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Polynomial Differences in Algebra Explained using Stirling Numbers of the Second Kind

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Earlier, we examined different polynomials with algebra to determine specific formula for a given degree. The table below is reproduced from this earlier page.

f(k)=a0k4+a1k3+a2k2+a3k+a4 [1.01]
x Δ0 or f(k) Δ1 or Δ Δ2 Δ3 Δ4
0 a4
1 a0+a1+a2+a3+a4 a0+a1+a2+a3
2 16a0+8a1+4a2+2a3+a4 15a0+7a1+3a2+a3 14a0+6a1+2a2
3 81a0+27a1+9a2+3a3+a4 65a0+19a1+5a2+a3 50a0+12a1+2a2 36a0+6a1
4 256a0+64a1+16a2+4a3+a4 175a0+37a1+7a2+a3 110a0+18a1+2a2 60a0+6a1 24a0
5 625a0+125a1+25a2+5a3+a4 369a0+61a1+9a2+a3 194a0+24a1+2a2 84a0+6a1 24a0

Using the table of values for Stirling Numbers of the Second Kind, to determine the factorial representation of 1.01
i
01234
k(0)k(1)k(2)k(3)k(4)
a4a4
a30a3
a20a2a2
a10a13a1a1
a00a07a06a0a0
Suma4a3+a2+a1+a0a2+3a1+7a0a1+6a0a0

By adding up the conversions for each factorial, we find:
f(k)=a4+a3k+a2k2+a1k3+a0k4≡a4+(a3+a2+a1+a0)k(1)+(a2+3a1+7a0)k(2)+(a1+6a0)k(3)+a0k(4) [1.02]

From this page, we presented the general formula for the mth differences of a polynomial factorial k, degree n:
16 [1.03]
In particular (from the same page) the first difference is:
∆k(n)=nk(n-1) [1.04]

Computing the First Difference

Using 1.04 with 1.02 we find:
∆f(k)=(a3+a2+a1+a0)+2(a2+3a1+7a0)k(1)+3(a1+6a0)k(2)+4a0k(3)[1.05]

k=0

When k=0, we find:
∆f(0)=a3+a2+a1+a0 which is what we found in the table

k=2

For k=2, k(3)=2(2-1)(2-2)=0, k(2)=2(2-1), and k(1)=2
∆f(2)=(a3+a2+a1+a0)+2(a2+3a1+7a0)k(1)+3(a1+6a0)k(2)
=(a3+a2+a1+a0)+4(3a1+a2+7a0)+6(a1+6a0)
=a3+5a2+19a3+65a0

Computing the Second Difference

Using 1.05:
∆f(k)=(a3+a2+a1+a0)+2(a2+3a1+7a0)k(1)+3(a1+6a0)k(2)+4a0k(3) [1.05 repeated]
And applying 1.03 for m=2
2f(k)=2(a2+3a1+7a0)+6(a1+6a0)k(1)+12a0k(2) [1.06]

k=0

For k=0,
2f(k)=2(a2+3a1+7a0)
=2a2+6a1+14a0

k=1

For k=1, k(3)=1(1-1)(1-2)=0, k(2)=1(1-1)=0, and k(1)=1

2f(1)=2(a2+3a1+7a0)+6(a1+6a0)
=2a2+12a1+50a0

Computing the Third Difference

2f(k)=2(a2+3a1+7a0)+6(a1+6a0)k(1)+12a0k(2) [1.06, repeated]

3f(k)=6(a1+6a0)+24a0k(1) [1.07]

k=0

For k=0, k(1) =0
So
3f(0)=6(a1+6a0)
=6a1+36a0





Ken Ward's Mathematics Pages


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