Fibonacci is here13世紀提出,A pair of rabbits starts breeding a month after birth,A pair of newborn rabbits are born every month
,It is assumed that rabbits only breed,沒有死亡,問第kHow many pairs of rabbits will there be in the month of birth：
解：
in pairs,The number of pairs of rabbits bred each month forms a series,This is the famous Fibonacci sequence,
1 , 1 , 2 , 3 , 5 , 8... 1,1,2,3,5,8... 1,1,2,3,5,8...,,this book F k F_k Fk,滿足條件：
F 0 = 1 , F 1 = 1 , F k + 2 = F k + 1 + F k F_0=1,F_1=1,F_{k+2}=F_{k+1}+F_k F0=1,F1=1,Fk+2=Fk+1+Fk
**解法1：**The eigenroot solution of difference equations
The characteristic roots of the difference equation are ：
λ 2 − λ − 1 = 0 \lambda^2-\lambda-1=0 λ2−λ−1=0
特征根 λ 1 = \lambda_1= λ1= 1 − 5 2 , λ 2 = \frac{1-\sqrt{5}}{2},\lambda_2= 21−5,λ2= 1 + 5 2 \frac{1+\sqrt{5}}{2} 21+5是互異的.所以通解為：
F k = c 1 ( 1 − 5 2 ) k + c 2 ( 1 + 5 2 ) k F_k=c_1\left( \frac{1-\sqrt{5}}{2} \right) ^k+c_2\left( \frac{1+\sqrt{5}}{2} \right) ^k Fk=c1(21−5)k+c2(21+5)k
Use initial value conditions： F 0 = 1 , F 1 = 1 F_0=1,F_1=1 F0=1,F1=1,得到方程組：
{ c 1 + c 2 = 1 c 1 ( 1 − 5 2 ) + c 2 ( 1 + 5 2 ) = 1 \begin{cases} c_1+c_2=1\\ c_1\left( \frac{1-\sqrt{5}}{2} \right) +c_2\left( \frac{1+\sqrt{5}}{2} \right) =1\\ \end{cases} { c1+c2=1c1(21−5)+c2(21+5)=1
解得： c 1 = 1 2 − 5 10 , c 2 = 1 2 + 5 10 c_1=\frac{1}{2}-\frac{\sqrt{5}}{10},c_2=\frac{1}{2}+\frac{\sqrt{5}}{10} c1=21−105,c2=21+105
於是：初值問題的解為：
F k = ( 1 2 − 5 10 ) ( 1 − 5 2 ) k + ( 1 2 + 5 10 ) ( 1 + 5 2 ) k F_k=\left( \frac{1}{2}-\frac{\sqrt{5}}{10} \right) \left( \frac{1-\sqrt{5}}{2} \right) ^k+\left( \frac{1}{2}+\frac{\sqrt{5}}{10} \right) \left( \frac{1+\sqrt{5}}{2} \right) ^k Fk=(21−105)(21−5)k+(21+105)(21+5)k
代碼:
import sympy as sp
sp.var("t,c1,c2")
t0=sp.solve(t**2-t-1)#Solve the eigenroot equation,
eq1=c1+c2-1
eq2=c1*t0[0]+c2*t0[1]-1
s=sp.solve([eq1,eq2])#求解線性方程組
print("c1=",s[c1],"\n""c2=",s[c2])#Output the solution of the linear system of equations
print("初值問題的解為：""\n",s[c1]*t0[0]+s[c2]*t0[1])#Output the solution to the initial problem
解法二：直接利用python軟件求解
import sympy as sp
sp.var("k")
y=sp.Function("y")
f=y(k+2)-y(k+1)-y(k)
s=sp.rsolve(f,y(k),{
y(0):1,y(1):1})#注意這裡是rsolve
print(s)
