Difference of Two Powers

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Theorem

Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.

Let $n \in \N$ such that $n \ge 2$.


Then for all $a, b \in \mathbb F$:

\(\displaystyle a^n - b^n\) \(=\) \(\displaystyle \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }\)


For convenience of applicability, these results are sometimes separated into two cases for odd and even indices:

Difference of Two Odd Powers

Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.

Let $n \in \Z_{\ge 0}$ be a positive integer.


Then for all $a, b \in \mathbb F$:

\(\displaystyle a^{2 n + 1} - b^{2 n + 1}\) \(=\) \(\displaystyle \paren {a - b} \sum_{j \mathop = 0}^{2 n} a^{n - j} b^j\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {a^{2 n} + a^{2 n - 1} b + a^{2 n - 2} b^2 + \dotsb + a b^{2 n - 1} + b^{2 n} }\)


Difference of Two Even Powers

Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.


Then for all $a, b \in \mathbb F$:

\(\displaystyle a^{2 n} - b^{2 n}\) \(=\) \(\displaystyle \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {a + b} \paren {a^{2 n - 2} + a^{2 n - 4} b^2 + a^{2 n - 6} b^4 + \dotsb + a^2 b^{2 n - 4} + b^{2 n - 2} }\)


General Commutative Ring

The result can also be extended into the general abstract algebraic context as follows:


Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$.

Let $a, b \in R$.

Let $n \in \N$ such that $n \ge 2$.


Then:

\(\displaystyle a^n - b^n\) \(=\) \(\displaystyle \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }\)


Proof 1

Let $\displaystyle S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$.

This can also be written:

$\displaystyle S_n = \sum_{j \mathop = 0}^{n - 1} b^j a^{n - j - 1}$


Consider:

$\displaystyle a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j$

Taking the first term (where $j = 0$) out of the summation, we get:

$\displaystyle a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$


Similarly, consider:

$\displaystyle b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}$

Taking the first term (where $j = 0$) out of the summation:

$\displaystyle b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$

This is equal to:

$\displaystyle b^n + \sum_{j \mathop = 1}^{n - 1} a^j b^{n - j}$

by Permutation of Indices of Summation.


So:

\(\displaystyle \paren {a - b} S_n\) \(=\) \(\displaystyle a \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j - b \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}\)
\(\displaystyle \) \(=\) \(\displaystyle a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - b^n\)
\(\displaystyle \) \(=\) \(\displaystyle a^n - b^n\)

$\blacksquare$


Proof 2

An instance of Difference of Two Powers in a General Commutative Ring.

$\blacksquare$


Proof 3

From Sum of Geometric Progression:


\(\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j\) \(=\) \(\displaystyle \frac {x^n - 1} {x - 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\dfrac a b}^n - 1\) \(=\) \(\displaystyle \paren {\dfrac a b - 1} \sum_{j \mathop = 0}^{n - 1} \paren {\dfrac a b}^j\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\dfrac {a^n - b^n} {b^n} }\) \(=\) \(\displaystyle \paren {\dfrac {a - b} b} \paren {\paren {\dfrac a b}^{n - 1} + \paren {\dfrac a b}^{n - 2} + \dotsb + \paren {\dfrac a b}^1 + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^n - b^n\) \(=\) \(\displaystyle \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + \dotsb + a b^{n - 2} + b^{n - 1} }\) multiplying both sides by $b^n$

$\blacksquare$


Proof 4

The proof will proceed by the Principle of Complete Finite Induction on $\Z_{>1}$.

Let $S$ be the set defined as:

$\displaystyle S := \set {n \in \Z_{>0}: a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j}$

That is, $S$ is to be the set of all $n$ such that:

$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$


Basis for the Induction

We have that:

$\displaystyle a^1 - b^1 = \paren {a - b} \sum_{j \mathop = 0}^0 a^{1 - 0 - 1} b^j$


So $1 \in S$.

This is the basis for the induction.


Induction Hypothesis

It is to be shown that if $j \in S$ for all $r$ such that $1 \le r \le k$, then it follows that $k + 1 \in S$.


This is the induction hypothesis:

$\forall r \in \Z_{>0}: 1 \le r \le k: \displaystyle a^r - b^r = \paren {a - b} \sum_{j \mathop = 0}^{r - 1} a^{r - j - 1} b^j$


It is to be demonstrated that it follows that:

$\displaystyle a^{k + 1} - b^{k + 1} = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j$


Induction Step

This is the induction step:


\(\displaystyle a^{k + 1} - b^{k + 1}\) \(=\) \(\displaystyle \paren {a + b} \paren {a^k - b^k} - a \paren {a^{k - 1} - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \)



So $\forall r \in S: 0 \le r \le k: r \in S \implies k + 1 \in S$ and the result follows by the Principle of Complete Finite Induction:

$\forall n \in \Z_{>0}: \displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$


Examples

Difference of Two Cubes

$x^3 - y^3 = \paren {x - y} \paren {x^2 + x y + y^2}$


Difference of Two Fourth Powers

$x^4 - y^4 = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}$


Difference of Two Fifth Powers

$x^5 - y^5 = \paren {x - y} \paren {x^4 + x^3 y + x^2 y^2 + x y^3 + y^4}$


Difference of Two Sixth Powers

$x^6 - y^6 = \paren {x - y} \paren {x + y} \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$


Sources