The Man who Knew Infinity
by djmathman, May 30, 2016, 10:37 PM
was actually not a bad movie. It definitely didn't feature Oscar-worthy performances, but it did get emotional at times. The mathematics also seemed pretty legit, which makes sense considering that Ken Ono was involved with that aspect of the film.
Although I had to explain a bunch of things to my parents afterward OOPS
Although I had to explain a bunch of things to my parents afterward OOPS
Putnam and Beyond #314 wrote:
Let 
be a positive integer and 
a positive real number. Prove that ![\[\lim_{n\to\infty}\binom nk\left(\dfrac\mu n\right)^k\left(1-\dfrac\mu n\right)^{n-k} = \dfrac{\mu^k}{e^\mu\cdot k!}.\]](//latex.artofproblemsolving.com/f/f/7/ff77599329fa34c29de0ee0ba3526ea3f2475897.png)
be a positive integer and 
a positive real number. Prove that ![\[\lim_{n\to\infty}\binom nk\left(\dfrac\mu n\right)^k\left(1-\dfrac\mu n\right)^{n-k} = \dfrac{\mu^k}{e^\mu\cdot k!}.\]](http://latex.artofproblemsolving.com/f/f/7/ff77599329fa34c29de0ee0ba3526ea3f2475897.png)
I now claim that ![\[\lim_{n\to\infty}\left(1 - \dfrac{\mu}n\right)^n = e^{-\mu}.\]](http://latex.artofproblemsolving.com/8/7/5/875144d3602a6e9be10f3a3cdd6194d020d38625.png)
If this is true, then the problem is solved: note that the limit of the first fractional expression is 
since the numerator and denominator are polynomials in 
with the same degree 
be the desired limit, and note that ![\[\ln L = \lim_{n\to\infty} n\ln\left(1 - \dfrac{\mu}n\right) = \lim_{n\to\infty}\dfrac{\ln(1-\mu/n)}{1/n}.\]](http://latex.artofproblemsolving.com/d/2/f/d2f69b78ae2934bc4ddc04324a9c01d4cd19b609.png)
Now L'Hopital's Rule gives ![\[\lim_{n\to\infty}\dfrac{\ln(1-\mu/n)}{1/n} = \lim_{n\to\infty}\dfrac{\mu/n^2}{1-\mu/n}\cdot\dfrac{1}{-1/n^2} = \lim_{n\to\infty}\dfrac{-\mu}{1-\mu/n} = -\mu.\]](http://latex.artofproblemsolving.com/7/8/c/78c5567dbcaf1d8f30ae6b08e649507a2e743156.png)
Raising both sides to the power of 
gives the desired.Putnam and Beyond #318 wrote:
Prove that ![\[\lim_{n\to\infty}n^2\int_0^{1/n}x^{x+1}\,dx = \dfrac12.\]](//latex.artofproblemsolving.com/8/3/7/8372c7addb172f162a09aa8113e11bdbd11111d0.png)
![\[\lim_{n\to\infty}n^2\int_0^{1/n}x^{x+1}\,dx = \dfrac12.\]](http://latex.artofproblemsolving.com/8/3/7/8372c7addb172f162a09aa8113e11bdbd11111d0.png)
.![$[0,\tfrac1n]$](http://latex.artofproblemsolving.com/6/1/2/612c2ff89ee29cf6b5041d08837f988008ae59c3.png)
,
.
is always positive, 
.
,![\[n^2\int_0^{1/n}x^{x+1}\,dx\leq n^2\int_0^{1/n}x\,dx = n^2\left[\dfrac12x^2\right|_0^{1/n} = \dfrac12.\]](http://latex.artofproblemsolving.com/3/6/a/36a5d2fd107d3487b917af58c33f41d0d54b950f.png)
Finding a lower bound is similar. Remark that 
,
.![\[n^2\int_0^{1/n}x^{x+1}\,dx \geq n^2\int_0^{1/n}x^{1+1/n}\,dx = n^2\left[\dfrac{1}{2+1/n}x^{2+1/n}\right|_0^{1/n} = \dfrac{n^{1/n}}{2+1/n}.\]](http://latex.artofproblemsolving.com/6/7/2/6721e3dd763998553503f9cced38d17ae9d6c5bc.png)
This means that ![\[\dfrac{n^{1/n}}{2+1/n}\leq n^2\int_0^{1/n}x^{x+1}\,dx\leq\dfrac12\]](http://latex.artofproblemsolving.com/7/9/6/796f5ad6604e2f77c7b6391fe8458f6deba7b61f.png)
for all 
.
and 
as 
.
as well, which is what we wanted.This post has been edited 1 time. Last edited by djmathman, May 30, 2016, 10:38 PM
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