During the discussion of 3rd targil (factorials) many unexpected ideas were
mentioned, mostly by Alexey (Gladkikh), Gal and Oded, several nice new
solutions and remarks, so I have decided to write an appendix to the
solutions.
1(b). Third solution (Alexey). Choose at random (with uniform probability
distribution) a point A, and independently, M+K more points. What is the
probability that the point A will hold place M+1, meaning, there will be M
points before and K points after?
We shall compute it in two ways. First way: well, there are
 M  K   M  K !
ways to choose divide M+K other points into M
 K 
M !K !


specific points and K specific points, and now multiply this number if ways
by the probability that M specific points will be before and K specific points
K
will be after. If point A is at place x then the probability is x M 1  x  , so in
general we get integral over probability measure of this dx. In the end we get
 M  K ! 1 x M 1  x K dx
 
M ! K ! 0
The second way to compute the same thing: all animals are equal, so the
probability for point A to be on each place, from 1 to M+K+1 is the same:
1
M  K 1
Hence
 M  K ! 1 x M 1  x K dx 
1


M !K ! 0
M  K 1
1
M !K !
 M  K  1!
0
Of course, this solution, as well as integration by parts, works for integer
values only.
So:
 x 1  x 
M
K
dx 
 1
2. Second solution (Gal & Dan) We shall compute   !, and
 2
dx
1  1 1
, hence 2udu  dx , so:
 !    ! . Let u  x , then du 
2 x
2  2 2




2
2
e x
eu
 1
dx  
2udu  2 eu du   eu du
  !  
u
 2 0 x
0
0

2
This integral is classical and equal to  . So the answer is

1  1 1
 !    ! 
2
2  2 2

u
 e du was explained in the second solution
2
Remark. How to compute

for problem 3. This function, eu , is called Gaussian, bell-shaped (‫)פעמון‬, or
normal distribution function, and it plays important role in probability
theory, partly because of central limit theorem.
Though we know the total integral under it, we can’t get its indefinite
integral as an elementary function (it is not arithmetical combination of
powers, exponents, logs and trigo), hence a special notation was invented, it
(up to some coefficients) is called erf(x).
2
3. Third solution (Oded). Like in the first solution, for the volume of the
1
ball we wrote, VN   VN 1 
1

1
1  z2
1

N 1
1
dz  
1


1 z
1  z2
 
N 1

N 1
dz . So all we need is
1 z

N 1
dz
Use a linear substitution which brings interval [-1,1] to interval [0,1]
u = (z+1)/2 , du = dz/2 , we get

1
1 z
1
1
2
N
2

N 1
2
N 1
dz  
 u 1  u 
0
1
1
N 1
2

1 z
 
N 1
1 z

N 1
1
dz    2u 
0
du 
N !2 N
 N 1 
  2 !


2
N 1
2
 2 1  u  
N 1
2
2du 

4. (a) I N   sin N  x  dx
0
eix  eix
Second solution (Alexey). sin x 
, when we raise it to the power N
2i
 N  N  K ix
N
we get sum of expressions of the kind   e  e  Kix all divided by  2i  .
K 

So, we have to compute  e
N  2 K ix
dx .
0
If K=2N then we get integral of 1, which is  . In other cases we get


e
 N  2 K ix
e
dx

0
 N  2K  i 0

N  2 K ix
this is 0 for even N  2 K , that is, for even N, and for odd N we get.

e
0
 N  2 K ix


e
2
dx 

 N  2K  i 0  N  2K  i
N  2 K ix
Hence the answer is, for even N:
N  N
 
 N 
K  N  K ix  Kix

1
e
e
dx



 N 2 

 
K 0 0  K 
N
I N   sin  x  dx 
  N
N
2
 2i 
0
For odd N

N  N
N
 
N
K  N  2 K ix
K
N  2 K ix
dx     1  e
dx

 K   1 e


K
0
I N   sin N  x  dx  K 0 0  
 K 0  

N
N
2
i
2
i
 
 
0
N 1
N 1
KN
K  N 
K  N 
2
2

2

1

2

1

1






K 
K  N
K 
N
N




 


  N 1
N
N
2  N  2K 
 N  2K 
K 0  2i   N  2 K  i
K 0
K 0 2
Second solution (Alexey). Let t  sin 2 x , then 1  t  cos2 x ,


1
2
I N   sin N  x  dx  2  sin N  x  dx   t N 2
0
0
0
dt
t 1  t 
 2dx
N
 !
dt
2

t 1  t   N  1! 


2

Because of 1(b).
 2N 
1

N  4 N 

N




2

N 
An elementary aproach, which gives a less good but similar inequality based
on a classical riddle, I recalled it two days after the meeting.
4(b).
22 N  N !   2 N !!
 2 N !!
N  2N 
4 




 2 N !
 2 N !  2 N  1!!
N 
 2 N !!  2  4  ...  2 N ,
Denote AN 
 2 N  1!! 1 3 2 N  1
2
BN 
2
 2 N  1!!  3  5  ...  2 N  1 .
2N
 2 N !! 2 4
2 3 4 5 6
     ...  1 we get, when we multiply such inequalities
1 2 3 4 5
2 BN 1  AN  BN .
Since
On the other hand, BN 1 AN  2 N , BN AN  2 N  1 hence 4 N  AN2  2 N  1 .
A more precise inequality was found by Alexey.
1

N  2N 
 N    4 
  N  1
4

N 
Start by taking the ln:
1
1


ln   ln  N  
ln   ln  N  
4



 N ln 4  2ln  N !  ln   2 N ! 
2
2
Let us denote the left LN, the middle MN, the right RN. Lets compute
LN+1 – LN
MN+1 – MN
RN+1 – RN
We get
1
1


ln  N  1    ln  N  
4
4


,
2
ln 4  2ln  N  1  ln  2 N  1  ln  2 N  2  ,
1
1


ln  N  1    ln  N  




2
Simplify it and multiply by 2
1 
 
1 
1



ln 1 
 , 2  ln 1 
  , ln 1 

 N  1/ 4 
 N  1/  
  2N  1  
If we would prove LN+1 – LN < MN+1 – MN < RN+1 – RN we would get the
statement by trivial induction. Unfortunately, the inequalities are vice versa.
1   
1 
1



Let’s prove ln 1 
>2
ln
1

>ln
1

  



 N  1/ 4    2 N  1  
 N  1/  
2
1
1 
1

1
> 1 
 >1 
N  1/ 4  2 N  1 
N  1/ 
1
2
1
1
>
+

2
N  1/ 4 2 N  1  2 N  1
N  1/ 
1
1
1
1
1




2
N  1/ 4 N  1/ 2  2 N  1
N  1/  N  1/ 2
1 1

1/ 4
1
2



2
 N  1/ 4  N  1/ 2   2 N  1  N  1/   N  1/ 2 
2
4
1
1



2 N  1/ 2 2 N  1 2 N  2 / 
This is easy.
So, we can’t apply standard induction from 1 to infinity, we can only hope to
do it vice versa – from infinity to 1. By the easy version of inequality,
 2N  N
N
 4 1
lim
N
N 


Hence LH  M H , M H  RH for huge H are very close to 0, closer than any
given positive epsilon, and we have proved that
 LN 1  LN    M N 1  M N  ,  M N 1  M N    RN 1  RN  is always negative,
hence
LK 1  M K 1  LH  M H 
M K 1  RK 1  M H  RH 
H 1
  L
N  K 1
N 1
 LN    M N 1  M N     ,
H 1
  M
N  K 1
N 1
 M N    RN 1  RN    
for each positive  hence LK 1  M K 1 , M K 1  RK 1  0 and so
LK  M K , M K  RK  0 ,
QED.
Especially for those who think that was not cool enough, Alexey had
improved this inequality even further three days later:
1
1
1 


N  2N 
N



4



N  N  
4
4 4N  3 




The idea is the following. Suppose we want to prove a good upper bound for
our expression. In the spirit of what was done before, we have to choose a
smaller number t, such that an inequality
 
1 
1


2  ln 1 
  >ln 1 
 will hold for all M ≥ N .
2
M

1
M

t

1/
4



 
He found out that when you choose any t, the expression
 
1 
1


DM  2  ln 1 
   ln 1 

 M  t  1/ 4 
  2M  1  
is negative for small M and positive after it passes certain point.
This point is N for which the original inequality can be proven.
To find this point he sets the derivative DM as a function of M to be 0,
and gets an expression for t.
 N N!  1
5. lim 

N 
N

 e
Gal asked, whether it is possible to prove it using Stirling’s formula. I didn’t
accept it, since this equality is a first step in proving or guessing Stirling’s
formula.
For the sake of those who couldn’t come to the meeting, here we shall give
the proof of Stirling’s formula.
Stirling’s formula is about approximating N!
The problem 5 gives a (false) hope, that
N
N! 1
e
that is, lim   N !  1.

N  N
N N eN
 
Let us check how far from truth is that.
  e N 
Denote AN  ln    N !  N  N ln N  ln  N ! , let
 N 



1
 N 1

BN  1  N ln N   N  1 ln  N  1  ln  N  1  1  N ln 

1

N
ln
1




N
 N 

We want AN to have a limit, it is equivalent to convergence of the series

. From the expression BN  1  N ln 1 
B
N 1
N

1
  0 we have a hope, but if
N
we look closer
1
1

1
 1
BN  1  N ln 1    1  N  
 O  N 3   
 O  N 2 
2
N

 N 2N
 2N

So
B
N
N 1
diverges even though BN goes to 0.
However, would we have
1 
1
1  1
1



BN  1   N   ln 1    1   N   
 O  N 3    O  N 2 
2
2 
N
2  N 2 N




Then
B
N 1
N
would converge and AN would have a limit.
This would happen if we’d have
1
1


BN  1   N   ln N   N  1   ln  N  1  ln  N  1
2
2


  e N N ! 
1

and AN  ln   
 N   N   ln N  ln  N ! .

 N  N 
2



N
 e  N!
So, the expression  
has a finite positive limit  , in other words
N
N
N
N
N!  N  
e
To find  take Wallis formula, of exercise 4(b) which is a more precise
version of Wallis formula.
2
N

N 
4  N   
2

4 N   N !
 e  

N 

N
 2 N !
 2N 
 2N 

 e 
Surprisingly, all powers of N and e and most powers of 2 cancel out and we
get   2 and Stirling’s formula is
N
N
N
N !  2 N  
e
It is on of the most beautiful and important formulas about factorials.
The most natural prove, which I showed above, I have read it in a book on
probability theory, by a famous French mathematician, Henri Poincare, who
lived about a 100 years ago. Stirling’s formula has also several other
classical proofs.

2 2 4 4 6 6
      follows
2 1 3 3 5 5 7

sin  x 

x2 
 x   1  2  .
from Euler’s formula for sine:

N 
N 1 
If you revert Wallis formula you get
2 1 3 3 5 5 7
1 3 3  5 5  7 7  9
22  1 42  1 62  1
        2  2  2  2   2  2  2  
 2 2 4 4 6 6
2
4
6
8
2
4
6
1 
1 
1

 1  2   1  2   1  2  
 2   4   6 
This is precisely what You get when You substitute ½ into Euler’s formula.
Gal has remarked, that Wallis formula

Another nice conclusion from Euler’s formula turns out when You compute

2
1
3
  2 .
the coefficient of x , on both sides: 
6
N 1 N
sin  x 
Intuitively, Euler’s formula presents
as a polynomial, whose roots

are all integer numbers, and hence it is a product over all integer numbers.
Gal has asked for the proof of Euler’s formula. Unfortunately, the only proof
I know uses complex functions (merukavot) so some of You may not
understand it. But anyway, here it comes.
Firstly, let us discuss convergence. In general when you have a product

 1  a  and the sequence a
N 1
N
N
tends to 0. We say that this product
converges well if it converges to a number, which is neither 0 nor infinity.
Lemma.


N 1
N 1
 1  aN  converges well iff  an converges.
(iff is short for “if and only if”)

Proof of lemma.
 1  a  converges well iff its log, which is
N 1
N

 log 1  a  converges. But for small numbers log 1  a   a . QED.
N 1
n
n
n
From this lemma, we see Euler’s product converges.
For every function f(z) there is a nice construction, called logarithmic
derivative f’(z)/ f(z). It is defined well and nicely even when definition of
logarithm is messy. A logarithmic derivative of a product is a sum of
logarithmic derivatives (this is precisely Leibniz rule).
Logarithmic derivative of Euler’s formula (of both sides) gives:
1   1/ N
1/ N  1   1
1 
 ctg  x     


   

x N 1  1  x / N 1  x / N  x N 1  x  N x  N 
It is as beautiful as the original formula itself, and is equivalent to it.
Take derivative of both parts, and change the signs, You will get a third
beautiful equivalent formula,
2
1

2
sin  x  NZ  x  N 2
It converges even better so the order of summation is not important.
We shall prove the third formula. Both left hand and right hand sides are
defined in all complex plane, periodic with period 1, and analytic
everywhere except integer points. At integer points they have precisely the
1
at all integer points, and far from real line both are
z2
bounded and tend to 0.
same singularity
So the difference of the 2 is an analytic bounded function on complex plane,
and it is constant by Liouville theorem, 0 in our case since it goes to 0 far
from real axis. QED.