Given the sum:
$\sum_{k=1}^{m} n_{k} log (p_{k}) + \lambda (1 - \sum_{k=1}^{m}p_k)$ ?
derive for p and $\lambda$, So far I've gotten it to this, but I'm unsure if my answers are correct, could someone clarify?
Assume $log(x)$ to base $e$
$f(p) = \sum_{k=1}^{m} n_{k} log (p_{k}) + \lambda - \lambda\sum_{k=1}^{m}p_k$
$=>\lambda + \sum_{k=1}^{m} n_{k} log (p_{k}) - \lambda\sum_{k=1}^{m}p_k$
$=> \lambda + \sum_{k=1}^{m} \frac{n_{k}}{p_k ln(e)} - \lambda\sum_{k=1}^{m}0$
$f'(p) = \lambda + \sum_{k=1}^{m} \frac{n_{k}}{p_k}$
is this correct?
$f(\lambda) = \lambda +\sum_{k=1}^{m} n_{k} log p_{k} - \lambda\sum_{k=1}^{m}p_k$
$f'(\lambda) = \lambda +\sum_{k=1}^{m} n_{k} log p_{k} - \lambda\sum_{k=1}^{m}p_k$
$=> 0 + \sum_{k=1}^{m} n_{k} log p_{k} - 0\sum_{k=1}^{m}p_k$
$f'(\lambda) = \sum_{k=1}^{m} n_{k} log p_{k}$
Could anyone clarify if these are correct or am I very off the mark? Any and all help wold be greatly appreciated