On the basis of today's lecture you can attempt Examples Sheet 1, #3, #4, #16. You might also look at #14 and think about how using a multinomial coefficient might help sort this one out.
The proof of Stirling's formula is not examinable. So why do we include it? Well, it would be a pity to so often use in our other work a result whose proof we did not understand. I also think it is interesting for you to see how a careful analysis of an asymptotic like this can be made.
Section 3.3 on Improved Stirling's formula (not examinable), which shows that $n!/\left(\sqrt{2\pi}n^{n+1/2}e^{-n}\right)$ lies between $e^{1/(12n+1)}$ and $e^{1/(12n)}$. This is from a paper, A remark on Stirling's formula, by Herbert Robbins.
Tim Gowers once lectured this course and wrote a blog post called Removing the magic from Stirling’s formula. He was looking for a more natural proof. He comments: It turned out to be hard to find a proof that didn’t look magic: the arguments had a “consider the following formula, do such-and-such, observe that this is less than that, bingo it drops out” sort of flavour.
I tried to explain in this lecture how we can understand the motivation behind most of the steps in this proof so that it feels less like pulling-a-rabbit-out-of-a-hat.
Why do we need to make the definition $I_n=\int_0^{\pi/2} \sin^n\theta\, d\theta$? Why not directly define $I_{2n}=\tfrac{1}{2}\tfrac{3}{4}\cdots \frac{2n-1}{2n}\frac{\pi}{2}$ and $I_{2n+1}=\tfrac{2}{3}\tfrac{4}{5}\cdots \frac{2n}{2n+1}$?
The answer is that we need to show that $I_n$ is decreasing in $n$, and this follows from the fact that $\sin^n\theta$ is decreasing in $n$ since $0<\sin\theta<1$.
Our calculation today of $I_{2n}/I_{2n+1}\to 1$ is related to a proof of Wallis's product. This is $$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots =\frac{\pi}{2}. $$ Perhaps the trickiest thing in the proof was the evaluation of the constant $\sqrt{2\pi}$. The following is a method of doing it in a less magical way. Suppose $n!\sim B n^{n+1/2}e^{-n}$. Then $p=2^{-2n}\binom{2n}{n}=(2n)!/(2^{n}n!)^2\sim \sqrt{2/n}/B$ is the probability that in $2n$ tosses of a fair coin there are $n$ heads and $n$ tails. If we knew the central limit theorem (Lecture 23), we could use it to estimate $p=1/\sqrt{n\pi}$, and from that find $B=\sqrt{2\pi}$.
The proof of Stirling's formula is not examinable. So why do we include it? Well, it would be a pity to so often use in our other work a result whose proof we did not understand. I also think it is interesting for you to see how a careful analysis of an asymptotic like this can be made.
Section 3.3 on Improved Stirling's formula (not examinable), which shows that $n!/\left(\sqrt{2\pi}n^{n+1/2}e^{-n}\right)$ lies between $e^{1/(12n+1)}$ and $e^{1/(12n)}$. This is from a paper, A remark on Stirling's formula, by Herbert Robbins.
Tim Gowers once lectured this course and wrote a blog post called Removing the magic from Stirling’s formula. He was looking for a more natural proof. He comments: It turned out to be hard to find a proof that didn’t look magic: the arguments had a “consider the following formula, do such-and-such, observe that this is less than that, bingo it drops out” sort of flavour.
I tried to explain in this lecture how we can understand the motivation behind most of the steps in this proof so that it feels less like pulling-a-rabbit-out-of-a-hat.
Why do we need to make the definition $I_n=\int_0^{\pi/2} \sin^n\theta\, d\theta$? Why not directly define $I_{2n}=\tfrac{1}{2}\tfrac{3}{4}\cdots \frac{2n-1}{2n}\frac{\pi}{2}$ and $I_{2n+1}=\tfrac{2}{3}\tfrac{4}{5}\cdots \frac{2n}{2n+1}$?
The answer is that we need to show that $I_n$ is decreasing in $n$, and this follows from the fact that $\sin^n\theta$ is decreasing in $n$ since $0<\sin\theta<1$.
Our calculation today of $I_{2n}/I_{2n+1}\to 1$ is related to a proof of Wallis's product. This is $$ \frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots =\frac{\pi}{2}. $$ Perhaps the trickiest thing in the proof was the evaluation of the constant $\sqrt{2\pi}$. The following is a method of doing it in a less magical way. Suppose $n!\sim B n^{n+1/2}e^{-n}$. Then $p=2^{-2n}\binom{2n}{n}=(2n)!/(2^{n}n!)^2\sim \sqrt{2/n}/B$ is the probability that in $2n$ tosses of a fair coin there are $n$ heads and $n$ tails. If we knew the central limit theorem (Lecture 23), we could use it to estimate $p=1/\sqrt{n\pi}$, and from that find $B=\sqrt{2\pi}$.
