Evaluate Tan Alpha for Sum of Inverse Tangent Series

To evaluate $\tan \alpha$, we first simplify the general term of the given summation.

The given equation is: $\alpha = \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right)$

Let $T_p$ denote the $p$-th term of the summation: $T_p = \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right)$

Notice that $2^{2p-1} = 2^p \cdot 2^{p-1}$ and $2^{p-1} = 2^p - 2^{p-1}$. We can rewrite $T_p$ as: $T_p = \tan^{-1}\left(\frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}}\right)$

Using the standard inverse trigonometric identity: $\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y \quad \text{for } x, y > 0$

we get: $T_p = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$

Now, summing $T_p$ from $p = 1$ to $11$ forms a telescoping series: $\sum_{p=1}^{11} T_p = \sum_{p=1}^{11} \left[ \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) \right]$

Expanding the sum: $\sum_{p=1}^{11} T_p = \left( \tan^{-1}(2^1) - \tan^{-1}(2^0) \right) + \left( \tan^{-1}(2^2) - \tan^{-1}(2^1) \right) + \dots + \left( \tan^{-1}(2^{11}) - \tan^{-1}(2^{10}) \right)$

All intermediate terms cancel out, leaving: $\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \tan^{-1}(2^0)$

Since $2^0 = 1$ and $\tan^{-1}(1) = \frac{\pi}{4}$: $\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \frac{\pi}{4}$

Substitute this result back into the expression for $\alpha$: $\alpha = \frac{\pi}{4} + \left( \tan^{-1}(2^{11}) - \frac{\pi}{4} \right)$ $\alpha = \tan^{-1}(2^{11})$

Taking the tangent on both sides: $\tan \alpha = \tan\left(\tan^{-1}(2^{11})\right) = 2^{11}$

Since $2^{11} = 2048$: $\tan \alpha = 2048$