漸化式:Recurrence relation †
In mathematics, a recurrence relation is an equation that recursively defines a sequence: each term of the sequence is defined as a function of the preceding terms.
A difference equation is a specific type of recurrence relation.
漸化式の例:フィボナッチ数列 †
- Example: Fibonacci numbers
Fn+2 = Fn+1 + fn with initial values F0=0,F1=0.
We obtain the sequence of Fibonacci numbers which begins:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
- 一般解の導出
The Fibonacci recursion
Fn+2 - Fn+1 - Fn =0
is similar to the defining equation of the golden ratio in the form
x^2 - X -1 = 0
which is also known as the generating polynomial of the recursion.
By definition α,β is a root of the equation
α = (1+√5)/2, β=(1-√5)/2.
Any root of the equation above satisfies x^2 - X -1 = 0 and multiplying by x^n shows:
x^(n+2)-x^(n+1)-x^n = 0
Both α^n and β^n are geometric series (for n = 1, 2, 3, ...) that satisfy the Fibonacci recursion.
Linear combinations of series α^n and β^n, with coefficients a and b, can be defined by
Fc(n)=a・α^n + b・β^n for any real a and b.
All thus-defined series satisfy the Fibonacci recursion
Fc(n+2)=a・α^(n+2) + b・β^(n+2)
=a・(α^(n+1)+α^n) + b・(β^(n+1)+β^n)
=a・α^(n+1)+ b・β^(n+1) + a・α^n+b・β^n
=Fc(n+1) + Fc(n).
Requiring that Fc(0)=0,Fc(1)=1 yields
Fc(0)=a・α^0 + b・β^0 = a + b =0
Fc(1)=a・α^1 + b・β^1 = a・α + b・β =0
Then a=1/√5 and b=-1/√5.
The solution of the Fibonacci recursion is
Fn=(α^n - β^n)/√5 ={(1+√5)/2}^n + {(1+√5)/2}^n
α = (1+√5)/2, β=(1-√5)/2.
- このことから 特性方程式の解の線形結合で、一般解が求められることが理解できる。
線形差分方程式の特性関数を用いた一般解法 †
An order linear homogeneous recurrence relation with constant coefficients is an equation of the form:
Fn = a1・Fn-1 + a2・Fn-2 + a3・Fn-3 + ・・・・+am・Fn-m
The characteristic polynomial is
p(x)=x^m-a1・xm-1 + a2・xm-2 + a3・xm-3 + ・・・・+am=0.
- m次元の線形漸化式は、特性方程式のm個の解と初期条件を使って、n番目のFnを解のn乗線形結合で表わされることが理解できる。
- このことは、線形差分方程式の解が明示的に求められることを意味する。言い換えれば、積分計算をしなくても解析解を示せることを意味する。
微分方程式の差分による解法:オイラー法の例 †
When solving an ordinary differential equation numerically, one typically encounters a recurrence relation. For example, when solving the initial value problem
y'(t)=f(t,y(t)) , y0=y(t0)
with Euler's method and a step size h, one calculates the values
y0=y(t0), y1=y(t0+h),y2=y(t0+2h),・・・・・・・・
by the recurrence
yn+1 = yn + h・f(tn,yn)
Systems of linear first order differential equations can be discretized exactly analytically using the methods shown in the discretization article.
- y'(t)=f(t,y(t))=r・y(t)の場合
- 厳密解は、y(t)=e^rt
積分区間[t0,tf]をn分割すると、h=(tf-t0)/nである。
yn=y(to+nh)=yo・e^(rnh) =y0・e^(r(tf-t0)) ・・・・・厳密解
- オイラー法では
yn+1 = yn + h・f(tn,yn)= yn + hr・yn=(1+hr)・yn
これを繰り返すので
yn = y0・(1+hr)^n ・・・・・オイラー法の近似解
刻み幅hを十分小さくするとyn= y0・(1+hr)^n は、指数eの定義より
Lim y0・(1+r(tf-t0)/n)^n = y0・e^(r(tf-t0))
となり、厳密解に収束する。