Math Problem Statement

  1. Show that the sequence {xn} defined by x1 = √ 7 and xn+1 = √ 7 + xn converges to the positive root of the equation x 2 − x − 7 =

Solution

Let's solve the problem by showing that the sequence {xn}\{x_n\} defined by x1=7x_1 = \sqrt{7} and xn+1=7+xnx_{n+1} = \sqrt{7 + x_n} converges to the positive root of the equation x2x7=0x^2 - x - 7 = 0.

Step 1: Find the roots of the equation

The given equation is: x2x7=0x^2 - x - 7 = 0

To find the roots, use the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=1b = -1, and c=7c = -7. Plugging in these values: x=(1)±(1)24(1)(7)2(1)x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-7)}}{2(1)} x=1±1+282x = \frac{1 \pm \sqrt{1 + 28}}{2} x=1±292x = \frac{1 \pm \sqrt{29}}{2}

So the roots are: x=1+292andx=1292x = \frac{1 + \sqrt{29}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{29}}{2}

Since 29>1\sqrt{29} > 1, the positive root is: α=1+292\alpha = \frac{1 + \sqrt{29}}{2}

Step 2: Show that the sequence converges to α\alpha

We need to show that limnxn=α\lim_{n \to \infty} x_n = \alpha.

Assume that the sequence {xn}\{x_n\} converges to some limit LL. Then, by the definition of the sequence: L=7+LL = \sqrt{7 + L}

Square both sides: L2=7+LL^2 = 7 + L

Rearranging terms gives: L2L7=0L^2 - L - 7 = 0

This is exactly the quadratic equation we solved earlier, so LL must be one of the roots of the equation. Since x1=7x_1 = \sqrt{7} is positive, and each xnx_n is defined as a square root (which is positive), the sequence {xn}\{x_n\} is positive, so the limit LL must be the positive root: L=1+292L = \frac{1 + \sqrt{29}}{2}

Step 3: Check if the sequence is increasing and bounded

Let's check whether the sequence {xn}\{x_n\} is increasing and bounded above by α=1+292\alpha = \frac{1 + \sqrt{29}}{2}.

Show that the sequence is increasing:

We will show that xn+1>xnx_{n+1} > x_n for all nn.

Assume xn<αx_n < \alpha. Then: xn+1=7+xnx_{n+1} = \sqrt{7 + x_n} Since f(x)=7+xf(x) = \sqrt{7 + x} is an increasing function (because its derivative f(x)=127+x>0f'(x) = \frac{1}{2\sqrt{7+x}} > 0), and xn<αx_n < \alpha, it follows that: xn+1=7+xn>7+α=αx_{n+1} = \sqrt{7 + x_n} > \sqrt{7 + \alpha} = \alpha So the sequence is increasing.

Show that the sequence is bounded:

Next, we check if the sequence is bounded above by α\alpha.

If xnαx_n \leq \alpha, then: xn+1=7+xn7+α=αx_{n+1} = \sqrt{7 + x_n} \leq \sqrt{7 + \alpha} = \alpha

Thus, the sequence is bounded above by α\alpha.

Conclusion

Since the sequence {xn}\{x_n\} is increasing and bounded above, by the Monotone Convergence Theorem, it converges. The limit of the sequence must satisfy the equation L2L7=0L^2 - L - 7 = 0, and since the sequence is positive, the limit is the positive root α=1+292\alpha = \frac{1 + \sqrt{29}}{2}.

Thus, the sequence {xn}\{x_n\} converges to the positive root of the equation x2x7=0x^2 - x - 7 = 0.


Would you like any further details or explanations?

Here are some related questions you might find helpful:

  1. What is the Monotone Convergence Theorem?
  2. How do you determine if a sequence is bounded?
  3. What are the general properties of quadratic equations?
  4. Can a sequence have more than one limit?
  5. How does the concept of limits apply in real analysis?
  6. What are other methods to solve quadratic equations besides the quadratic formula?
  7. How can you determine the rate of convergence of a sequence?

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Math Problem Analysis

Mathematical Concepts

Sequences
Limits
Quadratic Equations

Formulas

Quadratic formula
Monotone Convergence Theorem

Theorems

Monotone Convergence Theorem

Suitable Grade Level

Advanced High School / Undergraduate