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Method

Successive Differences of Powers

Consider a sequence that is given by the following explicit rule.

a_{n} = a_{1} \cdot n^{p}

In this rule

p

is an integer. The values of

a_{1}

and

p

can be found by calculating the differences between consecutive terms of the sequence. In the table, various consecutive terms of a sequence

a_{n}

are given. Note that

121.5

is not the first term of the sequence.

This sequence is given by the rule above. To determine the values of

a_{1}

and

p

in the explicit rule, there are five steps to follow.

1

Subtract Consecutive Terms

First, calculate the differences between consecutive terms.

$a_{n}$	$a_{n - 1}$	$a_{n} - a_{n - 1}$	Difference
$288$	$121.5$	$288 - 121.5$	$166.5$
$562.5$	$288$	$562.5 - 288$	$274.5$
$972$	$562.5$	$972 - 562.5$	$409.5$
$1543.5$	$972$	$1543.5 - 972$	$571.5$

2

Create a New Sequence

A new sequence can then be created using these differences.

3

Repeat Until the Differences Are Constant

Continue finding the differences until all the terms of the sequence are constant.

4

Identify

p

The exponent $p$ is the same as the number of differences needed. Since after the third differences the finite differences result in a nonzero constant, $p$ equals $3 .$

5

Calculate

a_{1}

The value of

a_{1}

can be calculated by using the below equation, where

d

is the constant found in the

p^{th}

differences.

d = a_{1} \cdot p!

In this case

d

is equal to

27,

and from the previous step the value of

p

is

3 .

Next, substitute these values into the equation and solve for

a_{1} .

d = a_{1} \cdot p!

$d = 27$ , $p = 3$

27 = a_{1} \cdot 3!

▼

Solve for

a_{1}

Write as a product

27 = a_{1} \cdot 3 \cdot 2 \cdot 1

Multiply

27 = a_{1} \cdot 6

$LHS / 6 = RHS / 6$

4.5 = a_{1}

Rearrange equation

a_{1} = 4.5

Therefore, the given data set can be represented by the following rule.

a_{n} = 4.5 n^{3}

When the $p^{th}$ differences of equally-spaced data are nonzero and constant, the data can be modeled by a polynomial function of degree $p .$

Why

Consider the difference between the

n^{th}

and the

(n - 1)^{th}

terms of the sequence

a_{n} .

a_{n} - a_{n - 1} = a_{1} n^{p} - a_{1} (n - 1)^{p} ⇓ a_{n} - a_{n - 1} = a_{1} [n^{p} - (n - 1)^{p}]

Notice that the expression in brackets is the difference of

p^{th}

powers, which can be factored as follows.

x^{p} - y^{p} = (x - y) (x^{p - 1} + x^{p - 2} y + \dots + x y^{p - 2} + y^{p - 1})

Using this rule, the right hand-side of the equation can be simplified.

a_{n} - a_{n - 1} = a_{1} [n^{p} - (n - 1)^{p}]

$x^{p} - y^{p} = (x - y) (x^{p - 1} + x^{p - 2} y + \dots + x y^{p - 2} + y^{p - 1})$

a_{n} - a_{n - 1} = a_{1} [(n - (n - 1)) (n^{p - 1} + n^{p - 2} (n - 1) + \dots + n (n - 1)^{p - 2} + (n - 1)^{p - 1})]

Distribute $- 1$

a_{n} - a_{n - 1} = a_{1} [(n - n + 1) (n^{p - 1} + n^{p - 2} (n - 1) + \dots + n (n - 1)^{p - 2} + (n - 1)^{p - 1})]

Subtract term

a_{n} - a_{n - 1} = a_{1} [(1) (n^{p - 1} + n^{p - 2} (n - 1) + \dots + n (n - 1)^{p - 2} + (n - 1)^{p - 1})]

$a \cdot 1 = a$

a_{n} - a_{n - 1} = a_{1} [n^{p - 1} + n^{p - 2} (n - 1) + \dots + n (n - 1)^{p - 2} + (n - 1)^{p - 1}]

This final equation shows that taking differences reduces the degree of the polynomial by

1 .

If taking the differences of consecutive terms continues, after the

p^{th}

differences the degree of the polynomial will be

0 .

Therefore, the polynomial will be a constant.

Conclusion If a polynomial has degree p, then the p^{th} differences are a nonzero constant .

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