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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
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 fixed-width 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| - p|b| >= 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.
Bob

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
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: 3y-5x= -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, one-dimensional. You can
simplify it by multiplying both sides by 15.

Algebryonic
Christian Bau

Joined: 01 May 2005
Posts: 77

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: JSH: Time is not on your side

jstevh@msn.com wrote:

 Quote: To me the basic surrogate factoring method, where you factor T and also j is not an elegant solution. However, it's the easiest to show, so I posted the equations, and I've shown I'm right. For even a beginning math student, the math is basically trivial. You people, however, don't seem to have a clue what is going on. Right now, someone can exploit the theory, and factor some really huge numbers. My guess is that the theory can be developed with a distributed solution to the point where you could factor a 2048 bit number in milliseconds. It's anybody's guess how long it would take develop the theory and implementiaon to that point, but I'd guess about two months. Right now, it probably would take someone a couple of days with the BASIC theory and some powerful workstations to factor a 2048 bit number, just with the very basic and inelegant set of equations I've posted the last two days.

We'll leave the honour to you.
Jim Ferry
science forum beginner

Joined: 29 Apr 2005
Posts: 20

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: JSH: Time is not on your side

jstevh@msn.com wrote:

 Quote: The more elegant solution I'm working on involves just factoring T.

Factoring a number by factoring a different number? Devious, yet
brilliant!

 Quote: And it's so elegant it allows a recursive solution, which potentially could factor a 4048 bit number on a single pc in a few hours.

Of course! You can factor T by factoring yet another number, and
factor that number by factoring another . . . the recursion makes my

 Quote: The surrogate factoring method in and of itself is just knowledge, and it's really important knowledge, which would have an impact anyway, but if it gets developed on the fringes and is acknowledged by the mainstream only when something really bad happens, then I assure you the world as you know it, the world of today, will be gone, replaced by who know's what.

It's heartening to hear a genius who's aware of the social implications
of his work. You've done with numbers what the Manhattan Project did
with atoms.

 Quote: Soon enough events will unfold very rapidly. I fear your decision, so I suggest to those who listen that they enjoy the world, whatever their piece of it may be, as much as they can over the next few days, as soon enough, it will pass away, thanks to people who call themselves "mathematicians".

All this is reminding me of a Dan Brown book I read called "Digital
Fortress". In this book, the NSA has fantastically advanced computers
which can easily crack codes with 64-bit keys in a few minutes. So
they can crack any code, because even if someone uses a 1024-bit key,
it takes only a day or so. I know what you're thinking: they must be
using James's algorithm. But no: they're using brute force -- they're
checking every possible key. Now here's the truly diabolical twist:
someone invents a code with a continuously morphing key, and even
though it's a trivial, 64-bit key, the code is uncrackable because even
if you guessed it, it wouldn't be right, because the key would have
changed. If you just enter the correct key it works though, and, if
you haven't guessed, the correct key is just "3" (or
"0000000000000000000000000000000000000000000000000000000000000011" if
you like) -- that's why the guy at the beginning was holding up three
fingers as he was dying under the cruel Spanish sun!

Now in that book, Dan Brown made it quite clear that the world would
essentially end if the code was cracked by The Bad Guys, and, as you
can probably tell, Dan Brown knows his crypto. So dismiss James at

 Quote: The tests were to be mathematical, and they were, given to those who called themselves mathematicians. You have ignored the three previous, but cannot ignore the final one. It is the final test for those who rejected the first three. You were given three chances. You failed them all.

Oh crap! I wasn't even paying attention! Can I take a make-up test?

 Quote: So now, The Hammer is here, and with it, the end of days. The world will be destroyed, and then remade, as foretold.

So no make-up test, then. This reminds me of

"There will be weeping and wailing and gnashing of teeth!"
"What if we have no teeth?"
"Teeth will be provided."

Ah, Reader's Digest . . . Humor *is* the Best Medicine. Will Reader's
Digest be destroyed? And, if so, what will it come back as when the
world is remade? I think it should come back as a sadomasochistic
platypus named Gertrude, but naturally these decisions are yours to
make, not mine.

 Quote: You will be lost, with your children, and then there will be others, and one day they will be tested, and will pass, but that is another story.

Oooh. Tell that other story.

The thing I worry about is that the San Andreas fault is, essentially,
a number, and that when Neferion codes up James's algorithm, he will
proceed to factor that number, and California will to fall into the
sea.

 Quote: James Harris

Oh, James Harris? Really? I kind of thought so.
Cathy Matthews
science forum beginner

Joined: 24 Mar 2005
Posts: 1

Arturo Magidin
science forum Guru

Joined: 25 Mar 2005
Posts: 1838

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Please help me solve this riddle

In article <eiitqe4igqa7@legacy>,
Cathy Matthews <CathyMMatthews@aol.com> wrote:
 Quote: There is a fork in the road. One road leads to heaven and one road leads to hell. Three men are standing at the fork, Bill Clinton, Dali Lami and Hitler. Each man looks and talks the same, they cannot be told apart by sound or appearance. Dali Lama always tells the truth, Hitler always lies and Bill Clinton could do either. You may ask three YES or NO questions. All three will answer the questions. What are the three questions necessary to ask to determine which road goes to heaven?

"The three questions necessary" seems to imply one and only one
solution. This is simply not the case. There are many solutions. In
fact, you can do it with two questions:

For example, you could ask first "Are you Bill Clinton?"

The Dalai Lama will certainly answer "no". Hitler will certainly
answer "yes". Bill Clinton will either answer "yes" or "no".

If you get two "yes" and one "no", then you have identified the Dalai
Lama (the only one who said "no"). Ask next "Which road leads to
Heaven?", and simply follow whichever road the Dalai Lama tells you.

If you get two "no" and one "yes", then you have identified Hitler
(the only one who said "yes"). Ask next "Which road leads to Heaven?"
and follow the road which Hitler does NOT point to.

--
======================================================================
"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
Arturo Magidin
science forum Guru

Joined: 25 Mar 2005
Posts: 1838

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Please help me solve this riddle

In article <cur5tu\$oom\$1@agate.berkeley.edu>,
Arturo Magidin <magidin@math.berkeley.edu> wrote:
 Quote: In article , Cathy Matthews wrote: There is a fork in the road. One road leads to heaven and one road leads to hell. Three men are standing at the fork, Bill Clinton, Dali Lami and Hitler. Each man looks and talks the same, they cannot be told apart by sound or appearance. Dali Lama always tells the truth, Hitler always lies and Bill Clinton could do either. You may ask three YES or NO questions. All three will answer the questions. What are the three questions necessary to ask to determine which road goes to heaven? "The three questions necessary" seems to imply one and only one solution. This is simply not the case. There are many solutions. In fact, you can do it with two questions:

In fact, you can get away with one question:

"If I were to ask someone whose answers are the opposite of yours
whether this is the Road to Heaven, what would he reply?"

Assume that you are indeed pointing to the Road to Heaven. Then:

(i) The Dalai Lama will say "No", since that is the response Hitler
would have given you.

(ii) Hitler will also say "No": the Dalai Lama would have said
"yes", but Hitler is lying to you about it.

(iii) Clinton would either say "Yes" or "No"; it does not matter.

Assume you are pointing to the Road to Hell. Then:

(i) The Dalai Lama will say "Yes", since Hitler would have told you
it is the Road to Heaven.

(ii) Hitler will also say "Yes", because the Dalai Lama would have
told you "no".

(iii) Clinton will either say "Yes" or "No", it does not matter.

In either case, you will get at least two "Yes"s if the road is the
road to Hell, and at least two "No"s if the road is the road to
Heaven. So simply take a tally, and do the opposite of what the
majority claims.

--
======================================================================
"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
henry

Joined: 24 Mar 2005
Posts: 56

Posted: Thu Mar 24, 2005 8:13 pm    Post subject: Re: Differencial calculus in R

On 9 Feb 05 17:59:36 -0500 (EST), Henry wrote:
 Quote: What will be the result of f(x)=squ(xE2-3x+2) study? An dhow i calcalute the derifative for g(x)=ln[tg(pi/4+x/2)] And if it worsens the best is h and i so that is needed the calculation of one primitive to [exsin x) and (x^2+x-3)Ee2x ..so do you have the answer?

g(x)= ln[tg(pi/4 + x/2)]

dg/dx = (sec^2(pi/4 + x/2)*1/2/tg(pi/4 + x/2))
dg/dx= 1/2 * ((1+tg^2(pi/4 + x/2)/tg (pi/4 + x/2)
dg/dx= 1/2[cotg (x/2 + pi/4)+tg(x/2+ pi/4)]
dg/dx= 1/2*(cos (pi/4+x/2)/sin (pi/4 + x/2)+
+(sin (pi/4+x/2)/cos (pi/4 + x/2)
dg/dx= 1/(2sin(pi/4+x/2)cos(pi/4+x/2)=
dg/dx=1/sin(pi/2 + x)
dg/dx=1/cos x

therefor,

d2^g/d^2x=sec x.tg x

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