mm-1719
 
===
Subject: Re: Standard symbol for the group of multiplicative characters?
> Is there a common symbol for the group of multiplicative characters of a
> certain group?  (In particular, I'm considering the group of
> multiplicative characters of Z_p.)
The (multiplicative) group of units of a ring R is notated R^*
or R^x tho the former preferable.  (Z_p)^* = (Z_p)0 when p prime.
(Z_4)^* = { 1,3 }
===
Subject: Re: Standard symbol for the group of multiplicative characters?
> Is there a common symbol for the group of multiplicative characters of 
a
> certain group?  (In particular, I'm considering the group of
> multiplicative characters of Z_p.)
> The (multiplicative) group of units of a ring R is notated R^*
> or R^x tho the former preferable.  (Z_p)^* = (Z_p)0 when p prime.
> (Z_4)^* = { 1,3 }
Yes, but OP didn't ask about the units of a ring, but about the 
characters of the ring - these are the homomorphisms into the 
complex numbers.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Standard symbol for the group of multiplicative characters?
>Is there a common symbol for the group of multiplicative characters of a
>certain group?  (In particular, I'm considering the group of
>multiplicative characters of Z_p.)
>>The (multiplicative) group of units of a ring R is notated R^*
>>or R^x tho the former preferable.  (Z_p)^* = (Z_p)0 when p prime.
>>(Z_4)^* = { 1,3 }
> Yes, but OP didn't ask about the units of a ring, but about the 
> characters of the ring - these are the homomorphisms into the 
> complex numbers.
Short answer:
IIUYC
You can use '(+)' and '(.)' respectively to represent the additive and 
multiplicative groups/axiom sets, that comprise, (along with the data 
elements, of course) a ring. [per the Penguin dictionary of Mathematics]
In books I've seen plus sign enclosed in hollow circle for '(+)' and X 
enclosed in a hollow circle--such as '(x) or (X)?'--for '(.)'
I don't know how one would render these typographically in a code set 
right?
*IIUYC If I Understand You Correctly
Ignore the blather that follows...
Boy, it sure is difficult to answer the simple questions, isn't it?
I know you know all this already, sorry, I'm organizing my thoughts here.
Relation := maps an n-tuple (ordered set) of elements to a value
            The value can, of course, be another n-tuple.
# could read these here as
# = -- is a
# + -- has a
# '{' -- collection of
Ring = {
   Elements  = set { number_system | chars }
   + A = group = relation + set { axioms }
   + M = group = relation + set { axioms }
}
The two groups that comprise a ring (among the instances of rings being 
those we know so fondly as C, R, Z), that is, the two collections A plus 
sign enclosed in a hollow circle is frequently used (Gullberg, Dubisch, 
etc.) to denote that generic/abstract operation (the group, that is) 
that would correspond in the specific instance of C, R, Z, whatever, to 
the additive properties of that ring. This lets you roll your own 
operation (group) or to refer in conversation to the set of properties 
without specifically talking about C's mapping this to an operation with 
an arbitrary sequence of characters, and X enclosed in a hollow circle 
to denote the operation (a group) that corresponds to mapping the 
multiplicative properties to arbitrary sequences of characters, numbers, 
what-have-you. Just as you say, homomorphisms of C's or R's additive and 
multiplicative, respectively, groups. Is that what you're referring to?
I've always thought that with abstract algebra we are heirs to an undue 
emphasis on the data, at the expense of the flexibility we'd have if we 
de-coupled operations from what they act on. type-less and type-ful 
seeming to be preferences that are more or less useful depending on the 
context. but I'm probably on crack, and overlooking majestic elegancies 
of axiomatic structure.
SL
===
Subject: Re: Standard symbol for the group of multiplicative characters?
> Is there a common symbol for the group of multiplicative characters of a 
> certain group?  (In particular, I'm considering the group of 
> multiplicative characters of Z_p.)
> By multiplicative character you mean a map h : Z_p -> C,
> the set of complex numbers, with h(x+y) = h(x)+h(y) and
> h(xy) = h(x)h(y) ?
> Or, if not that, then what?
I think multiplicative character just means h(xy) = h(x) h(y). 
You could call it the dual group of G and denote it G-dagger.
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: bus route
A bus route from town A to town D goes along a straight line and passes 
first through town B and then through town C. Jason has a positioning device 

that can tell him the distance from his current location to any of the four 

towns on the route. When he is somewhere between A and B, from the 
information supplied by the device Jason learns that the sum of the 
distances 
to A, to B and to D is greater than the distance to C by 5.9 km. When Jason 

is between B and C, he finds out that the sum of the distances to all four 
towns is 16.7 km. When he is between C and D, Jason discovers that the sum 
of 
the distances to B and to D is 8.1 km. What is the distance between C and D? 

This is what i think but im not sure if its right:
X is the point between A and B; Y between B and C; Z between C and D 
at X: AX + XB + XD = XC + 5.9 
or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9 
or AB + CD = 5.9 
at Y: AY + BY + YC + YD = 16.7 
or (AB + BY) + BC + (YC + CD) = 16.7 
or AB + 2xBC + CD= 16.7 
or with the first point: 2xBC = 16.7 - 5.9 
or BC = 5.4 
at Z: BZ + ZD = 8.1 
or (BC + CZ) + ZD = 8.1 
or BC + CD = 8.1 
with the above CD= 8.1 - 5.7 = 2.7 
this would make AB= 5.9 - 2.7 = 3.2
===
Subject: Re: bus route
>A bus route from town A to town D goes along a straight line and
>passes first through town B and then through town C. Jason has a
>positioning device that can tell him the distance from his current
>location to any of the four towns on the route.
Long lines are really hard to read; I've reformatted them for you.  
I've also made a sketch:
A--------X-----B----Y---------C----------Z----D
>When he is somewhere between A and B, from the information supplied 
>by the device Jason learns that the sum of the distances to A, to B 
>and to D is greater than the distance to C by 5.9 km.
>at X: AX + XB + XD = XC + 5.9 
>or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9 
>or AB + CD = 5.9 
Looks right so far. I think you've got the sense of the problem: 
that the distances from X need to be rewritten as distances between 
the points.
>When Jason is between B and C, he finds out that the sum of the 
>distances to all four towns is 16.7 km.
>at Y: AY + BY + YC + YD = 16.7 
 AY = AB+BY and BY+YC = BC and YD=YC+CD, and therefore as you said:
>or (AB + BY) + BC + (YC + CD) = 16.7 
Intermediate step that I took a while to see:
      AB + BY + BC + (BC-BY) + CD = 16.7
>or AB + 2xBC + CD= 16.7 
>or with the first point: 2xBC = 16.7 - 5.9 
>or BC = 5.4 
I agree.
>When he is between C and D, Jason discovers that the sum of the
>distances to B and to D is 8.1 km.
>at Z: BZ + ZD = 8.1 
True -- or BD = 8.1; CD = BD-BC = 8.1-5.4 = 2.7 km
>or (BC + CZ) + ZD = 8.1 
>or BC + CD = 8.1 
>with the above CD= 8.1 - 5.7 = 2.7 
>What is the distance between C and D? 
>this would make AB= 5.9 - 2.7 = 3.2
What you've done looks fine to me. I saw a slightly easier way at 
the very end, but if there's a really easy alternative methol for 
the whole problem I sure don't see it.
-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/
===
Subject: Re: bus route
===
Subject: regular polygon exists for all n>2?
Just want to make sure if I have n nodes(n>2), there always exists an 
alignment in which all edges and angles are equivalent.
gim
===
Subject: Re: regular polygon exists for all n>2?
For any n > 2, you can divide the angle of 360 degrees into n equal parts.
Draw a circle. Draw at its center the n lines (or rather rays emanating from 

the center) that form n 360/n angles. These will intersect the circle at the 

vertices of a regular n-gon.
A. Bogomolny
http://www.cut-the-knot.org
===
Subject: Parametric curve intersections.
How would I find where r = 3sin(theta) and r = 1 + sin(theta) intersect?  I 

tried doing this the normal system way, but I don't think that finds all the 

answers in a parametric system, does it?
===
Subject: Re: Parametric curve intersections.
<30229374.1115126280485.JavaMail.jakarta@nitrogen.mathforum.org>,
> How would I find where r = 3sin(theta) and r = 1 + sin(theta) intersect?  

I
> tried doing this the normal system way, but I don't think that finds all 
the
> answers in a parametric system, does it?
It does if you are careful. I think you meant to tell us this was in
polar cordinates not parametric (although, of course, there _is_ a
parameter).
3sin(t) = 1 + sin(t) has _many_ solutions since sine is periodic. It
only gives two _distinct_ points on both (polar) curves.
-- 
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Parametric curve intersections.
===
Subject: Re: Parametric curve intersections.
Intersection <=> encounter <=> idem locus.
FOR CARTESIAN Curves, say f(x) and g(x) 
     a point x* for which f(x*)=g(x*) ,
     ->same abscissa, same ordinate.
FOR r THETA /POLAR Curves   I use t for theta ,
     a point t*  r1(t*) =r2(t*),
     ->same radius, same angle:
     here r1(t)=3*sin(t) r2(t)=1+sin(t) (r1 and r2 not r),
     try solving it .
    F=> abscissa m(t) ,ordinate n(t), a common value t
    G=> abscissa p(t) ,ordinate q(t), a common value t ;
    name t1 the m and n parameter ,
         t2 the p and q parameter ,
    we obtain system {m(t1)= p(t2) ,n(t1)=q(t2)}
              same abcissa  ,  same ordinate
Hope it helps,
Alain.
===
Subject: Re: Parametric curve intersections.
>FOR r THETA /POLAR Curves   I use t for theta ,
>     a point t*  r1(t*) =r2(t*),
>     ->same radius, same angle:
>     here r1(t)=3*sin(t) r2(t)=1+sin(t) (r1 and r2 not r),
>     try solving it .
That's true, I think  if  r is constrained to be >= 0. But 
I believe that r = 3 sin t will have negative r for half the values 
of t.
I agree with the basic approach of solving simultaneous equations 
as you suggest, but it's complicated by the fact that say
     (2,75Á) and (-2,255Á)
are the same point.
-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/
===
Subject: Re: Parametric curve intersections.
>FOR r THETA /POLAR Curves   I use t for theta ,
>     a point t*  r1(t*) =r2(t*),
>     ->same radius, same angle:
>     here r1(t)=3*sin(t) r2(t)=1+sin(t) (r1 and r2 not r),
>     try solving it .
But Susan is correct that the above method does not find all the
points where the graphs intersect. The problem is that there is more
than one way to represent points in polar coordinates, particularly
the origin. In her example when t = 0 or Pi, r1 = 0 so the first curve
passes through the origin, and when t = 3 Pi/2, r2 = 0 so the second
curve passes through the origin. To the notion of same r, same t
doesn't find that point.
--Lynn
===
Subject: Re: I really need your help-probably simple to solve
> hi
> Could this be done with the use of horner's algorithm
> for polynomials ?
> ___ a[n]___a [n-1]___...__a[0]___________
>    |
> c  |       cb[n-1]        cb[0]
> ___|___________________________________
>    |b[n-1] b [n-2]       | remainder
> I can't seem to figure out how to solve it with the
> use of horner's algorithm
> thank you very much
I am not very familiar with Horner's method, but my understanding is that it 

is used to find roots of a polynomial given its coefficients.  What you want 

to do is the opposite.  You want to find the coefficients given the roots.  

Is there some particular reason you want to adapt Horner's method to your 
problem?
Here is the solution to your problem again using your earlier start at it.  

The problem is to find a third order polynomial, p(x), which has double 
roots 
at x = -3 and such that p(2) = 0 and p(-2) = 8.  You started with
p(x) = a(x+3)(x+3)(x+c)
Use p(2) = 0 to find c:
p(2) = a(2+3)(2+3)(2+c) = 50a + 25ac.
Divide through by 25a to get 
0 = 2 + c => c = -2.
So now we have
p(x) = a(x+3)(x+3)(x-2)
Next use p(-2) = 8 to find a:
p(-2) = a(-2+3)(-2+3)(-2-2) = -4a.
Divide through by -4 to get 
a = -2.  
The final solution is
p(x) = -2(x+3)(x+3)(x-2)
- MO
===
Subject: Convergence of Taylor Series
Suppose I have some function defined on the complex plane f(z) with a simple 

pole at some point z1. 
Consider F(z)= 1/(1-z) with the pole at z=1.
-Nate
===
Subject: Re: Convergence of Taylor Series
>Suppose I have some function defined on the complex plane f(z) with a 
simple pole at some point z1. 
>Consider F(z)= 1/(1-z) with the pole at z=1.
No, that doesn't follow. Uniform convergence is a funny thing...
You're sort of asking the wrong question, asking where it's not
uniformly convergent. Because uniform convergence is not something
that happens at a point, it's something that happens on a _set_.
The series is not uniformly convergent on the entire unit disk.
Suppose it _were_ uniformly convergent on the open disk. Let's
say the n-th partial sum is F_n. Then letting epsilon = 1 in
the definition of uniform convergence it would follow that there
exists N such that if n > N then
(*)  |F(z) - F_n(z)| < 1 for _all_ z with |z| < 1.
But there is no n such that (*) holds. Why not? Suppose that
z is very close to 1. (We can't let z = 1 because that would
not satisfy |z| < 1, so suppose that |z| < 1 but z is very
very close to 1.) Then F_n(z) is close to n (or n+1) or
whatever, but f(z) is much much larger than n. So (*)
does not hold for this n.
What's really happening: If r < 1 then the series converges
uniformly on the disk |z| <= r. But as r tends to 1 this
uniform convergence happens slower and slower.
>-Nate
************************
David C. Ullrich
===
Subject: Re: Convergence of Taylor Series
>Suppose I have some function defined on the complex plane f(z) with a 
simple pole at some point z1. 
>Consider F(z)= 1/(1-z) with the pole at z=1.
>-Nate
Just look at the real series.
Call sn =  1 + x^2 +... x^n = (1 + x^(n+1))/( 1 - x). The sum of the
series is s = 1 / (1 - x) and s - sn = x^(n+1)/( 1 - x )
If sn -> s uniformly on (0,1) then for e > 0 there is N such that
 |s - sn| = | x^(n+1)/( 1 - x ) | < e if n >= N for all x in (0,1).
But this can't be because no matter what m > N you pick, 
limit(x -> 1) | x^(m+1)/( 1 - x ) | = infinity so it isn't less than e
on (0, 1).
--Lynn
===
Subject: quotient of bounded functions
If f>0 and g>0 are continuous, smooth and bounded in a interval [a,b]. Can I 

say that there exist a k1 and K2, such that k1<=f(x)/g(x)<=K2 for all x in 
[a,b] ? If so, is this that the right way of stating it? If not, which other 

conditions would I need to reassure this result?
===
Subject: Re: quotient of bounded functions
 
>If f>0 and g>0 are continuous, smooth and bounded in a interval [a,b]. Can 

I say that there exist a k1 and K2, such that k1<=f(x)/g(x)<=K2 for all x in 

[a,b] ? 
Yes, but since you're not sure whether it's even true you should 
probably explain _why_ it's true. (Hint: f has a min and a max on
[a,b], and the min is _positive_.)
>If so, is this that the right way of stating it? If not, which other 
conditions would I need to reassure this result?
************************
David C. Ullrich
===
Subject: Re: quotient of bounded functions
>If f>0 and g>0 are continuous, smooth and bounded in a >interval [a,b]. Can 

I say that there exist a 0 < k1 < >K2, such that k1 <= f(x)/g(x) <= K2 for 
all x in [a,b] ? >If so, is this the right way of stating the result? If 
>not, which other conditions would I need to reassure >this result?
f,g > 0 and continuous on [a,b] -> there exist k, K > 0
such that k <= f(x)/g(x) <= K for all x in [a,b].
Best wishes
Torsten.
===
Subject: 2005 equation
Find all pairs of positive integers x and y that safisfy the equation: 
1/x - 1/y = 1/2005
I'm thinking cross multiply and solve....but i don't know what to do! Can 
someone help? Thamks!
===
Subject: Re: 2005 equation
> Find all pairs of positive integers x and y that safisfy the equation:
> 1/x - 1/y = 1/2005
> I'm thinking cross multiply and solve....but i don't know what to do!
> Can someone help? Thamks!
Think?  Don't think.  Do it.  What do you get?  Get rid of the 1/2005
also so you can deal with integers as you want, instead of fractions.
Then what?  Factor 2005?  Sure why not.  What do you get?
Put it all together into one equation without fractions.
Now you need know number theory theorems.  How many you know?
If p is prime and p | ab, then p | a or p | b (using | for divides).
I don't know where it goes from here, it's your problem and until you
soften up the equation, I'll abide my time before giving it more thought.
===
Subject: Limit of the supremum function
Let g(t,y), I am trying to compute 
Limit_{t->oo} Sup_y{ g(t,y) }.
I know that g(t,y)-> k (constant) as t->oo.
Can I simply exchange the Limit and Sup signs? If not, which 
conditions would I need on g(t,y) in order to allow it? I would appreciate 
any suggestion.
===
Subject: Re: Limit of the supremum function
>Let g(t,y), I am trying to compute 
>Limit_{t->oo} Sup_y{ g(t,y) }.
>I know that g(t,y)-> k (constant) as t->oo.
>Can I simply exchange the Limit and Sup signs? 
No.
>If not, which conditions would I need on g(t,y) in order to allow it? 
None spring to mind, sorry. As far as I can see you need to
know what g is, and then find those sups.
>I would appreciate any suggestion.
************************
David C. Ullrich
===
Subject: Re: Limit of the supremum function
> Let g(t,y), I am trying to compute
> Limit_{t->oo} Sup_y{ g(t,y) }.
> I know that g(t,y)-> k (constant) as t->oo.
> Can I simply exchange the Limit and Sup signs? If not, which
> conditions would I need on g(t,y) in order to allow it? I would
> appreciate any suggestion.
My first inclination would be that convergence needs be uniform
g(t,y) = y/t for |y| <= t
        = 1 for t < |y|
sup_y g(t,y) = 1
lim_t g(t,y) = 0
===
Subject: Re: Limit of the supremum function
>> Let g(t,y), I am trying to compute
>> Limit_{t->oo} Sup_y{ g(t,y) }.
>> I know that g(t,y)-> k (constant) as t->oo.
>> Can I simply exchange the Limit and Sup signs? If not, which
>> conditions would I need on g(t,y) in order to allow it? I would
>> appreciate any suggestion.
>My first inclination would be that convergence needs be uniform
That doesn't help here. It would help if, for example, he wanted
to interchange two lim's. But the limit of a sup simply has
nothing to do with the sup of the limit.
>g(t,y) = y/t for |y| <= t
>       = 1 for t < |y|
>sup_y g(t,y) = 1
>lim_t g(t,y) = 0
************************
David C. Ullrich