Re: Long division

Quentin Grady wrote (in part):

This has been a sensuous feast for me. What started out as
a single proof of a useful formula has become a focal point
for bringing together all manner of mathematical processes.

I added another method and posted all three algebraic methods
in an ap-calculus list. Below is the expanded version I posted
there, along with a reply I sent by e-mail to someone who
suggested (in an e-mail to me) using the chain rule to get
1/g(x) and then the product rule to get from there to f(x)/g(x).

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The Quotient Rule: 3 Algebraic Approaches
http://mathforum.org/kb/thread.jspa?messageID=5550331

Yesterday, I saw a post by Quentin Grady in sci.math
that suggested an interesting way to obtain the quotient
rule for differentiation. I wrote a reply yesterday
(METHODS #2 & 3 below). This morning, it occurred to
me that those in this group might also be interested.

I decided to add another method (METHOD #1 below) while
preparing this post for the ap-calculus listserv. I've
known about METHODS #1 & 2, but METHOD #3 (the method
suggested by Quentin Grady below) was new to me. It's
not new in the sense that I had previously tried but
failed to obtain the quotient rule in this way. Rather,
it's new in the sense that I'd never previously considered
trying this approach. In one of his replies since I made
my post, Quentin Grady said he saw the method in a
1925 calculus book, but he didn't give any additional
bibliographic information.

Incidentally, it appeared to me that Quentin Grady
may have just been asking how one carries out long
division on polynomials, but I decided to go ahead
and address the more interesting issue. I figured
others would explain polynomial long division to
him, or at least point him to an appropriate web
page. Also, it's often the case that what posters
literally ask about in their posts are not the real
issues they're having trouble with.

What follows is Quentin Grady's original post and my
reply, the latter modified to include METHOD #1 and
the labeling of the other two methods as METHOD 2
and METHOD 3.

sci.math thread "Long division" (28 February 2007)
http://groups.google.com/group/sci.math/msg/5a8909e2c18f72c0

Quentin Grady wrote:

I was reading a math book which had a proof of
differentiation of a quotient usually written as
u/v that I hadn't seen before.

In essence the proof relied on doing a long division
of the form a + b into c + d

As the division proceeded the terms became smaller
and smaller and so insignificant.

The catch is I've never seen a long division of the
form a+b into c+d so can't make sense of it.

Please can someone explain how one does such a long
division.

----------------------------------------------------

METHOD 1: WORKING DIRECTLY FROM (delta y/x) - y/x

(delta y/x) - y/x

[ y + (delta y) ] / [ x + (delta x) ] - y/x

{ x*[y + (delta y)] - y*[x + (delta x)] } / {x*[x + (delta x)]}

[ x*(delta y) - y*(delta x) ] / [ x^2 + x*(delta x) ]

We can ignore the x*(delta x) term, since it is
"infinitely smaller than" the x^2 term [or simply
divide numerator and denominator by (delta x)].
This gives us

[ x*(delta y) - y*(delta x) ] / x^2,

which is the quotient rule (in first-order finite form).

----------------------------------------------------

METHOD 2: RATIONALIZING THE DENOMINATOR OF (delta y/x)

One way that you can do this, a way that you can
sometimes find in older texts (such as those
published before 1940), is by rationalizing
the denominator.

Beginning with

(delta y/x) = [y + (delta y)] / [x + (delta x)],

which is the change in y/x,

multiply both the numerator and the denominator
by x - delta(x).

This gives you

[y + (delta y)]*[x - (delta x)] / [ x^2 - (delta x)^2 ]

[ xy + x*(delta y) - y*(delta x) - (delta x)(delta y)]

divided by

[ x^2 - (delta x)^2 ].

Keeping only the first order changes, which is what
one does for linear approximations, we get

[ xy + x*(delta y) - y*(delta x) ] / x^2

(xy)/(x^2) + [ x*(delta y) - y*(delta x) ] / x^2.

Hence, up to first-order terms, (delta y/x) - y/x
is equal to

[ x*(delta y) - y*(delta x) ] / x^2,

which is the quotient rule (in first-order finite form).

----------------------------------------------------

METHOD 3: LONG DIVISION APPLIED TO (delta y/x)

What you're [Quentin Grady] suggesting is that
we can get the same thing by long division. So
let's just do the long division and see what happens.

We want to divide x + (delta x) into y + (delta y).

Ideally, you'll want to set this up the usual
way when dividing polynomials: the column format,
where leading term is divided into leading term,
multiply and bring down, subtract, divide leading
term into leading term again, etc. I'll describe
what I'm doing, but I'm not going to attempt any
ASCII art to display things the usual way -- the
way that one would write this in a high school
algebra class.

Dividing x into y gives y/x.

Multiplying y/x by x + (delta x) gives y + (y/x)*(delta x).

Subtracting y + (y/x)*(delta x) _from_ y + (delta y)
gives (delta y) - (y/x)*(delta x).

Dividing x into (delta y) gives (delta y)/x.

Multiplying (delta y)/x by x + (delta x) gives
(delta y) + (delta x)(delta y)/x.

Subtracting (delta y) + (delta x)(delta y)/x _from_
(delta y) - (y/x)*(delta x) gives
(-y/x)*(delta x) - (delta x)(delta y)/x.

Dividing x into (-y/x)*(delta x) gives (-y/x^2)*(delta x).

Multiplying (-y/x^2)*(delta x) by x + (delta x) gives
(-y/x)*(delta x) - (y/x^2)*(delta x)^2.

Subtracting (-y/x)*(delta x) - (y/x^2)*(delta x)^2 _from_
(-y/x)*(delta x) - (delta x)(delta y)/x leaves only terms
of second order, and so to a first order approximation
we're done.

The quotient we've generated is

y/x + (delta y)/x - (y/x^2)*(delta x).

Hence, up to first-order terms, (delta y/x) - y/x
is equal to

(delta y)/x - (y/x^2)*(delta x),

which is equal to

[ x*(delta y) - y*(delta x) ] / x^2,

and again we have the quotient rule.

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Here's the e-mail reply I sent to someone who suggested
(in an e-mail to me) using the chain rule to get 1/g(x)
and then the product rule to get from there to f(x)/g(x),
suggesting that his method might be easier to teach.

Well, this does require the chain rule, but if that's
going to be covered anyway (and it is, of course), it's
certainly easier. Same with the differentiation of rational
powers of x. This can be done from scratch -- and it's not
really all that hard, but you have to know the (a-b)*(stuff)
factorization of a^q - b^q for a positive integer q (in order
to rationalize a difference of q'th roots) and the expansion
of (c + d)^p for a positive integer p (or at least the first
two terms and the fact that the there are finitely many later
terms, each of which has a greater-than-one-power-of-d factor)
to take care of the definition of a derivative for x^(p/q),
where p and q are positive integers (the case where p is
a negative integer and q is a positive integer can be
manipulated without too much trouble so that the same
methods that work for positive integers p & q can be applied
as well) -- but the algebraic details are beyond what most
calculus students are willing to go through. [I have done it
in rare cases with exceptional classes or as a supplementary
extra credit project (with hints) for students.] However,
rational powers of x are usually taken care of by implicit
differentiation and the chain rule, and all you need is
the f(x)-to-an-integer-power version of the chain rule.

My intent was to point out some old-fashioned methods
that others might not be aware of, for their own
enlightenment and entertainment, not for something I
would advocate doing in class (unless as a "filler
or supplementary" topic). Still, the first two methods
*are* the types of calculations/arguments one sees in
engineering and physics quite a bit, where the main
thrust is on obtaining approximations by using calculus
methods, not on proving calculus methods by a rigorous
analysis of approximations. It's the third method that
was new to me, and which I thought might be new to quite
a few others as well (even those with a physics or engineering
background).

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Dave L. Renfro

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