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1 Forward Difference Operator

Definition 1: Forward difference operator is a linear operator which is denotedby and is defined as follows:

f(x) = f(x + 1) f(x)

Following properties follows from definition:

(f(x) + g(x)) = f(x) + g(x)

If k is a constant then,(kf(x)) = kf(x)

Following is product rule due to Lebinitz:

(f(x)g(x)) = f(x)g(x) + g(x)f(x) (1)If one applies operator again on f(x) then one has,

2f(x) = f(x) = f(x + 1) f(x) = f(x + 2) 2f(x + 1) + f(x)

after which one has,

3f(x) = f(x+2)2f(x+1)+f(x) = f(x+3)3f(x+2)+3f(x+1)f(x)

Carrying out similarly one will note occurence of bionomial coefficients.Theorem 1: For n 0 one has,

nf(x) =n

j=0

n

j

(1)njf(x + j) (2)

Proof: If one assumes (2) then one has,

n+1f(x) =

nj=0

n

j

(1)nj(f(x + j + 1) f(x + j))

which after the property of bionomial coefficients,n

j

=

n + 1

j

n

j 1

produces,

n+1

f(x) =

n+1j=0

n + 1j

(1)n+1j

f(x + j)

Hence (2) is proved using induction on n.If one puts f(x) = xk in (2) then one has,

nxk =

nj=0

n

j

(1)nj(x + j)k (3)

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One would also be able to note that each time he applies on a polynomial he

gets a degree less than that of polynomial. Thereafter one has,For n > k,

nxk =

nj=0

n

j

(1)nj(x + j)k = 0 (4)

Theorem 2: For Bernoulli numbers Bn one has,

Bn =

nj=0

1

j + 1

jm=0

(1)m

j

m

mn (5)

Proof: One has,x = log(1 + ex 1)

and after infinite series expansion,

x

ex 1=

j=0

(1 ex)j

j + 1

j=0

(1 ex)j

j + 1=

j=0

1

j + 1

jm=0

j

m

(1)mexm

x

ex 1=

j=0

1

j + 1

jm=0

(1)m

j

m

n=0

xnmn

n!

j=0

1

j + 1

j

m=0

(1)mj

m

n=0

xnmn

n!

=

n=0

xn

n!

j=0

1

j + 1

j

m=0

(1)mj

mm

n

Therefore one has,

x

ex 1=

n=0

xn

n!

j=0

1

j + 1

jm=0

(1)m

j

m

mn

and after the view of (3) one has,

x

ex 1=

n=0

xn

n!

j=0

1

j + 1(1)j |jxn|x=0

Comparing both sides and after the view of (4) one has,

Bn =nj=0

1

j + 1(1)j |jxn|x=0

Definition 2: Stirling numbers of second kind will be denoted as S(n, j) andare defined as follows:

S(n, j) =jxn|x=0

j!

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It also follows from definition that for j > n, S(n, j) = 0.Theorem 3

: For 0 < j n one has,

S(n + 1, j) = jS(n, j) + S(n, j 1) (6)

Proof: (6) follows from definition itself.From (6) one can conlude that S(n, j) are integers.Stirling numbers of second kind are interesting combinatorial objects. 1 There-after we have,

Bn =nj=0

j!

j + 1(1)jS(n, j)

Theorem 4: Let p be a prime number then,If p-1 divides 2n(p = 2) (denoted by p 1|2n) then,

p1m=0

(1)m

p 1

m

m2n 1(modp) (7)

and if p-1 does not divide 2n then,

p1m=0

(1)m

p 1

m

m2n 0(modp) (8)

Proof: One has from Fermats little theorem,

mp1 1(modp)

for m = 1, 2,...,p 1If p 1|2n then one has,

m2n 1(modp)

for m = 1, 2,...,p 1and afterwards,

p1m=1

(1)m

p 1

m

m2n

p1m=1

(1)m

p 1

m

(modp)

and therefore,p1

m=0

(1)mp 1mm2n 1(modp)

If p-1 does not divide 2n then after Fermats theorem one has,

m2n m2n(p1)(modp)

1Refer to Combinatorics by Russell Merris

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for m = 1, 2,...,p 1,

If one lets = [

2n

p1 ] then after iteration one has,

2

m2n m2n(p1)(modp)

for m = 1, 2,...,p1 where one has 0 < 2n(p1) < p1. After summationon m one has,

p1m=0

(1)m

p 1

m

m2n

p1m=0

(1)m

p 1

m

m2n(p1)(modp)

which after the view of (4) gives (8).Theorem 5 For a > 2 and b > 2,

ab|(ab 1)! (9)

Proof: (9) is easy to see.

2 Von-Staudt Clausen Theorem

Von-Staudt Clausen theorem says:

B2n = In

p1|2n

1

p

where In is an integer.One already has,

B2n =2nj=0

1

j + 1

jm=0

(1)m

j

m

m2n (10)

and also,

B2n =2nj=0

j!

j + 1(1)jS(2n, j) (11)

where S(2n, j) are integers.If j + 1 is a composite number then, j + 1|(j)! for j > 3 by (9).

For j = 3 we have,

3m=0

(1)m

3

m

m2n = 3.22n 32n 3 0(mod4)

2[x] represents greatest integer less than or equal to x

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If j + 1 is prime, j is even except for j = 1

For j = 1 we have,

1

m=0

(1)m

1

m

m2n 1(mod 2)

Using above arguments and (7) and (8) one can derive Von-Staudt Clausentheorem.Exercise 1: Prove using induction:

n(f(x)g(x)) =nj=0

n

j

njf(x)jg(x)

Exercise 2: Use Von-Staudt Clausen theorem to show that denominators ofBernoulli numbers are divisible by 6.Exercise 3: Prove following using induction:

jk =

jm=0

(j)(j 1)....(j m + 1)S(k, m)

Exercise 4: Prove the following as a corrolary of Ex.3:

jk =

jm=0

j

m

m!S(k, m)

Define m!S(k, m) = A(k, m) and use:

j

m

=

j + 1

m + 1

j

m + 1

to obtain:M1j=N

jk =

jm=0

(

M

m + 1

N

m + 1

)A(k, m)

Exercise 5: Define Bernoulli polynomials Bk(x) as:

Bk(x) = k

jm=0

x

m + 1

A(k 1, m) + Ck

where Ck is a constant.Show the following:

jk =Bk+1(j + 1) Bk+1(j)

j + 1

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