Download binomial coefficient and the gaussian

BINOMIAL COEFFICIENT AND THE GAUSSIAN
The binomial coefficient is defined asC (n, k ) =
n!
k!(n − k )!
and can be written out in the form of a Pascal Triangle starting at the zeroth row with
element C(0,0)=1 and followed by the two numbers C(1, 0)=1 and C(1, 1)=1 in the first
row. The second row produces the numbers [1, 2, 1] and the third row the four numbers
[1, 3, 3, 1]. The row number is here given by the value of n, and the number of elements in
a row will be n+1. From these first few results one can state that-
C (n, k ) + C (n, k + 1) = C (n + 1, k + 1)
The Pascal Triangle thus reads1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
It is known that the numbers in the individual rows of this triangle also give the
coefficients in the binomial expansion of (a+b)^n. Thus, for example, for n=4, we
have (a + b) 4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4
One also notices that the binomial coefficients for fixed n are symmetric about
their central maximum value. Plotting their normalized values at n=72 with
-15<k<15 we get the graph-
We have superimposed on these values the Gaussian exp(-k^2/36). The results are
almost identical. Indeed as n is allowed to become large one has the result-
exp(−2k 2 / n) ≈
[(n / 2)!]2
[(n / 2)!]2
C (n, k + n / 2) =
n!
(n / 2 + k )!(n / 2 − k )!
with n>2k. At n=128 and k=8 one finds-
exp(1) ≈
128!
2
[64! C (128,72)]
= 2.704150286…
which is close to the correct value of 2.718281828.. An even better approximation
occurs when k is near zero since the two sides of the equation are normalized at k=0. So
keeping k=1 and k<<n, we find-
 N + 2
exp(1) ≈= 

N 

N
This yields the approximation 2.71801005.. for N=10,000.
Note that one can also use the Stirling Approximation for large n. It gives, for example,
the estimaten +1 / 2
C (n, k ) ≈
1
n
[
]
2π k k +1 / 2 (n − k ) n − k +1 / 2
for
n >> 1
which yields the approximation C(1000,500)≈0.270355…·10^300 compared to the exact value
of C(1000,5000)=0.270288…·10^300.
Going back to the Pascal Triangle given above one notices the following2
∑ C (2, k ) = 1 + 2 + 1 = 2
2
k =0
3 C (3, k ) = 1 + 3 + 3 + 1 = 23
∑
k =0
4
∑ C (4, k ) = 1 + 4 + 6 + 4 + 1 = 2
4
k =0
From this it follows that-
n
∑ C ( n, k ) = 2
k =0
n
.
One also notices that if C(n,1) is prime then C(n,k)/C(n,1) is an integer provided 0<k<n.
Furthermore a violation of this condition indicates that the number is composite . Thus one has
thatC ( n, k )
( N − 1)!
1 k −1
N is prime if T (k ) =
C (n,1)
=
k!( N − k )!
=
∏ ( N − n)
k! n =1
is an integer for all values of k ranging from k=1 through k=N/2. Any violation will
indicate the number N is composite. Consider the number N=177. On running the one
line MAPLE programfor k from 1 to 60 do {k,(1/k!)*product((177-n),n=1..k-1)}od;
one finds T(3),T(6),T(9),T(12),T(15),T(59) up to k=60 to be non-integer and hence the
number is composite. An extra benefit is that one actually finds the factors to be the
lowest values of k not multiples of a lower value. Thus k=3 and k=59 in this case, so that
177=3·59. Unfortunately this approach for determining the primeness of a large number
will be quite cumbersome and is actually more time consuming than just looking at the
ratio N/(2k+1) for 2k+1<sqrt(N). Running the one-linerfor k from 1 to 14 do {2*k+1,177/(2*k+1)}od;
already produces an integer for k=1 yielding 177/3=57 so that 177=3(57).
Going back to the Pascal Triangle we observe that the sequence of numbers lying on the vertical
line dividing the triangle in half is1-2-6-20-70-252-924- …
In terms of the C(n,k) coefficient this means the sequence goes asC(0,0)-C(2,1)-C(4,2)-C(6,3)-C(8,4)-…C(2n,n)…
Adding the terms together through the 2m th row, we find the Sum(m) to bem
m ( 2n)!
2
n = 0 n!
Sum(m) = ∑ C (2n, n) = ∑
n =0
For m=6 this sum equals Sum(6)=1+2+6+20+70+252+924=1275. One can also ask for the value
of the reciprocal series which converges to the finite value -
1 1 1
1
n!2
=1+ + +
+
+ ... = 1.73639985871871507790979516836…
∑
2
6
20
70
(
2
)!
n
n=0
∞
We can also work out the seriesn
S (n) = ∑ [C (n, k )]2
k =0
It yields S(1)=2, S(2)=6, S(3)=20, and S(4)=70. From this one sees that S(n) sums to C(2n,n).
Finally we point out that one can generalize the Pascal Triangle starting with m ones in the first
row and then adding the m nearest values in the (n-1) row to find the element in the nth row. For
m=3 this will lead to the triangle-
1
1 4
1 5 15
1 6 21 50
1
1
2
3
6
10 16
30 45
90 126
1
1
1
3
2 1
7
6 3 1
19 16 10 4 1
51 45 30 15 5 1
141 126 90 50 21 6 1
etc.
with the generating formula-
D(n, k ) = D(n − 1, k − 2) + D(n − 1, k − 1) + D(n − 1, k )
Thus D(5,5)=51=16+19+16 and D(6,7)=126=51+45+30. Note that for this m=3 case that the
number of elements in the nth row is 2n+1 and they sum to 3^n. The elements in the nth row
again have the character of a Gaussian distribution with very small discrepancies noticed near the
tails as shown on the following graph-
. Comparing this result and that for the classic Pascal triangle, we can conclude that-
m ones in the first row of a modified Pascal Triangle will lead to a
triangle whose elements in the nth row number [(m-1)n]+1 and add
up to m^n.
The element F(n,k) of such a modified Pascal Triangle will be given byF(n,k)= F(n-1,k-m+1)+F(n-1,k-m+2)+……..+F(n-1,k)
Also one notes that there is a generating function for the coefficients F(n,k) for a given value of
m . For example at m=3 the generating function is G=1+x+x^2 which has coefficients [1,1,1] and
corresponds to the first row of D(1,k). G^2=1+2x+3x^2+2x^3+x^4 has the coefficients
[1,2,3,2,1] and thus yields the coefficients D(2,k). For m=3 and n=6 we find12
(1 + x + x 2 ) 6 = ∑ D(6, k ) x k = 1 + 6 x + 21x 2 + 50 x 3 + 90 x 4 + 126 x 5 + 141x 6
k =0
+ 126 x 7 + 90 x 8 + 50 x 9 + 21x10 + 6 x11 + x12
December 2009