Typical example: a (n+1) = (a (an)+b)/(c (an)+d)
Note: I feel that this is the only thing that usually needs a fixed point. It is enough to remember its solution.
It is not impossible for us to solve this problem by general methods, but it is too complicated to determine the coefficient and ask for the reciprocal. If the fixed point method is used, the problem will be easy to solve. x=(ax+b)/(cx+d).
That is, cx2+(d-a)x-b=0.
Let the two roots of this equation be x 1, x2,
If x 1=x2
Then it is1/(a (n+1)-x1) =1/(an-x1)+p.
P can be solved by undetermined coefficient method, and then by arithmetic progression's general formula.
Note: If you have the ability, you can remember the expression of p, p=2c/(a+d).
If x 1≠x2, there is (a (n+1)-x1)/(a (n+1)-x2) = q ((an-x1)).
Q can be solved by the undetermined coefficient method, and then by the general term formula of geometric series.
Note: If you have the ability, you can remember the expression of q, q=(a-cx 1)/(a-cx2).
Simply put, substitute an=x in recursion.
A(n+ 1) is also equal to X.
Then construct a sequence. (However, it should be noted that the fixed point method is not omnipotent, and some recursive formulas have no fixed points, but other construction methods can be used to find the general term; Some can't be found out)
Let me give you a few specific examples:
1。 It is known that a (1) = m.a (n+1) = [a * a (n)+b]/[c * a (n)+d] to find the general term of an.
A(n) and a(n+ 1) respectively represent the nth item and the nth+1item of the sequence.
Solution: I have two solutions to this form of recursive formula, the undetermined coefficient method and the fixed point method. Here I use the fixed point method to solve this problem.
Replace both a[n] and a[n+ 1] in the original recursive formula with x, and get the equation x=(ax+b)/(cx+d).
Is that cx? +(d-a)x-b=0
Remember that the roots of the equation are x 1, x2 (for simplicity, assume that the equation has two real roots).
The original equation can be transformed into -x(a-cx)=b-dx.
So -x=(b-dx)/(a-cx), and replace x 1, x2.
-x 1 =(b-dx 1)/(a-CX 1)
-x2=(b-dx2)/(a-cx2)
Subtract x 1 from both sides of the recursive formula to get a [n-1]-x1= [(a-CX1) a [n]+b-dx1]/(ca [n]+d).
That is, a [n-1]-x1= (a-CX1) [a [n]+(b-dx1)]/(ca [n]
Substituting-x1= (b-dx1)/(a-CX1) gives:
a[n- 1]-x 1 =(a-CX 1)(a[n]-x 1)/(ca[n]+d)
Similarly: a [n-1]-x2 = (a-cx2) (a [n]-x2)/(ca [n]+d)
Divide the two formulas to get (a [n+1]-x1)/(a [n+1]-x2) = [(a-CX1)/(a-cx2)] * [
So {(a[n]-x 1)/(a[n]-x2)} is a geometric series.
(a[n]-x 1)/(a[n]-x2)=[(m-x 1)/(m-x2)]*[(a-cx 1)/(a-cx2)]^(n- 1)
So a [n] = {x2 * [(m-x1)/(m-x2)] * [(a-CX1)/(a-cx2)] (n-1)-x/kloc-0.
2。 an = 2/A(n- 1)+A(n- 1)/2
Find the general term of An
Solution: Find the general term by using the fixed point:
Let f(x)=2/x+x/2.
When f(x)=x
X =-2,2, and this point is a fixed point.
an-2=[a(n- 1)-2]^2/2a(n- 1)
an-(-2)=[a(n- 1)-(-2)]^2/2a(n- 1)
Two types of division
An-2 = [a (n-1)-2] 2
—— ——————
An+2 [a (n-1)+2] 2
Did you find a pattern?
Let {Bn}=(An-2)/(An+2).
B 1=(4-2)/(4+2)= 1/3
The recurrence formula is bn = b (n- 1) 2.
So bn = (1/3) [2 (n- 1)]
Relationship between Bn General Term and General Term
Solution: an = {2 * (1/3) [2 (n-1)]+2}/
{ 1-( 1/3)^[2^(n- 1)]}
Try to simplify yourself.
To add: fixed points are mostly used in limit processes. For example, the general forms of implicit function theorem and inverse function theorem in mathematical analysis, and the existence and uniqueness theorem of solutions of initial value problems of differential equations are all proved by fixed point theory.
You can refer to any book about combinatorial mathematics. Because the sequence is an iteration of fractional linear transformation, which can correspond to the power of the second-order matrix, it can also be solved by the eigenvalue of linear algebra to get the standard form, which is similar to the idea. -This is the mathematical content behind this question.
The specific content is probably very long to write. I suggest you look up books, books on combinatorial mathematics or books on mathematical competitions to talk about a part or series of combinatorial mathematics.
For high school students, of course, we can look at this problem from a more natural point of view: the recursive formula can be transformed into (one or two) geometric series solutions through appropriate transformation.
I found an article about linear recurrence and fractional linear recurrence series on the Internet, which is helpful to you:
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References:
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