If we knew, say, that the sum of the natural numbers was a quadratic equation, or polynomial of degree 2, then we know that the equation is ax

In the table below, n refers to the natural numbers, S

n | 1 | 2 | 3 | 4 | 5 |

S_{n} |
1 | 3 | 6 | 10 | 15 |

Δ^{1} |
2 | 3 | 4 | 5 | |

Δ^{2} |
1 | 1 | 1 |

We note that the Δ

The scheme mentioned here is highly simplified.

First we write out some values for analysis, using integral x-values beginning at 0. Then we compute the differences between these values (the first differences) and successively compute the differences between the differences until we reach a repeating constant value (successive differences will be zero).

The difference (first, second, etc) at which we reach this constant value is the degree of the polynomial generating the values. For instance, in the table below, we have some values for various n's. The differences have been computed in the table.

n | 0 | 1 | 2 | 3 | 4 |

f(n) | 7 | 12 | 29 | 70 | 147 |

Δ^{1} |
5 | 17 | 41 | 77 | |

Δ^{2} |
12 | 24 | 36 | ||

Δ^{3} |
12 | 12 |

Because the constant values appear at the third difference, we conclude that the equation producing our values is of degree 3. We can represent it as ax

One way to proceed is to substitute values in our equation (ax

d=7 (n=0) [1]

a+b+c+d=12 (n=1) [2]

8a+4b+2c+d=29 (n=2) [3]

27a+9b+3c+d=70 (n=3) [4]

First eliminate d from 2, 3, and 4:

a+b+c=5 [5]

8a+4b+2c=22 [6]

27a+9b+3c=63 [7]

Eliminate c from 6 and 7, using 5:

6a+2b=12 [8]

24a+6b=48 [9]

Eliminating b, we obtain:

6a=12, so a=2

Substituting a=2 in 8, we obtain:

12+2b=12, giving b=0

Substituting a=2 and b=0 in Equation 5:

2+0+c=5, so c=3

So the required equation is:

2x

n | 0 | 1 | 2 | 3 |

f(n) | 0 | 1 | 2 | 3 |

Δ^{1} |
1 | 1 | 1 |

In the following diagram, the red rectangles represent the differences between successive y-values. As they are of unit width, they also approximate the tangent to the curve in the area.

The tangent to a polynomial is always one degree less than the polynomial. So we would expect the differences to be represented by a curve which is one degree less than the actual curve.

Similarly, as we take more differences, approximating more tangents of tangents (or higher order differentials), these differences finally represent a constant value, a straight line parallel to the x-axis,.

From calculus, we note that the nth differential of a polynomial of degree n is the coefficient of the x

Actually, however, the constant term obtained through this scheme is always n! times the coefficient of the polynomial where n is the degree of the polynomial (and also the nth differences where the constant value appears).

This area concerning differences is quite fascinating and was also fascinating to Newton, who filled his notebooks with discoveries in this area.

We can note the following

- The nth, Δ
^{n}, difference, is n!a_{0} - The n+1 difference, Δ
^{n+1}, and higher differences, are zero

[2.01]

Suppose the steps along the x axis are h. The next f(x) value at x+h is:

[2.02]

We recall, by definition:

[2.03]

That is, the difference is equation 2.02 minus equation 2.01:

[2.03]

If we expand the left-hand parts of each term, we find:

[2.04]

The last term, a

[2.05]

Therefore for a polynomial of degree n, step h

[2.06]

This is reminiscent of:

[2.07]

Applying 2.06 again, we get:

[2.07]

If we apply the
formula
2.06 n
times, we have:

Or

Of course, because this is a constant (it is independent of x), the n+1 difference and further differences will be zero, so:

When h=1, we can write for a polynomial of degree n:

Or

Of course, because this is a constant (it is independent of x), the n+1 difference and further differences will be zero, so:

When h=1, we can write for a polynomial of degree n:

Ken Ward's Mathematics Pages

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