  # Ken Ward's Mathematics Pages

Series Contents

## Page Contents

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 0 1 2 3 4 k(0) k(1) k(2) k(3) k(4) a4 a4 a3 0 a3 a2 0 a2 a2 a1 0 a1 3a1 a1 a0 0 a0 7a0 6a0 a0 Sum a4 a3+a2+a1+a0 a2+3a1+7a0 a1+6a0 a0

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: [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

# Faster Arithmetic - by Ken Ward

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle:
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