mm-709
===
Subject: Re: Mystery equations ;-)
> ....
> r3 = (b1 + b2)*(1 + r1 + r2),
> r4 = b1*(r2 + r1*r2) + b2*(1 + r1 + r2 + r1*r2)
> r5 = b2*(b1*(1 + r2 + r1 + r1*r2) + r1 + r1*r2)
> I would think that anyone should be able to easily solve
for b1 and b2
> in terms of r1..r5 using any two of the three equations
above, but that
> does not appear to be possible....
Yes, itÕs possible over any [CapitalThorn]eld. The 
[CapitalThorn]rst two equations are
linear, and the usual theory gives you either no solution (if
theyÕre
inconsistent) or a unique solution or more than one. The
third equation
is best left till last.
Writing the [CapitalThorn]rst two equations in matrix form gives
(1 + r1 + r2 1 + r1 + r2 ) (b1) (r3)
( ) ( ) = ( )
((1 + r1)r2 (1 + r1)(1 + r2)) (b2) (r4)
These are consistent, i.e. some solution exists, if and only
if
the rank of
(1 + r1 + r2 1 + r1 + r2 )
( )
((1 + r1)r2 (1 + r1)(1 + r2))
is equal to the rank of
(1 + r1 + r2 1 + r1 + r2 r3)
( )
((1 + r1)r2 (1 + r1)(1 + r2) r4)
But I shanÕt carry through the general case because of what
you
said later.
> .... for my particular
> application (factoring), all variables (b1,b2,r1..r5) are
boolean with
> values restricted to either 0 or 1....
forming the Boolean ring {0,1} which is in fact just the [CapitalThorn]eld
of order 2. The above theory then gives different results
depending on
the values of r1 and r2. I shanÕt spell out all the details,
but
solving and then substituting in your third equation leads to
the
following list of solutions.
Case r1 = 0, r2 = 0.
Consistency requires r5 = (r3 + r4)r4, in which case the
solution is b1 = r3 + r4, b2 = r4.
Case r1 = 0, r2 = 1.
Consistency requires r3 = r5 = 0, in which case the solution
is
b1 = r4, b2 arbitrary (either 0 or 1).
Case r1 = 1, r2 = 0.
Consistency requires r3 = r4 = 0, in which case the solution
is
b1 arbitrary (either 0 or 1), b2 = r5.
Case r1 = 1, r2 = 1.
Consistency requires r4 = r5 = 0, in which case the solution
is
b1 arbitrary (either 0 or 1), b2 = r3 + b1.
Because the consistency conditions have such a strong effect,
you
canÕt expect solutions expressed much more simply than all
that.
Ken Pledger.
===
Subject: Re: Mystery equations ;-)
> r3 = (b1 + b2)*(1 + r1 + r2),
> r4 = b1*(r2 + r1*r2) + b2*(1 + r1 + r2 + r1*r2)
> r5 = b2*(b1*(1 + r2 + r1 + r1*r2) + r1 + r1*r2)
<snip matrix stuff the Boolean ring {0,1} which is in fact just the [CapitalThorn]eld of
order 2.
> The above theory then gives different results depending on
the
> values of r1 and r2.
> Case r1 = 0, r2 = 0.
> Consistency requires r5 = (r3 + r4)r4, in which case the
> solution is b1 = r3 + r4, b2 = r4.
IÕm not following. Coming directly from the equations.
r3 = b1 + b2; r4 = b2; r5 = b1 b2
ok, itÕs all the same
b2 = r4; b1 = r3 + r4; r5 = (r3 + r4)r4
> r3 = (b1 + b2)*(1 + r1 + r2),
> r4 = b1*(r2 + r1*r2) + b2*(1 + r1 + r2 + r1*r2)
> r5 = b2*(b1*(1 + r2 + r1 + r1*r2) + r1 + r1*r2)
> Case r1 = 0, r2 = 1.
> Consistency requires r3 = r5 = 0, in which case the
solution is
> b1 = r4, b2 arbitrary (either 0 or 1).
r3 = 0; r4 = b1; r5 = 0
> Case r1 = 1, r2 = 0.
> Consistency requires r3 = r4 = 0, in which case the
solution is
> b1 arbitrary (either 0 or 1), b2 = r5.
r3 = 0; r4 = 0; r5 = b2
> Case r1 = 1, r2 = 1.
> Consistency requires r4 = r5 = 0, in which case the
solution is
> b1 arbitrary (either 0 or 1), b2 = r3 + b1.
r3 = b1 + b2; r4 = 0 = r5; b1 = r3 + b2 (making b2 the
parameter)
> Because the consistency conditions have such a strong
effect, you
> canÕt expect solutions expressed much more simply than all
that.
Indeed, for what sets of r1, r2, r3, r4, r5 are there no
solutions.
===
Subject: Re: Mystery equations ;-)
> You need to make clear whether youÕre working in the 
Boolean
> _ring_ {0,1} in which 1 + 1 = 0, or the Boolean _algebra_
{0,1} in
> which 1 + 1 = 1.
Ok sure. IÕm working in the Boolean _ring_ in which 1 + 1 =
0. Having
been a computer programmer all my life, I wasnÕt even aware
that a
Boolean algebra _exists_ in which 1 + 1 = 1. Ha! I assume the
former
uses + to mean exclusive-or whereas the latter uses + to mean
inclusive-or?
-- Ron
===
Subject: Re: Mystery equations ;-)
> So does (1 + r1 + r2) mean (1 OR r1 OR r2)? In which case
it always
> evaluates to 1; Or for r1=1 and r2=2 does it mean (1+1+1)=3?
I already mentioned that the multiplication operator in my
particular
case is in fact the logical AND operator, but I didnÕt
mention that
the plus operator is actuallythen exclusive-or XOR operator.
So in
my _particular_ case, (1 + r1 + r2) means (1 XOR r1 XOR r2).
The fact
that I happen to be working with binary values however, is
totally
unrelated to my central question of not being able to solve
for b1 and
b2 in terms of the constants r1 - r5 (even if b1,b2,r1..r5
were real
numbers rather than binary numbers). Maybe I should have
phrased my
question as follows:
Why isnÕt it possible to solve for x and y in terms of the
constants c1
- c5 in the following set of three (overdetermined) equations:
c3 = (x + y)*(1 + c1 + c2),
c4 = x*(c2 + c1*c2) + y*(1 + c1 + c2 + c1*c2)
c5 = y*(x*(1 + c2 + c1 + c1*c2) + c1 + c1*c2)
-- Ron
===
Subject: Re: Mystery equations ;-)
> So does (1 + r1 + r2) mean (1 OR r1 OR r2)? In which case
it always
> evaluates to 1; Or for r1=1 and r2=2 does it mean
(1+1+1)=3?
> I already mentioned that the multiplication operator in my
particular
> case is in fact the logical AND operator, but I didnÕt
mention that
> the plus operator is actuallythen exclusive-or XOR
operator. So in
> my _particular_ case, (1 + r1 + r2) means (1 XOR r1 XOR
r2). The fact
> that I happen to be working with binary values however, is
totally
> unrelated to my central question of not being able to solve
for b1 and
> b2 in terms of the constants r1 - r5 (even if b1,b2,r1..r5
were real
> numbers rather than binary numbers). Maybe I should have
phrased my
> question as follows:
> Why isnÕt it possible to solve for x and y in terms of the
constants c1
> - c5 in the following set of three (overdetermined)
equations:
> c3 = (x + y)*(1 + c1 + c2),
> c4 = x*(c2 + c1*c2) + y*(1 + c1 + c2 + c1*c2)
> c5 = y*(x*(1 + c2 + c1 + c1*c2) + c1 + c1*c2)
> -- Ron
Dumping the third equation gives two linear equations in two
unknowns
which will have a unique solution unless (1 + c1 + c2) = 0 or
(1 + c1) = 0.
If that unique solution exists, it may be incompatible with
any solution
to the third equation, and if it does not exist, there is
most likely no
solution at all to the [CapitalThorn]rst two equations by themselves, much
less one
which will [CapitalThorn]t the third equation.
===
Subject: Re: Mystery equations ;-)
> You have 3-equations in 5-Unknowns according your statement
(R1,
> R2, R3, R4, R5).
> HOW do you isolate a solution to R1 - R5?
Please excuse me for failing to point out that r1 - r5 are
known
constants. For example, in one particular case they are:
r1=1
r2=1
r3=1
r4=0
r5=0
-- Ron
===
Subject: Re: Mystery equations ;-)
P.S.
As I said in my original question, I want to solve for b1 and
b2 in
terms of r1..r5.
===
Subject: S_3 is not decomposible
I would like to know following solution of S_3 is not
decomposible
is correct or not.
sol>
Since |S_3| = 6 ,
If S_3 is isomorphic to H x K (H,K are subgroup of G) and
H=/={1}=/=K,
then W.L.O.G it must be H=<(12)>, K=<(123)> or
H=<(23)>,K=<(123)> or
H=<(13)>, K=<(123)>.
Now, in all above three cases,
since H x K ~= C_6 (cyclic group of order 6) ,
H x K is cyclic group but S_3 is not a cyclic group.
therefore, S_3 has no subgroups H,K satisfying S_3 ~= H x K,
H=/={1}=/=K.
=> so, S_3 is not decomposible.
I read this solution at my book but, I think this solution is
false
because above H is not a normal subgroup of G.
I know the de[CapitalThorn]nition of decomposible group as follows :
Any group G is decomposible iff there are normal subgroups
H,K of G
satisfying
G~= H x K , |H|=/=1=/=|K|
Is my thinking false ?
===
Subject: S_3 is not decomposible
I would like to know following solution of S_3 is not
decomposible
is correct or not.
sol>
Since |S_3| = 6 ,
If S_3 is isomorphic to H x K (H,K are subgroup of G) and
H=/={1}=/=K,
then W.L.O.G it must be H=<(12)>, K=<(123)> or
H=<(23)>,K=<(123)> or
H=<(13)>, K=<(123)>.
Now, in all above three cases,
since H x K ~= C_6 (cyclic group of order 6) ,
H x K is cyclic group but S_3 is not a cyclic group.
therefore, S_3 has no subgroups H,K satisfying S_3 ~= H x K,
H=/={1}=/=K.
=> so, S_3 is not decomposible.
I read this solution at my book but, I think this solution is
false
because above H is not a normal subgroup of G.
I know the de[CapitalThorn]nition of decomposible group as follows :
Any group G is decomposible iff there are normal subgroups
H,K of G
satisfying
G~= H x K , |H|=/=1=/=|K|
Is my thinking false ?
===
Subject: Re: Algebraic solution to the following polynominal
equation?
> That is, t^a + t^b - x = 0.
Better to write
t^a + t^b = c
as commonly x is a variable and c is a constant.
> Is it logically possible to solve this problem
algebraically? Or
> should I conduct numerical simulations where x, c and d are
> numerically speci[CapitalThorn]ed?
No, use numerical method. There is however, a special
function,
(the Lambert W) more complicated than log thatÕs the 
solution
for
x e^x = c
which perhaps you could manipulate your equation into.
===
Subject: Re: Why do we teach curve sketching?
Excellent. I particularly like the words think for yourself, a
facility
which is all to often ignored nowadays.
> Here are couple of answer I can give to students who depend
too heavily
on
> graphic calculators.
> 1) I am here to teach you how to think for yourself, not to
use the
> machine. Graphing a graph and knowing what you graphed is
corrected are
two
> different things.
> Then you proceed to ask them graph something that a graphic
calculator
canÕt
> handle at all like
> f(x)= x where x is rational and zero everywhere
> or
> f(x)=sin(x)/x
> or the impulse function which is de[CapitalThorn]ned as a rectangle
whose area is
> de[CapitalThorn]ned as x * 1/x where x is the base and 1/x is the 
height.
> 2) Tell you student that their graphic calculator wonÕt be
with they
always
> but their brain will.
> Ben
===
Subject: Plot uncertainty?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5TC3bM25192;
the form of ellipses, from a given covariance matrix.
However, i canÕt
[CapitalThorn]nd layman information on it, can anyone guide me?
Other than drawing ellipse, can the above be visualised in
other
terms? Eg: volume based or simple 1D quantity based, so to be
plotted
as charts instead.
rdgs
===
Subject: Abstract/Modern Algebra
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i5TC3cM25249;
Does anyone know of a online/ distance learning course in
Abstract/Modern Algebra?