Fibonacci数列 Huffman树

Fibonacci数列

for k = 1:6k = int16(k);fprintf('f%d = %d\n', k, Fibonacci(k));
endfunction fk = Fibonacci(k)assert(isinteger(k) && numel(k)==1 && k>=1);if k > 2%fk = Fibonacci(k-1) + Fibonacci(k-2);dp = ones(k,1);for i = 3:kdp(i) = dp(i-1) + dp(i-2);endfk = dp(k);elsefk = 1;end
end

f1 = 1
f2 = 1
f3 = 2
f4 = 3
f5 = 5
f6 = 8

Fibonacci通项

最简单的证明的方法是使用数学归纳法.
f n = ϕ n − ψ n ϕ − ψ f_n = \frac{\phi^n - \psi^n}{\phi - \psi} fn=ϕ−ψϕn−ψn
f n + f n + 1 = ϕ n + 1 − ψ n + 1 + ϕ n − ψ n ϕ − ψ = ( ϕ + 1 ) ϕ n − ( ψ + 1 ) ψ n ϕ − ψ = ϕ 2 ϕ n − ψ 2 ψ n ϕ − ψ = f n + 2 f_{n} + f_{n+1} = \frac{\phi^{n+1} - \psi^{n+1} + \phi^{n} - \psi^{n}}{\phi - \psi} = \frac{(\phi+1)\phi^{n} - (\psi+1)\psi^{n}}{\phi - \psi} = \frac{\phi^2 \phi^{n} - \psi^2 \psi^{n}}{\phi - \psi} = f_{n+2} fn+fn+1=ϕ−ψϕn+1−ψn+1+ϕn−ψn=ϕ−ψ(ϕ+1)ϕn−(ψ+1)ψn=ϕ−ψϕ2ϕn−ψ2ψn=fn+2
ϕ = 1 + 5 2 \phi = \frac{1+\sqrt{5}}{2} ϕ=21+5 , ψ = 1 − 5 2 \psi = \frac{1-\sqrt{5}}{2} ψ=21−5 , ϕ + 1 = ϕ 2 \phi + 1 = \phi^2 ϕ+1=ϕ2, ψ + 1 = ψ 2 \psi + 1 = \psi^2 ψ+1=ψ2, 原因是他们是 x 2 − x − 1 = 0 x^2-x-1=0 x2−x−1=0的根.

使用初等数学方法求解Fibonacci通项主要通过化归到求解等比数列及其求和的通项.

f n = f n − 1 + f n − 2 f_{n} = f_{n-1} + f_{n-2} fn=fn−1+fn−2
g n − 1 = f n + B f n − 1 = A ( f n − 1 + B f n − 2 ) = A g n − 2 g_{n-1} = f_{n} + B f_{n-1} = A (f_{n-1} + B f_{n-2}) = A g_{n-2} gn−1=fn+Bfn−1=A(fn−1+Bfn−2)=Agn−2
g n − 1 = f n + − 1 − 5 2 f n − 1 = 1 − 5 2 ( f n − 1 + − 1 − 5 2 f n − 2 ) = 1 − 5 2 g n − 1 g_{n-1} = f_{n} + \frac{-1-\sqrt{5}}{2} f_{n-1} = \frac{1-\sqrt{5}}{2} \left( f_{n-1} + \frac{-1-\sqrt{5}}{2} f_{n-2} \right) = \frac{1-\sqrt{5}}{2} g_{n-1} gn−1=fn+2−1−5 fn−1=21−5 (fn−1+2−1−5 fn−2)=21−5 gn−1
g n − 1 = f n + − 1 + 5 2 f n − 1 = 1 + 5 2 ( f n − 1 + − 1 + 5 2 f n − 2 ) = 1 + 5 2 g n − 1 g_{n-1} = f_{n} + \frac{-1+\sqrt{5}}{2} f_{n-1} = \frac{1+\sqrt{5}}{2} \left( f_{n-1} + \frac{-1+\sqrt{5}}{2} f_{n-2} \right) = \frac{1+\sqrt{5}}{2} g_{n-1} gn−1=fn+2−1+5 fn−1=21+5 (fn−1+2−1+5 fn−2)=21+5 gn−1
{ g n = A n g 0 } \{g_n = A^n g_0\} {gn=Ang0}呈等比数列, 容易求解通项.
f n = ( − B ) n ( f 0 + ∑ k = 1 n g n − k ( − B ) n − k + 1 ) = ( − B ) n ( f 0 + ∑ k = 1 n A n − k g 0 ( − B ) n − k + 1 ) = ( − B ) n ( f 0 + g 0 − B ∑ k = 1 n ( A − B ) n − k ) f_n = (-B)^n \left( f_0 + \sum\limits_{k=1}^{n} \frac{g_{n-k}}{(-B)^{n-k+1}} \right) = (-B)^n \left( f_0 + \sum\limits_{k=1}^{n} \frac{A^{n-k} g_0}{(-B)^{n-k+1}} \right) = (-B)^n \left( f_0 + \frac{g_0}{-B} \sum\limits_{k=1}^{n} \left(\frac{A}{-B}\right)^{n-k} \right) fn=(−B)n(f0+k=1∑n(−B)n−k+1gn−k)=(−B)n(f0+k=1∑n(−B)n−k+1An−kg0)=(−B)n(f0+−Bg0k=1∑n(−BA)n−k)呈等比数列求和, 容易求解通项.

使用高等数学方法求解Fibonacci通项主要通过化归到频域的多项式真分式(proper polynomial fraction).

离散时间线性时不变系统
x [ n ] = x [ n − 1 ] + x [ n − 2 ] x[n] = x[n-1] + x[n-2] x[n]=x[n−1]+x[n−2]
连续化(上采样)
x ( t ) δ T ( t ) = ∑ n = 0 + ∞ x ( t ) δ ( t − n T ) = ∑ n = 0 + ∞ x ( n T ) δ ( t − n T ) x(t) \delta_T(t) = \sum\limits_{n=0}^{+\infty} x(t) \delta(t-nT) = \sum\limits_{n=0}^{+\infty} x(nT) \delta(t-nT) x(t)δT(t)=n=0∑+∞x(t)δ(t−nT)=n=0∑+∞x(nT)δ(t−nT)
Laplace变换
Z [ x [ n ] ] = L [ x ( t ) ] = ∑ n = 0 + ∞ x ( n T ) e − s n T = ∑ n = 0 + ∞ x [ n ] z − n \mathcal{Z}\left[x[n]\right] = \mathcal{L}\left[x(t)\right] = \sum\limits_{n=0}^{+\infty} x(nT) e^{-snT} = \sum\limits_{n=0}^{+\infty} x[n] z^{-n} Z[x[n]]=L[x(t)]=n=0∑+∞x(nT)e−snT=n=0∑+∞x[n]z−n
Z变换
X ( z ) = z − 1 ( X ( z ) − x [ − 1 ] z ) + z − 2 ( X ( z ) − x [ − 1 ] z − x [ − 2 ] z 2 ) = ( z − 1 + z − 2 ) X ( z ) + A 0 + A 1 z − 1 X(z) = z^{-1} \left( X(z) - x[-1]z \right) + z^{-2} \left( X(z) - x[-1]z - x[-2]z^{2} \right) = \left( z^{-1} + z^{-2} \right) X(z) + A_0 + A_1 z^{-1} X(z)=z−1(X(z)−x[−1]z)+z−2(X(z)−x[−1]z−x[−2]z2)=(z−1+z−2)X(z)+A0+A1z−1
逆Z变换
x [ n ] = L − 1 [ X ( z ) ] = L − 1 [ A 0 + A 1 z − 1 1 − z − 1 − z − 2 ] = L − 1 [ 5 − 1 2 5 A 0 + A 1 z − 1 1 − 1 − 5 2 z − 1 + 5 + 1 2 5 A 0 + A 1 z − 1 1 − 1 + 5 2 z − 1 ] x[n] = \mathcal{L}^{-1}\left[X(z)\right] = \mathcal{L}^{-1}\left[ \frac{A_0 + A_1 z^{-1}}{1 - z^{-1} - z^{-2}} \right] = \mathcal{L}^{-1}\left[ \frac{\sqrt{5} - 1}{2 \sqrt{5}} \frac{A_0 + A_1 z^{-1}}{1 - \frac{1-\sqrt{5}}{2} z^{-1}} + \frac{\sqrt{5} + 1}{2 \sqrt{5}} \frac{A_0 + A_1 z^{-1}}{1 - \frac{1+\sqrt{5}}{2} z^{-1}} \right] x[n]=L−1[X(z)]=L−1[1−z−1−z−2A0+A1z−1]=L−1[25 5 −11−21−5 z−1A0+A1z−1+25 5 +11−21+5 z−1A0+A1z−1]
= ( 1 − 5 2 ) n ( 5 − 1 2 5 A 0 + 5 − 1 2 5 A 1 1 − 5 2 ) + ( 1 + 5 2 ) n ( 5 + 1 2 5 A 0 + 5 + 1 2 5 A 1 1 + 5 2 ) = \left( \frac{1-\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} - 1}{2 \sqrt{5}} A_0 + \frac{\sqrt{5} - 1}{2 \sqrt{5}} \frac{A_1}{\frac{1-\sqrt{5}}{2}} \right) + \left( \frac{1+\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} + 1}{2 \sqrt{5}} A_0 + \frac{\sqrt{5} + 1}{2 \sqrt{5}} \frac{A_1}{\frac{1+\sqrt{5}}{2}} \right) =(21−5 )n(25 5 −1A0+25 5 −121−5 A1)+(21+5 )n(25 5 +1A0+25 5 +121+5 A1)
= ( 1 − 5 2 ) n ( 5 − 1 2 5 A 0 + A 1 5 ) + ( 1 + 5 2 ) n ( 5 + 1 2 5 A 0 + A 1 5 ) = \left( \frac{1-\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} - 1}{2 \sqrt{5}} A_0 + \frac{A_1}{\sqrt{5}} \right) + \left( \frac{1+\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} + 1}{2 \sqrt{5}} A_0 + \frac{A_1}{\sqrt{5}} \right) =(21−5 )n(25 5 −1A0+5 A1)+(21+5 )n(25 5 +1A0+5 A1)
= ( 1 − 5 2 ) n ( 5 − 1 2 A 0 ^ + A 1 ^ ) + ( 1 + 5 2 ) n ( 5 + 1 2 A 0 ^ + A 1 ^ ) = \left( \frac{1-\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} - 1}{2} \hat{A_0} + \hat{A_1} \right) + \left( \frac{1+\sqrt{5}}{2} \right)^n \left( \frac{\sqrt{5} + 1}{2} \hat{A_0} + \hat{A_1} \right) =(21−5 )n(25 −1A0^+A1^)+(21+5 )n(25 +1A0^+A1^)
边界条件
0 = x [ 0 ] = 5 A 0 ^ + 2 A 1 ^ 0 = x[0] = \sqrt{5} \hat{A_0} + 2 \hat{A_1} 0=x[0]=5 A0^+2A1^
1 = x [ 1 ] = 5 A 0 ^ + A 1 ^ 1 = x[1] = \sqrt{5} \hat{A_0} + \hat{A_1} 1=x[1]=5 A0^+A1^
求解待定值
A 0 ^ = 2 5 \hat{A_0} = \frac{2}{\sqrt{5}} A0^=5 2
A 1 ^ = − 1 \hat{A_1} = -1 A1^=−1
求解解析解
x [ n ] = − 1 5 ( 1 − 5 2 ) n + 1 5 ( 1 + 5 2 ) n = f n x[n] = - \frac{1}{\sqrt{5}} \left( \frac{1-\sqrt{5}}{2} \right)^n + \frac{1}{\sqrt{5}} \left( \frac{1+\sqrt{5}}{2} \right)^n = f_n x[n]=−5 1(21−5 )n+5 1(21+5 )n=fn

− 1 + 5 2 − 1 − 5 2 = − 1 \frac{-1+\sqrt{5}}{2} \frac{-1-\sqrt{5}}{2} = -1 2−1+5 2−1−5 =−1
z − 2 + z − 1 − 1 = ( − 1 − 5 2 − z − 1 ) ( − 1 + 5 2 − z − 1 ) = − 1 − 5 2 ( 1 − 1 − 1 − 5 2 z − 1 ) − 1 + 5 2 ( 1 − 1 − 1 + 5 2 z − 1 ) = − ( 1 + − 1 + 5 2 z − 1 ) ( 1 + − 1 − 5 2 z − 1 ) z^{-2} + z^{-1} - 1 = \left(\frac{-1-\sqrt{5}}{2} - z^{-1}\right) \left(\frac{-1+\sqrt{5}}{2} - z^{-1}\right) = \frac{-1-\sqrt{5}}{2} \left(1 - \frac{1}{\frac{-1-\sqrt{5}}{2}} z^{-1}\right) \frac{-1+\sqrt{5}}{2} \left(1 - \frac{1}{\frac{-1+\sqrt{5}}{2}} z^{-1}\right) = - \left(1 + \frac{-1+\sqrt{5}}{2} z^{-1}\right) \left(1 + \frac{-1-\sqrt{5}}{2} z^{-1}\right) z−2+z−1−1=(2−1−5 −z−1)(2−1+5 −z−1)=2−1−5 (1−2−1−5 1z−1)2−1+5 (1−2−1+5 1z−1)=−(1+2−1+5 z−1)(1+2−1−5 z−1)
1 ( 1 + − 1 + 5 2 z − 1 ) ( 1 + − 1 − 5 2 z − 1 ) = 5 − 1 2 5 1 1 + − 1 + 5 2 z − 1 + 5 + 1 2 5 1 1 + − 1 − 5 2 z − 1 = 5 − 1 2 5 1 1 − 1 − 5 2 z − 1 + 5 + 1 2 5 1 1 − 1 + 5 2 z − 1 \frac{1}{\left(1 + \frac{-1+\sqrt{5}}{2} z^{-1}\right) \left(1 + \frac{-1-\sqrt{5}}{2} z^{-1}\right)} = \frac{\sqrt{5} - 1}{2 \sqrt{5}} \frac{1}{1 + \frac{-1+\sqrt{5}}{2} z^{-1}} + \frac{\sqrt{5} + 1}{2 \sqrt{5}} \frac{1}{1 + \frac{-1-\sqrt{5}}{2} z^{-1}} = \frac{\sqrt{5} - 1}{2 \sqrt{5}} \frac{1}{1 - \frac{1-\sqrt{5}}{2} z^{-1}} + \frac{\sqrt{5} + 1}{2 \sqrt{5}} \frac{1}{1 - \frac{1+\sqrt{5}}{2} z^{-1}} (1+2−1+5 z−1)(1+2−1−5 z−1)1=25 5 −11+2−1+5 z−11+25 5 +11+2−1−5 z−11=25 5 −11−21−5 z−11+25 5 +11−21+5 z−11

Fibonacci求和通项

可以通过等比数列证明.
Fibonacci通项公式是等比数列之和, Fibonacci求和通项是等比数列求和之和.

可以通过递推关系证明.

f 2 + ∑ k = 1 n f k = f 2 + f 1 = f 3 f 3 + ∑ k = 2 n f k = ⋯ = f m + f m − 1 = f m + 1 f m + 1 + ∑ k = m n f k = ⋯ = f n − 1 + f n − 2 = f n f n + ( f n − 1 + f n ) = f n − 1 + f n = f n + 1 f n + f n + 1 = f n + f n + 1 = f n + 2 f n + 2 f_2 + \sum\limits_{k=1}^{n} f_k \stackrel{f_2+f_1 = f_3}{=} f_3 + \sum\limits_{k=2}^{n} f_k = \cdots \stackrel{f_{m}+f_{m-1} = f_{m+1}}{=} f_{m+1} + \sum\limits_{k=m}^{n} f_k = \cdots \stackrel{f_{n-1}+f_{n-2} = f_{n}}{=} f_{n} + \left( f_{n-1} + f_{n} \right) \stackrel{f_{n-1}+f_{n} = f_{n+1}}{=} f_{n} + f_{n+1} \stackrel{f_{n}+f_{n+1} = f_{n+2}}{=} f_{n+2} f2+k=1∑nfk=f2+f1=f3f3+k=2∑nfk=⋯=fm+fm−1=fm+1fm+1+k=m∑nfk=⋯=fn−1+fn−2=fnfn+(fn−1+fn)=fn−1+fn=fn+1fn+fn+1=fn+fn+1=fn+2fn+2

∑ k = 1 n f k = f n + 2 − f 2 = f n + 2 − 1 \sum\limits_{k=1}^{n} f_k = f_{n+2} - f_{2} = f_{n+2} - 1 k=1∑nfk=fn+2−f2=fn+2−1

∑ k = 1 n k f k \sum\limits_{k=1}^{n} kf_k k=1∑nkfk

频域微分

Z [ n x [ n ] ] = d X ( z ) d z − 1 \mathcal{Z}\left[nx[n]\right] = \frac{\mathrm{d} X(z)}{\mathrm{d}z^{-1}} Z[nx[n]]=dz−1dX(z)

时域卷积

Z [ ∑ k = 0 n k x [ k ] ] = Z [ n x [ n ] ∗ u [ n ] ] = d X ( z ) d z − 1 ⋅ 1 1 − z − 1 \mathcal{Z}\left[\sum\limits_{k=0}^{n} kx[k]\right] = \mathcal{Z}\left[nx[n] \ast u[n]\right] = \frac{\mathrm{d} X(z)}{\mathrm{d}z^{-1}} \cdot \frac{1}{1 - z^{-1}} Z[k=0∑nkx[k]]=Z[nx[n]∗u[n]]=dz−1dX(z)⋅1−z−11

∑ k = 1 n k f k = ∑ m = 1 n ∑ k = m n f k = n ∑ k = 1 n f k − ∑ m = 1 n ∑ k = 1 m − 1 f k = n ∑ k = 1 n f k − ∑ m = 1 n − 1 ∑ k = 1 m f k \sum\limits_{k=1}^{n} k f_{k} = \sum\limits_{m=1}^{n} \sum\limits_{k=m}^{n} f_{k} = n \sum\limits_{k=1}^{n} f_{k} - \sum\limits_{m=1}^{n} \sum\limits_{k=1}^{m-1} f_{k} = n \sum\limits_{k=1}^{n} f_{k} - \sum\limits_{m=1}^{n-1} \sum\limits_{k=1}^{m} f_{k} k=1∑nkfk=m=1∑nk=m∑nfk=nk=1∑nfk−m=1∑nk=1∑m−1fk=nk=1∑nfk−m=1∑n−1k=1∑mfk
= n ( f k + 2 − 1 ) − ∑ m = 1 n − 1 ( f m + 2 − 1 ) = n ( f k + 2 − 1 ) − ∑ m = 3 n + 1 ( f m − 1 ) = n f k + 2 − ∑ m = 3 n + 1 f m − 1 = n \left( f_{k+2} - 1 \right) - \sum\limits_{m=1}^{n-1} \left( f_{m+2} - 1 \right) = n \left( f_{k+2} - 1 \right) - \sum\limits_{m=3}^{n+1} \left( f_{m} - 1 \right) = n f_{k+2} - \sum\limits_{m=3}^{n+1} f_m - 1 =n(fk+2−1)−m=1∑n−1(fm+2−1)=n(fk+2−1)−m=3∑n+1(fm−1)=nfk+2−m=3∑n+1fm−1
= n f k + 2 − ( f n + 3 − 1 − f 1 − f 2 ) − 1 = n f k + 2 − f n + 3 + 2 = n f_{k+2} - \left( f_{n+3} - 1 - f_1 - f_2 \right) - 1 = n f_{k+2} - f_{n+3} + 2 =nfk+2−(fn+3−1−f1−f2)−1=nfk+2−fn+3+2

Fibonacci分布的Huffman树

因为
∑ k = 1 n f k = f n + 2 − 1 \sum\limits_{k=1}^{n} f_k = f_{n+2} - 1 k=1∑nfk=fn+2−1
所以( n ⩾ 3 n \geqslant 3 n⩾3)
f n + 1 < ∑ k = 1 n f k < f n + 2 f_{n+1} < \sum\limits_{k=1}^{n} f_k < f_{n+2} fn+1<k=1∑nfk<fn+2

因此 f k f n + 2 − 1 \frac{f_k}{f_{n+2}-1} fn+2−1fk的Huffman树是
c o d e ( f k ) = { 1 ⋯ 1 ⏟ ( n − k ) o n e s 0 k > 1 1 ⋯ 1 ⏟ ( n − 1 ) o n e s k = 1 \mathtt{code}(f_k) = \begin{cases} \underbrace{\mathtt{1 \cdots 1}}_{(n-k) ~ \mathrm{ones}} \mathtt{0} & k > 1 \\ \underbrace{\mathtt{1 \cdots 1}}_{(n-1) ~ \mathrm{ones}} & k = 1 \\ \end{cases} code(fk)=⎩⎪⎪⎨⎪⎪⎧(n−k) ones 1⋯10(n−1) ones 1⋯1k>1k=1

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{f_n, … ,f₁}(code=xxx11...)

f_n+1(code=xxx01)

f_n+2(code=xxx0)

f_n+2-1

f_n+3-1

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f₁ = 1 (code=1111)

f₂ = 1 (code=1110)

f₃ = 2 (code=110)

f₄ = 3 (code=10)

f₅ = 5 (code=0)

Fibonacci分布的Huffman树的期望编码长度

− 1 f n + 2 − 1 + ∑ k = 1 n f k f n + 2 − 1 ( n + 1 − k ) = − 1 f n + 2 − 1 + ( n + 1 ) ∑ k = 1 n f k f n + 2 − 1 − ∑ k = 1 n k f k f n + 2 − 1 - \frac{1}{f_{n+2}-1} + \sum\limits_{k=1}^{n} \frac{f_k}{f_{n+2}-1} (n+1-k) = - \frac{1}{f_{n+2}-1} + (n+1) \sum\limits_{k=1}^{n} \frac{f_k}{f_{n+2}-1} - \frac{\sum\limits_{k=1}^{n} kf_k}{f_{n+2}-1} −fn+2−11+k=1∑nfn+2−1fk(n+1−k)=−fn+2−11+(n+1)k=1∑nfn+2−1fk−fn+2−1k=1∑nkfk
= n + 1 − 1 f n + 2 − 1 − n f k + 2 − f n + 3 + 2 f n + 2 − 1 = 1 + f n + 3 − n − 3 f n + 2 − 1 = f n + 3 + f n + 2 − n − 4 f n + 2 − 1 = f n + 4 − ( n + 4 ) f n + 2 − 1 = n+1 - \frac{1}{f_{n+2}-1} - \frac{n f_{k+2} - f_{n+3} + 2}{f_{n+2}-1} = 1 + \frac{f_{n+3} - n - 3}{f_{n+2}-1} = \frac{f_{n+3}+f_{n+2} - n - 4}{f_{n+2}-1} = \frac{f_{n+4} - (n+4)}{f_{n+2}-1} =n+1−fn+2−11−fn+2−1nfk+2−fn+3+2=1+fn+2−1fn+3−n−3=fn+2−1fn+3+fn+2−n−4=fn+2−1fn+4−(n+4)

(g) Fibonacci分布的Huffman树的期望编码长度的极限

lim ⁡ n → + ∞ f n + 4 − ( n + 4 ) f n + 2 − 1 = lim ⁡ n → + ∞ ϕ n + 4 − ψ n + 4 ϕ − ψ − ( n + 4 ) ϕ n + 2 − ψ n + 2 ϕ − ψ − 1 = lim ⁡ n → + ∞ ϕ n + 4 ϕ n + 2 = ϕ 2 = ϕ + 1 = 3 + 5 2 \lim_{n \to +\infty} \frac{f_{n+4} - (n+4)}{f_{n+2}-1} = \lim_{n \to +\infty} \frac{\frac{\phi^{n+4} - \psi^{n+4}}{\phi - \psi} - (n+4)}{\frac{\phi^{n+2} - \psi^{n+2}}{\phi - \psi} - 1} = \lim_{n \to +\infty} \frac{\phi^{n+4}}{\phi^{n+2}} = \phi^2 = \phi + 1 = \frac{3+\sqrt{5}}{2} limn→+∞fn+2−1fn+4−(n+4)=limn→+∞ϕ−ψϕn+2−ψn+2−1ϕ−ψϕn+4−ψn+4−(n+4)=limn→+∞ϕn+2ϕn+4=ϕ2=ϕ+1=23+5

ϕ = 1 + 5 2 \phi = \frac{1+\sqrt{5}}{2} ϕ=21+5 , ψ = 1 − 5 2 \psi = \frac{1-\sqrt{5}}{2} ψ=21−5 , ϕ + 1 = ϕ 2 \phi + 1 = \phi^2 ϕ+1=ϕ2, ψ + 1 = ψ 2 \psi + 1 = \psi^2 ψ+1=ψ2, 原因是他们是 x 2 − x − 1 = 0 x^2-x-1=0 x2−x−1=0的根.

Fibonacci其他恒等式

f 2 k − 1 = f 2 k − f 2 k − 2 ⟹ ∑ k = 1 n f 2 k − 1 = f 2 n − f 0 f_{2k-1} = f_{2k} - f_{2k-2} \implies \sum\limits_{k=1}^{n} f_{2k-1} = f_{2n} - f_{0} f2k−1=f2k−f2k−2⟹k=1∑nf2k−1=f2n−f0
f 2 k = f 2 k + 1 − f 2 k − 1 ⟹ ∑ k = 1 n f 2 k = f 2 n + 1 − f 1 f_{2k} = f_{2k+1} - f_{2k-1} \implies \sum\limits_{k=1}^{n} f_{2k} = f_{2n+1} - f_{1} f2k=f2k+1−f2k−1⟹k=1∑nf2k=f2n+1−f1
∑ k = 1 n ( − 1 ) k f k = ( − 1 ) n f n − 1 − 1 \sum\limits_{k=1}^{n} (-1)^k f_k = (-1)^n f_{n-1} - 1 k=1∑n(−1)kfk=(−1)nfn−1−1
f k 2 = f k + 1 f k − f k f k − 1 ⟹ ∑ k = 1 n f k 2 = f n f n + 1 f_{k}^2 = f_{k+1}f_{k} - f_{k}f_{k-1} \implies \sum\limits_{k=1}^{n} f_k^2 = f_{n}f_{n+1} fk2=fk+1fk−fkfk−1⟹k=1∑nfk2=fnfn+1
f n = f k + 1 f n − k + f k f n − k − 1 ⟹ f n + 1 2 − f n − 1 2 = f 2 n f_{n} = f_{k+1}f_{n-k} + f_{k}f_{n-k-1} \implies f_{n+1}^2 - f_{n-1}^2 = f_{2n} fn=fk+1fn−k+fkfn−k−1⟹fn+12−fn−12=f2n
f n = ∑ k = 1 n − 1 ( n − 1 − k k ) f_{n} = \sum\limits_{k=1}^{n-1} \binom{n-1-k}{k} fn=k=1∑n−1(kn−1−k) ( ( N k ) = 0 \binom{N}{k} = 0 (kN)=0 if N < k N < k N<k)
∑ k = 1 + ∞ f k k ! = e ϕ − e ψ 5 \sum\limits_{k=1}^{+\infty} \frac{f_k}{k!} = \frac{e^{\phi} - e^{\psi}}{\sqrt{5}} k=1∑+∞k!fk=5 eϕ−eψ
∑ k = 1 n 1 f 2 k = 3 − f 2 n − 1 f 2 n \sum\limits_{k=1}^{n} \frac{1}{f_{2^k}} = 3 - \frac{f_{2^n-1}}{f_{2^n}} k=1∑nf2k1=3−f2nf2n−1
https://mathworld.wolfram.com/ReciprocalFibonacciConstant.html