PROOF OF THE DEPASCALISATION THEOREM

Amarnath Murthy, S.E.(E&T), WLS, Oil and Natural Gas Corporation Ltd., Sabarmati, Ahmedabad,-380005 INDIA.

In [1] we have defined Pascalisation as follows:

Let b1, b2, . . . be a base sequence. Then the Smarandache Pascal derived sequence

d1 , d2, . . . is defined as

d1 = b1

d2 = b1 + b2

d3 = b1 + 2b2 + b3

d4 = b1 + 3b2 + 3b3 + b4

. . .

n

dn+1 = S nCk .bk+1

k=0

Now Given Sd the task ahead is to find out the base sequence Sb . We call the process of extracting the base sequence from the Pascal derived sequence as Depascalsation. The interesting observation is that this again involves the Pascal's triangle , but with a difference.

On expressing bk 's in terms of dk's We get

b1 = d1

b2 = -d1 + d2

b3 = d1 - 2d2 + d3

b4 = -d1 + 3d2 - 3d3 + d4

. . .

which suggests the possibility of

n

bn+1 = S (-1)n+k. nCk .dk+1

k=0

This I call as Depascalisation Theorem.

PROOF: We shall prove it by induction.

Let the proposition be true for all the numbers l £ k+1. Then we have

bk+1 = kC0 (-1)k+2 d1 + kC1 (-1)k+1 d2 +. . . + kCk (-1)2

Also we have

dk+2 = k+1C0 b1 + k+1C1 b2 + . . . + k+1Cr br+1 + . . . + k+1Ck+1 bk+2 , which gives

bk+2 = (-1) k+1C0 b1 - k+1C1 b2 - . . . - k+1Cr br+1 - . . . + dk+2

substituting the values of b1, b2, . . .etc. in terms of d1, d2, . . . , we get the coefficient of d1 as

(-1) k+1C0 + (-k+1C1)(-1C0) +(-k+1C2)( 2C0) +. . .+ (-1)r . k+1Cr)(rC0) +. . .+ (-1) k+1(k+1Ck)(kC0)

- k+1C0 + k+1C1. 1C0 - k+1C2 . 2C0 +. . .+ (-1)r . k+1Cr . rC0 +. . .+ (-1)k+1. k+1Ck . kC0

similarly the coefficient of d2 is

k+1C1. 1C1 + k+1C2 . 2C1 +. . .+ (-1)r+1. k+1Cr . rC1 +. . .+ (-1) k+1 . k+1Ck . kC1

on similar lines we get the coefficient of dm+1 as

k+1Cm. mCm + k+1Cm+1 . m+1Cm - . . .+ (-1)r+m. k+1Cr+m . r+mCm +. . .+ (-1) k+m . k+1Ck . kCm

k-m

= S (-1)h+1 k+1Cm+h . m+hCm

h=0

(k+1)-m

S (-1)h+1 k+1Cm+h . m+hCm + (-1) k+m . k+1Ck+1 . k+1Cm -------------(1)

h=0

Applying theorem {4.2} of reference [2], in (1) we get

= k+1Cm { 1 + (-1)}k+1-m + (-1)k+m . k+1Cm

= (-1)k+m . k+1Cm

which shows that the proposition is true for (k+2) as well. The proposition has already been verified for k+1 = 3 , hence by induction the proof is complete.

In matrix notation if we write

[ b1, b2 , . . bn ]1xn * [pi,j ]'nxn = [ d1, d2 , . . dn ]1xn

where [pi,j ]'nxn = the transpose of [pi,j ]nxn and

[pi,j ]nxn is given by pi,j = i-1Cj-1 if i £ j else pi,j = 0

Then we get the following result

If [qi,j]nxn is the transpose of the inverse of [pi,j ]nxn Then

qi,j = (-1)j+i .i-1Cj-1

We also have

[ b1, b2 , . . bn ] * [qi,j ]'nxn = [ d1, d2 , . . dn ]

where [qi,j ]'nxn = The Transpose of [qi,j ]nxn

References:

[1] Amarnath Murthy, ' Smarandache Pascal Derived Sequences', SNJ, March ,2000.

[2] Amarnath Murthy, 'More Results and Applications of the Smarandache Star Function., SNJ, VOL.11, No. 1-2-3, 2000.