91影视

(a) Assume we get a paramagnet has $N$ magnets, and shut to half them are as in spin-up state, giving in $N鈫�=$$\frac{1}{2}N\text{,}$, where $N_{鈫�}$by signifies the energy units. This method provides quantity of microstates:

${\mathrm{惟}}_{max}=\left(\begin{array}{c}{N}_{鈫�}+N_{鈫�}+1\\ N_{鈫�}\end{array}\right)=\left(\begin{array}{c}\frac{N}{2}+\frac{N}{2}+1\\ \frac{N}{2}\end{array}\right)$

when N is solely an honest

${\mathrm{惟}}_{max}鈮�\left(\begin{array}{l}N\\ \frac{N}{2}\end{array}\right)=\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}$

For such design variables, we'll employ Stirling's approximation:

$n!鈮�\sqrt{2蟺n}{n}^n{e}^{鈭�n}$

$\begin{array}{r}{\mathrm{惟}}_{max}鈮�\frac{\sqrt{2蟺N}{N}^N{e}^{鈭�N}}{\left(\sqrt{2蟺\left(\frac{N}{2}\right)}{\left(\frac{N}{2}\right)}^{\left(\frac{N}{2}\right)}{e}^{{\left.鈭�\left(\frac{N}{2}\right)\right)}^2}\right.}\\ \end{array}$

${\mathrm{惟}}_{max}鈮�\frac{\sqrt{2蟺N}{N}^N{e}^{鈭�N}}{\left(2蟺\left(\frac{N}{2}\right){\left(\frac{N}{2}\right)}^N{e}^{鈭�N}\right)}$

$\begin{array}{c}{\mathrm{惟}}_{max}鈮�\frac{\sqrt{2蟺N}{N}^N}{蟺N{\left(\frac{N}{2}\right)}^N}\\ \end{array}$

$\begin{array}{c}{\mathrm{惟}}_{max}鈮�\frac{\sqrt{2蟺N}{2}^N}{蟺N}\end{array}$

$\begin{array}{c}{\mathrm{惟}}_{max}鈮�\frac{2^{N+1}}{\sqrt{2蟺N}}\\ \end{array}$

${\mathrm{惟}}_{max}鈮�\frac{2^{N+\frac{1}{2}}}{\sqrt{蟺N}}$

${\mathrm{惟}}_{max}鈮�2^N$

b) We derive its multitude solution within the location of the crest, let:

$x=N_{鈫�}鈭�\frac{N}{2}鈫�N_{鈫�}=\frac{N}{2}+x$

$x=N鈭�N_{鈫�}鈭�\frac{N}{2}鈫�N_{鈫�}=\frac{N}{2}鈭�x$

Therefore, while the subsequent is accurate:

$\mathrm{惟}=\left(\begin{array}{c}{N}_{鈫�}+N_{鈫�}+1\\ N_{鈫�}\end{array}\right)=\left(\begin{array}{c}\frac{N}{2}+x+\frac{N}{2}鈭�x+1\\ \frac{N}{2}鈭�x\end{array}\right)鈮�\left(\frac{N}{2}鈭�x\right)$

$\mathrm{惟}鈮�\left(\frac{N}{2}鈭�x\right)=\frac{N!}{\left(\frac{N}{2}鈭�x\right)!\left(\frac{N}{2}+x\right)!}$

As a conclusion, the redundancy think about units of $x$is:

$\mathrm{惟}鈮�\frac{\sqrt{2蟺N}{N}^N{e}^{鈭�N}}{\left(\sqrt{2蟺\left(\frac{N}{2}鈭�x\right)}{\left(\frac{N}{2}鈭�x\right)}^{\left(\frac{N}{2}鈭�x\right)}{e}^{鈭�\left(\frac{N}{2}鈭�x\right)}\right)\left(\sqrt{2蟺\left(\frac{N}{2}+x\right)}{\left(\frac{N}{2}+x\right)}^{\left(\frac{N}{2}+x\right)}{e}^{鈭�\left(\frac{N}{2}+x\right)}\right)}$

$e^{鈭�N}$with $e^{鈭�\left(\frac{N}{2}+x\right)}{e}^{鈭�\left(\frac{N}{2}鈭�x\right)}$

$\mathrm{惟}鈮�\frac{\sqrt{2蟺N}{N}^N}{\left(\sqrt{2蟺\left(\frac{N}{2}鈭�x\right)}{\left(\frac{N}{2}鈭�x\right)}^{\left(\frac{N}{2}鈭�x\right)}\right)\left(\sqrt{2蟺\left(\frac{N}{2}+x\right)}{\left(\frac{N}{2}+x\right)}^{\left(\frac{N}{2}+x\right)}\right)}$

$\sqrt{2蟺\left(\frac{N}{2}+x\right)}脳\sqrt{2蟺\left(\frac{N}{2}鈭�x\right)}=2蟺\sqrt{{\left(\frac{N}{2}\right)}^2鈭�x^2}$

$\mathrm{惟}鈮�\frac{\sqrt{2蟺N}{N}^N}{2蟺\sqrt{{\left(\frac{N}{2}\right)}^2鈭�x^2}\left({\left(\frac{N}{2}鈭�x\right)}^{\left(\frac{N}{2}鈭�x\right)}\right)\left({\left(\frac{N}{2}+x\right)}^{\left(\frac{N}{2}+x\right)}\right)}$

$\mathrm{惟}鈮�\frac{\sqrt{2蟺N}{N}^N}{2蟺\sqrt{{\left(\frac{N}{2}\right)}^2鈭�x^2}\left({\left(\frac{N}{2}鈭�x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}鈭�x\right)}^{鈭�x}\right)\left({\left(\frac{N}{2}+x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}+x\right)}^x\right)}$

${\left(\frac{N}{2}+x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}鈭�x\right)}^{\frac{N}{2}}={\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)}^{\frac{N}{2}}$

$\mathrm{惟}鈮�\frac{\sqrt{2蟺N}{N}^N}{2蟺\sqrt{{\left(\frac{N}{2}\right)}^2鈭�x^2}{\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)}^{\frac{N}{2}}{\left(\frac{N}{2}鈭�x\right)}^{鈭�x}{\left(\frac{N}{2}+x\right)}^x}$

$\mathrm{惟}鈮�\frac{\sqrt{N}{N}^N}{\sqrt{2蟺}{\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)}^{\frac{N}{2}+\frac{1}{2}}{\left(\frac{N}{2}鈭�x\right)}^{鈭�x}{\left(\frac{N}{2}+x\right)}^x}$

We also may function of notation and handle massive $N$applications without feeling worried concerning square root values. As a basis, here's a conservative calculation:

$\mathrm{惟}鈮�\frac{N^N}{{\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)}^{\frac{N}{2}}{\left(\frac{N}{2}鈭�x\right)}^{鈭�x}{\left(\frac{N}{2}+x\right)}^x}$

both on sides, have used the expectation andtake care of $\ln 鈦�\left(ab\right)=\ln 鈦�\left(a\right)+\ln 鈦�\left(b\right)\text{and}\ln 鈦�\left(\frac{a}{b}\right)=\ln 鈦�\left(a\right)鈭�\ln 鈦�\left(b\right)$

$\begin{array}{r}\ln 鈦�\left(\mathrm{惟}\right)鈮�\ln 鈦�\left(N^N\right)鈭�\ln 鈦�{\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)}^{\frac{蟺}{2}}鈭�\ln 鈦�{\left(\frac{N}{2}+x\right)}^x鈭�\ln 鈦�{\left(\frac{N}{2}鈭�x\right)}^{鈭�x}\end{array}$

$\begin{array}{r}\end{array}\begin{array}{r}\ln 鈦�\left(\mathrm{惟}\right)鈮�N\ln 鈦�\left(N\right)鈭�\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)鈭�x\ln 鈦�\left(\frac{N}{2}+x\right)+x\ln 鈦�\left(\frac{N}{2}鈭�x\right)\end{array}$

$\begin{array}{c}\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)=\frac{N}{2}\ln 鈦�\left[{\left(\frac{N}{2}\right)}^2\left(1鈭�{\left(\frac{2x}{N}\right)}^2\right)\right]\\ \end{array}$

$\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)=\frac{N}{2}\left[2\ln 鈦�\left(\frac{N}{2}\right)+\ln 鈦�\left(1鈭�{\left(\frac{2x}{N}\right)}^2\right)\right]$

$\ln 鈦�(1+x)=x\text{for}\left|x\right|<<1$

$\begin{array}{c}\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)=\frac{N}{2}\left[2\ln 鈦�\left(\frac{N}{2}\right)鈭�{\left(\frac{2x}{N}\right)}^2\right]\end{array}$

$\begin{array}{c}\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)=N\ln 鈦�\left(\frac{N}{2}\right)鈭�\frac{2x^2}{N}\\ \end{array}$

$\frac{N}{2}\ln 鈦�\left({\left(\frac{N}{2}\right)}^2鈭�x^2\right)=N\ln 鈦�N鈭�N\ln 鈦�2鈭�\frac{2x^2}{N}$

$\begin{array}{c}x\ln 鈦�\left(\frac{N}{2}+x\right)=x\ln 鈦�\left(\frac{N}{2}\right)+x\ln 鈦�\left(1+\frac{2x}{N}\right)\\ \end{array}$

$x\ln 鈦�\left(\frac{N}{2}+x\right)=x\ln 鈦�\left(\frac{N}{2}\right)+\frac{2x^2}{N}$

$\begin{array}{c}x\ln 鈦�\left(\frac{N}{2}鈭�x\right)=x\ln 鈦�\left(\frac{N}{2}\right)+x\ln 鈦�\left(1鈭�\frac{2x}{N}\right)\\ \end{array}$

$x\ln 鈦�\left(\frac{N}{2}鈭�x\right)=x\ln 鈦�\left(\frac{N}{2}\right)鈭�\frac{2x^2}{N}$

$\begin{array}{c}\ln 鈦�\left(\mathrm{惟}\right)鈮�N\ln 鈦�\left(N\right)鈭�N\ln 鈦�N+N\ln 鈦�2+\frac{2x^2}{N}鈭�x\ln 鈦�\left(\frac{N}{2}\right)鈭�\frac{2x^2}{N}+x\ln 鈦�\left(\frac{N}{2}\right)鈭�\frac{2x^2}{N}\end{array}$

$\begin{array}{c}\ln 鈦�\left(\mathrm{惟}\right)鈮�N\ln 鈦�2+\frac{2x^2}{N}鈭�\frac{2x^2}{N}鈭�\frac{2x^2}{N}\\ \end{array}$

$\ln 鈦�\left(\mathrm{惟}\right)鈮�\ln 鈦�\left(2^N\right)鈭�\frac{2x^2}{N}$

$\begin{array}{c}\mathrm{惟}鈮�2^N{e}^{鈭�\frac{2x^2}{N}}\end{array}$

$\begin{array}{c}鈫�\mathrm{惟}鈮�{\mathrm{惟}}_{max}{e}^{鈭�\frac{2x^2}{N}}\\ \end{array}$

$\mathrm{惟}鈮�{\mathrm{惟}}_{max}=2^N$

c) Oncethe top declines to $\frac{1}{e}$of its previous peak, the length of the crest is:

$\frac{{惟}_{\max }}{e}={惟}_{\max }{e}^{-\frac{2x^2}{N}}$

$鈫�e^{-1}=e^{-\frac{2x^2}{N}}$

$鈫�\frac{2x^2}{N}=1$

As a response, the peak's girth is increased.

d) If $1000000$ currencies were spun, the probability of getting 501000 heads is (let $N_{鈫�}$be the head):

Substitution giving in:

$\begin{array}{c}x=N_{鈫�}鈭�\frac{N}{2}\\ \end{array}$

$=501000鈭�\frac{1000000}{2}=1000$

$\begin{array}{c}\\ P=\frac{\mathrm{惟}}{{\mathrm{惟}}_{max}}\end{array}$

$P=e^{鈭�\frac{2x^2}{N}}$

$x=1000$and $N=1000000$are changed, giving in:

$P=e^{鈭�\frac{2(1000{)}^2}{1000000}}=0.1353$

It wouldn't be a very controversial outcome, given the big likelihood.

Obtaining a conclusion of $510000$heads seems to own a likelihood of:

$x=N_{鈫�}鈭�\frac{N}{2}$

$=510000鈭�\frac{1000000}{2}$

$=10000$

$P=e^{鈭�\frac{2(10000{)}^2}{1000000}}$

$=1.384脳{10}^{鈭�87}$