We can convert a polynomial to a factorial polynomial by dividing it by
k, k-1, etc.
k(0) is defined as 1.
Example
As an example, we will convert x2 to factorials.
h is taken to be 1. We suppose:
x2≡a0k(2)+a1k(1)+a2
In general successively divide the polynomial by x, x-1,
x-2...(x-n+1).
For x2 we divide by x and x-1.
We can divide how we please, but Synthetic Division
is perhaps preferred.
Write out the table as below. The second line is for the coefficients:
x 2
x 1
x 0
1
0
0
0
1
Divide by x:
x 2
x 1
x 0
1
0
0
0
0
0
1
0
0
1
In the normal fashion for synthetic division, we write down the 0 (of
x-0!) and write down the coefficient of x2, 1,
in the lower line.
We take 0x1 and write it below the coefficient of x, adding them.
Finally we take this sum, 0, multiply it by 0 and write it below the
coefficient of x0.
We are now finished with the last column, the value is 0, so there is
no k(0) term.
Next we divide by x-1:
x
2
x
1
x
0
1
0
0
0
0
0
1
0
0
1
1
1
1
Following the procedure for synthetic division, we divide by x-1. The
resulting factorial form is:
x2≡k(2)+k(1)
Next are the results for x3.
x
x
3
x
2
x
1
x
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
2
2
1
3
x3≡k(3)+3k(2)+k(1)
We can convert any polynomial to factorial form using this method.
As a final example, we convert the following:
x5+2x4+3x3+7x2+5x+19
x
x
5
x
4
x
3
x
2
x
1
x
0
1
2
3
7
5
19
0
0
0
0
0
0
1
2
3
7
5
19
1
1
3
6
13
1
3
6
13
18
2
2
10
32
1
5
16
45
3
3
24
1
8
40
4
4
1
12
Therefore:
x5+2x4+3x3+7x2+5x+19≡k(5)+12k(4)+40k(3)+45k(2)+18k(1)+19k
(0)
Using Synthetic Division, we can
comparitively easily determine the factorial form of a polynomial.
Later we shall see that using a table of Stirling Numbers
of the Second Kind, we can convert them even more easily.
Factorials
with Negative Powers
We can sometimes convert a fraction to negative factorials. For
instance, we wish to convert the following: [2.01]
We proceed by equating coefficients, after assuming: [2.02]
Multiplying throughout by (k+1)(k+2)(k+3): [2.03]
Note the coefficient of k2 on the LHS is 0, so
A=0
Equating coefficients of k:
1=B
And equating the constant:
1=3B+C
As B=1, C=-2
We can rewrite 2.02 supplying the values for the constants, A, B, C, so: [2.04]
Using the symbols for factorials, we can write 2.01 expressed as
factorial polynomials. [2.05] ■
[2.01]
[2.02]
[2.03]
[2.04]
[2.05] ■