If `x_(r) " is given by , " x_(r+1) = sqrt(1/2 ( 1 + x_(r)))" . Then, show: " cos^(-1) x_(0) = sqrt(1-x_(0)^(2))/(x_(1)x_(2)x_(3))` ......up to infinity .

Answer

To solve the problem, we need to show that:

\[ \cos^{-1}(x_0) = \frac{\sqrt{1 - x_0^2}}{x_1 x_2 x_3 \ldots} \]

where \( x_{r+1} = \sqrt{\frac{1}{2}(1 + x_r)} \).

### Step 1: Define \( x_0 \)
Let \( x_0 = \cos \theta \).

### Step 2: Find \( x_1 \)
Using the recursive formula:

\[ x_1 = \sqrt{\frac{1}{2}(1 + x_0)} = \sqrt{\frac{1}{2}(1 + \cos \theta)} \]

Using the identity \( 1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right) \):

\[ x_1 = \sqrt{\frac{1}{2} \cdot 2 \cos^2 \left(\frac{\theta}{2}\right)} = \cos \left(\frac{\theta}{2}\right) \]

### Step 3: Find \( x_2 \)
Continuing with the recursion:

\[ x_2 = \sqrt{\frac{1}{2}(1 + x_1)} = \sqrt{\frac{1}{2}(1 + \cos \left(\frac{\theta}{2}\right))} \]

Using the identity again:

\[ 1 + \cos \left(\frac{\theta}{2}\right) = 2 \cos^2 \left(\frac{\theta}{4}\right) \]

Thus,

\[ x_2 = \sqrt{\frac{1}{2} \cdot 2 \cos^2 \left(\frac{\theta}{4}\right)} = \cos \left(\frac{\theta}{4}\right) \]

### Step 4: Find \( x_3 \)
Continuing this process:

\[ x_3 = \sqrt{\frac{1}{2}(1 + x_2)} = \sqrt{\frac{1}{2}(1 + \cos \left(\frac{\theta}{4}\right))} \]

Using the identity again:

\[ 1 + \cos \left(\frac{\theta}{4}\right) = 2 \cos^2 \left(\frac{\theta}{8}\right) \]

Thus,

\[ x_3 = \sqrt{\frac{1}{2} \cdot 2 \cos^2 \left(\frac{\theta}{8}\right)} = \cos \left(\frac{\theta}{8}\right) \]

### Step 5: Generalize \( x_n \)
From the pattern, we can see that:

\[ x_n = \cos \left(\frac{\theta}{2^n}\right) \]

### Step 6: Compute the product \( x_1 x_2 x_3 \ldots \)
Now we need to compute the infinite product:

\[ x_1 x_2 x_3 \ldots = \prod_{n=1}^{\infty} x_n = \prod_{n=1}^{\infty} \cos \left(\frac{\theta}{2^n}\right) \]

### Step 7: Use the identity for sine
Using the sine double angle identity:

\[ \sin(2x) = 2 \sin(x) \cos(x) \]

We can express:

\[ \sin(\theta) = 2^n \sin\left(\frac{\theta}{2^n}\right) \prod_{k=1}^{n} \cos\left(\frac{\theta}{2^k}\right) \]

Taking the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} 2^n \sin\left(\frac{\theta}{2^n}\right) = \theta \]

### Step 8: Substitute back to show the result
Thus, we have:

\[ \frac{\sqrt{1 - x_0^2}}{x_1 x_2 x_3 \ldots} = \cos^{-1}(x_0) \]

### Conclusion
Therefore, we conclude that:

\[ \cos^{-1}(x_0) = \frac{\sqrt{1 - x_0^2}}{x_1 x_2 x_3 \ldots} \]