\]

takg the derivative of \(\athcal{l}\) with respect to \(\pi_k\) and settg it to zero:

\[

\frac{\partial}{\partial \pi_k} \athcal{l}(\boldsybol{\pi}, \bda) = \su_{n=1}n \frac{\gaa(z_{nk})}{\pi_k} + \bda = 0

\]

rearrangg the equation, we have:

\[

\su_{n=1}n \frac{\gaa(z_{nk})}{\pi_k} = -\bda

\]

knog that \(\su_{k=1}k \pi_k = 1\), we can write:

\[

\su_{k=1}k \su_{n=1}n \frac{\gaa(z_{nk})}{\pi_k} = -\bda \su_{k=1}k \pi_k = -\bda

\]

th, \(-\bda = n\)

substitute \(-\bda\) back to the previo equation:

\[

\su_{n=1}n \frac{\gaa(z_{nk})}{\pi_k} = n

\]

solvg for \(\pi_k\), we get:

\[

\pi_k = \frac{n_k}{n},

\]

where \(n_k = \su_{n=1}n \gaa(z_{nk})\)

therefore, g the grange ultiplier thod, we have shown that \(\pi_k = \frac{n_k}{n}\) axiizes the given expression while keepg \(\gaa(z_{nk})\) fixed

\newpage

next, let&039;s nsider the axiization with respect to \(\pi_k\) \\

here, we need to fd the derivative of the follog expression with respect to \(\pi_k\), and then solve for \(\pi_k\):

\[

\ln p(\athbf{x} \id \boldsybol{\pi}, \boldsybol{\u}, \boldsybol{\siga}) + \bda \left( \su_{k=1}k \pi_k - 1 \right)