Fast Fibonacci Transform Results

In How Many Transformations, Daniel came to the realization that linear recurrence of the form

{xn+1=a1xn+b1yn,yn+1=a2xn+b2yn,\left\{\begin{array}{l}x_{n+1}=a_1x_n+b_1y_n,\\ y_{n+1}=a_2x_n+b_2y_n,\end{array}\right.

could easily be solved by setting up the matrix interpretation, and then diagonalize the matrix (assuming that's possible) which would allow us quick exponentiation, and hence obtain the Nth term directly.

We've seen this in the context of the Fast Fibonacci Transform. Specifically, set xn=fn+1 x_n = f_{n+1} and yn=fn y_n = f_n, and you get the system of equations

{xn+1=1xn+1ynyn+1=1xn+0yn \begin{cases} x_{n+1} = 1x_n + 1y_n \\ y_{n+1} = 1x_n + 0 y_n \\ \end{cases}

Hence, (xnyn)=(1110)n(11) \begin{pmatrix} x_n \\ y_n \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}

Using this, show the following:
1. Performing the eigenvalue decomposition, prove Binet's formula.
2. Using only matrix properties, conclude that
f2n+1=fn+12+fn2. f_{2n+1} = f_{n+1} ^2 + f_n ^2.
Hint: A2n=An×An A^{2n} = A^n \times A^n .
3. Find a similar formula for f2n f_{2n} .
4. Express f3n f_{3n} in terms of fn,fn+1 f_{n}, f_{n+1}.

#Combinatorics #RecurrenceRelations

Note by Calvin Lin
7 years ago

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Comments

This kinda gave away the method one uses to solve my problem (if you don't use eigenvalue decomposition). However, I would still like to see how eigenvalue decomposition works, so it would be nice if a brilliant user could post a solution to the problems suggested in this note. Thanks.

Daniel Liu - 7 years ago

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Added a solution. It's pretty basic matrix manipulation, and would have been the approach that I thought you used.

Calvin Lin Staff - 7 years ago

Let ρ=1+52 \rho =\frac{1+\sqrt{5}}2 and ρ‾=1−52 {\overline \rho} = \frac{1-\sqrt{5}}2 . Let D=(ρ00ρ‾) D = \begin{pmatrix} \rho&0 \\ 0&{\overline \rho} \end{pmatrix} . Let A=(ρρ‾11) A = \begin{pmatrix} \rho&{\overline \rho} \\ 1&1 \end{pmatrix} . Then (1110)=ADA−1 \begin{pmatrix} 1&1\\1&0 \end{pmatrix} = ADA^{-1} .

Patrick Corn - 7 years ago

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(1) So (fn+1fn)=(ADA−1)n(10)=ADnA−1(10) \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} = (ADA^{-1})^n \begin{pmatrix} 1\\0 \end{pmatrix} = AD^nA^{-1} \begin{pmatrix} 1\\0 \end{pmatrix} . Some painful computations yield (fn+1fn)=15(ρn+1−ρ‾n+1ρn−ρ‾n) \begin{pmatrix} f_{n+1}\\f_n \end{pmatrix} = \frac1{\sqrt{5}} \begin{pmatrix} \rho^{n+1}-{\overline \rho}^{n+1} \\ \rho^n - {\overline \rho}^n \end{pmatrix} ; Binet's formula falls out of this.

Patrick Corn - 7 years ago

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(2) and (3): let M=(1110)=ADA−1 M = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = ADA^{-1} . Then Mn(10)=(fn+1fn) M^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} , and Mn(01)=Mn−1(10)=(fnfn−1) M^n \begin{pmatrix} 0 \\ 1 \end{pmatrix} = M^{n-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} f_n \\ f_{n-1} \end{pmatrix} . So Mn=(fn+1fnfnfn−1) M^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \end{pmatrix} .

Now then, (f2n+1f2n)=M2n(10)=Mn(fn+1fn)=(fn+1fnfnfn−1)(fn+1fn) \begin{pmatrix} f_{2n+1} \\ f_{2n} \end{pmatrix} = M^{2n} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = M^n \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \end{pmatrix} \begin{pmatrix} f_{n+1} \\ f_n \end{pmatrix} .

So we get (f2n+1f2n)=(fn+12+fn2fn+1fn+fnfn−1) \begin{pmatrix} f_{2n+1} \\ f_{2n} \end{pmatrix} = \begin{pmatrix} f_{n+1}^2 + f_n^2 \\ f_{n+1}f_n + f_n f_{n-1} \end{pmatrix} , and the formulas we want can be read off from there.

Patrick Corn - 7 years ago

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@Patrick Corn – (4) Similar computations give f3n=fn(3fn+12−3fnfn+1+2fn2) f_{3n} = f_n(3f_{n+1}^2-3f_nf_{n+1}+2f_n^2) . (Basically the same process as the previous, but I had to substitute fn−1=fn+1−fn f_{n-1} = f_{n+1}-f_n in some places.)

Patrick Corn - 7 years ago

I wish I saw this earlier

Agnishom Chattopadhyay - 6 years, 2 months ago

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Okay! Will go through this

Agnishom Chattopadhyay - 6 years, 2 months ago

For Fast Fibonacci Transform, here's a python code for calculating fibonacci numbers:

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# (Public) Returns F(n).
def fibonacci(n):
    if n < 0:
        raise ValueError("Negative arguments not implemented")
    return _fib(n)[0]


# (Private) Returns the tuple (F(n), F(n+1)).
def _fib(n):
    if n == 0:
        return (0, 1)
    else:
        a, b = _fib(n // 2)
        c = a * (b * 2 - a)
        d = a * a + b * b
        if n % 2 == 0:
            return (c, d)
        else:
            return (d, c + d)

Pranjal Jain - 6 years, 1 month ago
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