# Interesting mathematical statements (2 Viewers)

#### leehuan

##### Well-Known Member
$\bg_white \lim_{m \to \infty}\frac{(2m)!/m!}{2^{m+\frac{1}{2}}m^me^{-m}}=1$

##### -insert title here-
$\bg_white \lim_{m \to \infty}\frac{(2m)!/m!}{2^{m+\frac{1}{2}}m^me^{-m}}=1$
that's not very interesting.

it's a fairly trivial consequence of Stirling's Approximation... and Limits...

#### leehuan

##### Well-Known Member
Too bad, cause even simple things like the power series for exp are interesting when you first see it.

(Also it was not a consequence in my homework. It was an intermediate step to GET to Stirling's approximation)

##### -insert title here-
Too bad, cause even simple things like the power series for exp are interesting when you first see it.

(Also it was not a consequence in my homework. It was an intermediate step to GET to Stirling's approximation)
I know. But I'm not the one doing it, so ¯\_(ツ)_/¯

#### Sy123

##### This too shall pass
$\bg_white \\ Almost all real numbers have a (roughly; aka in the limit) equal proportion of 0s and 1s in their binary expansion.$

$\bg_white \\ To see this, note that if \ \{B_n\}_{n\geq 1} \ are a sequence of iid Bernoulli(0.5) trials, then the random variable \ Z_{0.5} = \sum_{n=1}^{\infty}\frac{B_n}{2^n} \sim U(0,1), \ is uniformly distributed$

$\bg_white \\ Moreover, whenever we do the same thing with Bernoulli(p) trials instead, to obtain \ Z_p \ it turns out that Z_p is not absolutely continuous (i.e. it has no density), so that all of its probability mass accumulates on a set of measure 0$

#### sida1049

##### Well-Known Member
$\bg_white \\ Almost all real numbers have a (roughly; aka in the limit) equal proportion of 0s and 1s in their binary expansion.$

$\bg_white \\ To see this, note that if \ \{B_n\}_{n\geq 1} \ are a sequence of iid Bernoulli(0.5) trials, then the random variable \ Z_{0.5} = \sum_{n=1}^{\infty}\frac{B_n}{2^n} \sim U(0,1), \ is uniformly distributed$

$\bg_white \\ Moreover, whenever we do the same thing with Bernoulli(p) trials instead, to obtain \ Z_p \ it turns out that Z_p is not absolutely continuous (i.e. it has no density), so that all of its probability mass accumulates on a set of measure 0$
It is as if you're trying to make our boys Uri and Daniel proud at the same time

#### mrbunton

##### Member
the expression:
( (i0*(1*(1%(((i0 + 1) / (i1 + 1))+1))*(1%(((i0 + 1) / (i2 + 1))+1))*(1%(((i0 + 1) / (i3 + 1))+1))*(1%(((i0 + 1) / (i4 + 1))+1)))) + (i1*(1*(1%(((i1 + 1) / (i0 + 1))+1))*(1%(((i1 + 1) / (i2 + 1))+1))*(1%(((i1 + 1) / (i3 + 1))+1))*(1%(((i1 + 1) / (i4 + 1))+1)))) + (i2*(1*(1%(((i2 + 1) / (i0 + 1))+1))*(1%(((i2 + 1) / (i1 + 1))+1))*(1%(((i2 + 1) / (i3 + 1))+1))*(1%(((i2 + 1) / (i4 + 1))+1)))) + (i3*(1*(1%(((i3 + 1) / (i0 + 1))+1))*(1%(((i3 + 1) / (i1 + 1))+1))*(1%(((i3 + 1) / (i2 + 1))+1))*(1%(((i3 + 1) / (i4 + 1))+1)))) + (i4*(1*(1%(((i4 + 1) / (i0 + 1))+1))*(1%(((i4 + 1) / (i1 + 1))+1))*(1%(((i4 + 1) / (i2 + 1))+1))*(1%(((i4 + 1) / (i3 + 1))+1)))))/(0+1*(1%(((i0 + 1) / (i1 + 1))+1))*(1%(((i0 + 1) / (i2 + 1))+1))*(1%(((i0 + 1) / (i3 + 1))+1))*(1%(((i0 + 1) / (i4 + 1))+1))+1*(1%(((i1 + 1) / (i0 + 1))+1))*(1%(((i1 + 1) / (i2 + 1))+1))*(1%(((i1 + 1) / (i3 + 1))+1))*(1%(((i1 + 1) / (i4 + 1))+1))+1*(1%(((i2 + 1) / (i0 + 1))+1))*(1%(((i2 + 1) / (i1 + 1))+1))*(1%(((i2 + 1) / (i3 + 1))+1))*(1%(((i2 + 1) / (i4 + 1))+1))+1*(1%(((i3 + 1) / (i0 + 1))+1))*(1%(((i3 + 1) / (i1 + 1))+1))*(1%(((i3 + 1) / (i2 + 1))+1))*(1%(((i3 + 1) / (i4 + 1))+1))+1*(1%(((i4 + 1) / (i0 + 1))+1))*(1%(((i4 + 1) / (i1 + 1))+1))*(1%(((i4 + 1) / (i2 + 1))+1))*(1%(((i4 + 1) / (i3 + 1))+1)))

is equal to the largest positive integer(and 0) out of i0,i1,i2,i3,i4.
"a/b" denotes floor(a/b) or integer division
"a%b" denotes a mod b
"a+b" and "a*b" is obvious

generated through a python script. I realised that it's not the shortest way to express such a thing, but still. The most efficient expression would still take up acouple lines(i would think). I encourage other people to try to write an expression that is equal to the largest of two numbers using only these operations.