Intereting Posts

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In Spivak’s *Calculus 3rd Edition*, there is an exercise to prove the following:

$$x^n – y^n = (x-y)(x^{n-1} + x^{n-2} y + … + x y^{n-2} + y^{n-1})$$

I can’t seem to get the answer. Either I’ve gone wrong somewhere, I’m overlooking something, or both. Here’s my (non) proof:

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$$\begin{align*}

x^n – y^n &= (x – y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\

&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\

&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\

&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} – x^{n-1}y – y^2 x^{n-2} – \cdots- x y^{n-1} – y^n \\

&= x^n + x^2 y^{n-2} – x^{n-2} y^2 – y^n \\

&\neq x^n – y^n

\end{align*}$$

Is there something I can do with $x^n + x^2 y^{n-2} – x^{n-2} y^2 – y^n$ that I’m not seeing, or did I make a mistake early on?

EDIT:

I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:

- Associate law for addition
- Existence of an additive identity
- Existence of additive inverses
- Commutative law for additions
- Associative law for multiplication
- Existence of a multiplicative identity
- Existence of multiplicative inverses
- Commutative law for multiplication
- Distibutive law

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You have everything right except the last line.

Maybe it is easier to do in this order:

$$(x−y)\left(x^{n−1}+x^{n−2}y+\cdots+xy^{n−2}+y^{n−1}\right)=\\

=x\cdot x^{n-1}-y\cdot x^{n-1} +x\cdot x^{n−2}y-

y\cdot x^{n−2}y+x\cdot x^{n−3}y^2-\cdots\\ \cdots -y\cdot x^2y^{n-3} +x\cdot xy^{n-2}-y \cdot y^{n-1}$$

The second term $y\cdot x^{n-1}$ is the same as the third term $x\cdot x^{n−2}y$ except the sign, similarly the 4th and the 5th terms are canceled… So the only terms left are: $x\cdot x^{n-1}$ and $y\cdot y^{n-1}$.

Here is the inductive step, presented more conceptually

$$\rm\frac{x^{n+1}-y^{n+1}}{x-y}\: =\ x^n\: +\ y\ \frac{x^n-y^n}{x-y}$$

So, *intuitively,* proceeding inductively yields

$$\rm\:x^n + y\: (x^{n-1} + y\: (x^{n-2} +\:\cdots\:))\ =\ x^n + y\: x^{n-1} + y^2\: x^{n-2} + \:\cdots $$

Use this intuition to compose a *formal* proof by induction.

I think it would be easier for you to recall

$$\left(1+x+x^2+\cdots+x^{n-1}\right)(x-1) = x^n-1$$

and put $x=\dfrac{b}{a}$

$$\eqalign{

& \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n – 1}}}}{{{a^{n – 1}}}}} \right)\left( {\frac{b}{a} – 1} \right) = \frac{{{b^n}}}{{{a^n}}} – 1 \cr

& \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n – 1}}}}{{{a^{n – 1}}}}} \right)\left( {\frac{{b – a}}{a}} \right) = \frac{{{b^n} – {a^n}}}{{{a^n}}} \cr

& {a^{n – 1}}\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n – 1}}}}{{{a^{n – 1}}}}} \right)\left( {b – a} \right) = {b^n} – {a^n} \cr

& \left( {{a^{n – 1}} + b{a^{n – 2}} + {b^2}{a^{n – 3}} + \cdots + {b^{n – 1}}} \right)\left( {b – a} \right) = {b^n} – {a^n} \cr} $$

A little bit “tidier”, so that we know what happens in between the dots…

$$\eqalign{

& {x^n} – 1 = \left( {x – 1} \right)\sum\limits_{k = 0}^{n – 1} {{x^k}} \cr

& \frac{{{b^n}}}{{{a^n}}} – 1 = \left( {\frac{b}{a} – 1} \right)\sum\limits_{k = 0}^{n – 1} {\frac{{{b^k}}}{{{a^k}}}} \cr

& \frac{{{b^n} – {a^n}}}{{{a^n}}} = \left( {\frac{{b – a}}{a}} \right)\sum\limits_{k = 0}^{n – 1} {\frac{{{b^k}}}{{{a^k}}}} \cr

& {b^n} – {a^n} = \left( {b – a} \right)\sum\limits_{k = 0}^{n – 1} {{b^k}{a^{n – k – 1}}} \cr} $$

Your method is sound, you just made a sort of arithmetic mistake. When cancelling or otherwise combining two sequences, try explicitly lining things up to make sure you do it right:

$$ \begin{align} x^n &+& x^{n-1} y &+& x^{n-2} y^2 &+& \cdots + x y^{n-1} &

\\ &-& x^{n-1} y &-& x^{n-2} y^2 &+& \cdots – x y^{n-1} &+& y^n

\end{align}

$$

I’ve found that, when shorthand starts becoming awkward and/or error prone, that it really is helpful to switch to summation notation. So, you are trying to prove

$$ x^n – y^n = (x-y) \sum_{k=0}^{n-1} x^k y^{n-1-k} $$

and the first stem of your work would be

$$ \cdots = \left( \sum_{k=0}^{n-1} x^{k+1} y^{n-1-k} \right)

– \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$

and now, we can change the index to line things up: I’m substituting k = j-1:

$$ \cdots = \left( \sum_{(j-1)=0}^{n-1} x^{(j-1)+1} y^{n-1-(j-1)} \right)

– \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$

and simplifying

$$ \cdots = \left( \sum_{j=1}^{n} x^{j} y^{n-j} \right)

– \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$

and now replacing $j$ with $k$.

$$ \cdots = \left( \sum_{k=1}^{n} x^{k} y^{n-k} \right)

– \left( \sum_{k=0}^{n-1} x^{k} y^{n-k} \right)$$

(can you take it from here?)

The $x^2 y^{n-2}$ term from $x \cdot x y^{n-2}$ is cancelled by the term from $(-y) \cdot

x^2 y^{n-3}$. Similarly, the $(-y) \cdot x^{n-2} y$ is cancelled by the $x \cdot x^{n-3} y^2$.

Since powers of x and y is always greater than or equal to zero, You can prove it by mathematical induction.

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