Vtgyjfy

I am solving problems from Spivak's calculus book's 1st chapter. Basically Spivak wants me to prove this:

$$x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + ldots + xy^{n-2} + y^{n-1})$$

Question 1: Is my proof correct?

Question 2: How to make it perfect? Any advise?

My proof:

I guess Spivak wants to say:
$$
x^n-y^n = (x-y)left(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
$$

Let:
$$
f(x,y,n) = left(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
$$

Then we can say (is this sub-proof perfect?):
$$begin{split}
x^{n+1}-y^{n+1} &= (x-y)left(sum_{i=1}^{i=n+1} x^{(n+1)-i}y^{i-1}right)\
&= (x-y)left(
begin{split}
&x^{(n+1)-(n+1)} y^{(n+1)-1}\
&+ sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
end{split}
right)\
&= (x-y)left(
x^{0} y^{n}
+ sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
right)\
&= (x-y)left(
y^{n}
+ sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
right)\
&= (x-y)left(
y^{n}
+ xsum_{i=1}^{i=n} x^{n-i}y^{i-1}
right)\
&= (x-y)left(
y^{n}
+ xf(x,y,n)
right)\
end{split}$$

We can also rewrite what Spivak wants into:
$$
x^n-y^n = (x-y)Big(y^{n-1} + xf(x,y,n-1)Big)
$$

We've already proven in [spivak_calc_probs.1.1.2]:
$$begin{split}
x^2-y^2 &= (x-y)(x+y)\
&= (x-y)left(sum_{i=1}^{i=2} x^{2-i}y^{i-1}right)\
&= (x-y)Big(y^{2-1} + xf(x,y,2-1)Big)\
end{split}$$

Then, by induction, we prove that:
$$begin{split}
x^2-y^2 &= (x-y)Big(y^{2-1} + xf(x,y,2-1)Big)\
x^3-y^3 &= (x-y)Big(y^{3-1} + xf(x,y,3-1)Big)\
vdots\
x^n-y^n &= (x-y)Big(y^{n-1} + xf(x,y,n-1)Big)\
end{split}$$

And since $(x-y)(y^{n-1} + xf(x,y,n-1))$ is only a rewrite of what Spivak
wants, i.e. $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $ldots$ $+$ $xy^{n-2})$,
therefore Q.E.D already.

alternative proof:

Using the distributive axiom:
$$begin{split}
& (x-y)left(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)\
&= xleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
-yleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)\
&= xleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
-yleft(x^{n-n}y^{n-1} + sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}right)\
&= xleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
-yleft(x^{0}y^{n-1} + sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}right)\
&= xleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
-yleft(y^{n-1} + sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}right)\
&= xleft(sum_{i=1}^{i=n} x^{n-i}y^{i-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= xleft(x^{n-1}y^{1-1} + sum_{i=2}^{i=n} x^{n-i}y^{i-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= xleft(x^{n-1}y^{0} + sum_{i=2}^{i=n} x^{n-i}y^{i-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= xleft(x^{n-1} + sum_{i=2}^{i=n} x^{n-i}y^{i-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= left(x^{n} + sum_{i=2}^{i=n} x^{n-i+1}y^{i-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= left(x^{n} + sum_{i=1}^{i=n-1} x^{n-(i+1)+1}y^{(i+1)-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= left(x^{n} + sum_{i=1}^{i=n-1} x^{n-i-1+1}y^{i+1-1}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= left(x^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)
-left(y^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= x^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - sum_{i=1}^{i=n-1} x^{n-i}y^{i}\
end{split}$$

Then using additive associative axiom:
$$begin{split}
& x^{n} + sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - sum_{i=1}^{i=n-1} x^{n-i}y^{i}\
&= x^{n}
- y^{n}
+ left(sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- sum_{i=1}^{i=n-1} x^{n-i}y^{i}right)\
&= x^{n}
- y^{n}
+ left(0right)\
&= x^{n}
- y^{n}
end{split}$$

A proof by induction (although in this case it's not the simplest approach, as pointed out in comments) would proceed as follows:

1) Base case - show the the expansion is correct for $n=1$. In this case the sum $sum_{i=1}^{i=n} x^{n-i}y^{i-1}$ has only one term which is $x^0y^0=1$ so this is straightforward.

2) Assume the expansion is correct for a specifc value of $n$, say $n=k$.

3) Find an expression that relates the $n=k+1$ case to the $n=k$ case. You could try:

$quad x^{k+1}-y^{k+1} = x(x^k-y^k) + (x-y)y^k$

4) Use (2) to expand your expression from (3):

$quad x^{k+1}-y^{k+1} = x(x-y)sum_{i=1}^{i=k} x^{k-i}y^{i-1} + (x-y)y^k\ quad=(x-y)left( xsum_{i=1}^{i=k} x^{k-i}y^{i-1} + y^kright) \quad =(x-y)left(sum_{i=1}^{i=k+1} x^{k+1-i}y^{i-1}right)$

So now you have shown that if the expansion is true for $n=k$ then it is also true for $n=k+1$. Together with your base case $n=1$, this proves by induction that the expansion is true for all $n in mathbb{N}$.

More simply, in your first proof, for brevity let $S(n)$ be the sentence $(x-y)f(x,y,n)=x^n-y^n.$ Then, using your calculation that $f(x,y,n+1)=y^n+xf(x,y,n),$ we have $$S(n)implies (x-y)f(x,y,n+1)=$$ $$=(x-y)(y^n+xf(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(( x-y)f(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(x^n-y^n)=$$ $$=x^{n+1}-y^{n+1}.$$ That is, we have $$S(n)implies S(n+1).$$ And $S(1)$ is true because $f(x,y,1)=1.$ So by induction on $n$ we have $S(n)$ for all $nin Bbb N.$

$(a - 1)(a^{n-1} + a^{n-2} + ... + 1) =$
$a^n + a^{n-1} + ... + a - (a^{n-1} + a^{n-2} + ... + 1) =$
$a^n - 1$

Set a = x/y and multiply by y$^n$.