# Exercise C.1.14

$\star$ By differentiating the entropy function $H(\lambda)$, show that it achieves its maximum value at $\lambda = 1/2$. What is $H(1/2)$?

Phew, it took me a while to figure out that I've forgotten calculus:

H(\lambda) = -\lambda\lg{\lambda} - (1 - \lambda)\lg(1 - \lambda) \\ \begin{align} H'(\lambda) &= -\lg{\lambda} - \frac{\lambda 1}{\lambda \ln2} + \lg(1 - \lambda) - \frac{(1-\lambda)}{(1-\lambda)(-1)\ln2} \\ &= \lg\frac{1 - \lambda}{\lambda} - \lg{e} + \lg{e} \\ &= \lg\frac{1 - \lambda}{\lambda} \end{align}

Let's find a critical point:

$$H'(\lambda) = 0 \\ \Downarrow \\ \lg\frac{1 - \lambda}{\lambda} = 0 \\ \Downarrow \\ \frac{1-\lambda}{\lambda} = 1 \\ \Downarrow \\ \lambda = 1/2$$

Because $H'(1/4) = \lg{3} > 0$ and $H'(3/4) = \lg(1/3) < 0$, we know that it is a maxima.

$$H(1/2) = - \lg(1/2)/2 -\lg(1/2)/2 = - \lg{1/2} = 1$$