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IVPL@Y science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
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Ladnor Geissinger science forum beginner
Joined: 04 May 2005
Posts: 18

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Linear diophantine equation question



(After a description of the method he finally arrived at, following
his professor's suggestion, for extending the Euclidean Algorithm)

On Thu, 03 Feb 2005 00:01:40 GMT, Jeff Heikkinen wrote:
Quote:  For the purposes of undergraduate level number theory, this method
slices, dices, makes Julienne fries, *and* deciphers the lyrics to
Louie Louie; it can help you with nearly any nontrivial problem, I am
told. 

It's nice that he is so happy about the method, but I'd like to point
out that he was very close to getting even more in a slick package
that allows for easy proofs as well as computation.
I had never heard of the Blankinship algorithm, but it seems to be
almost identical to a form of the Extended Euclidean Algorithm (EEA)
which I devised many years ago and which I have been teaching to my
students whenever the subject is elementary number theory  in a
number theory or abstract algebra or discrete math class. A brief
description of this algorithm follows.
EEA: Suppose given integers A > B > 0.
Form a sequence of triples as follows.
Initialize: first triple (A,1,0), second triple (B,0,1).
Now divide A by B and write A=Q1*B+R1 with Q1 and R1 integers, and
0<=R1<B. Then the third triple is (A,1,0)Q1*(B,0,1)=(R1,1,Q1).
Repeat this process until the first entry in a computed triple is 0.
(And DO compute the second and third entries in this last triple!)
To be a bit clearer about what exactly is to be done repeatedly, let's
look at some stage in this process where the previous and current
computed triples are (R,S,T) and (U,V,W). By our construction 0<=U<R.
Suppose that U is not zero. Then we will divide R by U to get R=Q*U+X
with Q and X integers and 0<=X < U.
Then the next triple will be (R,S,T)Q*(U,V,W)=(X,Y,Z).
These triples were constructed to have the property R=S*A+T*B and
U=V*A+W*B, and it is easy to check that then also X=Y*A+Z*B. This is
a loop invariant. Notice that this property is equivalent to saying
that each triple is orthogonal to the vector (1,A,B).
Also note that the computation in this step can be described in matrix
language as follows:
Let E=[0,1; 1,Q], then E*[R,S,T; U,V,W]=[U,V,W; X,Y,Z]. Note that E
has determinant 1. Each previous step in the loop can be carried out
by multiplying by such a matrix, so it follows that if (U,V,W) is the
nth computed triple, and if F is the product of all these previous
matrices like E, then det(F)=(1)^n and
F*[A,1,0; B,0,1]= [R,S,T; U,V,W].
Which of course means that F= [S,T; V,W], so we were keeping track of
this matrix all along without noticing!
Finally, suppose that U=0. Then as usual we can easily show that R is
the greatest common divisor GCD(A,B), and we have the desired result
that GCD(A,B)=S*A+T*B from the penultimate triple. The last triple
now yields 0=V*A+W*B, and we can easily show that
V*A=W*B=LCM(A,B), that is, V=+B/R and W=+A/R. (see below)
Moreover, for any integers K and L such that R=K*A+L*B, it must be the
case that (K,L)=(S,T)+m*(V,W) for some integer m.
Some observations on the resulting sequence of integer triples:
1. The column of first entries is strictly decreasing and ends in 0.
2. After the initial triples, the entries in columns 2 and 3 are
nonzero, they alternate in sign, and are strictly increasing in
absolute value.
3. Suppose that we replace the first entry in each triple by the
quotient when that entry is divided by R, i.e.
newcolumn1=oldcolumn1/R. Also let A=A'*R and B=B'*R, then A' and B'
are coprime. And if we carry out our algorithm beginning with A' and
B' we will get this newcolumn1 along with the SAME old columns 2 and
3. Which means that F*[A',1,0; B',0,1]= [1,S,T; 0,V,W] at the end
(when U=0). And the inverse F^(1)=(1)^n*[W,T; V,S] so when you
multiply it on the left and check the first column you see that
(1)^n*[W; V]=[A'; B'].
By the way, note that this algorithm requires no confusing
backtracking of the sort that appears in most elementary number
theory discussions in order to get the S and T such that
GCD(A,B)=S*A+T*B. Also, we really only need 6 memory locations to
store numbers used in the algorithm  we could print out the initial
triples (so A and B are recorded), and then at each step we need only
the previous triple and the current triple to compute the next triple,
which we could then print out if we want it, or just wait until the
end and print out the final 2 triples. And finally, you can carry
this out easily in Excel or any other spreadsheet. 

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Bob science forum addict
Joined: 30 Apr 2005
Posts: 80

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: help with stats problems



On 02 Feb 2005 07:21:40 GMT, Jack <barqs100@hotmail.com> wrote:
Quote:  barqs100@hotmail.com (Jack) wrote in news:cec0p51s8psm@legacy:
I have this stats problem and I have no idea how to solve it.
In a certain small city, traffic accidents occur independently at a
constant rate of 30 per year. What is the probability that 20 or fewer
accidents will occur in the year 2005, followed by 40 or more in 2006.
All I know so far is that I have to add up P(x <= 20) with P(x >= 40).
Any help is appreciated.
Ok, let's change it to average rate instead of constant rate.
Would the solution be some what like this using the poisson distribution
formula.
P(X <= 20) * P(X >= 40)
= ( P(X = 1) + P(X = 2) + ... P(X = 20) ) * ( 1  P(X < 40) )
If so, the solution would require a lot of calculation then...

A spreadsheet would do it in a wink!
But also... the function is well behaved. Calculate a few of those
points, and sketch what it looks like. Beyond some point, p will
obviously be negligible, and you don't need to calculate them all. And
I bet it is damn near symmetrical at average of 30.
bob 

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Algebryonic science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: I have some intermediate algelbra questions!!!



charmel822@yahoo.com asks on 02/02/2005:
Quote:  3y5x= 8
Y/5 + 1/3= 7/15

Those are elementary level algebra.
The first one is just a linear equation which easily permits x and y intercepts
by inspection; and slope also by inspection.
The second one is just a point on a number line, onedimensional. You can
simplify it by multiplying both sides by 15.
Algebryonic 

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Bob science forum addict
Joined: 30 Apr 2005
Posts: 80

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: conditional prob



On Wed, 2 Feb 2005 12:48:56 +0000 (UTC), jiejing_15@yahoo.com (irrix)
wrote:
Quote:  hi can someone help me solve this question thank you.
2 cards are randomly selcted from a deck of 52 playing cards.
a)what is the probability they constitute a pair
b) what is the conditional probability they constitute a pair given
that they are of different suits
isnt it a pair come from different suit?

yes
To see the point of that second part, imagine a particular case. First
card is a 2D (two of diamonds). Now you draw second card. What is p
that it is another 2?
Among the possible second cards are 3D, 4 D etc. But in part b, these
are excluded as allowed (or considered) outcomes. In part b, if the
second card is a D, we just reject that trial. So, now knowing that we
will reject any D for the second card, what is the desired p that we
get another 2. It is higher than in part a.
bob 

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Jeff Heikkinen science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Linear diophantine equation question



Q: Are we not men? A: We are Virgil!
Quote:  I finally figured that out (using a method my prof showed the class
where the fact that you're using matrices is less obvious) at about the
same time you posted, but I was still hoping there was a way that
involved fewer steps. For three digit a and b values one of which I
think was a prime number (they were definitely relatively prime), it
took nearly half an hour and filled an entire page to do the sample
problem I was working on under exam conditions (and that's with being
allowed to use a calculator). Is there a way to speed up the "work your
way back up" part using similar matrix operations?
It can be done without using the matrix format, but the number of steps
involved is about the same with any method I am aware of, except lucky
guessing.

Actually, I've since realized, thanks to a hint from my prof, that the
matrix format (or rather, a version of it where you're not explicitly
writing out the matrices) works best and cuts the number of steps by
about half, more if you count trivial gathering of terms, compared to
what I describe above. It's similar to running the Euclidean Algorithm
backwards.
So to answer my own original question:
You make a chart with four columns, and fill the first two rows in like
so (this will be best viewed in a fixedwidth font, and I am assuming
that a > b):
 a 1 0
p b 0 1
Where the dash is, you don't need to put anything; what I'm calling p
is the largest integer such that a  pb >= 0.
Then you multiply the rest of the second row (the second, third and
fourth columns) by p, then subtract it from the first row and write down
the results in the third row, second through fourth columns. Obviously
this is just standard row operations, but the chart is quicker than
writing them out "properly". Then you figure out a new p value fitting
the above criterion and write it down in the first column of your new
row. Repeat this process until you get a zero in the second column
(don't bother writing out the rest of that row).
The number immediately above the zero is the GCD of a and b, but for
this particular problem that's not the relevant bit (especially if a and
b are relatively prime). The relevant bit is the third and fourth
columns. Let's call the number in the second column r (for remainder),
the third column u and the fourth column v. Each row is in fact, the
coefficients in a correct equation of the form
ua + vb = r
using the original a and b you were given.
Once you have a version of this equation where r is equal to the GCD (1
in this case, by my own stated assumptions), you are one trivial step
away from the answer; you just multiply through by c (or more generally
c/r, but in this case r is known to be 1). x = cu and y = cv, and either
you're done or you're one trivial step away from the general solution,
depending on which type of problem you're given.
For the purposes of undergraduate level number theory, this method
slices, dices, makes Julienne fries, *and* deciphers the lyrics to Louie
Louie; it can help you with nearly any nontrivial problem, I am told. 

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Jim Spriggs science forum Guru
Joined: 24 Mar 2005
Posts: 761

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Probability distribution function



Jack wrote:
Quote: 
Hi,
Can anyone help me solve the following problem:
Suppose that X is a continuous random variable with probability
distribution function:
f(x) = xe^(x)  bx^2 for 0 <= x <= 1
What is the value of b? Is this a valid probability distribution
function?

You must define f over all R not just on [0, 1].
The integral of f over all R must equal 1.
If you're lucky f = 0 outside [0, 1] and integrating it allows you to
determine b. 

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Stan Brown science forum Guru Wannabe
Joined: 06 May 2005
Posts: 279

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Question on probability distribution function



"Torsten Hennig" wrote in alt.math.undergrad:
Quote:  if the mass of the distribution is concentrated on [0,1], a
necessary condition on f to be a probability distribution function
is f(0)=0 and f(1)=1.

Huh? How do you get that? I agree that the cumulative function
F(x) = INT[ 0, x, f(t) dt ]
must be 1 at x=1, but I can't see any reason why the f(1) must be
1.

Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/ 

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Stan Brown science forum Guru Wannabe
Joined: 06 May 2005
Posts: 279

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Simplifying Equation



"mortalfunk" wrote in alt.math.undergrad:
Quote:  I can't seem to get the right answer... either the answer is wrong or
I'm doing something wrong.
The equation is:
1(1/(x+3))/(x+(2/(x+3)))

No, it's not: it's an expression.
Calling a random collection of letters and symbols an "equation" is
like fingernails on a blackboard to anyone who knows anything about
math.
I would have got past that, but you gave us no indication at all
what you're supposed to do with this expression, nor what you've
tried already.

Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/ 

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Jim Spriggs science forum Guru
Joined: 24 Mar 2005
Posts: 761

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: GCSE past papers



Sara wrote:
Quote: 
im lookin for the same thing so if sume one replys to you will you let
me know please

The op asked on 14 August 2004. What is the likelihood that they will
be looking for replies now? 

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henry science forum addict
Joined: 24 Mar 2005
Posts: 56

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Trigonomentry equatrions



On 10 Sep 01 09:57:21 0400 (EDT), Saleem Ahmed wrote:
Quote:  Would appreciate if could help me solve the following. I tried it a
few times but was stuck.
Solve
1.
cotx * cosx = 1+sinx
2.
7 secx = 3(cosxtanx)
Thanks,
Saleem

2)
7sec x=3(cos xtan x)
7/cos x=3((cos^2xsin x)/cos x)
7=3cos x(1sin^2 xsin x)
1=7/3sin(pi/2x)(1sin^2 xsin x)
7/3 sin (90x)=1 or sin x(sin x1)=0
x=kpi,kEz or x=3pi/2 +2kpi.KEz
and 7/3 = 2.33333333
bye... 

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ashok kumar science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: IIT screening test question



On 12 Feb 05 03:20:19 0500 (EST), pH wrote:
Quote:  Q: A bear is chasing a man, who reaches the edge of a cliff and jumps
down. The man reaches the ground in 5s. The height of the cliff is
123
metres. What is the colour of the (fur of the) bear?
Note: The above question is not a joke. whoever gets it is a true
genius (max time: 5 minutes). Ideally, the question will not take
more
than 2 minutes to solve.
pH

white because the value of g=9.8 which is at north pole and south pole
which is polar region 

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Peter Webb science forum Guru Wannabe
Joined: 05 May 2005
Posts: 192

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: IIT screening test question



"Stan Brown" <the_stan_brown@fastmail.fm> wrote in message
news:38fq6gF5o1ffoU1@individual.net...
Quote:  "ashok kumar" wrote in alt.math.undergrad:
On 12 Feb 05 03:20:19 0500 (EST), pH wrote:
Q: A bear is chasing a man, who reaches the edge of a cliff and jumps
down. The man reaches the ground in 5s. The height of the cliff is
123 metres. What is the colour of the (fur of the) bear?
white because the value of g=9.8 which is at north pole and south pole
which is polar region
Nonsense. The time of 5 s contains one significant digit. The
variation in the acceleratikon of gravity over the earth's surface
is FAR too small to show at that level of precision.

If you want to get picky, then there aren't likely to any 123 metre cliffs
at the North Pole (as it is a floating ice mass), and the nearest bears to
the South Pole are koala "bears", 5,000 kms away.
However, as is well known, the answer to all questions which involve the
color of bears is "white", just as all questions which require you to "spell
it" are answered "i", "t". 

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abhishek science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: IIT screening test question



On 12 Feb 05 03:20:19 0500 (EST), pH wrote:
Quote:  Q: A bear is chasing a man, who reaches the edge of a cliff and jumps
down. The man reaches the ground in 5s. The height of the cliff is
123
metres. What is the colour of the (fur of the) bear?
Note: The above question is not a joke. whoever gets it is a true
genius (max time: 5 minutes). Ideally, the question will not take
more
than 2 minutes to solve.
pH

You said it came in jee screening. When?
In good spirit.


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Paul Sperry science forum Guru
Joined: 08 May 2005
Posts: 371

Posted: Thu Mar 24, 2005 8:13 pm Post subject:
Re: Sum of series



In article <o8qnm4tivish@legacy>, john <johnmc@aol.com> wrote:
[...]
John conjectured that sum(k/2^k,k = 1..oo) = 2 which was confirmed by
Ashutson and myself. Ever since, I have had the nagging thought that I
should have recognized the problem.
Sure enough.
A sequence of the form
a + (a + d)*r + (a + 2*d)*r^2 + ... + (a + n*d)*r^n + ... is an
"artihmeticgeometric" series and converges to
a/(1  r) + r*d/((1  r)^2) for r < 1.
So for sum(k/2^k,k = 1..oo) take a = 0, d = 1, r = 1/2 and get
a sum of 2.
So there!

Paul Sperry
Columbia, SC (USA) 

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