How to prove the limit of sequence with definition method Please give the steps to solve the problem. Judge whether the sequence has limit. If so, please write Xn=cos（1/n）

The limit of the sequence is 1. The proof is as follows: for any ε > 0, let
　　　　|cos(1/n)-1| = |-2{sin[(1/n)/2]}^2]| < 2*[(1/n)/2]}^2 < 1/n^2 < 1/n< ε,
If n > 1 / ε and N = [1 / ε] + 1, then when n > N, there is
　　　　|cos(1/n)-1| < 1/n< 1/N

How to find the limit of sequence with the method of indefinite point
Let X1 = 2 xn + 1 = 2 + 1 / xn find the limit of sequence

Let the limit be a, then a = 2 + 1 / A, a ^ 2-2a-1 = 0
The solution is a = √ 2 + 1

How to find the general term in the fixed point method of sequence limit

Generally, in order to find the general term of recursive sequence a [n + 1] = (CA [n] + D) / (EA [n] + F) [C, D, e, f are not all zero constants, C, e are not all zero at the same time], we can use the fixed point method to solve it. If the sequence {a [n]} satisfies a [n + 1] = f (a [n]), we call x = f (x) the fixed point equation of function f (x), Its root is called the fixed point of function f (x). As for why the general term of recursive sequence can be solved by the fixed point method, it is enough to write a book. But the general understanding is that when n tends to infinity, if there is a limit of sequence {a [n]}, there is no difference between a [n] and a [n + 1]
Firstly, it should be noted that not all recursive sequences have corresponding fixed point equations, such as a [n + 1] = a [n] + 1 / a [n]. Secondly, fixed points have different fixed points and coincident fixed points
The following is a few specific examples combined with various methods of solving general term by fixed point method
Example 1: given a [1] = 2, a [n + 1] = 2 / (a [n] + 1), find the general term
Note: This is an example of "different fixed points."
Find the fixed point first
∵a[n+1]=2/(a[n]+1)
Let x = 2 / (x + 1), the fixed points are: x = 1 and x = - 2 [different fixed points]
(a [n + 1] - 1) / (a [n + 1] + 2) [use fixed point]
=(2/(a[n]+1)-1)/(2/(a[n]+1)+2)
=(2-a[n]-1)/(2+2a[n]+2)
=(-a[n]+1)/(2a[n]+4)
=(-1/2)(a[n]-1)/(a[n]+2)
∵a[1]=2
∴(a[1]-1)/(a[1]+2)=1/4
{(a [n] - 1) / (a [n] + 2)} is an equal ratio sequence with the first term of 1 / 4 and the common ratio of - 1 / 2
∴(a[n]-1)/(a[n]+2)=1/4(-1/2)^(n-1)
The solution is a [n] = 3 / [1 - (- 1 / 2) ^ (n + 1)] - 2
Example 2: given that the sequence {a [n]} satisfies a [1] = 3, a [n] a [n-1] = 2A [n-1] - 1, find the general term
Note: This is an example of "coincident fixed point". The method of taking the reciprocal is often used in "coincident fixed point"
∵a[n]=2-1/a[n-1]
The fixed point method is used to make x = 2-1 / X
Namely: x ^ 2-2x + 1 = 0
‖ x = 1 [coincident fixed point]
∵a[n]=2-1/a[n-1]
A [n] - 1 = 2-1 / a [n-1] - 1 [use fixed point]
a[n]-1=(a[n-1]-1)/a[n-1]
Take the reciprocal on both sides to get: 1 / (a [n] - 1) = a [n-1] / (a [n-1] - 1)
That is: 1 / (a [n] - 1) - 1 / (a [n-1] - 1) = 1
∵a[1]=3
{1 / (a [n] - 1)} is an arithmetic sequence with the first term of 1 / (a [1] - 1) = 1 / 2 and tolerance of 1
That is: 1 / (a [n] - 1) = 1 / 2 + (n-1) = (2n-1) / 2
∴a[n]=2/(2n-1)+1=(2n+1)/(2n-1)
Example 3: given that the sequence {a [n]} satisfies a [1] = 1 / 2, s [n] = a [n] n ^ 2-N (n-1), find the general term
Note: the coefficients of the fixed point equation obtained in the above two examples are all constants. Now let's look at an example where the coefficients of the fixed point equation contain n
∵S[n]=a[n]n^2-n(n-1)
∴S[n+1]=a[n+1](n+1)^2-(n+1)n
By subtracting the above two expressions, we get:
a[n+1]=a[n+1](n+1)^2-a[n]n^2-(n+1)n+n(n-1)
(n^2+2n)a[n+1]=a[n]n^2+2n
(n+2)a[n+1]=na[n]+2
a[n+1]=a[n]n/(n+2)+2/(n+2) 【1】
Using fixed point method, let x = xn / (n + 2) + 2 / (n + 2)
The solution is: x = 1 [coincidence fixed point]
Let: a [n] - 1 = B [n], then: a [n] = B [n] + 1 [use fixed point]
Substituting into formula [1], B [n + 1] + 1 = (B [n] + 1) n / (n + 2) + 2 / (n + 2)
b[n+1]=b[n]n/(n+2)
That is: B [n + 1] / B [n] = n / (n + 2)
So: [because the right is divided into several lines, write a few more lines to see clearly]
B [n] / B [n-1] = (n-1) / (n + 1) [denominator is reserved here]
B [n-1] / B [n-2] = (n-2) / n [here the denominator is reserved]
b[n-2]/b[n-3]=(n-3)/(n-1)
b[n-3]/b[n-4]=(n-4)/(n-2)
.
b[5]/b[4]=4/6
b[4]/b[3]=3/5
B [3] / B [2] = 2 / 4
B [2] / B [1] = 1 / 3
By multiplying the above items on the left and right, we get the following results:
b[n]/b[1]=(1*2)/[n(n+1)]
∵a[1]=1/2
∴b[1]=a[1]-1=-1/2
∴b[n]=-1/[n(n+1)]
The general term a [n] = B [n] + 1 = 1-1 / [n (n + 1)]
Example 4: given that the sequence {a [n]} satisfies a [1] = 2, a [n + 1] = (2a [n] + 1) / 3, find the general term
Note: this example shows that some problems can be solved by fixed point method or other methods
∵a[n+1]=(2a[n]+1)/3
Find the fixed point: x = (2x + 1) / 3, get: x = 1 [coincidence fixed point]
A [n + 1] - 1 = (2a [n] + 1) / 3-1 [use fixed point]
Namely: a [n + 1] - 1 = (2 / 3) (a [n] - 1)
{a [n] - 1} is an equal ratio sequence with the first term of a [1] - 1 = 1 and the common ratio of 2 / 3
That is: a [n] - 1 = (2 / 3) ^ (n-1)
∴a[n]=1+(2/3)^(n-1)
[also] ∵ a [n + 1] = (2a [n] + 1) / 3
∴3a[n+1]=2a[n]+1
In this case, the undetermined coefficient method can be used, or even the observation method can be used directly
3a[n+1]-3=2a[n]-2
∴a[n+1]-1=(2/3)(a[n]-1)
Same as above
Example 5: given that the sequence {x [n]} satisfies x [1] = 2, x [n + 1] = (x [n] ^ 2 + 2) / (2x [n]), find the general term
Note: let's take an example where the fixed point is an irrational number, in which the logarithm method is also used
∵x[n+1]=(x[n]^2+2)/(2x[n])
Using the fixed point method, let y = (y ^ 2 + 2) / (2Y)
y^2=2
The fixed point of the solution is y = ± √ 2
(x [n + 1] - 2) / (x [n + 1] + √ 2) [use fixed point]
={(x[n]^2+2)/2x[n]-√2}/{(x[n]^2+2)/2x[n]+√2}
=(x[n]^2-2√2x[n]+2)/(x[n]^2+2√2x[n]+2)
={(x[n]-√2)/(x[n]+√2)}^2
∵x[n+1]=(x[n]^2+2)/2x[n]=x[n]/2+1/x[n]≥2/√2=√2
Ψ ln {(x [n + 1] - 2) / (x [n + 1] + √ 2)} = 2ln {(x [n] - 2) / (x [n] + √ 2)} [logarithm]
∵x[1]=2>√2
∴(x[1]-√2)/(x[1]+√2)=3-2√2
{ln ((x [n] - 2) / (x [n] + √ 2))} is an equal ratio sequence with LN (3-2 √ 2) as the first term and 2 as the common ratio
That is: ln {(x [n] - 2) / (x [n] + √ 2)} = 2 ^ (n-1) ln (3-2 √ 2)
(x[n]-√2)/(x[n]+√2)=(3-2√2)^[2^(n-1)]
x[n]-√2=(3-2√2)^[2^(n-1)](x[n]+√2)
x[n]-x[n](3-2√2)^[2^(n-1)]=√2(3-2√2)^[2^(n-1)]+√2
∴x[n]=√2{1+(3-2√2)^[2^(n-1)]}/{1-(3-2√2)^[2^(n-1)]}
Example 6: given that the sequence {a [n]} satisfies a [1] = 2, a [n + 1] = (1 + a [n]) / (1-A [n]), find the general term
Note: let's take an example where the fixed point is an imaginary number to show that some problems can be solved by the fixed point method, but other solutions may be more convenient
Find the fixed point: x = (1 + x) / (1-x), that is: x ^ 2 = - 1, get:
X [1] = I, x [2] = - I
(a [n + 1] - I) / (a [n + 1] + I) [use fixed point]
={(1+a[n])/(1-a[n]-i}/{(1+a[n])/(1-a[n]+i}
=(1+a[n]-i+a[n]i)/(1+a[n]+i-a[n]i)
={(1+i)/(1-i)}{(a[n]-i)/(a[n]+i)}
=i(a[n]-i)/(a[n]+i)
∵a[1]=2
{(a [n] - I) / (a [n] + I)} is an equal ratio sequence with the first term of (a [1] - I) / (a [1] + I) = (2-I) / (2 + I) and the common ratio of I
Namely: (a [n] - I) / (a [n] + I) = [(2-I) / (2 + I)] I ^ (n-1)
(a[n]-i)(2+i)=(a[n]+i)(2-i)i^(n-1)
2a[n]-2i+ia[n]+1=(2a[n]+2i-ia[n]+1)i^(n-1)
{2+i-(2-i)(i)^(n-1)}a[n]=2i-1+(2i+1)i^(n-1)
a[n]=[2i-1+(2i+1)i^(n-1)]/[2+i-(2-i)i^(n-1)]
∴a[n]=[2i-1+(2-i)i^n]/[2+i-(2-i)i^(n-1)]
Let's use "triangle substitution" to see if it's more ingenious
∵a[n+1]=(1+a[n])/(1-a[n])
Let a [n] = Tan θ, then a [n + 1] = [Tan (π / 4) + Tan θ] / [1-tan (π / 4) Tan θ] = Tan (π / 4 + θ)
∵θ=arctan(a[n]),π/4+θ=arctan(a[n+1])
By subtracting the above two expressions, we get arctan (a [n + 1]) - arctan (a [n]) = π / 4
∵a[1]=2
{arctan (a [n])} is an arithmetic sequence whose first term is arctan (a [1]) = arctan2 and tolerance is π / 4
That is: arctan (a [n]) = arctan2 + (n-1) π / 4
∴a[n]=tan[(n-1)π/4+arctan2]

Ask a proof about similar matrix (linear algebra)
Let a and B be matrices of order n, and a be similar to B, and E be identity matrices of order n. It is proved that te-a is similar to te-b for any constant t

In this way, for any constant T, we have: P * (te-b) * P ^ (- 1) = P * te * P ^ (- 1) - p * b * P ^ (- 1) = t (p * e * P ^ (- 1)) - a = t (p * P ^ (- 1)) - a = te-a, then te-a = P * (te-b) * P ^ (- 1)

Let matrix a be invertible, and prove that (a *) - 1 = |a-1|a

If the matrix is invertible, then a & # 732; &# 185; = 1 / | a | x a *, equivalent to AA * = | a | e, (E is the unit matrix of order n)
Because | A & # 732; & # 185; | = 1 / | a |, so | A & # 732; & # 185; | × AA * = e,
So the inverse matrix of a * is: | A & # 732;, # 185; | × a

On the proof of invertible matrix
Let p be an invertible matrix of order n, if B = P ^ (- 1) AP, it is proved that B ^ m = P ^ (- 1) a ^ MP, where m is any integer
M is a positive integer

This proves that:
B^m=P^(-1)A^mP=BB… B (multiplication of m b) = (P ^ (- 1) AP) * (P ^ (- 1) AP) （p^(-1)AP）=p^(-1)AP*p^(-1)AP*p^(-1)AP*… p^(-1)AP
Because p ^ (- 1) * P = e, the above formula becomes B ^ m = P ^ (- 1) a * e * a * e * *A*P=P^(-1)A^mP
B ^ m = P ^ (- 1) a ^ MP,
This question is proved

It is proved that the matrix is always invertible
It is proved that ((a ^ t) a + λ I) is always an invertible matrix, where λ is always positive

Consider the system of linear equations [(a ^ t) a + λ I] x = 0, so (a ^ t) AX = - λ x, that is, X is the eigenvector of (a ^ t) a corresponding to negative eigenvalue - λ. Because (a ^ t) a is a semi positive definite matrix, its eigenvalues are nonnegative, so x = 0, so the matrix (a ^ t) a + λ I is invertible

On a linear matrix
Given that the cubic power of a is equal to e, then how much is the square of a plus a plus E and whether it is zero
A3 = e, then whether A2 + A + e is zero, please write according to
In fact, I want to prove whether a + e is reversible, so I have to launch a ^ 2-A + e first, but I can't figure out what kind of situation it is. Please be more clear

A ^ 3 = e, then the eigenvalues of a are cubic unit roots
If a does not contain eigenvalue 1, then A-E is reversible, and a ^ 2 + A + e = 0 is obtained from 0 = a ^ 3-E = (A-E) (a ^ 2 + A + e)
If a contains eigenvalue 1, the conclusion is not correct
Add: you see the sign clearly first, if it is a + e, then it must be reversible, just look at the eigenvalue

If matrix A is similar to B, then ()
Let's help
1. If matrix A is similar to B, then () a. | a | = | B|
b. A and B are similar to a diagonal matrix
c. For the same eigenvalues, matrices A and B have the same eigenvectors
2. It is known that the eigenvalues of the third-order matrix A are 0, - 1, + 1
A. Matrix A is irreversible B. the sum of main diagonal elements of matrix A is 0
C. The eigenvectors corresponding to - 1 and - 1 are orthogonal
D. The fundamental solution system of AX = 0 consists of a vector
E. Matrix A-E is an irreversible matrix
F. Matrix A + e is similar to diagonal matrix
Choose what ah, the best is to give a reason for the master ah, thank you

1.A
The same eigenvalue / determinant / trace of similar matrix
2.C
Eigenvectors belonging to different eigenvalues are linearly independent and not necessarily orthogonal
Orthogonality of real symmetric matrices

How to prove the limit of sequence with definition method Please give the steps to solve the problem. Judge whether the sequence has limit. If so, please write Xn=cos（1/n）