Long division
Quentin Grady
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.
Thank you,
Best wishes,
--
Quentin Grady ^ ^ /
New Zealand, >#,#< [
/ \ /\
"... and the blind dog was leading."
user923005
Consider (for instance) that all the math we do is essentialy with
polynomials operating against polynomials.
e.g.:
1234 * 5678
That could also be considered as
(1000 * 1 + 100 * 2 + 10 * 3 + 4) * (1000 * 5 + 100 * 6 + 10 * 7 + 8)
Of course, that grouping is arbitrary. We can collect the terms
however we like and still get the same result.
If you give a pointer to the proof in question, I guess that someone
can help you to see it clearly.
Virgil
Quentin Grady <que...@paradise.net.nz> wrote:
> G'day G'day Folks,
>
> 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.
>
> Thank you,
>
> Best wishes,
Perhaps if you could reproduce a few lines of that long division, we
might have a better idea of what is going on.
Dave L. Renfro
> 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.
-----------------------------------------------------------------
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).
-----------------------------------------------------------------
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.
-----------------------------------------------------------------
Dave L. Renfro
Quentin Grady
On 28 Feb 2007 10:10:51 -0800, "Dave L. Renfro" <renf...@cmich.edu>
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.
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.
Quentin Grady
wrote:
>Perhaps if you could reproduce a few lines of that long division, we
>might have a better idea of what is going on
G'day G'day Virgil,
Thanks for the offer. Dave has already helped me out. Perhaps
there is a fifth approach yet to be considered.
Quentin Grady
wrote:
>Consider (for instance) that all the math we do is essentialy with
>polynomials operating against polynomials.
>
>e.g.:
>1234 * 5678
>That could also be considered as
>(1000 * 1 + 100 * 2 + 10 * 3 + 4) * (1000 * 5 + 100 * 6 + 10 * 7 + 8)
>Of course, that grouping is arbitrary. We can collect the terms
>however we like and still get the same result.
G'day G'day,
Thank you for this insight.
>If you give a pointer to the proof in question, I guess that someone
>can help you to see it clearly.
Take a look at Dave's reply. This helped me considerably. I wouldn't
be surprised to find there are other approaches.
Han de Bruijn
> Take a look at Dave's reply. This helped me considerably. I wouldn't
> be surprised to find there are other approaches.
At a somewhat higher level, did you try Logarithmic Differentiation?
ln(u/v) = ln(u) - ln(v)
[ ln(u/v) ]' = [ ln(u) ]' - [ ln(v) ]'
(u/v)' / (u/v) = u'/u - v'/v
(u/v)' = ( u'.v - v'.u ) / v^2
Han de Bruijn
Virgil
Nice! It avoids all that hassle with limits.
It also works for the product rule, at least when u > 0 and v > 0.
ln(u*v) = ln(u) + ln(v)
(u*v)'/(u*v) = u'/u + v'/v
(u*v)' = u'*v + u*v'.
Lee Rudolph
>In article <3f127$45e68d1b$82a1e228$30...@news1.tudelft.nl>,
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> wrote:
>
>> Quentin Grady wrote:
>>
>> > Take a look at Dave's reply. This helped me considerably. I wouldn't
>> > be surprised to find there are other approaches.
>>
>> At a somewhat higher level, did you try Logarithmic Differentiation?
>>
>> ln(u/v) = ln(u) - ln(v)
>>
>> [ ln(u/v) ]' = [ ln(u) ]' - [ ln(v) ]'
>>
>> (u/v)' / (u/v) = u'/u - v'/v
>>
>> (u/v)' = ( u'.v - v'.u ) / v^2
>>
>> Han de Bruijn
>
>Nice! It avoids all that hassle with limits.
Not to mention all that hassle with functions u and v that take on
non-positive values somewhere. Down with pessimism!
Lee Rudolph
Quentin Grady
<Han.de...@DTO.TUDelft.NL> wrote:
>Quentin Grady wrote:
>
>> Take a look at Dave's reply. This helped me considerably. I wouldn't
>> be surprised to find there are other approaches.
>
>At a somewhat higher level, did you try Logarithmic Differentiation?
G'day G'day Han,
I hope that is the correct greeting.
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.
When I said I wouldn't be surprised to find there are other approaches
I did so without thinking it likely there would be any within my
pretty basic math knowledge. Seeing the use of conjugates and now
natural logs (both things I've met) has been most enlightening. My
maths had in the past been largely restricted to the pragmatic needs
of electrical students taking a certificate course. Now I get do
maths for the joy and relaxation it brings me.
>ln(u/v) = ln(u) - ln(v)
>
>[ ln(u/v) ]' = [ ln(u) ]' - [ ln(v) ]'
>
>(u/v)' / (u/v) = u'/u - v'/v
>
>(u/v)' = ( u'.v - v'.u ) / v^2
>
>Han de Bruijn
Best wishes,
Quentin Grady
wrote:
>Nice! It avoids all that hassle with limits.
>
>It also works for the product rule, at least when u > 0 and v > 0.
>
>ln(u*v) = ln(u) + ln(v)
>
>(u*v)'/(u*v) = u'/u + v'/v
>
>(u*v)' = u'*v + u*v'.
G'day G'day Virgil,
Thank you. I'm delighted that such neat proofs flow from a asking
such "simple" question.
Dave L. Renfro
> 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).
**************************************************************
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.
**************************************************************
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).
**************************************************************
Dave L. Renfro
Quentin Grady
wrote:
>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.
G'day G'day Dave,
At the time I wasn't interested in the division of polynomials.
Method 3 dealt with exactly what I wanted to achieve. However, after
having worked through Method 3, I did take a look at converting octal
based numbers to denary using a nested format for the base eight
polynomial. eg 372(8) = 3*8^2 + 7*8 + 2
Using nested format for evaluation,
3*8 = 24
24+7 = 31
31*8 = 248
248+2= 250 (10)
That lead me on to the remainder theorem, the factor theorem and
surprise, surprise, the division of polynomials.
To put it another way, I didn't intend to perform any division of
polynomials but ended up there anyway on my latest maths meander.
Best wishes and thank you,