Re: Long division
- From: Quentin Grady <quentin@xxxxxxxxxxxxxxx>
- Date: Thu, 01 Mar 2007 15:38:11 +1300
This post not CC'd by email
On 28 Feb 2007 10:10:51 -0800, "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
wrote:
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.
G'day G'day Dave,
Thanks. The book I saw the proof of the quotient rule performed by
division was originally published in 1925. The proofs I was familiar
with were the rearrangement of the product rule and one involving a
reciprocal.
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).
This is the first time I've seen the proof by rationalizing the
denominator, though I'm familiar with the concept of rationalizing the
denominator from doing simple complex number calculations.
Thank you illustrating it.
-----------------------------------------------------------------
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.
Thank you. That is exactly what I was looking for. At first it had
seemed impossible to do the long division with only unknown symbols eg
a, b, c and d. or U + deltaU and V + deltaV which fit the actual
proof I desired to perform. I'd used a, b and c, d to make the ASCII
easier.
-----------------------------------------------------------------
Dave L. Renfro
Thank you Dave. There is such a difference between having A proof of
something like the product rule for differentiation and have four or
perhaps more so that one approach it from different starting points.
Best wishes,
--
Quentin Grady ^ ^ /
New Zealand, >#,#< [
/ \ /\
"... and the blind dog was leading."
http://homepages.paradise.net.nz/quentin
.
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