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Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate

Statement

Let $a, b \in R$ and let $n \in N$ with $n \geq 1$ , with powers as in Integer powers $a^{m}$ and finite sums as in Finite sums and finite products, by recursion. Then

$b^{n} - a^{n} = (b - a) \sum_{k = 0}^{n - 1} a^{k} b^{n - 1 - k} .$

Lipschitz estimate. If $0 \leq a \leq b \leq M$ then

$0 \leq b^{n} - a^{n} \leq n M^{n - 1} (b - a),$

where $n$ denotes the canonical field image $n \cdot 1_{R}$ of the natural n (Canonical naturals are positive and strictly increasing, The unique embedding of ℚ into an ordered field). This estimate is the form in which the factorisation is used, later on this page, to prove that $n$ -th roots exist.

Facts & Assumptions

Given: Elements $a, b, M$ of $R$ and a natural $n \geq 1$ . Define $c : N \to R$ by $c_{k} = a^{k} b^{n - k}$ for $k \leq n$ and $c_{k} = a^{n}$ for $k > n$ ; only the values $c_{0}, \dots, c_{n}$ occur below.

[L1]

Definition of powers (Integer powers $a^{m}$ ): $a^{0} = 1$ and $a^{j + 1} = a^{j} a$ .

[L2]

Laws of finite sums (Laws of finite sums and finite products, themselves proved by induction, The principle of mathematical induction): telescoping $\sum_{k < n} (c_{k + 1} - c_{k}) = c_{n} - c_{0}$ ; scaling $\sum_{k < n} λ d_{k} = λ \sum_{k < n} d_{k}$ and the constant sum $\sum_{k < n} λ = n λ$ ; and monotonicity, $d_{k} \leq e_{k}$ for all $k < n$ implies $\sum_{k < n} d_{k} \leq \sum_{k < n} e_{k}$ .

[L3]

Laws of integer exponents (Laws of integer exponents): $M^{k} M^{n - 1 - k} = M^{n - 1}$ for $k \leq n - 1$ .

[L4]

Monotonicity of powers (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ): $0 \leq x \leq y$ implies $0 \leq x^{j} \leq y^{j}$ .

[L5]

Multiplying inequalities of nonnegatives: $0 \leq x \leq y$ and $0 \leq u \leq v$ imply $x u \leq y v$ . Multiplying inequalities of positives states exactly this nonstrict form, alongside the strict one, and it is the only product-of-inequalities move used below. Two instances occur: in step 1.3 with the four nonnegatives $a^{k} \leq M^{k}$ and $b^{n - 1 - k} \leq M^{n - 1 - k}$ , and in step 4.1 with $0 \leq \sum_{k < n} a^{k} b^{n - 1 - k} \leq n M^{n - 1}$ and $0 \leq b - a \leq b - a$ , which yields $(\sum_{k < n} a^{k} b^{n - 1 - k}) (b - a) \leq n M^{n - 1} (b - a)$ . No instance with a possibly negative factor and no strict instance is needed, so the multiplicative order claims of Sign rules for products and monotonicity of multiplication, which are strict and require a nonzero multiplier, are not what carries either move.

Proof

technique · direct

1.1L1algebra

The consecutive differences of $c$ factor: for $k < n$ , $c_{k + 1} - c_{k} = a^{k + 1} b^{n - k - 1} - a^{k} b^{n - k} = a^{k} b^{n - 1 - k} a - a^{k} b^{n - 1 - k} b = a^{k} b^{n - 1 - k} (a - b)$ , using $a^{k + 1} = a^{k} a$ and $b^{n - k} = b^{n - 1 - k} b$ .

1.2L1algebra

The endpoints of $c$ are $c_{0} = a^{0} b^{n} = b^{n}$ and $c_{n} = a^{n} b^{0} = a^{n}$ .

1.3L3L4L5

Now assume $0 \leq a \leq b \leq M$ ; then for every $k < n$ , $0 \leq a^{k} \leq M^{k}$ and $0 \leq b^{n - 1 - k} \leq M^{n - 1 - k}$ , so $0 \leq a^{k} b^{n - 1 - k} \leq M^{k} M^{n - 1 - k} = M^{n - 1}$ .

1.4L4L5

Also $b - a \geq 0$ and $M^{n - 1} \geq 0$ , and $b^{n} - a^{n} \geq 0$ because $0 \leq a \leq b$ gives $a^{n} \leq b^{n}$ .

2.1step 1.2L2

Telescoping the differences: $\sum_{k < n} (c_{k + 1} - c_{k}) = c_{n} - c_{0} = a^{n} - b^{n}$ .

2.2step 1.1L2

Pulling the constant factor out of the same sum: $\sum_{k < n} (c_{k + 1} - c_{k}) = \sum_{k < n} a^{k} b^{n - 1 - k} (a - b) = (a - b) \sum_{k < n} a^{k} b^{n - 1 - k}$ .

2.3step 1.3L2

Summing the bound of step 1.3 over $k < n$ and using monotonicity and the constant sum: $\sum_{k < n} a^{k} b^{n - 1 - k} \leq \sum_{k < n} M^{n - 1} = n M^{n - 1}$ .

3.1step 2.1step 2.2algebra

Comparing the two evaluations gives $a^{n} - b^{n} = (a - b) \sum_{k < n} a^{k} b^{n - 1 - k}$ , and multiplying both sides by $- 1$ gives the stated factorisation $b^{n} - a^{n} = (b - a) \sum_{k = 0}^{n - 1} a^{k} b^{n - 1 - k}$ .

4.1step 3.1step 1.3step 1.4step 2.3L2L5∎

Multiplying the inequality of step 2.3 by the nonnegative factor $b - a$ and substituting into the factorisation gives $b^{n} - a^{n} = (b - a) \sum_{k < n} a^{k} b^{n - 1 - k} \leq n M^{n - 1} (b - a)$ ; the multiplication is the instance of [L5] with $0 \leq \sum_{k < n} a^{k} b^{n - 1 - k} \leq n M^{n - 1}$ and $0 \leq b - a \leq b - a$ , the left-hand nonnegativity holding because every term is nonnegative by step 1.3 and finite sums are monotone. Together with $b^{n} - a^{n} \geq 0$ this is the Lipschitz estimate.

Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Factorisation of bn−anbn−an, and the resulting Lipschitz estimate

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate