Question 4 (1.5 points)
The Fibonacci sequence 1, 2, 3, 5, 8, 13, 21, 34, 55, ... is usually defined recursively via
$c_0 = 1$, $c_1 = c_0$, $c_n = c_{n-1} + c_{n-2}$ for $n \ge 2$.
If we use this sequence as coefficients of a Maclaurin Series (= Taylor Series around 0)
$T(z) = \sum_{n=0}^{\infty} c_n z^n = 1 + z + 2z^2 + 3z^3 + 5z^4 + 8z^5 + 13z^6 + 21z^7 + \dots$
-- what happens?
Select 4 correct answer(s)
Multiplying out and comparing coefficients shows that
$(1 + z + z^2) \cdot T(z) = 1$, so wherever the series converges, it is identical to
$f(z) = \frac{1}{1 + z + z^2}$.
Multiplying out and comparing coefficients shows that
$(1 - z - z^2) \cdot T(z) = 1$, so wherever the series converges, it is identical to
$f(z) = \frac{1}{1 - z - z^2} = \frac{-1}{z^2 + z - 1}$.
Multiplying out and comparing coefficients shows that
$(1 - z + z^2) \cdot T(z) = 1$, so wherever the series converges, it is identical to
$f(z) = \frac{1}{1 - z + z^2}$.
The function $f$ identified above has two singularities, at the zeros of its denominator. The Maclaurin Series of $f$ thus converges on the disk around 0 whose radius is the smaller of the absolute values of these roots: This number is $\sqrt{5} - 1 \approx 1.23$.
The function $f$ identified above has two singularities, at the zeros of its denominator. The Maclaurin Series of $f$ thus converges on the disk around 0 whose radius is the smaller of the absolute values of these roots: This value is $\frac{1}{2}(\sqrt{5} - 1) \approx 0.618$.
The function $f$ identified above has two singularities, at the zeros of its denominator. The Maclaurin Series of $f$ thus converges on the disk around 0 whose radius is the smaller of the absolute values of these roots: This value is $\frac{1}{2}(\sqrt{5} + 1) \approx 1.618$.
If $\varphi$ and $\psi$ denote the two singularities of our $f$, one can now use a partial fraction decomposition to write $f(z) = \frac{A}{z - \varphi} + \frac{B}{z - \psi}$ for certain numbers A, B.
Each of the summands in the partial fraction decomposition can further be expanded (e.g. writing $\frac{A}{z - \varphi} = -\frac{A}{\varphi} \cdot \frac{1}{1 - z/\varphi}$) into a geometric series. One of them has radius of convergence $|\varphi|$, the other $|\psi|$, so their sum converges on the smaller of the disks around 0 with these radii. Collecting the coefficients for this sum, and using the Identity Theorem, provides us with a (not particularly beautiful, but) explicit i.e. non-recursive formula for each $c_n$.
