Proving that Shannon entropy is maximised for the uniform distribution I know that Shannon entropy is defined as $-\sum_{i=1}^kp_i\log(p_i)$. For the uniform distribution, $p_i=\frac{1}{k}$, so this becomes $-\sum_{i=1}^k\frac{1}{k}\log\left(\frac{1}{k}\right)$.
Further rearrangement produces the following:
$-\sum_{i=1}^k\frac{1}{k}\log(k)^{-1}$
$\sum_{i=1}^k\frac{1}{k}\log(k)$
This is where I am stuck. I need the solution to come to $\log(k)$. What is the next step?
 A: Using Lagrange multipliers we have the equation:
$$\mathcal{L} = \left \{ -\sum_i^k p_i \log p_i - \lambda\left ( \sum_i^k p_i - 1 \right )\right \}$$
Maximizing with respect to the probability,
$$\frac{\partial \mathcal{L}}{\partial p_i} = 0 = -\log p_i - 1 - \lambda \implies $$
$$p_i = e^{-(1+\lambda)}\tag{1}$$ 
Maximizing with respect to $\lambda$:
$$\frac{\partial \mathcal{L}}{\partial \lambda} = 0 = - \sum_i^k p_i + 1 \implies$$
$$ \sum_i^k p_i = 1 \tag{2}$$
Substituting equation (1) into equation (2):
$$\sum_i^k e^{-(1+\lambda)} = 1 \implies$$
$$k e^{-(1+\lambda)} = 1 $$
Since $p_i = e^{-(1+\lambda)}$
$$p_i = \frac{1}{k}$$
The Shannon Entropy formula now becomes
$$ H = - \sum_i^k \frac{1}{k}\log \frac{1}{k}$$
Since $k$ does not depend on the summation,
$$H = \frac{k}{k} \log k = \log k$$
A: This is a constrained maximization problem in $k$ variables $p_1,p_2,...p_k$. The objective function is 
$$-\sum_{i=1}^kp_i\log(p_i)$$
and the constraint is 
$$\sum_{i=1}^kp_i = 1$$
Form the Lagrangean and I guess you can proceed from here.
