这是一份GLA格拉斯哥大学MATHS4076_1作业代写的成功案例

To find the best probability assignment subject to the constraint of given mean, we use the MaxEnt method. I.e. maximse
$$
S[p]=-k_{\mathrm{B}} \sum_{k \geq 0} p(k) \ln p(k)
$$
subject to the constraint
$$
\sum_{k \geq 0} k p(k)=\mu .
$$
using the method of Lagrangian multipliers to take constraints into account.
To simplify notation we can assume $k_{\mathrm{B}}=1$; this amounts to redefining Lagrangian multipliers in units of $k_{\mathrm{B}}$ (can you see why?). Thus define
$$
\mathcal{L}[p]=-\sum_{k \geq 0} p(k) \ln p(k)+\lambda_{0}\left(\sum_{k \geq 0} p(k)-1\right)+\lambda_{1}\left(\sum_{k \geq 0} k p(k)-\mu\right)
$$

MATHS4076_1 COURSE NOTES ：

$$
\mu=\sum_{k \geq 0} k p(k)=\left(1-\mathrm{e}^{\lambda_{1}}\right) \sum_{k=1}^{\infty} k \mathrm{e}^{\lambda_{1} k}=\frac{\mathrm{e}^{\lambda_{1}}}{1-\mathrm{e}^{\lambda_{1}}}
$$
required by the constraint of the given mean. Hence
$$
\mathrm{e}^{\lambda_{1}}=\frac{\mu}{1+\mu} \quad \Longrightarrow \quad p(k)=\frac{1}{1+\mu}\left(\frac{\mu}{1+\mu}\right)^{k}
$$
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