The idea on the first floor is right.
(a 1-√3)*(a2-√3)
=(a 1-√3)*( 1+2/( 1+a 1)-√3)
=(a 1-√3)*(2+( 1-√3)*( 1+a 1))/( 1+a 1)
=(a 1-√3)*(3-√3+( 1-√3)* a 1)/( 1+a 1)
=(a 1-√3)*( 1-√3)*(a 1-√3)/( 1+a 1)
=-(√3- 1)*(a 1-√3)^2/( 1+a 1)<; 0
So the root number 3 is between a 1 and a2.
(2)
Separable a 1
Set a 1
Just prove A2-√ 3.
After substituting a2, the inequality can be sorted as follows
( 1-√3)*(a 1-√3)/( 1+a 1)+(a 1-√3)& lt; 0,
According to a 1-√ 3
Similarly, when a1>; √3,
It proves that √ 3-A2
After substituting a2, the inequality can be sorted as follows
( 1-√3)*(a 1-√3)/( 1+a 1)+(a 1-√3)>0,
According to a1-√ 3 >; 0,a 1 & gt; 0 inequality is clearly established.
(3)
Synthesis (1) and (2),
Let b(n) be a sequence of all even terms of a(n)
Let c(n) be the sequence of all odd terms of a(n)
It is easy to know that B (n) and C (n) approach √3 from the left and right respectively.
B (n) and c (n) are monotone bounded sequences,
Therefore, the limit of b(n) exists and is set to x; C(n) limit exists and is set to y.
Whether B (n) and C (n) approach √3 from the left or right depends on whether a 1 is on the left or right of √3.
It is known that A (n+2) =1+2/(1+A (n+1)) is used instead of A (n+1) =1+.
The recursive formula for finding a(n+2) from a (n) is as follows:
a(n+2)=(3+2a(n))/(2+a(n))
Therefore, the recurrence formula of the sequence b(n) is: b(n+ 1)=(3+2b(n))/(2+b(n)).
Because the limit of b(n) is x, you can get it by replacing both b(n+ 1) and b(n) in the above formula with X.
X=(3+2X)/(2+X),
Where a 1 is the approximate value of √3 and a 1 >; 0, according to the conclusions in (1) and (2), it is not difficult to judge that a(n) is greater than 0.
So b(n) is greater than 0, and natural x cannot be equal to a negative number.
The above equation about x can be arranged as follows
X 2 = 3, so X=√3.
Similarly, you can get Y=√3.
Both the odd and even terms of the sequence a(n) are restricted by √3, so a(n) must also be restricted by √3.
The first term of a(n) is a rational number. By observing the recurrence formula of a(n), we can draw the conclusion that a(n) is rational.
Therefore, the sequence a(n) constitutes a method to find a rational near quasi-value with root number 3.