实数序列 x1,x2,x3,…x_{1}, x_{2}, x_{3}, \dots 满足递推关系 xn+1=5xn+22xn+3对于所有 n≥1.x_{n+1} = \dfrac{5x_{n} + 2}{2x_{n} + 3}\quad\text{对于所有 } n \ge 1.

  1. (a)

    已知该序列收敛于 ll,求 ll 的所有可能的精确值。

  2. (b)

    描述当 x1=3x_{1} = 3 时该序列的性态。

  3. (c)

    已知 x5=35032158x_{5} = \dfrac{3503}{2158},求 x1x_{1} 的值。

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完整解法

4(a)

1. Set up the equation for the convergence limit.

Since the sequence converges to ll, we can take the limit as n→∞n \to \infty on both sides of the recurrence relation, setting xn+1=lx_{n+1} = l and xn=lx_n = l:
l=5l+22l+3l = \frac{5l + 2}{2l + 3}

2. Rearrange the equation into a standard quadratic form.

l(2l+3)=5l+2l(2l + 3) = 5l + 2
2l2+3l=5l+22l^2 + 3l = 5l + 2
2l2−2l−2=02l^2 - 2l - 2 = 0
l2−l−1=0l^2 - l - 1 = 0

3. Solve the quadratic equation to find the exact values of $l$.

Using the quadratic formula:
l=−(−1)±(−1)2−4(1)(−1)2(1)l = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}
l=1±1+42l = \frac{1 \pm \sqrt{1 + 4}}{2}
l=1±52l = \frac{1 \pm \sqrt{5}}{2}

答案

l=1±52l = \frac{1 \pm \sqrt{5}}{2}

4(b)

1. Calculate the first few terms to observe the sequence's behaviour.

x1=3x_1 = 3
x2=5(3)+22(3)+3=179≈1.889x_2 = \frac{5(3)+2}{2(3)+3} = \frac{17}{9} \approx 1.889
x3=5(179)+22(179)+3=10361≈1.689x_3 = \frac{5(\frac{17}{9})+2}{2(\frac{17}{9})+3} = \frac{103}{61} \approx 1.689
x4=5(10361)+22(10361)+3=637389≈1.638x_4 = \frac{5(\frac{103}{61})+2}{2(\frac{103}{61})+3} = \frac{637}{389} \approx 1.638

The values are strictly decreasing.

2. Determine the value the sequence converges to by observing long-term behaviour.

As n→∞n \to \infty, using the calculator to evaluate subsequent terms shows that xnx_n approaches approximately 1.618031.61803, which corresponds to 1+52\frac{1+\sqrt{5}}{2} (the positive possible exact value of ll found in part (a)).

Therefore, the sequence decreases and converges to 1+52\frac{1+\sqrt{5}}{2}.

答案

The sequence decreases and converges to 1+52\frac{1+\sqrt{5}}{2}.

4(c)

1. Make $x_n$ the subject of the recurrence relation.

xn+1=5xn+22xn+3x_{n+1} = \frac{5x_n + 2}{2x_n + 3}
xn+1(2xn+3)=5xn+2x_{n+1}(2x_n + 3) = 5x_n + 2
2xnxn+1+3xn+1=5xn+22x_nx_{n+1} + 3x_{n+1} = 5x_n + 2
xn(2xn+1−5)=2−3xn+1x_n(2x_{n+1} - 5) = 2 - 3x_{n+1}
xn=2−3xn+12xn+1−5x_n = \frac{2 - 3x_{n+1}}{2x_{n+1} - 5}

2. Iterate backwards from $x_5$ to find $x_1$ using a graphing calculator.

Using a graphing calculator to generate the previous terms sequentially:
x4=2−3(35032158)2(35032158)−5=563344x_4 = \frac{2 - 3\left(\frac{3503}{2158}\right)}{2\left(\frac{3503}{2158}\right) - 5} = \frac{563}{344}
x3=2−3(563344)2(563344)−5=9154x_3 = \frac{2 - 3\left(\frac{563}{344}\right)}{2\left(\frac{563}{344}\right) - 5} = \frac{91}{54}
x2=2−3(9154)2(9154)−5=158x_2 = \frac{2 - 3\left(\frac{91}{54}\right)}{2\left(\frac{91}{54}\right) - 5} = \frac{15}{8}
x1=2−3(158)2(158)−5=2910x_1 = \frac{2 - 3\left(\frac{15}{8}\right)}{2\left(\frac{15}{8}\right) - 5} = \frac{29}{10}

答案

x1=2910x_1 = \frac{29}{10}

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