2m+1}+\phi^{-2m-1}}{2}. $$
By equation (14),
$$ F_{2m+1} =\frac{1}{\sqrt5}\left(\phi^{2m+1}+\phi^{-2m-1}\right), $$
because
$$ \hat\phi^{,2m+1}=-\phi^{-2m-1}. $$
Therefore
$$ \sin nz =(-1)^m\frac{\sqrt5}{2}F_{2m+1}. $$
Dividing by (1),
$$ \frac{\sin nz}{\sin z} =(-1)^mF_{2m+1}. $$
Since
$$ i^{,1-(2m+1)} =i^{-2m} =(-1)^m, $$
we obtain
$$ \frac{\sin nz}{\sin z}=i^{,1-n}F_n $$
when $n$ is odd.
The identity therefore holds for all integers $n$:
$$ \boxed{\frac{\sin nz}{\sin z}=i^{,1-n}F_n }. $$
This completes the proof.
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