Re: Long division

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.

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One way that you can do this, a way that is often 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).

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What you're 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 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, y/x - (delta 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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Dave L. Renfro

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