Math Problem Statement

  1. Two sequences {xn} and {yn} are defined by xn+1 = 1 2 (xn + yn), yn+1 = p (xnyn) for n ≥ 1 and x1 > 0, y1 > 0. Prove that both sequences converge to a common liR

Solution

Let's consider the sequences {xn}\{x_n\} and {yn}\{y_n\} defined by the recursive formulas:

xn+1=12(xn+yn)x_{n+1} = \frac{1}{2}(x_n + y_n) yn+1=xnyny_{n+1} = \sqrt{x_n y_n} with the initial conditions x1>0x_1 > 0 and y1>0y_1 > 0.

We need to prove that both sequences converge to the same limit.

Step 1: Monotonicity

Let's first prove that {xn}\{x_n\} is decreasing and {yn}\{y_n\} is increasing.

  1. Sequence {xn}\{x_n\} is decreasing:

    Since xn+1=12(xn+yn)x_{n+1} = \frac{1}{2}(x_n + y_n), and by the arithmetic-geometric mean inequality (AM-GM inequality):

    xn+yn2xnyn\frac{x_n + y_n}{2} \geq \sqrt{x_n y_n}

    This implies:

    xn+1yn+1x_{n+1} \geq y_{n+1}

    Therefore, xnxn+1x_n \geq x_{n+1} for all nn, so {xn}\{x_n\} is decreasing.

  2. Sequence {yn}\{y_n\} is increasing:

    Similarly, yn+1=xnyny_{n+1} = \sqrt{x_n y_n}, and by the same AM-GM inequality:

    yn+1=xnynxn+yn2=xn+1y_{n+1} = \sqrt{x_n y_n} \leq \frac{x_n + y_n}{2} = x_{n+1}

    Thus, ynyn+1y_n \leq y_{n+1} for all nn, so {yn}\{y_n\} is increasing.

Step 2: Boundedness

Next, we show that both sequences are bounded.

  1. {xn}\{x_n\} is bounded below:

    Since xnx_n is decreasing and positive (as x1>0x_1 > 0), it is bounded below by 00.

  2. {yn}\{y_n\} is bounded above:

    Since yny_n is increasing and xnynx_n \geq y_n, the sequence {yn}\{y_n\} is bounded above by x1x_1.

Step 3: Convergence

Since {xn}\{x_n\} is decreasing and bounded below, it converges to some limit LxL_x.

Similarly, since {yn}\{y_n\} is increasing and bounded above, it converges to some limit LyL_y.

Step 4: Common Limit

Now, we show that Lx=LyL_x = L_y. Taking limits in the recursive relations:

  1. From the equation for xn+1x_{n+1}:

    Lx=12(Lx+Ly)L_x = \frac{1}{2}(L_x + L_y)

    Simplifying this:

    2Lx=Lx+Ly2L_x = L_x + L_y Lx=LyL_x = L_y

    So, both sequences converge to the same limit, say LL.

  2. From the equation for yn+1y_{n+1}:

    L=L×L=LL = \sqrt{L \times L} = L

Thus, both sequences converge to a common limit LL.

Conclusion

The sequences {xn}\{x_n\} and {yn}\{y_n\} both converge to a common limit LL.


Would you like more details or have any other questions?

Related Questions:

  1. What is the significance of the AM-GM inequality in this proof?
  2. How can we determine the actual value of the common limit LL?
  3. Are there other methods to prove the convergence of these sequences?
  4. What happens if x1x_1 and y1y_1 are not positive?
  5. Can we generalize this approach to other types of recursive sequences?
  6. How does the bounded monotonic sequence theorem apply here?
  7. What if the initial conditions were such that x1=y1x_1 = y_1?
  8. How does the concept of fixed points relate to the convergence of sequences?

Tip:

Always check for monotonicity and boundedness when proving the convergence of sequences. These are powerful tools that simplify the proof process.

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Math Problem Analysis

Mathematical Concepts

Recursive sequences
Convergence of sequences
Arithmetic-geometric mean inequality

Formulas

Recursive formulas for sequences

Theorems

Monotonic sequence theorem
Bounded sequence theorem

Suitable Grade Level

Advanced Undergraduate