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server
science forum beginner

Joined: 24 Mar 2005
Posts: 26

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Trigonometric equations

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Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

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

In article <MPG.1c6a322ece80edb498a070@news.easynews.com>,
Jeff Heikkinen <no.way@jose.org> wrote:

 Quote: Q: Are we not men? A: We are Virgil! In article , Jeff Heikkinen wrote: Greetings. As most of you probably know, linear diophantine equations are (at their simplest) of the form ax + by = c where all five variables must be integers. Normally you want to find one or more solutions for x and y, for given values of a, b and c. It can be shown that if the greatest common denominator of a and b divides c, such a problem has infinitely many solutions, while if this is not the case, it has none. (For example, no integers x and y satisfy 3x + 9y = 4.) I understand all that. I also understand why once you have one solution, you have all of them. That is simple high school algebra, if you have one solution for x and y, you can generate an arbitrarily large number of other ones by adding b to x and subtracting a from y or vice versa: a(x+b) + b(y+a) = ax + ab + by -ab = ax + by (look familiar?) = c What I don't understand is how, in general, to get one solution in the first place! What is a good general-purpose method for doing this? I would guess it would involve the Euclidean algorithm but the application escapes me. Assume that a and b are relatively prime, and that the a, b and c are too large to just solve by inspection (at least three digits). See, for example, http://mathworld.wolfram.com/BlankinshipAlgorithm.html So basically, you solve for the GCD (which is one, given my above assumption), then using the matrices you generated along the way, work your way back up until you've expressed the number one in terms of the original a and b, then multiply through by c? 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.
Guest

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Parametric equations

Jo wrote:
 Quote: Thank you Matt I'm starting to understand the mechanics of it. But take xy =16 How do I go about this one? x=16/y = t > y = 16/t y= 16/x = t > x = 16/t But this is wrong from the book's answer. Sorry, to go on... Jo

Just to recap on the basics, given xy = 16, or y = 16/x, there are, as
always, infinitely many parameterisations. The simplest (trivial) one
is

x = t
y = 16/t

And then in general

x = f(t)
y = 16/f(t)

will do, where f is (almost) ANY function of t. (The only caveat is
that you need to be alert to the ranges/domains of the functions
involved. For example, if the original function allows x < 0, and you
set x = t^2 then you won't be able to get the negative values of x
without allowing t to go imaginary. Nevertheless, the "spirit" of the
statement is true. You can choose any old function you like.)

The next question is what your book thinks is the "right answer".
Generally speaking there IS no "right" answer, but for certain special
well-known curves there is a "standard parameterisation". For example,
the standard parameterisation for the unit circle is x = cos(t), y =
sin(t). To get the "right" answer according to the book you need to
learn what these standard parameterisations are.

xy = 16 is a rectangular hyperbola, and my best guess is that the book
is looking for x = kt, y = k/t, for some k to be found. If I'm wrong

Assuming that this is correct, simply find from x = kt, y = k/t that xy
= k^2. Comparing with xy = 16 it's obvious that k = 4. This then gives
the parameterisation x = 4t, y = 4/t.

I'm not sure what you're doing with this:

 Quote: x=16/y = t > y = 16/t y= 16/x = t > x = 16/t

Each of these gives a valid (though trivial) parameterisation
individually, but you can't combine them. You can have EITHER

x = 16/y = t => x = t, y = 16/t
OR

y = 16/x = t => x = 16/t, y = t

but not both at the same time.
Rich Carreiro
science forum beginner

Joined: 03 May 2005
Posts: 28

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Parametric equations

hello@there.com (Jo) writes:

 Quote: Thank you Matt I'm starting to understand the mechanics of it. But take xy =16 How do I go about this one?

I would start simple:
x = t

then
y = 16/x = 16/t

So a parametric version of xy=16 is:
x = t
y = 16/t

And just to point out there's no one right answer,
you could do:
x = 4t
y = 16/x = 16/(4t) = 4/t

Thus
x = 4t
y = 4/t

--
Rich Carreiro rlcarr@animato.arlington.ma.us
Guest

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

Torsten Hennig wrote:
 Quote: On 02 Feb 2005 04:07:50 GMT, Ali G wrote: Hi, Can someone 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? Any help is appreciated. Hi, 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. So you get 1 = 1*e^(-1) - b*1^2 as an equation to determine b. If the resulting function f is a valid distribution function depends on whether it is monotonically increasing on [0,1] or not. Best wishes Torsten.

You're talking about a cumulative distribution function here, whereas I
think the OP meant a probability density function.
Guest

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

Ali G wrote:
 Quote: Hey, thanks for reply. I solved the integral and got the same answer as your b. But how do you know if this is a valid probability function? What makes it valid?

For f(x) to be a valid (continuous) probability density function, the
integral of f(x) over the stated domain must be equal to one, and we
must also have f(x) >= 0 everywhere in this domain. I think that's
pretty much it unless you want to get massively technical.
Arturo Magidin
science forum Guru

Joined: 25 Mar 2005
Posts: 1838

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

In article <e5pb1aq7sgh8@legacy>, Steve <thethreemorgans@cox.net> wrote:
 Quote: Barry, our best dog groomer, noticed that Bert and Bess both had fleas. If one flea jumps from Bert to Bess, then both will have the same amount. However, if one flea jumps from Bess to Bert, then Bert will have five times the amount as Bess. How many fleas does each dog have before jumping takes place?

Let r be the number of flees on Bert, s the number of flees on Bess.

First statement says that r+1 = s-1.

Second statement says that r-1 = 5(s+1).

Solve for r and s.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu
William Elliot
science forum Guru

Joined: 24 Mar 2005
Posts: 1906

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

On Wed, 2 Feb 2005, 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

(13*12/2) / 52*51

 Quote: 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. If two cards of different suits are drawn, when are they a pair?

3 / 3*13 = 1/13
Sara
science forum beginner

Joined: 24 Mar 2005
Posts: 1

 Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: GCSE past papers im lookin for the same thing so if sume one replys to you will you let me know please
Dave L. Renfro
science forum Guru

Joined: 29 Apr 2005
Posts: 570

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Imaginary numbers...

THeSkeptic <get1@aol.com>
[alt.math.undergrad: February 19, 2005 00:19:12 -0500 (EST)]

wrote:

 Quote: I have always wondered why they invented imaginary numbers for negative square roots, when it seemed simpler to me to simply define a negative square number as a the product of a number and its inverse. A negative square root would then have two solutions, a number and its inverse. What is wrong with this idea?

I think you mean "square root of a negative", not "negative
of a square root". Also, the product of a number and its
inverse which, if you are, you should have been explicit
would dictate you meant the multiplicative inverse. So let's
assume you meant additive inverse. Then by the "negative
square of 3" you mean (+3)*(-3) = -9. I don't see how this
helps. This process just spits out real numbers again,
and so equations like x^2 = -1 still have no solution.

Dave L. Renfro
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

Google's first hit searching for "GCSE past papers":
http://www.gcsemathspastpapers.com/.

Since you're asking in a mathematics group I assume it's mathematics
papers that you want. But tell me, are you also studying English?
Virgil
science forum Guru

Joined: 24 Mar 2005
Posts: 5536

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

In article <d187uj1kx7x0@legacy>, 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

What is the probability, after drawing a card, that the second card has
the same face value?

Count outcomes! How many cards are left? How many of those cards will
produce a pair?

 Quote: b) what is the conditional probability they constitute a pair given that they are of different suits

What is the probability, after drawing a card and removng all other
cards of the same suit from the deck, that the second card will have the
same face value?

Count outcomes! How many cards are left? How many of those cards will
produce a pair?
Guest

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

Jeff Heikkinen wrote:
 Quote: Q: Are we not men? A: We are Virgil! In article , Jeff Heikkinen wrote: So basically, you solve for the GCD (which is one, given my above assumption), then using the matrices you generated along the way, work your way back up until you've expressed the number one in terms of the original a and b, then multiply through by c?

Is there a way to speed up the "work your
 Quote: way back up" part using similar matrix operations?

You don't need to find the GCD first, then work back.
Say you wanted to solve
128X + 62 Y = 4.

Left matrix multiplication is basically
and row swapping. So you can dispense with
the actual matrices. If you multiply the
identity matrix by the matrices you find,
you are just performing the same row operations on
it.
You can write out the calculations like this:

128 1 0
62 0 1 R1 - 2*R2

4 1 -2
62 0 1 swap

62 0 1
4 1 -2 R1 -15*R2

2 -15 31
4 1 -2 swap

4 1 -2
2 -15 31 R1 - 2*R2

0 31 -64
2 -15 31 Swap

2 -15 31
0 31 -64

2 divides 4 so you know there
are solutions. Othewise you
know there is no solution.

If the determinant of the matrix is
not equal to +/-1 at any stage, you
know you have made a mistake.
Ari
science forum beginner

Joined: 24 Mar 2005
Posts: 30

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: uniqeness proofs are totally useless

"fishfry" <BLOCKSPAMfishfry@your-mailbox.com> wrote in message
news:BLOCKSPAMfishfry-B4C831.23114927012005@comcast.dca.giganews.com...
 Quote: In article , "david k" wrote: What is the point of proving that something is unique? If a certain mathematical object does what it is supposed to do, what is the point of proving that there is only one of it? Every group theory book starts with the axioms and then goes on to prove that the identity element is unique. I fail to see the point of doing this. Who cares if there is more than one identity as long as they act like the identity what should it matter? If there were two different identities, how would you define inverses?

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

Perhaps inverses are totally useless, too.

Aristotle Polonium

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Ari
science forum beginner

Joined: 24 Mar 2005
Posts: 30

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: repeated eigenvalues

"Jeremy Watts" <jwatts1970@hotmail.com> wrote in message news:gs1Ld.795\$vC.500@newsfe3-win.ntli.net...
 Quote: Stumped> wrote in message news:omsov018e3r507ub95csgouvle57fpl9h6@4ax.com... On Sat, 29 Jan 2005 11:55:22 GMT, "Jeremy Watts" jwatts1970@hotmail.com> wrote: is there any way of telling whether a matrix will have repeated eigenvalues? and if so, which of the eigenvalues found are the repeated ones? maybe see if it has a common row sum or collumn sum then compare that to its trace? yeah thats the sort of thing i've been looking at - i've noticed that some matrices with linearly dependent rows or columns have repeated eigenvalues, so i was wondering whether there is some underying theory there, but cant seem to find one in books or out there in net land yet... :)

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

It can be shown that if a matrix has distinct eigenvalues then its
eigenvectors are linearly independent.

The contrapositive of this statement is: if the eigenvectors are linearly
dependent then the eigenvalues are not distinct.

Of course, there are matrices with repeated eigenvalues that have
linearly independent eigenvalues (e.g., the identity matrix).

http://www.ma.iup.edu/projects/CalcDEMma/JCF/jcf.html

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

Aristotle Polonium

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +

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