[quote="SM4RT"]not sure if this is right..btw this is just the answer if its right i think i'll post my solution
[hide="5"]
$\frac{5^{4}}{3^{5}2^{6}}$
[/hide][/quote]

No...that's not right.

[quote="SM4RT"]for Question 5..lets say we have:
1,2,3,4,4,5 and then we roll
1,2,4,3,4,5...i did my calculations as so they were different..are we assuming that it doesnt matter which two die shows the two 4's...[/quote]

Those two are considered different.

okay..then i made an error in my calculations or my idea is just wrong..would someone mind posting the answer!

[hide="Here is how to approach it"]
First of all, there are $6^{6}$ total outcomes.

Assume that the two 4s come first: $4, 4, ?, ?, ?, ?$.

This can happen in $1\cdot 1\cdot 5\cdot 5\cdot 5\cdot 5 = 625$ ways.

The case where the two 4s are the first two dice is one of $\dbinom{6}{2}=15$ ways the two 4s can come.

Thus, there are $625\cdot 15 = 9375$ total ways for exactly two 4s to appear.

The probability, then, is $\frac{9375}{6^{6}}$.[/hide]

hmm thats what i did too..ya my answer before it shouldve been $5^{5}$ instead of $5^{4}$...

Well for 2
[hide]
the numbers can be..
$1***5$
$2***4$
$3***3$
$4***2$
$5***1$
$6***0$

i think its pretty easy from there..
[/hide]

[hide="#4"]
5 of the integers are even, and 5 are odd.  For the sum to be even, either both chosen integers are even or both are odd.  So the total number of possibilities is $2\binom{5}{2}=20$.
[/hide]

for 3...i thought the vowels were to appear in like A E I O U..

[quote="ch1n353ch3s54a1l"][quote="rnwang2"]Have fun!

1. Suppose a family of four goes for a drive in the five-passenger car. How many ways can they sit in the car if

a) anyone can drive?
b) there are only two licensed drivers?

2. How many five-digit numbers are there in which the sum of the first and the last digits is six?

3. Consider the arrangements of the letters of FACETIOUS. How many have the vowels in their natural order (i.e. A, E, I, O, U)?

4. Given ten consecutive integers, in how many ways can two of them be chosen so that their sum will be even?

5. Suppose you roll six dice. What is the probability of getting exactly two 4s?

6. In crew, an "eight" is a boat with four rowers on the port side and four on the starboard side. If two of the eight rowers can only row port and three can only row starboard, in how many ways can the eight rowers of the First Boat be assigned to the shell?[/quote]

[hide="1"]
(a) Four ways to choose a driver, $\binom{4}{3}$ ways to choose the seats, $3!$ ways to arrange the people in the seats. $4\cdot \binom{4}{3}\cdot 3!=\boxed{96}$

(b) Two ways to choose a driver, $\binom{4}{3}$ ways to choose the seats, $3!$ ways to arrange the people in the seats. $2\cdot \binom{4}{3}\cdot 3!=\boxed{48}$
[/hide]

[hide="2"]
$10^{3}=1000$ ways to choose the middle three digits, 6 cases where the first and last digit sum to 6, $6\cdot 1000=\boxed{6000}$
[/hide]
[hide="3"]$6^{4}$ ways to choose what vowels to put the consonants in between, $4!$ ways to then arrange the consonants (because they're all different), $6^{4}\cdot 4!=\boxed{31104}$[/hide]
[hide="4"]$\binom{5}{2}$ ways to pick two odd numbers, you can pick two odds or two evens, so $2\binom{5}{2}=\boxed{20}$[/hide]
[hide="5"]$\left(\frac{1}{6}\right)^{2}$ probability of getting two 4's, $\left(\frac{5}{6}\right)^{4}$ probability of not getting four 4's, $\binom{6}{2}$ ways to get the fours, $\left(\frac{1}{6}\right)^{2}\cdot \left(\frac{5}{6}\right)^{4}\cdot \binom{6}{2}=\boxed{\frac{3125}{15552}}$[/hide]
[hide="6"]The two port rowers and three starboard rowers have to row there, and there are $\binom{3}{1}=\boxed{3}$ ways to choose one of the three people who can row both to row starboard[/hide][/quote]

3 and 6 are not correct.

[quote="SM4RT"][hide="3"]
i think this is the answer

$= \dbinom{11}{5}\cdot 4! = \boxed{11088}$[/hide][/quote]

[hide="???"]Umm...doesn't FACETIOUS only have 9 letters? Then, your answer is correct. Choose 5 of the 9 places for the vowels in their correct order, and then you can permute the consonants.[/hide]

[hide="#3 in another way"] There are 9 letters with 5 vowels and 4 consonants. If the vowels have to remain in their correct order you can treat them as if they are 5 repetitions of the same letter.

$\frac{9!}{5!}=3024$.[/hide]

[hide="#6"] 2 of the 8 rowers can only row port. The first has 4 choices, the second has 3 choices.

3 of the 8 rowers can only row starboard. The first has 4 choices, the second has 3 choices, the third has 2 choices.

There are now 3 remaining rowers.

The first has 3 choices, the second has 2 choices, and the last has only 1 choice.

The total is: $(4\cdot 3)\cdot (4\cdot 3\cdot 2)\cdot (3\cdot 2\cdot 1) =1782$.[/hide]

Yes it does have 9 letters...lol i dont know why i had 11...


geometry
3D geometry
sphere
algorithm
quadratics
vector
LaTeX

I've been trying for some time to solve for the points created by the intersection of three spheres (as part of a vehicle dynamics simulation).  Because the problem is based on real-world physics, I can assume that I will never have a case where any of the spheres are coincident or do not interesect.  I know the center points of each of the spheres in 3-space, and I know the radius of each sphere.  My most successful method so far has been the Newton-Raphson method, and I usually get convergence within 7 or 8 iterations (seeking less than 5e-7 error in position).  My problem is the roughly 5% of the time that I do not get convergence.  I can't determine a pattern as to when/why the algorithm fails to converge.  I've also tried using a gradient descent algorithm, which is less likely to explode, but is also less likely to converge, in fact, I have not found any cases where Newton's method doesn't converge and the gradient descent algorithm will.

My question is this:  Does anyone have reccommendations for other algorithms that might help me out?  Any modifications to the algorithms I have?

Thanks for your help!

Why do you need anything like that at all? :? The problem reduces to solving a few linear equations and one quadratic one. Indeed if $ |x \minus{} c_j|^2 \equal{} r_j^2$ ($ j \equal{} 1,2,3$) are your equations, you can subtract the second and the third one from the first and get $ \minus{} 2(x,c_j \minus{} c_1) \equal{} \minus{} |c_1|^2 \plus{} |c_j|^2 \plus{} r_1^2 \minus{} r_j^2$ ($ j \equal{} 1,2$) which is a pair of linear equations with 3 unknowns. Then you can easily find the line that is the intersection of these 2 planes and solve one quadratic equation to find the intersection point of this line with the first sphere. The whole thing is about as time consuming as the extraction of the square root in the quadratic formula and gives you the exact solution.

If I understand you correctly, you're saying I should solve for the circle created by the intersection of two spheres, then find the plane on which the circle resides.  Doing this twice for two different pairs of spheres gives two planes, their intersection defines a line, and the intersection of the line with any of the spheres should be my solution.

I've tried working through this way, and I'm still getting stuck.  Everything works except for determining the plane on which the circle resides.  Do you have any advice on this?  (It's possible that my problem is solved anyway, if I use an iterative technique to solve this problem, maybe I'll have better results?).  I can get the plane's unit normal by finding the vector connecting the centers of the two spheres, but I still need to know the location of some point on the circle (or in the plane) to complete the equation of the plane.  Suggestions?

Thanks!

The equation fedja wrote (choosing values for $ i$ and $ j$) is an explicit equation for one of the planes. There are three such planes, but their intersection is a line rather than a point (the radical axis).

The method also generalizes easily to arbitrarily many dimensions.

:what?: Didn't I just write the linear equations of the planes for you? :? Let me write them again:
\[ 2(x,c_j \minus{} c_1) \equal{} \minus{} |c_1|^2 \plus{} |c_j|^2 \plus{} r_1^2 \minus{} r_j^2 (j \equal{} 2,3)
\]
($ c_j$ are centers, $ r_j$ are radii, $ x$ is a point on the plane, $ (,)$ is the scalar product in $ \mathbb R^3$, $ j \equal{} 2,3$) I admit that I made a couple of misprints last time (wrong $ \minus{}$ on the left, for example) when I typed in a hurry :blush:, which might confuse you a bit.

Edit: Ah, jmerry was faster :)

I guess it's just not obvious to me that this equation is what I need.  Can you reccommend a web site where I can find a derivation?  I can kind of make out that it is the equation of a plane (I'm used to seeing something like (x , c_j - c_i) = 0, so setting it equal to a non-zero number would just change your 4th constant if you expand it into the form A * x + B * y + C *z + D = 0), but I don't understand how it works.

I appreciate all the help you guys have given me, I really think I'm close to getting it!  Thanks again!

-Kerry

edit:  also... how do you format your equations so nicely?

edit 2:  I made D (in the expanded form of the plane equation above) equal to (-norm(c_1)^2 + norm(c_j)^2 + R_1^2 - R_j^2) / 2 and it worked!!!  I would only like to understand why...

[quote="ETCkerry"]I guess it's just not obvious to me that this equation is what I need.  Can you reccommend a web site where I can find a derivation?  [/quote]
I just subtracted two equations of the spheres to get it. What other derivation are you looking for? :?
[quote]
edit:  also... how do you format your equations so nicely?[/quote]
$ \text{\LaTeX}$ is functional on this forum. If you are familiar with it, just put formulae inside 2 dollar signs and you are done :).

Thanks for your responses.

OK.  I've convinced myself that your equation is correct (I'm not used to seeing the equations of circles/planes in vector form), and I wrote a simple MATLAB script to test it.  I was surprised to find that it doesn't alwasy work.  I plotted the spheres and planes to help diagnose the problem, and it appears that my plane equation is wrong for cases where the distance between the sphere centers is less than or equal to the larger radius, or in other words, the center of the smaller sphere is inside the volume encompassed by the larger sphere.

Since the math works out, I have to believe that the error is in my code, but I've been over it a dozen times and I keep missing it.  The MATLAB function I'm using to compute the plane constants is below:
[code]function [A, B, C, D] = SpherePlane(C1, C2, R1, R2)

A = C1(1) - C2(1);
B = C1(2) - C2(2);
C = C1(3) - C2(3);
D = (norm(C2)^2 - norm(C1)^2 - R2^2 + R1^2) / 2;

return[/code]

Can anyone identify the problem?

Thanks!

-Kerry

edit:  Rather than speculate about why it doesn't work... here's an example:
C1 = [1 1 5], R1 = 4.3589
C2 = [2 2 2], R2 = 3.4641

Radii were chosen so the intersection of the spheres contains point [4 4 4].

I think you noticed the missing factor of $ 2$ yourself by now :) Remember, I wrote $ 2(x,c_1 \minus{} c_j)$ on the left, and your $ A,B,C$ were corresponding to $ (,)$ without $ 2$ :roll: That explains your remark in "edit 2". The last post is, probably outdated by now...

I'm not sure what you mean... can't I just divide both sides of the equation by 2?  Then I get the dot product on the left by itself and everything else on the right (all over 2).

I was under impression that you corrected some error and now everything works. Do you still have a problem? :? If so, post the full code of your program. The fragment you posted seems fine, so the error must be somewhere else.

Everything works out on paper, but it doesn't return the correct answer.  I think I've narrowed the problem down to solving for the plane constants, but as you said above, that code looks OK.  Below is the full code for my program, and some separate programs that I hoped would aid with debugging.

The main script (Each of these is it's own file)
[code]% File:  ThreeSpheres.m
% Auth:  K. Loux
% Date:  2/1/2008
% Desc:  Analytical solution for the two points created by the
%        intersection of three spheres.
clear;
clc;

% Given:
%  The sphere center points:
P1 = [1 1 5]';
P2 = [2 4 4]';
P3 = [3 1 1]';

% Just to make sure there is a solution (and to check our answer)
Answer = [4 4 4]';% Remember this is only one of the two solutions!!

%  The radii:
R1 = norm(Answer - P1);
R2 = norm(Answer - P2);
R3 = norm(Answer - P3);

% Plot the spheres
hold off;
PlotSphere(P1, R1, 12, 60, 'blue');
PlotSphere(P2, R2, 12, 60, 'red');
PlotSphere(P3, R3, 12, 60, 'green');

% The method:
%  1.  The intersection of two spheres creates a circle.  That circle
%      lies on a plane.  Determine this plane for any two spheres. A
%      function was created to handle this, more explaination is
%      contained in that file (SpherePlane.m).
%  2.  Determine the same plane as in step 1 for a different set of two
%      spheres.
%  3.  Find the line created by the intersection of the planes found in
%      steps 1 and 2.  Lines only have one degree of freedom, so this
%      will either be two equations in the same variable, i.e. y(x) and
%      z(x), or the second equation will depend on the first, i.e. y(z)
%      and x(y(z), z).  The equations used here are of the latter type
%      and take the form y(z) = mzy * z + by and
%      x(y(z), z) = myx * y(z) + mzx * z + bx.
%  4.  The intersection of the line and any sphere will yeild two
%      points (unless the spheres don't intersect or they intersect at
%      only one point).  These points are the solutions.

% The A coefficient is the denomonator in several divisions below, so
% the spheres in steps 1 and 2 will be selected to avoid this.

% Step 1 (Plane defined by intersection of first set of spheres)
% And step 2 (Plane defined by intersection of second set of spheres)
if P1(1) - P2(1) ~= 0 & P1(1) - P3(1) ~= 0
    % First set (spheres 1 and 2)
    [A1 B1 C1 D1] = SpherePlane(P1, P2, R1, R2);
    % Second set (spheres 1 and 3)
    [A2 B2 C2 D2] = SpherePlane(P1, P3, R1, R3);
    % Plot the planes
    PlotPlane([A1 B1 C1 D1], P1 + (P2 - P1) / 2, 2 * (R1 + R2), 10, 'magenta');
    PlotPlane([A2 B2 C2 D2], P1 + (P3 - P1) / 2, 2 * (R1 + R3), 10, 'cyan');
elseif P1(1) - P2(1) ~= 0 & P2(1) - P3(1) ~= 0
    % First set (spheres 1 and 2)
    [A1 B1 C1 D1] = SpherePlane(P1, P2, R1, R2);
    % Second set (spheres 2 and 3)
    [A2 B2 C2 D2] = SpherePlane(P2, P3, R2, R3);
    % Plot the planes
    PlotPlane([A1 B1 C1 D1], P1 + (P2 - P1) / 2, 2 * (R1 + R2), 10, 'magenta');
    PlotPlane([A2 B2 C2 D2], P3 + (P2 - P3) / 2, 2 * (R2 + R3), 10, 'cyan');
elseif P1(1) - P3(1) ~= 0 & P2(1) - P3(1) ~= 0
    % First set (spheres 1 and 3)
    [A1 B1 C1 D1] = SpherePlane(P1, P3, R1, R3);
    % Second set (spheres 2 and 3)
    [A2 B2 C2 D2] = SpherePlane(P2, P3, R2, R3);
    % Plot the planes
    PlotPlane([A1 B1 C1 D1], P1 + (P3 - P1) / 2, 2 * (R1 + R3), 10, 'magenta');
    PlotPlane([A2 B2 C2 D2], P2 + (P3 - P2) / 2, 2 * (R2 + R3), 10, 'cyan');
else
    % If we make it all the way down here, there is only one solution,
    % and we'll need special code to find it.  This will never happen in
    % the vehicle dynamics simulation.
    fprintf('Error:  All spheres share common x-coordinate!!\n\n');
    return;
end

% Step 3 (Line defined by intersection of planes from steps 1 and 2)
%  Coefficients for y(z):
mzy = (A1 * C2 - A2 * C1) / (A2 * B1 - A1 * B2);
by = (A1 * D2 - A2 * D1) / (A2 * B1 - A1 * B2);
%  Coefficients for x(y(z), z):
myx = -B1 / A1;
mzx = -C1 / A1;
bx = -D1 / A1;

% Step 4 (Intersection of line and sphere 1)
%  The quadratic
a = 1 + mzy^2 + (myx * mzy + mzx)^2;
b = 2 * (mzy * (by - P1(2)) + (by * myx + bx - P1(1)) * (myx * mzy + mzx) + P1(3));
c = P1(3)^2 + (by - P1(2))^2 + (by * myx + bx - P1(1))^2 - R1^2;
%  First solution
ans1(3) = (-b + sqrt(b^2 + 4 * a * c)) / (2 * a);
ans1(2) = mzy * ans1(3) + by;
ans1(1) = myx * ans1(2) + mzx * ans1(3) + bx;
%  Second solution
ans2(3) = (-b - sqrt(b^2 + 4 * a * c)) / (2 * a);
ans2(2) = mzy * ans2(3) + by;
ans2(1) = myx * ans2(2) + mzx * ans2(3) + bx;

% Display the solutions
fprintf('\nSolution 1:\nX: %7.3f\nY: %7.3f\nZ: %7.3f\n', ans1(1), ans1(2), ans1(3));
fprintf('\nSolution 2:\nX: %7.3f\nY: %7.3f\nZ: %7.3f\n', ans2(1), ans2(2), ans2(3));[/code]

Now the function that I think is the problem (same as I posted above)
[code]function [A, B, C, D] = SpherePlane(C1, C2, R1, R2)
% File:  SpherePlane.m
% Auth:  K. Loux
% Date:  2/4/2008
% Desc:  Returns the four constants in the plane equation of the
%        form A * x + B * y + C * z + D = 0.  The plane is defined
%        by the intersection of two spheres, centered at C1 and C2
%        with radii R1 and R2, respectively.
%
% Usage: [A B C D] = SpherePlane(C1, C2, R1, R2)
%
% The plane constants are determined by subtracting the equations
% for two planes from each other.  While substitution would result
% in the equation of the circle formed by the intersection of two
% spheres, subtraction cancels out the higher order terms, leaving
% just the four plane constants.
%
A = C1(1) - C2(1);
B = C1(2) - C2(2);
C = C1(3) - C2(3);
D = (norm(C2)^2 - norm(C1)^2 - R2^2 + R1^2) / 2;

return[/code]

The above two are all that are required to solve the intersection of three spheres problem, but I added these next two for help with visualizing the problem.
[code]function PlotSphere(C, r, numCirc, numPts, color)
% File:  PlotSphere.m
% Date:  2/6/08
% Auth:  K. Loux
% Desc:  Plots points on the surface of a sphere centered
%        around C with radius r.  PlotSphere will draw
%        numCirc circles around each of the three axis,
%        with numPts points on each circle, in the color
%        chosen with the fourth argument.
%
% Usage: PlotSphere(C, r, numCirc, numPts, color)
%

% Determine circle parameters
CircSpacing = 2 * r / (numCirc + 1);
ArcSpacing = 2 * pi / numPts;

% Draw circles around x-axis
for i = 1:numCirc
    local_r = r * sin(acos(abs(r - i * CircSpacing) / r));
    for j = 1:numPts + 1
        X(j) = C(1) + r - i * CircSpacing;
        Y(j) = C(2) + local_r * sin((j - 1) * ArcSpacing);
        Z(j) = C(3) + local_r * cos((j - 1) * ArcSpacing);
    end
    % Plot the circles as they're calculated
    plot3(X, Y, Z, 'color', color);
    hold on;
end[/code]

And for the planes:
[code]function PlotPlane(Coefficients, Center, Size, numLines, color)
% File:  PlotPlane.m
% Date:  2/6/08
% Auth:  K. Loux
% Desc:  Plots points on the surface of a plane with
%        coefficients A, B, C, and D.  It will plot
%        numLines lines along two axis (forming a hatch
%        pattern) centered around a specified point with
%        line length Size.  The color argument specifies
%        the color to plot each line.
%
% Usage: PlotPlane(Coefficients, Center, size, numLines, color)
%
%  Coefficients is a vector containing the four plane
%  constants:  Coefficients = [A B C D]
%

% Make sure our center point is a column vector
if size(Center, 1) < size(Center, 2)
    Center = Center';
end

% Determine the line parameters and extract plane constants
LineSpacing = Size / (numLines - 1);
A = Coefficients(1);
B = Coefficients(2);
C = Coefficients(3);
D = Coefficients(4);

% To determine our first direction for lines, we'll invert
% the normal vector.  The second direction will be the cross
% product of the normal and our first vector.  The if..else
% structure is to avoid divide by zero errors.
if A == 0
    if B == 0 % A and B are zero
        Dir1 = [1 1 0]';
    elseif C == 0 % A and C are zero
        Dir1 = [1 0 1]';
    else% Only A is zero
        %fprintf('Error:  Zero componenet in plane normal!\n\n');
        %return
        Dir1 = [1 0 0]';
    end
elseif B == 0
    if C == 0 % B and C are zero
        Dir1 = [0 1 1]';
    else% Only B is zero
        %fprintf('Error:  Zero componenet in plane normal!\n\n');
        %return
        Dir1 = [0 1 0]';
    end
elseif C == 0 % Only Z is zero
    %fprintf('Error:  Zero componenet in plane normal!\n\n');
    %return
    Dir1 = [0 0 1]';
else % No component is zero
    Dir1 = [-1/A -1/B -1/C]';
end
Dir2 = cross([A B C]', Dir1);
Dir1 = Dir1 / norm(Dir1);
Dir2 = Dir2 / norm(Dir2);

% Find the point on the plane closest to the center point.
% To do this, we'll use the plane equation to find the third
% coordinate, given two coordinates from the given center point.
% Those two coordinates are chosen from the three based on the
% slope of the plane.  We ignore the coordinate associated with
% the steepest direction (i.e. if the normal vector is [0 0 1],
% we use the x and y from the given center point and calculate z
% to satisfy the plane equation).
if max(abs([A B C])) == abs(A)% Calculate X
    Center(1) = -(D + B * Center(2) + C * Center(3)) / A;
elseif max(abs([B C])) == abs(B)% Calculate Y
    Center(2) = -(D + A * Center(1) + C * Center(1)) / B;
else % Calculate Z
    Center(3) = -(D + A * Center(1) + B * Center(2)) / C;
end

% Do the calculations for lines along Dir1
for i = 1:numLines
    Point1 = Center + Size / 2 * Dir2 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir1;
    Point2 = Center - Size / 2 * Dir2 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir1;
    plot3([Point1(1) Point2(1)], ...
          [Point1(2) Point2(2)], ...
          [Point1(3) Point2(3)], 'color', color);
      hold on;
end

% Do the calculations for lines along Dir2
for i = 1:numLines
    Point1 = Center + Size / 2 * Dir1 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir2;
    Point2 = Center - Size / 2 * Dir1 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir2;
    plot3([Point1(1) Point2(1)], ...
          [Point1(2) Point2(2)], ...
          [Point1(3) Point2(3)], 'color', color);
      hold on;
end

return[/code]

And this last one is what makes me think the problem is with the plane constants.  This plots two spheres and the plane defined by their intersection.  You can see that the plane is not correct...
[code]% File:  SphereTest.m
% Date:  2/6/2008
% Auth:  K. Loux
% Desc:  Tests the algorithm that solves for the plane
%        defined by the intersection of two spheres.
clear;

% Given:
%  The sphere center points:
P1 = [1 1 5]';
%P2 = [2 4 4]';% This one fails
P2 = [3 1 1]';% This one works
%P2 = [15 1 5]';
%P2 = [2 2 2]';% This one fails

% Just to make sure there is a solution
Answer = [4 4 4]';

%  The radii:
R1 = norm(Answer - P1);
R2 = norm(Answer - P2);

% I think we can identify the problem: -> wrong - this isn't always true!!
%if norm(P1-P2) <= R1 | norm(P1 - P2) <= R2
%    fprintf('Failure predicted\n\n');
%end

% Plot the spheres
hold off;
PlotSphere(P1, R1, 12, 60, 'blue');
PlotSphere(P2, R2, 12, 60, 'red');

[A B C D] = SpherePlane(P1, P2, R1, R2);
PlotPlane([A B C D], P1 + (P2 - P1) / 2, 2 * (R1 + R2), 10, 'magenta');

axis equal;[/code]

Do you agree that the plane is not correct?  I don't think it's an error with my plotting function (unless there are two errors above) because the top program doesn't return the correct answer either (one of the points it returns should be [4 4 4]).

Thanks for your help!

-Kerry

For anyone who doesn't want to go through the trouble of making files and running the above scripts, here's the output from ThreeSpheres.m (the main one up top):

Solution 1:
X:   5.253
Y:   3.791
Z:   4.627

Solution 2:
X:  -8.210
Y:   6.035
Z:  -2.105

And the plots from three different runs of SphereTest.m with different points for P2:

I did find a mistake with the code that plots the planes.  The result was a plane that wasn't normal to the correct vector.  The correct version of that code is below, but this was not the problem with the solution to the intersection of three spheres.  I have to believe the problem is in the calculation of the constant term in the plane equation.  Any ideas?

Thanks a lot!

-Kerry

[code]function PlotPlane(Coefficients, Center, Size, numLines, color)
% File:  PlotPlane.m
% Date:  2/6/08
% Auth:  K. Loux
% Desc:  Plots points on the surface of a plane with
%        coefficients A, B, C, and D.  It will plot
%        numLines lines along two axis (forming a hatch
%        pattern) centered around a specified point with
%        line length Size.  The color argument specifies
%        the color to plot each line.
%
% Modified:
%   2/8/2008 - Fixed bug that plotted planes at odd angles
%              with the normal vector, KRL.
%
% Usage: PlotPlane(Coefficients, Center, size, numLines, color)
%
%  Coefficients is a vector containing the four plane
%  constants:  Coefficients = [A B C D]
%

% Make sure our center point is a column vector
if size(Center, 1) < size(Center, 2)
    Center = Center';
end

% Determine the line parameters and extract plane constants
LineSpacing = Size / (numLines - 1);
A = Coefficients(1);
B = Coefficients(2);
C = Coefficients(3);
D = Coefficients(4);

% To determine our first direction for lines, we'll cross
% the normal vector with any other vector (being careful to
% avoid producing a zero vector), thus giving us a vector that
% is perpendicular to the normal vector.  The second direction
% will be the cross product of the normal and our first vector.
if max([A B C]/norm([A B C])) == 1% There are two zeros
    if A == 0
        Dir1 = cross([A B C]', [1 0 0]');
    else
        Dir1 = cross([A B C]', [0 1 0]');
    end
else % There are one or fewer zeros
    if A == 0
        Dir1 = cross([A B C]', [1 0 0]');
    elseif B ==0
        Dir1 = cross([A B C]', [0 1 0]');
    else
        Dir1 = cross([A B C]', [0 0 1]');
    end
end
Dir2 = cross([A B C]', Dir1);
Dir1 = Dir1 / norm(Dir1);
Dir2 = Dir2 / norm(Dir2);

% Find the point on the plane closest to the center point.
% To do this, we'll use the plane equation to find the third
% coordinate, given two coordinates from the given center point.
% Those two coordinates are chosen from the three based on the
% slope of the plane.  We ignore the coordinate associated with
% the steepest direction (i.e. if the normal vector is [0 0 1],
% we use the x and y from the given center point and calculate z
% to satisfy the plane equation).
if max(abs([A B C])) == abs(A)% Calculate X
    Center(1) = -(D + B * Center(2) + C * Center(3)) / A;
elseif max(abs([B C])) == abs(B)% Calculate Y
    Center(2) = -(D + A * Center(1) + C * Center(1)) / B;
else % Calculate Z
    Center(3) = -(D + A * Center(1) + B * Center(2)) / C;
end

% Do the calculations for lines along Dir1
for i = 1:numLines
    Point1 = Center + Size / 2 * Dir2 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir1;
    Point2 = Center - Size / 2 * Dir2 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir1;
    plot3([Point1(1) Point2(1)], ...
          [Point1(2) Point2(2)], ...
          [Point1(3) Point2(3)], 'color', color);
      hold on;
end

% Do the calculations for lines along Dir2
for i = 1:numLines
    Point1 = Center + Size / 2 * Dir1 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir2;
    Point2 = Center - Size / 2 * Dir1 + ...
             (Size / 2 - (i - 1) * LineSpacing) * Dir2;
    plot3([Point1(1) Point2(1)], ...
          [Point1(2) Point2(2)], ...
          [Point1(3) Point2(3)], 'color', color);
      hold on;
end

return[/code]

Since the line through your two points passes through $ (4,4,4)$, the linear part is correct. The error is, therefore, in the quadratic part. I suspect that you put $ \plus{}$ instead of $ \minus{}$ in the discriminant of the quadratic equation ;)

Anyway, the code as written is numerically unstable if the first coordinates of two centers are close to each other or in other cases when the line is almost parallel to the $ xy$ plane. If you are going to use it in something "real", I strongly recommend to revise it.

Ah yes, I did put the wrong sign in the discriminant.  I corrected it, and I now get these points:

Solution 1:
X:  -1.478
Y:   4.913
Z:   1.261

Solution 2:
X:  -1.478
Y:   4.913
Z:   1.261

I am aware that this code is not very robust, and will surely re-write it before I put it in my simulation.  Thank you for checking, though.  :lol:

Let me ask this question:  If I subtract the equations of two spheres, what am I physically doing?  I mean to say that I can visualize that the intersection of two spheres give you a circle, and that circle has to lie on some plane, which is the plane I am searching for.  When you subtract the equations of two spheres, the higher order terms drop out and you are left with the equation of a plane.  How do I know that this plane is in fact the plane that the circle lies one?  Judging from the output of the SphereTest.m file above, it would appear that this plane is sometimes the plane containing the circle, but not always.

Thanks for your help!

-Kerry

On second thought, while I still have your attention  :oops: , I think I would be better off asking this question than to just dismiss your warning about the numerical instabilities with my code:  You are referring to Step 3 and beyond?  Steps 1 and 2 should capture all cases, yes?  Thank you!


Let $P(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_{1}x+a_{0}$ be a polynomial with grade $n \geq 0$, integer coefficients and $a_{n}\neq 0$. Let $p = n+\left|a_{0}\right|+\left|a_{1}\right|+\cdot\cdot\cdot+\left|a_{n}\right|$.
How many $P(x)$ are there such that $p = 6$?

We know that n cannot be greater than 5.  We can just do casework.

[hide]
If n is 5, then we have only one possibility:  the coefficient of $x^{5}$ is 1.  However, we multiply by 2 because the coefficient can be positive or negative.

If n is 4, then we can either have two coefficients of 1 or one coefficient of 2.  The first situation contributes $4(4)$, or 16 polynomials (because the $x^{4}$ coefficient must be 1, leaving us 4 other terms which can have the other coefficient of 1).  There are only 2 possibilities if we have one coefficient of 2.

If n is 3, then we can have three coefficients of 1, one 1 and one 2, or one 3.  The first situation contributes 24 polynomials (we have two coefficients of 1 to distributed among 4 terms).  The second situation gives us $4(2)(3)$, or 24 polynomials because we can have either the 2 or the 1 in front of $x^{3}$.  Again, there are only 2 possibilities for a coefficient of 3.

If n is 2, then we can have at most 3 coefficients, meaning one of 4, two of 2, one of 1 and one of 3, or two of 1 and one of 2.    Having one 1 and one 3 contributes $4(2)(2)$, or 16 polynomials to our total.  Having two of 2 contributes $4(2)$, or 8 polynomials to our total.  Having two of 1 and one of 2 contributes 24 polynomials to our total.  Having one of 4 contributes two polynomials to our total.

If n is 1, then we can have at most 2 coefficients, meaning one of 5, one of 4 and one of 1, or one of 2 and one of 3.  Having one of 4 and one of 1 contributes $4(2)(1)$ or 8 polynomials to the total.  Having one of 3 and one of 2 contributes 8 polynomials to our total. Having only a 5 contributes two polynomials to our total.

If n is zero, we have only two polynomials where $a0=6$ or $a0=-6$

The total is $2+18+50+50+18+2 = 140$

Is that right?     :maybe:[/hide]


I don't have the results from 2003, but here is what I can find:

04 top scores:  37, 41, 42 (18th at nationals)
05 top scores:  38, 40. 42 (6th at nationals)
06 #4 score:  31 (24 at nationals)

Notice that Oklahoma does not always score among the "weak" states!

Even though I'm not from OK, a 40 should definitely get you into Nats in a state like Oklahoma.

But you do need to realize that fact that since 2003 is an old state year...it seems to be a little easier than this current day's tests.

It really does seem like the years are getting harder eh? Last year, it was kind of hard in chapter. Our top scorer was this DonkeyKong and he got a 35? Somewhere in the low 30's. But I guess since all the awesome states like Michigan and Texas and Colorado and Missouri (Just kind of listing out the private states) it was probably easy for them.  But math888: I took that test and I got a 42!!!! But I think I already told you over gmail. But redirecting AIME15, a 40 will definitely get you into nats.

I made Nationals from Oklahoma last year with just a 24 total at states.
I was under the impression that the highest score in OK at states last year was a 29, but I might be wrong about that.

Who are you number dance? You can just PM me or something.. Because I'm from OK... Are you from Whittier or Jenks?

Hello Jimmy.

Jonathan -- Neither, actually. While I was in middle school I was homeschooled, but now I'm at public school.

[quote="Eragon"][quote="jimmyw8888"]Hi everyone!!! My first post!![/quote]
$ \equal{}$ :spam:
 :moose:[/quote]

Um.. That's spam too. Don't revive dead topics...

[quote="yaofan"][quote="math888"]OK, I just did the 2003 state competition which I've never done before. I got a 26 on sprint round and a 7 on target round. What's the average score to get into top 3 at state? I think this year is more competitive than last year. :D[/quote]

how do u get a 7 on the target round?[/quote]
LOL. He meant 7 probs.  :D
You going to Nats this year, yaofan?

[/quote]
LOL. He meant 7 probs.  :D
You going to Nats this year, yaofan?[/quote]

no... :(

Congrats to the state team and good luck at Nats!

If you're still interested, a 40 would have been 3rd in 2003.

Well older tests were actually easier. Haven't seen you around that much Chinaboy!


abstract algebra
function
Ring Theory
superior algebra
superior algebra unsolved

$A$ is a ring, $A((X))$ the ring of formal series with coefficient on $A$ (i.e. infinite polynomial).
1/ if $A$ is a field, is $A((X))$ a principal ring.
2/ if $A$ is noetherian, is $A((X))$ noetherian ?
3/ what are the irreducible element of $\mathbb{Z}((X))$ ? (I haven't thought to this one yet)

[quote="alekk"]2/ if $A$ is noetherian, is $A((X))$ noetherian ?
[/quote] Yes, this is an extension of the Hilbert's basis theorem.

Shouldn't you use $A[[X]]$?

1. It is easy to prove that $K[[X]]$ is euclidian by using the function $o: K[[X]]\to \mathbb{N}$, where $o(S)$ is the least nonnegative integer $k$ such that $x^ k|P$.

2.  $A[X]$ is noetherian by Hilbert's basis theorem and $A[[X]]$ is the completion of $A[X]$ with respect to the maximal ideal $(X)$. A completion of a noetherian ring is noetherian. There may be some other conditions other than noetherianity, but I am sure that they are satisfied in this case :)

hello Amfulger, I must admit my proof for 2/ was not as straightforward as yours  ...  because I don't know what is the completion of a ring  :D  so what is it ?

regards.


[img]http://img297.imageshack.us/img297/3475/problem006wv9.png[/img]

Hallar $x$

[img]http://img456.imageshack.us/img456/2907/problem011pp7.png[/img]

Hallar $x$ nuevamente.

[img]http://img401.imageshack.us/img401/2774/problem013cr0.png[/img]

Disfrútenlos  :)

P.D.: cuando vayan a resolverlos, se ruega citar el problema dentro del post para así no tener que subir a cada rato para ver la imagen.

P.D.: por favor en cada uno de ellos evitar el uso de la trigonometría.

Para el problema 2, reflejemos D sobre AB y llamemos al punto reflejado N. Por congruencia vemos que ANB=ADB, entonces ANB+ACD=90 de manera que NAC es recto. Ademas por hipotesis DC=2DB=NC+DB=ND, de manera que D es punto medio de NC. Ahora como NAC es recto y AD es mediana, entonces AD=DC, de donde DAC=ACD=BAD=x, entonces 3x=90--> x=30.

Otra forma, reflejemos A sobre BC y sea M su reflejado. Vemos que CB es mediana de ACM y como BD=2DC entonces D es su gravicentro. Entonces MD corta a AC en su punto medio. Pero ademas, ADB=BDM=90-x, entonces vemos que MD es perpendicular a AC, de manera que el triangulo ACM es equilatero, y concluimos nuevamente que x=30.

La respuesta es correcta.

Aunque te pediría que vieras el 1er. post-data que coloqué por favor.

[hide="Adjunto mi solución"]

[img]http://img456.imageshack.us/img456/2907/problem011pp7.png[/img]

Puede verse sin problemas que el $\Delta ABD \sim \Delta CBA$, luego es directo que $\frac{{\overline{AB}}}{{\overline{BD}}}= \sqrt 3$, y como el $\Delta ABD$ es rectángulo, podemos inferir que corresponde a la mitad de un triángulo equilátero, por tanto $\measuredangle \ ADB = 60^\circ$, para finalmente concluir que $x=30^\circ$[/hide]

Saludos.

P.D.: no usas $\LaTeX$?

Para el problema 1, tomemos un punto M sobre AC tal que AD=MC.Es facil ver que DB=BC (por traslacion de angulos).  Aora como DAB=2ABD entonces se cumple que DB^2=DA(DA+AB)  (es un resultado conocido y es posible de probar sin trigonometria). Lo anterior por hipotesis y por construccion es equivalente a BC^2=MC(DA+DC)=MC(AC); de manera que los triangulos CMB y CBA son semejantes ---> CBM=CAB=2x. Aora como por construccion AD=MC , entonces AM=DC=AB. De aqui que AMB=ABM=90-x  --->BMC=90+x.
Para concluir observamos que el triangulo BMC tiene angulos de 2x. 90+x, 3x. Entonces 90+x+2x+3x=180--> x=15.

[quote="conejita"]$\overline{BD}^{2}=\overline{AD}(\overline{AD}+\overline{AB})$[/quote]

Interesante relación, no la conocía...

Me puedes hablar más de ella?, en qué circunstancias se aplica?.

Te lo agradecería, saludos.

Tu respuesta es correcta.

Ah pues mira, es una propiedad que tienen todos los triangulos que tienen dos angulos tales que uno es el doble del otro.
El enunciado asi dice:
Sea ABC un triangulo. ABC=2ACB  si y solo si AC^2=AB(AB+BC).

Una prueba facil (no trigonometrica) es prolongando BC por B y tomar un punto D sobre esta linea, tal que AB=BD. De aqui encontrando unos isosceles y una semejanza. Te dejo la prueba de tarea. jeje

Muchas gracias, en verdad no conocía tal relación.

Saludos.

P.D.: queda un problema  :maybe:

prolongemos DB hasta un punto E de tal manera que  DEC = 4x vemos tb que BCE = x , ahora vemos que el triangulo DEC es isosceles , por ello DC = ED = AB , ahora trazemos DF paralela a AB donde F pertenece a EC ahora vemos la congruencia entre los triangulos ADB y EFD  por ello BD = DF , ahora vemos que el triangulo DFC ; por ello DF = FC  ahora trazemos BF ahora vemos queel triangulo BDF es isosceles y tambien BFC ( completando angulo se puede ver ello ) por ello BDF es equilatero por ello x = 10

[quote="Geómetra"]Solución correcta  :ninja:[/quote]

Este comentario algunos lo considerarían como "spam", o dicho de una forma mejor que no colaboran mucho con el foro o con la calidad de él para ser específicos, no te apresures por postear lo importante no es que tengas muchos "posts" ( y que el número de estrellitas aumente)  sino la calidad de ellos y qué tanto colaboran con el foro. Es mi opinión. Me  animé a decirla pues he visto que varios usuarios "postean por postear"  cosas como:

*  Gracias!
*  Nadie lo ha intentado? (cuando han pasado unas horas)
*  Ese problema es fácil.

con el fin de aumentar sus posts, me refiero no solo a Spanish Communities, y tienen cientos  de posts vacíos (en contenido)  y la verdad es que varias veces me he sentido incomodo leyendo ese tipo de posts. Espero que este comentario no lo tomen a mal y como dije antes es solo una opinión.

$Tipe$

[quote="tipe"]varias veces me he sentido incomodo leyendo ese tipo de posts.[/quote]
En 1er. lugar no veo que tengas algún cargo, como moderador o algo.

Lo otro, es que si no te gusta ver estos pots, limítate a postear, es todo lo que tienes que hacer.

3ero., inicié yo la discusión y si postié de esa forma, fue para afirmar que su solución fue correcta.

Y 4to., no tengo ningún interés en acumular posts.

En resumen, mientras algún mod. no me diga que esté haciendo algo mal, pues todo bien.

Como dije es una opinión, no es necesario que sea moderador para que emita mi opinión. Cuando dije que me sentía incomodo es porque es así, me siento así cuando leo posts vacíos.

$Tipe$

Point D is chosen on the side AC of a triangle ABC such that AD = BC and BD = DC. Angle BAC =3x, angle BCA = 2x. Find x.


Find all functions $ f: R\to R$ satisfying $ f(xf(x) \plus{} f(y)) \equal{} f^2(x) \plus{} y$ .


here are 2 special funct equation :
               1)    f: from R+ to R+ satisfy
for all x: f(f(x))=6x-f(x)
               2)    f: from R to R satisfy
for all x: f(f(x))= x^2-2
                could anyone tell me how to find solution to these problem of this kind:one variable.Some notes,advices from you is meaningful to me.Thanks a lot.
                                      :D  :D

Some hints:

1) Fix the value of $x$. Study the (linear recursive) sequence $x$, $f(x)$, $f(f(x))$, $f(f(f(x)))$, ...

2) What do you know about the fixed points of $f$?

It may be helpful:
http://eqworld.ipmnet.ru/en/education/li.pdf


[quote="warut_suk"]Find all primes $p\leq q \leq r$ such that
$pq+r,pq+r^2,qr+p,qr+p^2,rp+q,rp+q^2$ are primes[/quote]

Partitial solution. Last step must be very simple but I don't see it right away.

[hide]
Obviously $p \neq q \neq r$ (suppose $p=q$, then: $rp+q^2 = rp+p^2 = p(r+p)$, and similar reasonings)

Suppose $p, q, r$ are all odd. The product of two odd numbers is odd, and the sum of two odd numbers is even. All expressions in the problem will be composite (even and $> 2$). Hence, $p, q$ or $r$ must be an even prime, so $p = 2$.

$qr+p,qr+p^2$ gives $qr+2,qr+4$. Both expressions must be prime, so both can't be divisible by $3$. Hence, $qr$ must be divisible by $3$, so $q = 3$.
...
[/hide]

Finished solution

[hide]
So we have (see post above): $2r+3, 2r+9, 3r+2, 3r+4, 6+r, 6+r^2$ must all be prime.

Lets look at the problem $\mod 10$
Suppose $r \equiv\& 1 \Rightarrow 2r+3 \equiv\& 5  \Rightarrow 5 \mid 2r+3$
Suppose $r \equiv\& 3 \Rightarrow 2r+9 \equiv\& 5  \Rightarrow 5 \mid 2r+9$
Suppose $r \equiv\& 7 \Rightarrow 3r+4 \equiv\& 5  \Rightarrow 5 \mid 3r+4$
Suppose $r \equiv\& 9 \Rightarrow 6+r \equiv\& 5  \Rightarrow 5 \mid 6+r^2$

Only possibility left for a prime $r$ is $r = 5$.
We check this solution: $pq+r = 11, pq+r^2 = 31, qr+p = 17, qr+p^2 = 19, rp+q = 13, rp+q^2 = 19$

So the only solution is $p = 2, q = 3, r = 5$.
[/hide]


limit trigonometry integration logarithms ratio

Try to post a formula for pi different from all the others posted already.

I'll start this thread:

1)$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$

2)$\pi=4\left(\sum_{i=0}^{\infty}(-1)^{i}\left(\frac{1}{2n+1}\right)\right)$

$\frac{Circumference}{Diameter}$ (circle)  :D

[quote="I"]$\frac{\text{Area of a circle}}{\text{radius}^{2}}$[/quote]
HaHaHa that's funny :rotfl:

[quote="lotrgreengrapes7926"]These are some I made up.
$\lim_{n\to\infty}{\frac{n}{2}\sqrt{2-2\cos{\frac{2\pi}{n}}}}=\pi$.
$\lim_{n\to\infty}{n\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}}=\pi$. :D

Except, they have $\pi$ in it. Of course, you could always replace it with $180^\circ$.[/quote]

your second one can be simplivied to $\lim_{n\to\infty}{n\sin{\frac{\pi}{n}}=\pi}$ because ${\lim_{n\to\infty}\cos{\frac{\pi}{n}}}=1$.   :wink:

[quote="lotrgreengrapes7926"]These are some I made up.
$\lim_{n\to\infty}{\frac{n}{2}\sqrt{2-2\cos{\frac{2\pi}{n}}}}=\pi$.
$\lim_{n\to\infty}{n\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}}=\pi$. :D

Except, they have $\pi$ in it. Of course, you could always replace it with $180^\circ$.[/quote]

For the second one, it can be reduced taking out the cosine but then $\lim_{n\to\infty}{n\sin{\frac{\pi}{n}}}$ is essentially asking for $\lim_{n\to\infty}{\pi}$ because as $n\to \infty$, sin x becomes like x so stuff cancels.

http://mathworld.wolfram.com/PiFormulas.html

I'll post an amusing one.

$\pi = \sqrt[4]{ 90 \cdot \left( \sum_{k=1}^{\infty}\frac{1}{k^{4}}\right) }$

:)

Edit:  Nonono, here's my favorite.

$\pi = \frac{22}{7}-\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}\, dx$

:D

[quote="Kent Merryfield"]Do you guys know about the Buffon needle problem? (Dropping a needle randomly on a floor with parallel floorboards and counting how often it lies across a joint?)[/quote]

Here it is Kent: http://mathworld.wolfram.com/BuffonsNeedleProblem.html

Here is a way to APPROXIMATE $\pi$:

Inscribe a unit circle in a square of side length 2.  Now, randomly choose points inside the square and get the ratio of the number of points inside the circle to the total number of points and multiply by 4.

I think this is similar to Kent MerryField's


[quote="Mathsniper"]it's minkowski's inequality. trivial.[/quote]
It's not plain Minkowsky, but yeah $2\sqrt{x^{2}+y^{2}}\geq \sqrt{x^{2}+y^{2}}$, so we just have to prove that \[\sqrt{(x+n)^{2}+y^{2}}+\sqrt{(x-n)^{2}+y^{2}}\geq \sqrt{4x^{2}+4y^{2}},\] which is just a special case of the inequality proven [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103460]here[/url] in 4 different ways. (Set $a=x-n, b=x+n, c=d=y$, then $a+b=2x$ and $c+d=2y$.)

sry i didn't see godel's post, but I was gonna say 'square the whole thing', which is how they proved minkowski two variable in the linked post.

[quote="1234567890"]Prove that for all real $x,y,n$,

$\sqrt{x^{2}+y^{2}}\le \sqrt{(x+n)^{2}+y^{2}}+\sqrt{(x-n)^{2}+y^{2}}$

Find a clever way to prove this.[/quote]
I think here is the slick solution [b]1234567890[/b] was probably referring  to.

Consider a triangle with vertices $A(0,0),\ B(x-n,y)$ and $C(x+n,y).$ Then, label $(x,y)$ as $D.$ It is clear that $AD$ is the median line to $BC.$ Then, we have $AD < AB+AC$ from which our result immediately follows.

Yeah it is $2$.
if $n=0$ we get $LS=k\sqrt{x^{2}+y^{2}}$ $RS=2\sqrt{x^{2}+y^{2}}$ so $k\le 2$
Then just square both sides and do some algebra, and you are done.

[quote="1234567890"]All really nice solutions. Thanks.

[b]Question[/b]:
If the inequality below holds for all real $k,x,y,n$,
$k(\sqrt{x^{2}+y^{2}})\le \sqrt{(x+n)^{2}+y^{2}}+\sqrt{(x-n)^{2}+y^{2}}$,
what is the maximum value of $k$? Is it $2$?  :|[/quote]
If you look at my geometric solution above, you will see that $AD \le \frac{1}{2}(AB+AC)$ from which it follows $k=2.$


linear algebra
matrix
group theory
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ GL(n,q)$ be the matrix group on $ GF(q)$. Let $ p$ be a prime divisor of $ q^n \minus{} 1$ and $ n$ is the smallest number such that $ q^n \equiv 1(\bmod p)$. Now suppose that $ \langle a\rangle$ and $ \langle b\rangle$ are both cyclic subgroups of order $ q^n \minus{} 1$. Whether or not $ p\nmid |\langle a\rangle \cap\langle b\rangle|$?

Suppose that $ p$ divides $ \langle a\rangle \cap\langle b\rangle$. An element of order $ p$ acts irreducibly on $ F_q^n$ by your assumption on $ p$. So the intersection $ C$ of your two cyclic subgroups is irreducible. By Schur's Lemma, the centralizer $ D$ of $ C$ in $ End(F_q^n)$ is a divison subring of $ End(F_q^n)$, so $ |D|\le q^n$. But $ a,b\in D$ have order $ q^n\minus{}1$, so $ \langle a\rangle$ and $ \langle b\rangle$ coincide and equal the multiplicative group of $ D$.

These cyclic subgroups of order $ q^n\minus{}1$ in $ GL(n,q)$ are called Singer cycles. In B. Huppert, Endliche Gruppen I, Satz II.7.3(a) shows that if $ p$ satisfies your conditions, and $ P$ is a subgroup of order $ p$ in a Singer cycle $ S$, then the normaliser of $ P$ in $ GL(n,q)$ is the normaliser of $ S$ (of order $ n(q^n\minus{}1)$). This shows that $ P$ is contained in at most one Singer cycle, so $ p$ divides $ |\langle a\rangle \cap\langle b\rangle|$ if and only if $ \langle a\rangle \equal{} \langle b\rangle$.


number theory
greatest common divisor
LaTeX
function
number theory proposed

If x^2 - 1 = 0 (mod n) proove that

x = a^( phi(n) - lambda(n) )/2 ( mod n ).

gcd (a, n) = 1.

In general if a^g = 1 and a^h = 1 (mod n), x = a^(g - h)/2 (mod n)

Let me put it another way. Assume g = h (mod 2) g > h

Assume a^g = 1 (mod n) and a^h = 1 (mod n)

You are looking for x^2 = 1 (mod n)

I say x = a^(g - h)/2 (mod n)

(Sorry, I can't us latex, I'm in an internet cafe.)

You can still use latex, just put dollar signs...?
If $x^{2}-1 \equiv 0 \mod{n}$ prove that
\[x \equiv a^{(\phi(n)-\lambda(n))/2}\mod{n}\]
Where $gcd(a, n) \equiv 1$

In general if $a^{g}\equiv 1$ and $a^{h}\equiv 1 \mod{n}$, $x \equiv a^{(g-h)/2}\mod{n}$

What do you mean by $\lambda(n)$? Is it the Carmichael function, which returns the "actual" period?

Assume $g \equiv h \mod{2}, g > h$

Assume $a^{g}\equiv 1 \mod{n}$ and $a^{h}\equiv 1 \mod{n}$

You are looking for $x^{2}\equiv 1 \mod{n}$

I say $x \equiv a^{(g-h)/2}\mod{n}$


Princeton college Support articles inequalities

There has been some interesting polling on another website recently on this topic. What do you think the minimum voting age should be?

[quote="CheesyP18"]When you show competence of the workings of government.  With a test.  A hard one.  Yeah, that kind of kills the voter bloc, since a lot of people are stupid, but who needs stupid people to vote?

 :ninja:[/quote]

Yeah.... like a literacy test. And a poll tax. And only allow people of a certain color to pass it. Gosh, where've we seen that before?!

More like an IQ test, not a literacy test.  10 year olds can take an IQ test and rock it.  Poll tax?  Did I say poll tax?  Noooooo.  Poll tax would suck.  I wouldn't spend money to vote, thats retarded.

[quote="Bobtheball"]when you're knowledgeable about the political system?[/quote]

Which is????

If we administer a test, what's going to be asked on the test?  History?  Knowing what was the 5th state to enter the union, for example, has nothing to do with whether or not one should vote TODAY.  I don't believe you should take away the right to vote of 'stupid people'.  I imagine many of them don't vote anyways.  And, really, defining 'stupid' would have to be done before you could do anything.

Depriving people of the right to vote is never a good idea.  And these merit tests/IQ tests/etc. are only going to end up decreasing the amount of people voting.  If you take the right to vote away from people, you will end up undoubtedly end up with a revolution on your hands.

If one doesn't know who the current President is, what Congress does, or what the Supreme Court is for, then I believe that person should not have the right to vote.  To have an opinion on a subject matter, such as politics, one must be at least slightly knowledgeable.  I agree that creating objective testing would be very difficult, and subjective testing would never work.  However, I believe such a system, if ever practical or enacted, would benefit the country (at the expense of the stupid people, I know; but who cares about them?  All we need are immigrants anyway)

 :ninja:

First, I think we need to decide what role the voter plays.  What is the purpose of the voter?

Sometimes the voter chooses the electors who choose the officers, sometimes the voter chooses the officer himself, and then there are times the voter voters on referendums.

So now, we need to decide if the voters should be educated.  Do they?  The educated and wealthy are the ones running our country, right?  So why is it a requirement for the people voting to be smart.  [i]They[/i] are the ones being governed.  So they deserve the right to at least voice their opinion and vote for the people who will control the government.

Ideally everyone who votes should be well informed of the policy stances of the candidates and of the major issues being debated amongst the candidates.  It's disheartening to know that many voters will vote blindly without knowing enough about the candidates or the issues.  But we still should not deprive them of the right to vote.

And so you believe, that Joe Bloggs (taken from Princeton Review, not a pun on you by any means) should have the right to vote even though when he goes into the poll, he's going to choose the name which sounds the coolest?

CheesyPi, you're supposing without evidence that happens.

Those of you who are suggesting that vast swaths of the population should lose sufferage: wouldn't it make more sense to work to improve the quality of education?

[quote="bleumoose"]I do have to say, though, that it's a little worrisome that the "educated" people that thinks we should have a test don't even know how to learn from history.[/quote]

That's very nicely put!

Take a test that includes wheter you are going to vote democrat or republican...if you choose republican than you are not allowed to vote...JK

I think the voting age should be 16.

Letting only the educated people vote wouldnt necessarily be a good thing because then its not the whole country...

I do wish there was a simple test to decide whether a person was capable of voting smartly or not existed, but it doesn't. (Although, to be honest, this isn't as bad as driving. I don't understand why you have to be 16 when there is a TEST to determine your driving ability.)

I don't think that a person has to be educated to be able to make a just decision when voting. But I don't believe there is some magic age when a person becomes suddenly able to understand politics and make reasonable decisions. (The government believes this to be 18, it seems.) I'm more politically aware than plenty of adults I know, and yet it's them that can vote. What a just society we live in.

[quote]It's taken a long time to get for everyone...[/quote]

That's incorrect. We do not have voting for everyone; that's the point of this topic.

I don't recall ever hearing any scientific study about the sudden intellectual boost one gains on his or her eighteenth birthday, do you?

Minors have an appalling lack of rights. And this indirect assertion of "the under-18 crowd is too politically unaware to vote" is just one example.

[quote="JellyWorld"]I think by education he means political awareness, not mathematical, scientific or literary ability.[/quote]

bingo.

No, there is no magical age where you become politically aware. Do you know why the voting age is 18 instead of 21? Do you? There's a very important reason. As important as that reason is, fewer than 20% of 18-21 year-olds exercise their right to vote. So what's the point of extending the voting population even younger? Voting is a responsibility just as much as it is a right; it's a responsibility that I am scared sometimes to think that I have, especially with such a major primary coming up.

I think it has to be an age where the voter is experienced enough to know what he or she wants, to be able to make his or her own decision. That age is probably unclear, but w/e. say...above 12 years old? Actually, I'm fine with 18 because i get to vote for president for my first voting experience.

The question put, but ignored in the discussion is, "What is the purpose of voting?"

Is it to make a decision?  Not really.

One could suggest that the reason for voting is to have the majority of those who can upset the political system through citizen action feel that they have sufficient input into the political process to decide not to attempt to make changes through upsetting the system (violence, disruptive behavior).  This provides stability to those not involved.  Why should I not have mail today or have to cross a picket line if I want to do business with a given firm, just because someone is upset about their pay?

If this is true, voting is needed for each family.  Until one goes out on their own and supports themselves, their family, if satisfied, will control most antisocial behavior of their children.

The discussions on line focus on the ability of the voters to make informed decisions.  This is not really a role of the voters.  The Presidential canidates were narrowed down to two viable canidates, neither of which was a disaster.  To make a decision between the two canidates (last time or most any time) is driven by one's basic political stands.  Is a left leaner, or out right communist going to decide to vote for anyone but the Democrat?  Only if they are willing to throw their vote away.  Similarly, is a right winger going to abandon the Republician because they are "too center?" or "too left?"  Again, not a "rational" decision.

The two parties have enough party voters, or those that freely decide, but have an amazing consistency to deny anyone else any political power across the nation.  There is the occasional exception, but it is not sufficiently large to be meaningful across the country.  (For example, the wife of a dead canidate, or a really rich person who spends his money to buy the office, or a public person who is willing to spend his fame to buy the office.)

The actual decision is made by the few true independents.  (Each side tries to get the canidate that meets two criteria:  1)  can get elected and 2) appeals to the interest of the party loyalists.  The party loyalists are  too far from the mainstream in both parties to have their real choice be close to being preferable to a majority of voters.  Each side wants the party loyalists happy (more money and time donated), but also want to win by 1 vote.  (This is the reason that the latest voting has been very close, no one wants to work for the political home run.  The potential office holders that can do that aren't acceptable to party loyalists.)

Given that those in office set the rules, one shouldn't be amazed that running against an incumbant is harder than it has ever been, looking at results.

SO . . . getting back to the question at hand, one really needs to have agreement on the reasons for voting or at least clarity as to what the potential reasons are for voting, before meaninful discussions can occur.

Basically I don't quite get the system, and I also don't get why Belgium has such a different one.  We have a king that is utterly obsolete, and every four years we have parliamentary elections.  The government is formed by a set of parties whoe percentage has a sum over 50%, and they make up the ministers and prime ministers.  The king is theoretically the head of state, but strictly ceremonial actually, and the prime minister is the true head of state.

Why is it so different in other countries, so in USA there are municipal, congress, and presidential elections??

And my second question:
What would happen  if in USA, in 2008 :
Adam gets 30%
Bob gets 29 %
Charles gets 31%
David gets 10%

Does Charles become the president?

I don't think we go for plurality. The top three candidates (or is it two?) are sent to the House of Representatives and they decide. This happened in 1800 and again in 1876. And probably some more.

[quote="fredbel6"]What would happen if in USA, in 2008...[/quote]
There's no way to know without knowing how the regional support of the candidates broke down, since the system is not based on a direct popular vote. Each state has some number of electoral votes, which are generally all awarded to the state's winner. These electoral votes are approximately proportional to state population, although low-population states have more votes than this. It is possible to win the election while losing the popular vote; this has happened four times, most recently in 2000.
If no candidate gets a majority in the electoral vote, the system gets really weird.
An [url=http://en.wikipedia.org/wiki/U.S._presidential_election,_1860]explanation[/url] of the most fractured presidential election in US history. Note that the second-place candidate by popular vote won only two states and came in fourth in the electoral vote.

In the modern USA system, there are no regional parties, and two major parties dominate by virtue of winner-take-all elections at every level. In order for a third party to get any representation, that party has to win outright somewhere.

I love how almost everyone I know votes on the basis of republican/democrat, without even considering the issues. I also absolutely love how almost no one can argue well anymore. It's just "Hey, the other guy will let terrorist eat your babies! Protect the children!" nonsense.

This is of course a gross overgeneralization, but I hope some people will start caring about their country. It's ok to admit the other side has a good argument! Dissent is good. Everyone should read slashdot when the good civil liberties articles are up. Some really bad disscussion takes place.

Also, I love how everyone railroads EVERY SINGLE DISSCUSSION into a contest to see who's philosophy is better.

Libertarian: I knew it! Government overstepping it's bounds!  Let's go get some guns, privitize everything, and weaken federal government!

Socialist hippie: No way! We can't have lazy johnny getting F's because the government doesn't give him lots of money for his motivation classes! It might ruin his teamwork skills! Free healthcare and cable to everyone!

Republican: But the terrorists might invade!

Democrat: And terrorists might not like gay people!

Libertarian: But everyone is entitled to their own opinions! And, by the way, guns don't hurt people, people hurt people.

Democrat Spin-meister: No they don't your study is faulty

Republican: No way!

Democrat: Yes !

Republican: No way!

Democrat: Yes !

Republican: No way!

Democrat: Yes !

Capitalist: You wouldn't have any of these problems if we had a true free market!

[quote="gr8sk8r"]yea, i'd rather leave the decision making to the slightly...more educated.[/quote]

This would be okay, if everyone had the exact same right to the exact same education in our country.  Obviously, that's not true.

Let's say there's this person, person A.  He grows up in inner-city New York, and is repeatedly led to believe the best he can do is maybe graduate from high school.  Many of his teacher do not even have college degrees, and his parent(s) don't care about his education.  He is sorrounded by the image of gangs, and deals with teen pregnancy at a very young age, and his parents might have even had to deal with those things.

And there's another person, person B.  Student B live in a very wealthy, upper-class neighborhood, and high parents pay 15,000 dollars a year for his education in one of the top private schools in the country.  His parents continuouslly push him to do well in school, and he is expected to get at least a bachelor's degree in college.  HIs mom might be a stay-at-home one, and he lives a very comfortable lifestyle, and if he ever has to deal with teen pregnancy (which he probably won't with the money to buy birth-control), his, or the girlfriends parents can probably get an abortion, and there won't be any consequences.

You can't say that person A shouldn't be able to vote, while person B can, just because person A is "dumber" than person B.  It's not really fair, is it?

Upon reading your person A and person B example, I believe that you are off-topic.

We're not discussing the discrimination of the less-educated, in terms of school education as you have presented.  I was commenting upon the lack of political awareness and knowledge.  It's not really fair [that person A, who has no idea about any presidential candidates, carries the same weight as person B, who makes a conscious decision about which candidate she or he likes],[sic] is it?

I understand why many of us want "smarter" people to run the country. Surely we won't have to work and they will take care of all our confilcts and make the US a utopia, right? Just because a person is creative, or intelligent by IQ test standards, or even wise, doesn't mean many people would agree with their policies. 'Dumb' people have to live in the US also.

Having only "smart" people vote would really skew the US's policies.

It is helpful to understand that similar issues were present during the formation of the U.S. government.

I think it might be safe to say that social differences were greater then. You had wealthy educated people leading the country, (George Washington or John Hancock, for instance), while the majority of the people were farmers.

Obviously, the leaders were concerned about the "uneducated vote" of old Farmer Joe who only had time to grow his crops. One thing they came up with, was the electoral college. Member of the college represents their state, cast their vote for the president. The majority vote of the electoral college for the candidate becomes president. If the majority of the people of a state vote for what is considered a bad candidate for president (let's say a full-blown communist), then the member of the electoral college can cast his or her vote for a better candidate.

One important thing about knowing history, particularly when debating political issues, etc., is that it gives you perspective. Being able to compare events now to events later is important. We can look at a similar event and say, "It may be bad now, but is was worse then". Or vise versa, "It was bad then, but it is worse now." And thus we can learn, or at least have a better understanding, of the issue when looking at it in such a way.

On the topic of the electoral college: did you know that in some states, if you are an electorate and don't vote for the person that the majority voted for, you can be fined? Learned that in gov't class too.

[quote="bleumoose"]So wait, you think it's "fair" that rich people are the only ones that would get to vote for candidates, because they are more "educated"? Have you thought maybe that rich people might vote for candidates that support things that RICH PEOPLE want, and in effect rule out, oh, I dunno, social programs of any sort?[/quote]

If you think about it, people definitely need to have at least $\$$1 million in their savings in order to understand that the popular vote means nothing in the election and that there are two major political parties in the US.

on the original topic i think at least in MA the voting age should be lowered to 17 because that is the age you are tried as an adult so if your resposible enough and knowledgeable enough about the rules to be convicted as an adult you should have their rights

[quote="bleumoose"]Fewer than 20% of 18-21 year-olds exercise their right to vote. So what's the point of extending the voting population even younger?[/quote]
Extending the voting age to a younger age (say 16) will allow those younger citizens who want to, and feel a need to, make a choice in politicians that will affect them [b]personally[/b] in the future.  For those who don't care, or can't be troubled to vote, the >80%, then it doesn't matter for them.  Politicians ignore younger people because, guess why, we can't vote against them!  And by the time we can vote, they won't be running for an office anymore, or they know that the majority of us won't even bother to vote.  So what harm will it do to lower the voting age?  Only those who truely care about their, our, future will take advantage of it, those who hope to make a differerence.  Really, when you think about it, can you complain about the president if you had the option, even the responsibility, to make your tiny voice heard by voting and you didn't vote?  Do you have the moral right to complain then?  If you didn't care enough to vote then, should you really be allowed to complain now?

If you lower the voting age to 16, then people will say the same thing about 14 year olds, and the age would just keep going lower lower until anyone could vote.

So I'm with 18.

[quote="h_s_potter2002"]If you lower the voting age to 16, then people will say the same thing about 14 year olds, and the age would just keep going lower lower until anyone could vote.[/quote]

[b]Could[/b] is the keyword there.  Shouldn't anyone who wants to vote, and is physically capable of voting (able to cast a ballot themselves, not a parent casting it for an infant) be allowed to vote, since the people in office are making decisions that affect all of us?  Even if you were to allow everyone to vote, not everyone would - in fact, the great majority of us under 20 wouldn't.  But at least give us a chance.

In my opinion, a person should at least have had an opportunity to have had a high-school education before voting.  You won't be young forever.  However, an age of 18 is unfair in some ways.  For example, in my senior year, I will be able to vote in the presidential election, whereas most of my classmates will have to wait a year to vote, two years for a major election, and four for a presidential.  However, similar problems would exist for any voting age.

Due to the unfortunate fact that education is largely based on wealth in this country, it would be unfair to base voting on education.  Also, any sort of voting test would by its very nature be subject to corruption.  ("How many bubbles are there in a bar of soap?").

In the U.S., I think that realistically, we should ditch the electoral college.  In Alaska or Wyoming, a given person's vote counts much more than it would in California.  This isn't fair.  Also, it is possible to lose the popular vote completely and still win a presidential election.  It would also make manipulation of votes more difficult.

Idealistically, I think that candidates, rather than voters, should be chosen selectively, as in Han China, at least on the federal and guberantorial levels.  (Of course, before this happened, we would have free college education.)  People wishing to be eligible for public office would first have to submit letters of recommendation and references of good character.  They would then be admitted to a central university for such people, and then have to undergo some sort of marathon exam, probably lasting for several days, like the Han exams, and somewhat less difficult than Olympiad questions, on subject of math, the sciences, politics, government, history, diplomacy, etc.  However, after performing sufficiently well there, the candidates would have to give a dissertation on a political subject.  This way, even if the voters chose the wrong candidate, it wouldn't be a complete disaster, since all candidates would be somewhat good.  There would not be the problem of random billionaires waltzing in and becoming senators.  However, to be sure that the system would not become dominated by an élite class concerned exclusively with its own benefit, a general vote could potentially overturn any decision reached by the government.  However, this is not likely to happen any time soon. ;)

By the way, I think it is good for us to debate our respective philosophies.

[quote="Boy Soprano II"]Idealistically, I think that candidates, rather than voters, should be chosen selectively, as in Han China, at least on the federal and guberantorial levels.  (Of course, before this happened, we would have free college education.)  People wishing to be eligible for public office would first have to submit letters of recommendation and references of good character.  They would then be admitted to a central university for such people, and then have to undergo some sort of marathon exam, probably lasting for several days, like the Han exams, and somewhat less difficult than Olympiad questions, on subject of math, the sciences, politics, government, history, diplomacy, etc.  However, after performing sufficiently well there, the candidates would have to give a dissertation on a political subject.  This way, even if the voters chose the wrong candidate, it wouldn't be a complete disaster, since all candidates would be somewhat good.  There would not be the problem of random billionaires waltzing in and becoming senators.  However, to be sure that the system would not become dominated by an élite class concerned exclusively with its own benefit, a general vote could potentially overturn any decision reached by the government.  However, this is not likely to happen any time soon. ;) [/quote]

Ah, but would it not be easy to  bias the exam significantly? And then we would have our rulers puppets of the professors of that university - rule by the educated elite. Not good.

It should not be an age; it should be determined based on a persons majority.  If you are a legal minor (which you are if you are under 18 and have not been emancipated) then you should not be allowed to vote.  We should have a different system to determine who is ans who isn't a legal adult, however, an age of majority really doesn't cut it either.

And it does seem to me that removing certain people from the voting pool would be a "good thing" (TM)......... but I'm not a believer in the supreme awesomeness and justice of democracy; in other words, appoint me dictator.  I won't abuse my power.  The world will be a better place.......

you get the idea.

right?

How about an upper age limit?  If the idea is that people under a certain age aren't contributing significantly to the tax base and don't have an understanding of politics sophisticated enough to make informed decisions, we could apply the same argument to retired people on social security, right?

The discussions so far are beating around the central problem of democracy, namely the fact that most people don't know enough to make informed decisions about how the country should be run.

But I don't mean "most people" and "informed decisions" like you think I do.

Consider the economy, for instance. The domestic and international economies (plural or singular? that is the question) are so huge and complex and convoluted that certainly a lot of special training is needed to understand how they work, if any amount of education can actually foster an understanding over such a beast. Very, very, very few people actually have the education to know what is actually a good economic strategy. But since more than just economics Ph.Ds can vote, our economic-political dialogue is reduced to such rediculous oversimplifications as "cutting taxes allows people to put money into the market and stimulate the economy" or "holding taxes decreases inflation and increases the value of the dollar." These phrases include one or two "economics" terms that allow people to feel like they know what they're talking about, but in reality this sort of dialogue does not even approach the complexity that must be the national economy.

I'm not saying only economics Ph.Ds should be able to vote, but I am pointing out that a permanent shortcoming of our electoral system is the fact that it is oriented towards the average person, and the average person doesn't know enough to appreciate the complexity of political affairs.

So we're talking about voting age and ability to comprehend the political system. 18 is fine. Changing the age will neither keep the system better oriented towards the average person-- as it is-- nor will it solve the problem of politicians rejecting a much-needed mental complexity in favor of a "simple" posture that appeases the average voter.

[quote="bthings_2000"]I am pointing out that a permanent shortcoming of our electoral system is the fact that it is oriented towards the average person, and the average person doesn't know enough to appreciate the complexity of political affairs. [/quote]

It's not clear to me at all that this is a "permanent shortcoming."  In the Athenian democracy (and in many modern communes, kibbuzim, and other similar organizations) the political unit was small enough that every citizen should have been able to understand all of the issues facing the state and to take an active part in the political system.  Certainly, the knowledge required to be a similarly empowered individual in the modern democratic state is much greater, but we've also demonstrated an ability to grasp enormous amounts of knowledge.  After all, nearly everyone on this website knew (or will know) nearly all of ancient Greek mathematical knowledge by the time they were (or will be) 15 years old, plus a not-insubstantial collection of mathematical knowledge from the following two thousand years.  It's not at all obvious that we can't provide a sufficiently good education to the average American to allow her to have a decent grasp of the economic, political, environmental, social, legal, scientific and other issues of the day.  The fact that we're not doing this (and in fact doing the opposite)* is problematic for this vision, but it does not mean this vision is impossible, just that it isn't imminent.

*One of the problems with the current emphasis on testing in American education, embodied by No Child Left Behind, is that the focus has been almost exclusively on achieving a basic level of mathematics and reading skill.  The result is that more complex sorts of knowledge and skills within those fields, and all other fields in their entirety, are increasingly de-emphasized in the education of the average citizen.

[quote="bthings_2000"]The problem is that a "sufficiently good" education by any reasonable definition cannot reasonably empower the average members of a society to grasp the complexities of the economy, international relations, etc. They can gain a rudimentary understanding, for sure, but not the sort needed to truly understand causality and so forth. Modern democracy is not only "much" greater than Athens or a kibbutz, but upwards of 1,000 times greater.[/quote]

The first three sentences are your unsupported opinion -- you give me no reason whatsoever to believe that they are true.  Your last sentence is rather odd (what does the word "greater" mean in that context?) and is also unsupported and also also doesn't do anything to skewer my point.

Back to the original question:
[quote="bleumoose"]Do you know why the voting age is 18 instead of 21? Do you?[/quote]
The voting age was originally dropped during the Vietnam War when there were protests about how people could be drafted, and forced to go fight and die for the country without the right to pick their elected leaders.  There are no such problems now.  Yes, we may wish to change the way the government is run, but realistically most people under 18 will either not vote or vote the way their parents vote.  Just because a very small group set of those under 18 are well informed enough to vote does not mean that the voting age should be dropped.

My school system requires US Gov to graduate, but most people just sleep through the class.  So, while the people who are smart enough to find this site and spend time here may be smart enough to vote, most people are not ready at 16, and there is no fair way to determine who is ready and who is not: no test can be truely nonbiased.  In addition, while we may pretend otherwise, inner-city public schools will not teach the required information to the majority of their students, so most of those who decreasing the voting age with a required test would add would come from, if not wealthy, at least middle class families, and much less from the poor.  Until the inequality in schooling is resolved, this unbalance will remain.


A friend of mine whose on the western mass arml team told me that you just have to go to the practices and the coach chooses the team from performance on homework. Is this the case? And where and when are the practices for 2006-2007?

The first practice for Western Mass ARML this year is October 7 at Hudson High School from 9 AM to 12:30 PM.

Tentative future dates are Nov 4, Dec 16, Jan 13, Feb 10, Mar 17, Apr 15, and May 12 or 19.  Probably half of those will change, but it gives you an idea of the commitment (because the number of practices is unlikely to change).

As for team selection, I wouldn't say it's primarily based on homework.  It's mainly based on your performance on sample questions in practice, which is where you show your mettle.  But if you don't do your homework, you risk being dropped to a lower team or not being brought to Pennsylvania at all.

Rick Yanco
Worcester Academy

Wow, homework? You western MA guys are hardcore.


group theory
abstract algebra
Euler
algebra
polynomial
Galois Theory
superior algebra

The parts of this problem form a proof of the fact that if $ G$ is a finite subgroup of $ F^*$, where $ F$ is a field (even if $ F$ is infinite), then $ G$ cyclic. Assume $ |G|\equal{}n$.
(a) If $ d$ divides $ n$, show $ x^d\minus{}1$ divides $ x^n\minus{}1$ in $ F[x]$, and explain why $ x^d\minus{}1$ has $ d$ distinct roots in $ G$.
(b) For any $ k$ let $ \psi(k)$ be the number of elements of $ G$ having order $ k$. Explain why $ \sum_{c|d}\psi(c)\equal{}d\equal{}\sum_{c|d}\phi(c)$, where $ \phi$ is the Euler $ \phi$-function.
(c) Use Mobius Inversion to conclude $ \psi(n)\equal{}\phi(n)$. Why does this tell us $ G$ is cyclic?

a)  The first part is trivial.  If $ |G| \equal{} n$ then the roots of $ x^n \minus{} 1$ are precisely the elements of $ G$, which are distinct.  (Note how strong this result is.  Specifying the order of $ G$ specifies $ G$ uniquely.)  Since the roots of $ x^d \minus{} 1$ are a subset of the roots of $ x^n \minus{} 1$, they must also consist of $ d$ distinct elements of $ G$.

b)  $ \sum_{c | d} \psi(c)$ is the number of elements having order dividing $ d$, and those are precisely the roots of $ x^d \minus{} 1$.  On the other hand, $ x^d \minus{} 1 \equal{} \prod_{c | d} \Phi_c(x)$ where $ \text{deg } \Phi_c \equal{} \phi(c)$.

c)  The first part is trivial.  Since $ \psi(n) > 0$, $ G$ has elements of order exactly $ n$.

This means that multiplicative groups of (possibly infinite) fields share the property with cyclic groups that there is at most one subgroup of a given order. The latter property can be used to show, via the FTGT, that there is a unique finite field of a given order. So, in infinite Galois theory, could this have interesting implications for infinite extensions whose Galois group is the multiplicative group of some field?

[b]t0rajir0u[/b], thanks for your help.  I have a few questions.  Why is the first part in #1a trivial [is there a way to show this]?  Also, why is the first part in #3c also trivial?  The last question I had was: is $ \Phi_c(x)$ just a polynomial in $ F[x]$?

The reason why I ask is for #1a first part is because is not necessarily true. If $ \text{char}(F) \equal{} p>2$ and $ x^p \minus{} 1$ divides $ x^n \minus{} 1$ then $ x^p \minus{} 1 \equal{} (x\minus{}1)^p$ has just one root.  Also, I am still not seeing why #3c first part is trivial.  Thanks for the help.

[quote="numbertheory12"]The reason why I ask is for #1a first part is because is not necessarily true. If $ \text{char}(F) \equal{} p > 2$ and $ x^p \minus{} 1$ divides $ x^n \minus{} 1$ then $ x^p \minus{} 1 \equal{} (x \minus{} 1)^p$ has just one root.  Also, I am still not seeing why #3c first part is trivial.  Thanks for the help.[/quote]

You're talking about the [i]second[/i] part of 1a.  But what you have is not a counterexample.  You would need $ p$ to divide $ n$, (look again at the question!) but if you glance at the definition of $ n$, you'll find that this cannot happen in a field of characteristic $ p$.

For the first part of 1a, do you know why $ y\minus{}1$ divides $ y^k \minus{} 1$?  Are there values you can choose for $ y$ and $ k$ that will give you the result you want?

For the first part of 3c, well, if you know what Möbius inversion is, you'll know that you can use it to uniquely extract the value of $ \psi$ from your equation.  But the equation is the same for $ \phi$, no?


Ok, this is a problem I've known about for years but I've never been able to solve, or prove that it is impossible. Maybe you know it.

You have a sheet of paper with these symbols on it.

X   X   X

Y   Y   Y

You need to connect each X to each Y, for a total of 9 lines. You can lift your pencil and everything, but the thing is you cannot cross any lines.

My friend insists that it's possible. Is it?

It is impossible. I'm not sure how to explain it on here, but I can prove it on a piece of paper.

This is impossible. It is famous, I know its name is utility graph. I know you can connect with 8 lines but with 9 lines you must cross it.

[Edit] Here I try and have picture you can do with 8 but last line must cross one line to connect top left X to bottom right Y

Your question is:

Is the bipartite graph $K_{3,3}$ planar?

Answer: No.

Why? Kuratowski's theorem: A graph is planar iff it does not contain within it any graph that is a graph expansion of the complete graph $K_5$ or  $K_{3,3}$.

[quote="10000th User"][quote="mathnerd314"]Of course, since it's a piece of paper and not a blackboard, you can always cheat and draw the ninth line on the other side of the paper. LOL

Maybe that was what your friend had in mind.[/quote]
 :rotfl: Back of paper!  :rotfl: Your friend is cheating!  :D[/quote]

Is he cheating? The problem never specified that you could only use the one side of the paper, so he wasn't breaking any rules!  :lol:

yep, its impossible. you could do it with 8 lines though.

I don't see how the back of the paper helps... :? unless you mean poking a hole through or something.

A simple proof that this graph is not planar:

For planar graphs, we have Euler's formula $V+F=E+2$. This graph has 6 vertices and 9 edges, so it would have to have 5 faces (including the outside) if it were planar. Since the graph is bipartite, every face must have at least 4 edges. Since each edge is on at most two faces, we must have at least 10 edges for five faces. There are only 9 edges, so this is impossible and the graph is not planar.

Arguing rigorously from a drawing is very difficult.

I know this problem as the Three Wells Problem... and it has something to do with the Jordanian Curve Theorem. But it is certainly impossible.

This is not impossible:

[hide][img]http://img369.imageshack.us/img369/2046/impossible3dc.jpg[/img][/hide]

You might want to view it as its full size, oh and that line from the left X that curves around the left side that ends up at the middle Y is supposed to go to the left Y.

You didn't connect the top right X to the bottom middle Y.

[quote]oh and that line from the left X that curves around the left side that ends up at the middle Y is supposed to go to the left Y. [/quote]

In that case, we would have two lines connecting the left X and left Y, which we don't need.

This is impossible  :P If you are still not able to form a confirmed view, here is a website for you http://www.mathforum.org/dr.math/faq/faq.3utilities.html

This is also known as the "Three houses and three utilities" problem.

Have a look and enjoy :D

wow, that is really cool!  I didn't think it was possible to do that problem, and it seemed like those before had proved its impossibility.  Thanks gammapurpose!

Umm... gammapurpose was already proved wrong, in case you hadn't noticed.

Hexaditidom's friend seems to think it is possible.

My cheap solution relies on my crazy assumption that the points show through the paper but the lines don't. So draw the first 8 lines on one side and the 9th on the other side, thus connecting the 9 points. But that wouldn't work.

What does your friend say?


Hi, This math will keep u awake all time..... :lol:

Five girls have a little store that opens from Monday through Friday. Since two people are always enough for taking care of it, they decide to do a work plan for the week, specifying who will work each day, and fulfilling the following conditions:

a) Each girl will work exactly two days a week
b) The 5 assigned couples for the week must be different
In how many ways can the girls do the work plan?

Please search before posting anything; this was recently posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208697]here[/url].

Swt la.....

Me juz register yesterday...... Didnt see.....

Sryy...... :blush:


So in my document I used:
\begin{document}
\lhead{My header}

After that, I go on writing my document using sections and subsections. However, the heading of my sections appear in the top right corner of every page for some reason. How can I take that out?

On the line after "\lhead{My header}", type "\rhead{}".


ratio
analytic geometry
graphing lines
slope

A line goes through point $ A(9, 1)$, point $ B(19, k)$ and point $ C(7, 0)$.
What is the value of $ k$?

Taking $ \plus{} 1$ for the $ y$-term with a $ \plus{} 2$ going from $ 7$ to $ 9$ in the $ x$-term, we find that $ \plus{} 10$ (going from $ 9$ to $ 19$) in the $ x$-term will cause $ \plus{} 5$ in the $ y$-term, and $ 1 \plus{} 5 \equal{} \boxed{6}$. The ratio of change stays the same as a line has constant slope.

[hide] You can find the equation of the line. Since a line goes through all three points, they all must be collinear. Determine the equation of line AC.

slope = 1/2,
thus, y=1/2(x-7)

Point B (19,k) lies on this line as well.

k=1/2(19-7)=1/2(12)=6.
[/hide]


Find all $ f$ $ \mathbb R\to \mathbb R$satisfying
$ f(f(x)\minus{}y)\plus{}f(f(x)\plus{}y)\equal{}2x\minus{}f(x)f(y)\plus{}xy$

Replace $ y$ with $ \minus{} y$. $ LHS$ remains unchanged so we get
$ 2x \minus{} f(x)f(y) \plus{} xy \equal{} 2x \minus{} f(x)f( \minus{} y) \minus{} xy$
$ \Longrightarrow f(x)(f(y) \minus{} f( \minus{} y)) \equal{} 2xy \minus{} \minus{} \minus{} \minus{} (1)$
Taking $ x \equal{} 0$ in the last equation we get
$ f(0)(f(y) \minus{} f( \minus{} y)) \equal{} 0$
Here we have 2 cases
[b]Case1.[/b] If $ f(y) \minus{} f( \minus{} y) \equal{} 0$
Now substitude $ x$ with $ \minus{} x$ in the main equation
$ 2x \minus{} f(x)f(y) \plus{} xy \equal{} f(f(x) \minus{} y) \plus{} f(f(x) \plus{} y) \equal{} f(f( \minus{} x) \minus{} y) \plus{} f(f( \minus{} x) \plus{} y) \equal{} \minus{} 2x \minus{} f( \minus{} x)f(y) \minus{} xy \equal{} \minus{} 2x \minus{} f(x)f(y) \minus{} xy$
$ \Longrightarrow 2x \equal{} \minus{} xy$ for all $ x,y \in R$, which is impossible
[b]Case2.[/b] If $ f(0) \equal{} 0$.
By taking $ x \equal{} 0$ in the main equation we get
$ f( \minus{} y) \plus{} f(y) \equal{} 0 \Longrightarrow f( \minus{} y) \equal{} \minus{} f(y)$
Now using this in the equation $ (1)$ we get
$ f(x)f(y) \equal{} xy \minus{} \minus{} \minus{} \minus{} (2)$
$ x \equal{} y \Longrightarrow f(x)^2 \equal{} x^2 \Longrightarrow f(x) \equal{} x$ or $ f(x) \equal{} \minus{} x$.
If there is nonzero $ t \in R$ such that $ f(t) \equal{} \minus{} t$ then equation $ (2)$ implies $ f(x) \equal{} \minus{} x$ for all $ x\in R$.
So there are two equations which satisfy our problem: $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{} x$.


For any positive integers $a,b,c,d,n$, it is given that $n$ is composite, such that $n=ab=cd$ and $S$ as \[S=a^2+b^2+c^2+d^2\]Prove that $S$ is never a prime number

I'm sorry, I've edited the question. There are some conditions missing. :P

[hide=Hint]Prove that one of the number is 1[/hint]

$a,b,c,d$ are positive integers with $ab=cd$. Show that $a^2+b^2+c^2+d^2$ is not prime.
Find and prove also a generalization.

$a = \frac{cd}{b}$. So $c = hm$, $d = km$ and $b = hk$. $a^2 + b^2 + c^2 + d^2 = m^2n^2 + h^2k^2 + h^2m^2 + k^2n^2 = (k^2 + m^2)(h^2 + n^2)$.

The generalization: $a, b, c, d$ are positive integers such that $ab = cd$, then $a + b + c + d$ is composite.

Choose positive integers $w$, $x$, $y$, $z$ with $a=wx$, $b=yz$, $c=wy$, $d=xz$. Then $S=(w^2+z^2)(x^2+y^2)$, so $S$ is composite.


real analysis
calculus
integration
function
algebra
domain
real analysis unsolved

I am studying 'ANALYSIS by Lieb and Loss '...

usually lebesgue integral is defined in terms of simple function

But

In this book, integral is defined in terms of Riemann Integration !!!

$ \int f d\mu : \equal{} \int_0^{\infty} \mu (\{x \in X : f(x) > t \}) dt$
of course, $ \mu$ is measure, f is measurable.

LHS -> general (lebesgue) integration
RHS -> Riemann integration

Have you ever seen this definition in any other books?

If so, which book ? I need Reference .. HELP ME PLEASE!!

-------------------------------
P.S. Sorry for poor english..

Two points:

1. You left out a vital hypothesis: this only works for nonnegative $ f.$

2. That's not, strictly speaking, a Riemann integral. The Reimann integral is defined only for bounded functions on bounded domains. $ [0,\infty)$ is not a bounded domain, and $ \mu(\{f>t\})$ may be unbounded as $ t\to0^\plus{}.$ The integral from $ a$ to $ b$ makes sense for $ 0<a<b<\infty$ as a Riemann integral since we're integrating a monotone function, but to extend it to all of $ [0,\infty)$ is to interpret it as an [i]improper[/i] integral.

As for the characterization: it is well known, and it appears in some form or another in most textbooks. I recently taught the course using the Wheeden and Zygmund textbook; this characterization of the Lebesgue integral is certainly in that book.

[quote="Kent Merryfield"]Two points:

1. You left out a vital hypothesis: this only works for nonnegative $ f.$

2. That's not, strictly speaking, a Riemann integral. The Reimann integral is defined only for bounded functions on bounded domains. $ [0,\infty)$ is not a bounded domain, and $ \mu(\{f > t\})$ may be unbounded as $ t\to0^ \plus{} .$ The integral from $ a$ to $ b$ makes sense for $ 0 < a < b < \infty$ as a Riemann integral since we're integrating a monotone function, but to extend it to all of $ [0,\infty)$ is to interpret it as an [i]improper[/i] integral.

As for the characterization: it is well known, and it appears in some form or another in most textbooks. I recently taught the course using the Wheeden and Zygmund textbook; this characterization of the Lebesgue integral is certainly in that book.[/quote]

I've just read the book. but i think it is different from above definition..

and I want a book that properties and thorems is derived using this definition.

Is there exists another books?


calculus
derivative
geometry
3D geometry
sphere
rectangle
Pythagorean Theorem

I need some help on Derivatives because I have a test coming up and these two problems are stumping me =/

1. Find the meximum volume of a cylinder inscribed in a sphere with radius 10.

I got that the cylinder's radius was 5. And then for the height I got sqrt(300). Am I correct?

2. Point B is 100km east of Point A. At noon a truck leaves A and travels north at 60km/h, while a car leaves B and travels west at 80km/h.

a) Write an expression for the distance d between the car and the truck after t hours of travel.
b) when will the distance between the vehicles be a minimum?
c) What is the minimum distance between the vehicles?

This question totally has me baffled.

For 1 you can slice the sphere such that you get a 2-dimensional picture, namely a circle with an inscribed rectangle, where one of the rectangle sides is the height $h$ of cylinder and the other half
the radius $r/2$. By Pythagoras you can show $r^{2}+\frac{h^{2}}{4}=R^{2}$ where $R$ is the sphere radius.
Now the volume is given by
\[V(r)=2\pi r^{2}\frac{h}{2}=2\pi r^{2}\sqrt{R^{2}-r^{2}}\]
differentiate
\[V'=4\pi r\sqrt{R^{2}-r^{2}}-\frac{2\pi r^{3}}{\sqrt{R^{2}-r^{2}}}\]
Now set $V'=0$ and find realistic solutions $\ldots$

[quote]2. Point $B$ is $100 \text{ km}$ east of Point $A$. At noon a truck leaves $A$ and travels north at $60 \text{ km/h}$, while a car leaves $B$ and travels west at $80 \text{ km/h}$.

a) Write an expression for the distance d between the car and the truck after t hours of travel.
b) when will the distance between the vehicles be a minimum?
c) What is the minimum distance between the vehicles?[/quote]
2a)
[hide]By the formula $distance = rate \cdot time$ at any time $t$, the car traveling west is $80 \text{ km/h}\cdot t \text{ h}=80t \text{ km}$ away from Point $B$.  Thus it's horizontal distance from Point $A$ is $100-80t ~~\forall~~ 0\leq t\leq \frac{100}{80}$ or $0\leq t\leq \frac{5}{4}$. (Also note that the distance from $A ~~\forall t>\frac{5}{4}$ is $80t-100$.)  The distance the truck is going north by the same is $60t$.

A much more simple diagram:
[img]http://img396.imageshack.us/img396/8112/triok1.jpg[/img]

Thus by the pythagorean theorem the distance between the car and the truck is:
$d^{2}=(100-80t)^{2}+(60t)^{2}~~\forall~~ 0\leq t\leq \frac{5}{4}$ and $d^{2}=(80t-100)^{2}+(60t)^{2}~~\forall~~ t>\frac{5}{4}$
$d=20\sqrt{5(5t^{2}-8t+5)}~~\forall~~ t\geq 0$; Notice we may reduce it down to one equation because in the one above $(100-80t)^{2}=(80t-100)^{2}$

Thus $\boxed{d=20\sqrt{5(5t^{2}-8t+5)}}~~\forall~~ t\geq 0$.[/hide]

2b)
[hide]Take a derivative with respect to $t$ and get:
$\frac{dd}{dt}=\frac{20\sqrt{5}(5t-4)}{\sqrt{5t^{2}-8t+5}}$

Set this equal to zero and eliminate the bottom part and constants:
$5t-4=0$
$t=\frac{4}{5}$

Set up a derivative line:

[img]http://img171.imageshack.us/img171/540/derivtx2.jpg[/img]
Because the derivative changes from negative to positive at $t=\frac{4}{5}$, by the First Derivative Test we can conclude that $\boxed{t=\frac{4}{5}}$ produces the abscissa of an absolute minimum.[/hide]

2c)
[hide]Plug $t=\frac{4}{5}$ into your distance equation and get $d|_{t=\frac{4}{5}}=\boxed{60}$[/hide]

stop I may have done something wrong since a quick calculation gave me $r=\sqrt{80}$
but the principles are correct

Thank you skimmc :)

I got r=8.165 on my calculator

intuitively it feels a bit much but .. :huh:


[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$

Prove that $ a^2bc \plus{} ab^2c \plus{} abc^2\leq \frac 13$.[/quote]
$ a^2bc\plus{}b^2ca\plus{}c^2ab \leq \frac{1}{3}(ab\plus{}bc\plus{}ca)^2 \leq \frac{1}{3}$

[quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$

Prove that $ a^2bc \plus{} ab^2c \plus{} abc^2\leq \frac 13$.[/quote]

$ a^2bc \plus{} ab^2c \plus{} abc^2\equal{}abc(a\plus{}b\plus{}c)\le\left(\frac{a\plus{}b\plus{}c}{3}\right)^3(a\plus{}b\plus{}c)\equal{}3\left(\frac{a\plus{}b\plus{}c}{3}\right)^4\le3\left(\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}}\right)^4\equal{} \frac{1}{3}$ by AM-GM and AM-QM

[quote="CSS-MU"][quote="kunny"]Let $ a,\ b,\ c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$

Prove that $ a^2bc \plus{} ab^2c \plus{} abc^2\leq \frac 13$.[/quote]

$ a^2bc \plus{} ab^2c \plus{} abc^2 \le a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \le \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{3} \equal{} \frac {1}{3}$[/quote]

Nice Solution! :lol:

\[ \boxed{\boxed{3(x^2 \plus{} y^2 \plus{} z^2)\geq (x \plus{} y \plus{} z)^2\geq 3(xy \plus{} yz \plus{} zx)}}\]

[quote="xzlbq"]How do you prove stronger:
Let $ a,\ b,\ c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$

$ bc \plus{} ca \plus{} ab \geq 3abc(b \plus{} c \plus{} a).$[/quote]

$ bc \plus{} ca \plus{} ab \equal{}(bc \plus{} ca \plus{} ab)(a^2\plus{}b^2\plus{}c^2) \ge (ab\plus{}bc\plus{}ca)^2 \geq 3abc(b \plus{} c \plus{} a).$

 :wink:

[quote="xzlbq"]How do you prove stronger:
Let $ a,\ b,\ c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1.$

$ bc \plus{} ca \plus{} ab \geq 3abc(b \plus{} c \plus{} a).$[/quote]
$ ab\plus{}bc\plus{}ca\equal{}\frac{(ab\plus{}bc\plus{}ca)^2}{ab\plus{}bc\plus{}ca}\ge\frac{(ab\plus{}bc\plus{}ca)^2}{a^2\plus{}b^2\plus{}c^2}\ge 3abc(a\plus{}b\plus{}c)$