Derivation of softmax function I'm reading Bishop's book on Pattern Recognition and machine learning and I wanted to reproduce a calculation for the softmax function, also known as normalized exponential. Basically, the calculation requires to get the multinomial distribution into its form as a member of the exponential family:
$$p(x|\eta) = h(x)g(\eta)\exp{\{\eta^{T}u(x)\}}$$
Starting from $\exp{\{\sum_{k=1}^{M}x_{k}\ln{\mu_{k}}\}}$ and after a few steps, we recognize that $\eta_{k}$ is given by:
$$\ln{\left[{\frac{\mu_{k}}{1-\sum_{j}^{M-1}{\mu_{j}}} }\right]} = \eta_{k}$$
then it says:
which we can solve for $\mu_{k}$ by first summing both sides over $k$ and then rearranging and back-substituting to give:
$$\mu_{k}=\frac{\exp{\{\eta_{k}\}}}{1+\sum_{j}\exp{\{\eta_{j}\}}}$$
But that's not what I get. Instead, I obtained (assuming $\sum_{k}\mu_{k}=1$)
$$\mu_{k}=\frac{\exp{\{\eta_{k}\}}}{\sum_{j}\exp{\{\eta_{j}\}}}$$
Wikipedia seems to agree with my answer but I'd like to get a confirmation or correct the derivation I did.
 A: The question was posted a long time ago but it could be useful for anyone else who is working through Bishop's book to note that both forms of the softmax function are equivalent since  \begin{equation}1+\sum_{j=1}^{M-1}\exp{\{\eta_{j}\}}={\sum_{j=1}^M\exp{\{\eta_{j}\}}}\end{equation}
A: As another user above has said, the crux is the following relation
$$
    \sum_{j=1}^{K} \exp(\eta_j) = \sum_{j=1}^{K-1} \exp(\eta_j) + 1,
$$
since from this equation the equality of the two stated definitions of the softmax functions follows directly.
As it is not obvious from the section in the Bishop where this relation comes from, I will provide a derivation here.
We can split up the sum
$$
    1 = \sum_{j=1}^{K} \pi_j = \sum_{j=1}^{K-1} \pi_j + \pi_K.
$$
By rearranging this we get
$$
    \pi_K = 1 - \sum_{j=1}^{K-1} \pi_j,
$$
and therefore
$$
    \eta_i =\log\left(\frac{\pi_i}{1-\sum_{j=1}^{K-1} \pi_j}\right)= \log\left(\frac{\pi_i}{\pi_K}\right).
$$
Thus,
$$
    \eta_K = \log\left(\frac{\pi_K}{\pi_K}\right) = \log(1) = 0.
$$
Finally, we split up the following sum, as we did before,
$$
    \sum_{j=1}^{K} \exp(\eta_j) = \sum_{j=1}^{K-1} \exp(\eta_j) + \exp(\eta_K),
$$
which is then nothing else as the answer above stated:
$$
    \sum_{j=1}^{K} \exp(\eta_j) = \sum_{j=1}^{K-1} \exp(\eta_j) + 1.
$$
