Partial Factor Decomposion
G Patel
When decomposing a proper rational function into simplest partial
factors, we set the rational function equal to the simplest factors
with unknown coefficients in the numerator of the factors.
Example:
1 / (x^2 -9) = A/(x-3) + B/(x+3) [x <> 3, -3]
Then we multiply through by (x-3)(x+3):
1 = A(x+3) + B(x-3)
This equation is only valid for all x not equal to 3 or -3, because the
equation/identity it came from had this restriction (we add the
extraneous 3, and -3 to the solution set of the identity by multiplying
by (x-3)(x+3), but it really shouldn't be there).
Hereon, one method to find A/B is to substitute key values for x into
the last equation.
x = 3 :: 1 = 6A therefore A = 1/6
BUTTT, x can not equal 3, because of the restriction (happens to cause
the original function to be undefined). So why does this method ask
for us to set x to these key numbers, which happened not to be in
domain of our original function?
What am I missing?
tanQ
Gerry Myerson
"G Patel" <gaya....@gmail.com> wrote:
Continuity.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
William Elliot
The method works for
1 = A(x+3) + B(x-3)
which you are trying to solve.
Otherwise, be old fashion ringing R[x]
a + b = 0
3a - 3b = 1
b = -a
3a + 3a = 1
a = 1/6, b = -1/6
G Patel
The original rational function has discontinuities at 3 and -3. Can
you explain what you mean? I know you are giving me a hint, but I fear
I can't take it over from there.
G Patel
Ringing x?
> a + b = 0
> 3a - 3b = 1
> b = -a
> 3a + 3a = 1
> a = 1/6, b = -1/6
Yes, this is the other method of equating coefficients. But why does
the other method work even though we are substituting in x values that
cause the original rational function to be undefined? This is just an
extra question I have after learning to mechanically do the
decomposition.
David Moran
"G Patel" <gaya....@gmail.com> wrote in message
news:1141964434....@j52g2000cwj.googlegroups.com...
Because once you multiply both sides by the LCD, you don't have any
fractions so you can use whatever values you want.
Dave
Alex. Lupas
==========
Suppose that we want to solve following problems :
[A] To determine real numbers A and B such that
(1) 1= A(x+3) +B(x-3) for all x in R .
[B] To determine real numbers A and B such that
(2) 1 =A(x+3)+B(x-3) forall x in M .
It's clear that if [A] is solved, then also [B] is solved.
To solve [A] put x= -3 and then x=3 , in (1) .
It follows A =1/6 , B= -1/6.
Because
(1.1) 1= (x+3)/6 -(x-3)/6 forall real x
we have also (select M:= R\{-3,3} )
(2.1) 1= (x+3)/6 -(x-3)/6 forall x in R\{-3,3} .
Further, divide (2.1) by (x+3)(x-3) .....perhaps help .
Bill Dubuque
> "G Patel" <gaya....@gmail.com> wrote:
>>
>> When decomposing a proper rational function into simplest partial
>> factors, we set the rational function equal to the simplest factors
>> with unknown coefficients in the numerator of the factors.
>>
>>
>> Then we multiply through by (x-3)(x+3):
>>
>> 1 = A(x+3) + B(x-3)
>>
>> This equation is only valid for all x not equal to 3 or -3, because the
>> equation/identity it came from had this restriction (we add the
>> extraneous 3, and -3 to the solution set of the identity by multiplying
>> by (x-3)(x+3), but it really shouldn't be there).
>>
>> Hereon, one method to find A/B is to substitute key values for x into
>> the last equation.
>>
>> x = 3 :: 1 = 6A therefore A = 1/6
>>
>> BUTTT, x can not equal 3, because of the restriction (happens to cause
>> the original function to be undefined). So why does this method ask
>> for us to set x to these key numbers, which happened not to be in
>> domain of our original function?
>>
>> What am I missing?
>
> Continuity.
SIMPLER If A(x+3)+B(x-3)-1 were nonzero at only finitely
many rationals, then it would be a nonzero polynomial with
infinitely many rational roots. But a nonzero polynomial
over a field (or domain) has no more roots than its degree.
Here one knows *a priori* that the solutions A,B must be
polynomials, so one can appeal to the properties of such.
Similarly other problems elegantly succumb to universal
properties of polynomials, for example see my prior posts
http://google.com/groups?selm=y8zwtm6mnpv.fsf%40nestle.csail.mit.edu
http://google.com/groups?selm=y8zznru8v1d.fsf%40nestle.ai.mit.edu
--Bill Dubuque
Brian VanPelt
wrote:
>Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>> "G Patel" <gaya....@gmail.com> wrote:
>>>
>>> When decomposing a proper rational function into simplest partial
>>> factors, we set the rational function equal to the simplest factors
>>> with unknown coefficients in the numerator of the factors.
>>>
>>> Example: 1/(x^2 -9) = A/(x-3) + B/(x+3) [x <> 3, -3]
>>>
>>> Then we multiply through by (x-3)(x+3):
>>>
>>> 1 = A(x+3) + B(x-3)
>>>
>>> This equation is only valid for all x not equal to 3 or -3, because the
>>> equation/identity it came from had this restriction (we add the
>>> extraneous 3, and -3 to the solution set of the identity by multiplying
>>> by (x-3)(x+3), but it really shouldn't be there).
>>>
>>> Hereon, one method to find A/B is to substitute key values for x into
>>> the last equation.
>>>
>>> x = 3 :: 1 = 6A therefore A = 1/6
>>>
>>> BUTTT, x can not equal 3, because of the restriction (happens to cause
>>> the original function to be undefined). So why does this method ask
>>> for us to set x to these key numbers, which happened not to be in
>>> domain of our original function?
>>>
>>> What am I missing?
>>
There is a second way to do these problems that also appeal to the use
of polynomials and their properties.
You have
1 = A ( x + 3 ) + B ( x - 3 )
OR
1 = ( A + B ) x + 3A - 3B
By comparing the coefficients on each side of the equation, you can
determine the values of A and B. Just to be more straightforward, we
have
0 x + 1 = ( A + B ) x + 3A - 3B
which yeilds the equations
0 = A + B
1 = 3A - 3B
In this situation, you are required to solve a system of equations
A + B = 0
3A - 3B = 1
OR
3A + 3B = 0
3A - 3B = 1
Add the two equations to get
6A = 1 ---> A = 1/6
and then plug that value back in to any of the previous two equations
to get the value of B.
Brian
Gerry Myerson
"G Patel" <gaya....@gmail.com> wrote:
A and B depend continuously on x.
William Elliot
> William Elliot wrote:
> > On Thu, 9 Mar 2006, G Patel wrote:
> >
> > > Example:
> > > 1 / (x^2 -9) = A/(x-3) + B/(x+3) [x <> 3, -3]
> > >
> > > Then we multiply through by (x-3)(x+3):
> > > 1 = A(x+3) + B(x-3)
> > >
f(x) = a(x + 3) + b(x - 3)
defined for all x /= 3,-3. Continuously extend that function to
g over all the reals. Now you can solve g(x) = 1, for a,b using
x = 3,-3. Checking the answer is an easy matter of formal algebra
of rational functions R(x)
> > > This equation is only valid for all x not equal to 3 or -3, because the
> > > equation/identity it came from had this restriction (we add the
> > > extraneous 3, and -3 to the solution set of the identity by multiplying
> > > by (x-3)(x+3), but it really shouldn't be there).
> > >
> > > Hereon, one method to find A/B is to substitute key values for x into
> > > the last equation.
> > >
> > > x = 3 :: 1 = 6A therefore A = 1/6
> > >
> > > BUTTT, x can not equal 3, because of the restriction (happens to cause
> > > the original function to be undefined). So why does this method ask
> > > for us to set x to these key numbers, which happened not to be in
> > > domain of our original function?
> > >
> > > What am I missing?
> > >
> > The method works for
> > 1 = A(x+3) + B(x-3)
> > which you are trying to solve.
> >
> > Otherwise, be old fashion ringing R[x]
> Ringing x?
>
R[x] is a ring. The below comes from working within the ring R[x], using
(a + b)x + 3(a - b) = 1 = 1 + 0x
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>>>"G Patel" <gaya....@gmail.com> wrote:
>>>>
>>>> When decomposing a proper rational function into simplest partial
>>>> factors, we set the rational function equal to the simplest factors
>>>> with unknown coefficients in the numerator of the factors.
>>>>
>>>> Example: 1/(x^2 -9) = A/(x-3) + B/(x+3) [x <> 3, -3]
>>>>
>>>> Then we multiply through by (x-3)(x+3):
>>>>
>>>> 1 = A(x+3) + B(x-3)
>>>>
>>>> This equation is only valid for all x not equal to 3 or -3, because the
>>>> equation/identity it came from had this restriction (we add the
>>>> extraneous 3, and -3 to the solution set of the identity by multiplying
>>>> by (x-3)(x+3), but it really shouldn't be there).
>>>>
>>>> Hereon, one method to find A/B is to substitute key values for x into
>>>> the last equation.
>>>>
>>>> x = 3 :: 1 = 6A therefore A = 1/6
>>>>
>>>> BUTTT, x can not equal 3, because of the restriction (happens to cause
>>>> the original function to be undefined). So why does this method ask
>>>> for us to set x to these key numbers, which happened not to be in
>>>> domain of our original function?
>>>>
>>>> What am I missing?
>>>
>>
>> SIMPLER If A(x+3)+B(x-3)-1 were nonzero at only finitely
>> many rationals, then it would be a nonzero polynomial with
>> infinitely many rational roots. But a nonzero polynomial
>> over a field (or domain) has no more roots than its degree.
>>
>> polynomials, so one can appeal to the properties of such.
>> Similarly other problems elegantly succumb to universal
>> properties of polynomials, for example see my prior posts
>> http://google.com/groups?selm=y8zwtm6mnpv.fsf%40nestle.csail.mit.edu
>> http://google.com/groups?selm=y8zznru8v1d.fsf%40nestle.ai.mit.edu
>
> to the use of polynomials and their properties.
>
> You have 1 = A ( x + 3 ) + B ( x - 3 )
>
> OR 1 = ( A + B ) x + 3A - 3B
>
> By comparing the coefficients on each side of the equation,
But you haven't justified why one can compare coefficients.
The heart of the problem stems from a confusion between
formal objects and functional objects. One shouldn't view
A/(x-3) as a function, but instead as a quotient of formal
polynomials, i.e. as an elt of the field Q(x). Therefore
1/(xx-9) = A/(x-3) + B/(x+3) in Q(x)
<-> 1 = A (x+3) + B (x-3) in Q(x)
is true simply because multiplying an equation by a unit
always produces an equivalent equation (over any ring).
For less trivial examples see my prior posts linked above.
Here are some other examples of function vs. form confusion
http://google.com/groups?selm=y8zaccqdqcj.fsf%40nestle.csail.mit.edu
http://google.com/groups?&q=author%3Adubuque+formal+polynomial
--Bill Dubuque
G Patel
I thought it would be a => relationship since we're multiply the first
equation by (x-3)(x+3), which is not a reversible operation.
>
> is true simply because multiplying an equation by a unit
> always produces an equivalent equation (over any ring).
>
So the second equation is not redundant (having a greater solution set)
relative to the first?
>
> For less trivial examples see my prior posts linked above.
> Here are some other examples of function vs. form confusion
> http://google.com/groups?selm=y8zaccqdqcj.fsf%40nestle.csail.mit.edu
> http://google.com/groups?&q=author%3Adubuque+formal+polynomial
>
Ok thanks.
The Qurqirish Dragon
After multiplying, you have removable discontinuities at the points
where the original function was undefined (which is where your question
comes from, if I understand you correctly). However, since at this
point we are working with a polynomial (rather than the original
rational function), we can remove the discontinuities. If you really
want to avoid doing this, consider the following limit process instead:
You have the equation:
1=A(x+3) + B(x-3); x=/= 3,-3
Take the limits as x approaches +/- 3. Since the discontinuities are
removable, these limits exist. so, for example, if x=3+eps, we have:
1=A(6+eps)+B(eps), which in the limit eps->0 obviously yields 1=6A
(which is, of course, what substituting 3 in for x originally would
have done)
Similar reasoning for x=3-eps establishes the left-side limit, if you
are worried about 1-sided limits.
Virgil
"Alex. Lupas" <alex....@gmail.com> wrote:
> If R:=(-infty,infty) let M be a subset of R .
> ==========
> Suppose that we want to solve following problems :
>
> [A] To determine real numbers A and B such that
>
> (1) 1= A(x+3) +B(x-3) for all x in R .
>
> [B] To determine real numbers A and B such that
>
> (2) 1 =A(x+3)+B(x-3) forall x in M .
If M contains at least two different values, say u and v, then one can
find A and B by considering the system of simultaneous equations
1 =A(u+3)+B(u-3) and
1 =A(v+3)+B(v-3)
with A and B as unknowns
Peter L. Montgomery
If you pick the right A and B, then
1 = A(x+3) + B(x-3)
will hold for all real x except possibly x = 3 and x = -3. The polynomial
g(x) = 1 - A(x+3) - B(x-3)
has infinitely many roots x. This is possible only
if g is identically zero. In particular, g(3) = g(-3) = 0.
A polynomial identity (in one variable) which holds for
infinitely many values of the variable must hold everywhere.
[I assume the polynomial is over a field -- if you don't
know what a field is, ignore this comment.]
--
VP Cheney Burr-ed his gun as a bird flew past The nation responds "burr"
as we await bird flu shots and fight a real cold war.
pmon...@cwi.nl Microsoft Research and CWI Home: Bellevue, WA
Brian VanPelt
<qurqi...@aol.com> wrote:
This, to me, is the closest explanation that I'd give to calc II
students as to why plugging in 3 and -3 produce the values for A and
B.
You have to remember that using words like ring, field, unit, etc. are
beyond the topics covered in the course the OP is taking. While I
(we) understand what a polynomial in an indeterminate X is, there is
almost no possibility that a calc student will understand that.
By the way, in my earlier post, I didn't try to give reasons why what
I did was correct, because I knew it was beyond what the OP has most
likely had in his course of study. It was just offered as an
alternative to the methods that were given. In fact, you can use the
limit idea to explain that technique, in a reasonable fashion, to calc
II students.
By the way, the word formal, and the word form, are typically beyond
calc II students as well.
Just my 2 cents.
Brian
G Patel
You are right. I'm definitely Calc 2 level (actually, not even that -
self study). The explanations were good and I thank you all for
helping. Now I can use the techniques with some understanding of why
it works, instead of just memorizing the computational steps.
William Elliot
> You are right. I'm definitely Calc 2 level (actually, not even that -
> self study). The explanations were good and I thank you all for
> helping. Now I can use the techniques with some understanding of why
> it works, instead of just memorizing the computational steps.
>
Thank you for wanting to understand math instead of
accepting the notion that math is algorithmic rules.