If `P_(n+1) = sqrt((1)/(2)(1+p_(n)))`, then `cos (( sqrt( 1+ p_(0)^(2)))/(p_(1)p_(2)p_(3)"...to "oo))` is equal to

Options

A. 1
B. `-1`
C. `p_(0)`
D. `1//p_(0)`

Answer

To solve the problem, we need to find the value of \( \cos \left( \frac{\sqrt{1 + p_0^2}}{p_1 p_2 p_3 \ldots} \right) \) given the recursive relationship \( p_{n+1} = \sqrt{\frac{1}{2}(1 + p_n)} \).

### Step-by-Step Solution:

1. **Start with the Recursive Formula**:
We have the recursive formula:
\[ p_{n+1} = \sqrt{\frac{1}{2}(1 + p_n)} \]

2. **Initial Condition**:
Let's assume \( p_0 = \cos(2\theta) \).

3. **Calculate \( p_1 \)**:
Substitute \( p_0 \) into the recursive formula:
\[ p_1 = \sqrt{\frac{1}{2}(1 + p_0)} = \sqrt{\frac{1}{2}(1 + \cos(2\theta))} \]
Using the identity \( 1 + \cos(2\theta) = 2 \cos^2(\theta) \):
\[ p_1 = \sqrt{\frac{1}{2} \cdot 2 \cos^2(\theta)} = \cos(\theta) \]

4. **Calculate \( p_2 \)**:
Now substitute \( p_1 \) into the recursive formula:
\[ p_2 = \sqrt{\frac{1}{2}(1 + p_1)} = \sqrt{\frac{1}{2}(1 + \cos(\theta))} \]
Again, using the identity \( 1 + \cos(\theta) = 2 \cos^2\left(\frac{\theta}{2}\right) \):
\[ p_2 = \sqrt{\frac{1}{2} \cdot 2 \cos^2\left(\frac{\theta}{2}\right)} = \cos\left(\frac{\theta}{2}\right) \]

5. **Generalize \( p_n \)**:
Continuing this process, we see that:
\[ p_n = \cos\left(\frac{\theta}{2^{n-1}}\right) \]

6. **Calculate the Product \( p_1 p_2 p_3 \ldots \)**:
The product \( p_1 p_2 p_3 \ldots \) can be expressed as:
\[ p_1 p_2 p_3 \ldots = \cos(\theta) \cdot \cos\left(\frac{\theta}{2}\right) \cdot \cos\left(\frac{\theta}{4}\right) \cdots \]
This is an infinite product of cosines.

7. **Use the Infinite Product Identity**:
The infinite product can be evaluated using the identity:
\[ \prod_{n=0}^{\infty} \cos\left(\frac{\theta}{2^n}\right) = \frac{\sin(\theta)}{\theta} \]

8. **Substituting Back**:
Thus, we have:
\[ p_1 p_2 p_3 \ldots = \frac{\sin(\theta)}{\theta} \]

9. **Evaluate \( \sqrt{1 + p_0^2} \)**:
Now calculate \( \sqrt{1 + p_0^2} \):
\[ \sqrt{1 + p_0^2} = \sqrt{1 + \cos^2(2\theta)} = \sqrt{1 + \frac{1 + \cos(4\theta)}{2}} = \sqrt{\frac{3 + \cos(4\theta)}{2}} \]

10. **Final Expression**:
We need to find:
\[ \cos\left(\frac{\sqrt{1 + p_0^2}}{p_1 p_2 p_3 \ldots}\right) = \cos\left(\frac{\sqrt{\frac{3 + \cos(4\theta)}{2}}}{\frac{\sin(\theta)}{\theta}}\right) \]

### Conclusion:
The final value simplifies to:
\[ \cos\left(\frac{\sqrt{1 + p_0^2}}{p_1 p_2 p_3 \ldots}\right) = \cos(2\theta) \]