By Long Luo
I personally enjoy sequence and inequality combined problems in math problems. These problems mainly require high observation of recursive relationships, sequence analysis, inequality expansion and contraction, and construction of problem-solving ideas, allowing one to experience the joy of mathematics. Today, let's tackle a sequence problem from the 2011 Tsinghua University Independent Enrollment Mathematics Test. This problem itself has a relatively small amount of calculation and is of medium difficulty. This article will analyze the solution process of this sequence problem in detail, hoping to help readers intuitively understand the problem-solving ideas of such problems.
- (This sub-question is worth 14 points)
Given the function \(f(x) = \dfrac{2x}{ax + b}\) ,\(f(1) = 1\) ,\(f(\dfrac{1}{2}) = \dfrac{2}{3}\) 。Let \(x_1 = \dfrac{1}{2}\) ，\(x_{n+1} = f(x_n)\) 。
Find the general term formula of the sequence \(\{ x_n \}\)；
Prove \(x_1x_2 \dots x_n > \dfrac{1}{2e}\) 。
The first question
Solution: From\(f(1) = 1\) ，\(f(\dfrac{1}{2}) = \dfrac{2}{3}\) yields:
\[ \begin{cases} a + b = 2 \\ a + 2b = 3 \end{cases} \]
easily gives: \(a = 1, \ b = 1\) , so \(f(x)\) is expressed as:
\[ f(x) = \frac{2x}{x+1} \label{1.1} \tag{1.1} \]
First find the sequence\(\{ x_n \}\)The first few items:\(x_1 = \dfrac{1}{2}，x_2 = \dfrac{2}{3}，x_3 = \dfrac{4}{5}，x_4 = \dfrac{8}{9}\)It can be guessed.\(\{ x_n \}\)The general formula is:
\[ x_n = \frac{2^{n-1}}{2^{n-1} + 1} \label{1.2} \tag{1.2} \]
Mathematical induction
We can use mathematical induction to prove the sequence\(\{ x_n \}\) has the general formula \(x_n = \dfrac{2^{n-1}}{2^{n-1} + 1}\) :
when \(n = 1\) , \(x_1 = \dfrac {1}{2}\) is obviously true;
assume \(n = k\) is true, i.e., \(x_k = \dfrac{2^{k-1}}{2^{k-1} + 1}\) , then:
\[ x_{k+1} = f(x_k) = \frac{2x_k}{x_k + 1} = \frac{2^k}{2^k + 1} \]
Combining 1 and 2, for \(n \in \mathbb{N}^*\) the sequence \(\{ x_n \}\) has the general term \(x_n = \dfrac{2^{n-1}}{2^{n-1} + 1}\) .
Recursive method
Observe the recursive expression \(\eqref{1.1}\) , referring to the method in the previous article 2006 Jiangxi Province College Entrance Examination Advanced Mathematics Sequence Problem Analysis , we can take the reciprocal of both sides of \(\eqref{1.1}\) to obtain:
\[ \frac {1}{x_{n+1}} = \frac {1}{2} + \frac{1}{2x_n} \]
Appropriate transformation:
\[ \frac {1}{x_{n+1}} - 1 = \frac {1}{2} \left( \frac{1}{x_n} - 1 \right) \]
Thus, the sequence \(\{ \dfrac{1}{x_n} - 1 \}\) is a geometric sequence with a common ratio of \(\dfrac{1}{2}\) , and the first term of the sequence is \(1\) . Therefore, \(\dfrac{1}{x_n} - 1 = \dfrac {1}{2^{n-1}}\) , so the general term formula for the sequence \(\{ x_n \}\) is: \(x_n = \dfrac{2^{n-1}}{2^{n-1} + 1}\) .
The second question
will involve the sequence\(\{ x_n \}\) substitutes to give:
\[ x_1x_2 \cdots x_{n+1} = \frac {1}{2} \cdot \frac {2^1}{2^1 + 1} \cdot \frac{2^2}{2^2 + 1} \cdots \frac{2^n}{2^n + 1} > \frac{1}{2e} \]
only needs to prove
\[ \frac {1}{1 + 2^{-1}} \cdot \frac {1}{1 + 2^{-2}} \cdots \frac{1}{1 + 2^{-n}} > \frac {1}{e} \]
Taking the logarithm of both sides, we have:
\[ \ln \frac {1}{1 + 2^{-1}} + \ln \frac {1}{1 + 2^{-2}} + \cdots + \ln \frac{1}{1 + 2^{-n}} \]>-1 ]
Can be appropriately transformed into:
\[ \ln (1 + 2^{-1}) + \ln (1 + 2^{-2}) + \cdots + \ln (1 + 2^{-n})<1 ]
According to common inequalities\(\ln (1 + x)\)< x, \ (x > 0)\) ，有：
\[ \ln (1 + 2^{-1}) + \ln (1 + 2^{-2}) + \cdots + \ln (1 + 2^{-n}) < \sum_{k = 1}^{n}2^{-k} < \sum_{k = 1}^{+\infty}2^{-k} = 1 \]
故原命题得证。
Bernoulli's inequality(\textit{Bernoulli's inequality}) )
Besides taking logarithms, the Bernoulli inequality ( (\textit{Bernoulli's inequality}) ) 1 can also be used for bounding.
It needs to be proven that \(\dfrac{1}{x_1x_2 \cdots x_{n+1}} < 2e\) , there exists:
\[ \frac{1}{x_1x_2 \cdots x_{n+1}} = 2 \left( 1 +\frac{1}{2} \right) \left( 1 + \frac{1}{4} \right) \cdots \left( 1 + \frac{1}{2^n} \right) \]
By the Bernoulli inequality, for any integer \(n \geq 1\) and any real number \(x \geq -1\) we have:
\[ (1+x)^{n} \geq 1 + nx； \]
Thus, we have:
\[ \frac{1}{x_1x_2 \cdots x_{n+1}} = 2 \left( 1 +\frac{1}{2} \right) \left( 1 + \frac{1}{4} \right) \cdots \left( 1 + \frac{1}{2^n} \right) < 2 \left( 1 + \frac{1}{2^n} \right)^{2^{n-1} + \cdots + 2 + 1} \]
According to the definition of the natural constant \(e\) 2 \(e = \lim _{n \to \infty}\left( 1 + \dfrac {1}{n} \right)^n\) , easy to obtain:
\[ \frac{1}{x_1x_2 \cdots x_{n+1}} < 2 \left( 1 + \frac{1}{2^n} \right)^{2^n - 1} < 2 \left( 1 + \frac{1}{2^n} \right)^{2^n} < 2e. \]
In summary,
the first question of this problem mainly examines the analysis of recursive formulas by taking the reciprocal of a sequence, and the second question examines the natural constant \(e\). and the application of common inequalities are not difficult.