# Ken Ward's Mathematics Pages

Series Contents

## Determining Polynomials: Algebra

On this page we explore polynomials of degrees 2, 3, 4 and 5 to search for patterns using algebra.

## Preamble

Previously we mentioned the results for the sum of the natural numbers, where n is the number in the series and Sn is the sum of these n:
 n 1 2 3 4 5 Sn 1 3 6 10 15 Δ1 2 3 4 5 Δ2 1 1 1

We noted the series is based on a quadratic because the second difference, Δ2, is a constant. So Sn is of the form a0n2+a1n+a2.
We note that the constant, a2, is 0 because the sum of 0 terms is 0, S0=0.
Because Δn=a0n!, when the steps are 1, we know:
Δ2=a02!, so:
a02!=1, and a0=1/2!=1/2
As the polynomial is a quadratic, and we would have needed 3 simultaneous equations to solve for the 3 unknowns, but we have already identified 2 of them, we can find the remaining coefficient, a1, from an equation with one unknown, for example:
S1=1=(1/2)n2+a1n
As n is 1, we have 1=(1/2)+a1
And so the equation is:
(a1n2+a2n)/2, or
n/2(n+1)

## Generalizing Polynomials Of Degree 2 With Algebra

Consider the equation:
f(x)=a0x2+a1x+a2 [1.01]

 x Δ0 or f(x) Δ1 or Δ Δ2 0 a2 1 a0+a1+a2 a0+a1 2 4a0+2a1+a2 3a0+a1 2a0 3 9a0+3a1+a2 5a0+a1 2a0
With the previous example, we can write the coefficients down immediately. So:
Δ2 =1=2a0, so a0=1/2
Δ =1=a0+a0, so a1=1/2
And a0=0.
So, writing n for x, the equation is:
Sn=(1/2)n2+(1/2)n

We can note that:
f(0)=a2, and generally, f(0)=an because this is the constant term, independent of x [1.02]
Δf(1)=a0+a1, and generally:
[1.03]
because this is just replacing the x's with 1, leaving the coefficients.
Δf(0), which is f(1)-f(0), is a0+a1 and generally:
[1.04]

because Δf(0)=f(1)-f(0), f(0)=an, and f(1) is the sum of the coefficients (by substituting 1 for x in a general polynomial).

## Example

 x Δ Δ2 Δ3 0 5 1 12 7 2 29 17 10 3 56 27 10 0
Because we find that Δ2 is a constant, 10, we conclude the polynomial is one of degree 2.
f(x)=a0x2+a1x+a2
Using our results for a polynomial of degree 2, we note:
a2=5
2!a0=10, so a0=5
Δf(0)=a0+a1=7=5+a1 so, a1=2
The underlying equation is therefore:
5x2+2x+5

## Generalizing Polynomials Of Degree 3 With Algebra

We can do the same algebra we did before with a degree 2 polynomial with a degree 3 (n=3):
f(x)=a0x3+a1x2+a2x+a3
 x Δ0 or f(x) Δ1 or Δ Δ2 Δ3 0 a3 1 a0+a1+a2+a3 a0+a1+a2 2 8a0+4a1+2a2+a3 7a0+3a1+a2 6a0+2a1 3 27a0+9a1+3a2+a3 19a0+5a1+a2 12a0+2a1 6a0 4 64a0+16a1+4a2+a3 37a0+7a1+a2 18a0+2a1 6a0

We note that what we have previously concluded applies here too (as it will to any polynomial) That is:
f(0)=an, which is a3 here.
f(1) is the sum of all the coefficients, a0+a1+a2+a3
Δf(0)=the sum of the coefficients, except the last a0+a1+a2
The constant term is n!a0

Tantalisingly,  Δ2 =6a0+2a1 and Δ3 =6a0 which suggests we see 3! appearing in the second difference. This appears to be the case when n=2. However, as we shall see later, it is not that simple.

## Example

 Δ Δ2 Δ3 0 5 1 17 12 2 47 30 18 3 107 60 30 12 4 209 102 42 12
At the third difference we obtain a constant so this is a polynomial with n=3
f(x)=a0x3+a1x2+a2x+a3
We note a3=5
a0=12/3!, so a0=2
From the general table above, we note that Δ2 f(0)=6a0+2a1
So 18=12+2a1
So, a1=3
Δf(0)=sum of coefficient=a0+a1+a2=12, 2+3+a2=12
So a2=7
The equation is therefore:
f(x)=2x3+3x2+7x+5

## Generalizing Polynomials Of Degree 4 With Algebra

We can do the same algebra we did before with a degree 3 polynomial with a degree 4 (n=4):
f(x)=a0x4+a1x3+a2x2+a3x+a4
 x Δ0 or f(x) Δ1 or Δ Δ2 Δ3 Δ4 0 a4 1 a0+a1+a2+a3+a4 a0+a1+a2+a3 2 16a0+8a1+2a2+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
Looking at Δ3  we note that n! does not appear in a simple way, but we also note that 6a1 looks like (n-1)!a1 and so do the other Δn-1 f(0) -- in the tables above. It seems there are patterns emerging, but we will not tax the algebra further and wait until we look at factorial polynomials for clarification. In fact, factorial polynomials help us greatly with difference equations, and regular polynomials can be converted to factorial polynomials. (We finally discover how to compute these values on this page).

## Generalizing Polynomials Of Degree 5 With Algebra

We can do the same algebra we did before with a degree 4 polynomial with a degree 5 (n=5):
f(x)=a0x54+a1x4+a2x3+a3x2+a4x+a5

 x Δ0 or f(x) Δ1 or Δ Δ2 Δ3 Δ4 Δ5 0 a5 1 a0+a1+a2+a3+a4+a5 a0+a1+a2+a3+a4 2 32a0+16a1+8a2+4a3+2a4+a5 31a0+15a1+7a2+3a3+a4 30a0+14a1+6a2+2a3 3 243a0+81a1+27a2+9a3+3a4+a5 211a0+65a1+19a2+5a3+a4 180a0+50a1+12a2+2a3 150a0+36a1+6a2 4 1024a0+256a1+64a2+16a3+4a4+a5 781a0+175a1+37a2+7a3+a4 570a0+110a1+18a2+2a3 390a0+60a1+6a2 240a0+24a1 5 3125a0+625a1+125a2+25a3+5a4+a5 2101a0+369a1+61a2+9a3+a4 1320a0+194a1+24a2+2a3 750a0+84a1+6a2 360a0+24a1 120a0 6 7776a0+1296a1+216a2+36a3+6a4+a5 4651a0+671a1+91a2+11a3+a4 2550a0+302a1+30a2+2a3 1230a0+108a1+6a2 480a0+24a1 120a0
Again, patterns seem to be suggested, but they aren't particularly clear yet. One approach we can take is to transform the polynomials into another form that might yield their patterns more clearly. We will look at factorial polynomials next to try to understand difference behaviour better.

## Example

Finally, on this page, we will look at an example of a polynomial degree 5

 Δ0 Δ1 Δ2 Δ3 Δ4 Δ5 0 11 1 22 11 2 95 73 62 3 452 357 284 222 4 1723 1271 914 630 408 5 5186 3463 2192 1278 648 240 6 13007 7821 4358 2166 888 240

This is, or appears to be a polynomial of degree 5:
f(x)=a0x54+a1x4+a2x3+a3x2+a4x+a5
a5=11 (f(0) )
a0=240/5!=2
Δ4f(0)=408=240a0+24a1
So, 24a1=408-2*240=-72
a1=-3
Δ3f(0)=150a0+36a1+6a2
Substituting known values,
150*2+36(-3)+6a2=222
a2=5
Δ2f(0)=30a0+14a1+6a2+2a3
Substituting known values:
30a0+14a1+6a2+2a3=62=30*2+14(-3)+6*5+2a3
a3=7

Δf(0)=a0+a1+a2+a3+a4
Substituting known values,
2-3+5+7+a4=11
So a4=0
So the equation is:
f(x)=2x54-3x4+5x3+7x2+11

Ken Ward's Mathematics Pages

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