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 Science, Math, and Philosophy Discussions regarding science, math, and/or philosophy.

09-24-2009, 09:01 AM   #101
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 I figured it out I think. I'm getting 3 weighings, which is consistent with log base 2 of 8. So, if the balls are numbered 1-8, take 1-3 and weigh it against 4-6. If they are equal, weigh 7v8. If these are equal, we're done. If they are different, weigh one against ball 1. Then you can figure out which one is the odd one out, and whether or not its heavier or lighter. If the 3-3 weighing is different, without loss of generality, assume 123 is the heavier group of balls. Now, weigh 1 and 2 against 3 and 7 (since we know 7 is a normal ball). If they are different, (say 1 and 2 are heavier), we can weigh them against each other to find the wrong ball. If they are the same, weigh 4 and 5 (two of the lighter balls). If they're the same, 6 is the light ball. Otherwise, you will know which ball is lighter. If we generalize this to n balls, with n-1 identical, would the number of weighings be the smallest integer greater than or equal to log base 2 of n?
Instead, once you know that either 1, 2, or 3 is the heavy ball, just weigh 1 v 2.

edit: sorry we're solving different problems should read more carefully.

What you wrote looks reasonable.

 09-24-2009, 08:46 PM #102 jono adept   Join Date: Jan 2005 Location: West Coast Posts: 705 Re: The Official Math/Physics/Whatever Homework questions thread Trying to help a friend out with this, but I'm not sure how to standardize this data (ie comparing two normal distributions if they have different arithmetic means and/or standard deviations) we have this data on 2 types of bacteria: -Initial Number of Bacteria at Time Zero -Number of Human Cells at Time Zero (Cells that will be infected) -Multiplicity of Infection (Initial Number of Bacteria divided by Number of Human Cells) -Number of Bacteria after 3 hours (Time 3-hr Supernatant) -Number of Bacteria attached and invaded into Human Cells after 3 hours (Time 3-hr Cells) -Number of Bacteria invaded after 6 hours (GEN) -Number of Bacteria attached (Time 3-hr Cells minus GEN) We are looking at: 1) Adhesiveness- how well the bacteria sticks to the surface of the human cell 2) Invasiveness- how well the bacteria gets inside the human cell I'm somewhat confused because if this was a straight stats problem it'd be cake but I don't see how I can compare/configure any means, standard deviations, or normal-distribution graphs with this info. thanks
 09-25-2009, 06:48 AM #103 myammy enthusiast   Join Date: Jul 2007 Location: epdog on cr Posts: 52 Re: The Official Math/Physics/Whatever Homework questions thread Stats 410 question Suppose X and Y are independent exponentially distributed random variable with mean 5 and 2 respectively. Find the pdf of w=X+Y Calculate P[w>10] we have just covered convolution, so this problem can be done using the cdf method or convolution. My integrals are comign out weird. Thanks edit: Sorry I missed the sticky. Here is my attempt i used convolution to get Integral from 0 to Infinity e^(-x/2) e^(x-w)/5) w.r.t. x this gives 10/3 e^(-w/5) as a pdf. What do you guys think? Last edited by myammy; 09-25-2009 at 06:52 AM. Reason: work
 09-26-2009, 01:49 PM #104 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread Can someone tell me how tricky this is? ---- Prove that the function on the open unit disk... f(b, a) = ∥(b - a) / (1 - āb) ∥ ...is a metric. ---- I don't want to spend a lot of time on it, but I'd like to understand the basic idea. (Viz.---how to show f(b,c) ≤ f(b,a) + f(a,c). Inequalities own my soul.)
09-26-2009, 03:56 PM   #105
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Pyromantha Neither. There is no 'right e' here. The point is that for *any* choice of e, there exists an N for which the inequality is satisfied by a_N, and all subsequent terms, if the sequence does indeed converge to l. Perhaps a concrete example would help. Take a simple sequence a_n = 1/n, and we suspect that the limit is 0. Suppose we choose e arbitrarily to start with, lets say we choose e = 1. Well then, the inequality |a_n-l| < e becomes |1/n| < 1, which is satisfied for n >= 2. If we chose e = 1/10, then we need |1/n| < 1/10, which is satisfied for n>=11. For *any* choice of positive e, we need |1/n| < e, which is satisfied for n >= 1/e. So all the terms after a_1/e satisfy the inequality regardless of the choice of e, and so a_n converges to zero.
if you are trying to prove that a sequence doesn't converge, then you can pick an epsilon and show the inequality can't be satisfied. since the definiton says the inequality holds for all epsilon, finding an epsilon that doesn't work is enough to show that the sequence doesn't converge

 09-26-2009, 09:47 PM #106 DrQian grinder   Join Date: Sep 2008 Location: Bed Stuy Posts: 406 Re: The Official Math/Physics/Whatever Homework questions thread Hello. I am doing a project on the construction of the real numbers from the rationals, using Cauchy sequences. More info here: http://en.wikipedia.org/wiki/Constru...e_real_numbers I'm currently trying to show that the operation of division (or equivalently, taking the reciprocal) is well defined for all non-zero equivalence classes. I think I need to show that given (a_n) -> (b_n), (a_n)-/-> 0, we can conclude that 1/(a_n) -> 1/(b_n). I'm sure this is fairly easy, but I just can't see how to do it without (a_n) or (b_n) having a limit. Any help much appreciated, thanks.
 09-27-2009, 04:58 AM #107 Pyromantha veteran   Join Date: Dec 2007 Posts: 2,227 Re: The Official Math/Physics/Whatever Homework questions thread Are the a_n and b_n sequences? What is the equivalence relation '->' (assuming that was an equivalence relation?). I thought initially it might be 'converge to the same limit', but this doesn't make sense after you said you couldn't do it without them having a limit ah wait nm should have checked the link: Two Cauchy sequences are called equivalent if and only if for every rational ε > 0, there exists an integer N such that |xn -yn|<ε for all n > N. Back to thinking about it edit: why can't we drop all the terms for which a_n or b_n are zero and do a term-by-term division? Last edited by Pyromantha; 09-27-2009 at 05:06 AM.
09-27-2009, 05:06 AM   #108
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Pyromantha Are the a_n and b_n sequences? What is the equivalence relation '->' (assuming that was an equivalence relation?).
They're Cauchy sequences of rationals. Two such sequences (a_i), (b_i) are equivalent if for any rational e > 0 there is some n such that for all m > n, |a_m - b_m| < e.

The question is how do you prove that (a_i/b_i) is also a Cauchy sequence, etc.

 09-27-2009, 11:48 AM #109 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread Let me re-frame that question. Suppose ϕa(z) is the Mobius transformation ϕa(z) = (z-a) / (1-āz). Then why does... |ϕz(x)| ≤ |ϕy(x)| + |ϕz(y)| ...for all x, y, z ∈ↁ, the unit disk?
 09-27-2009, 11:56 AM #110 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread In general, can anything interesting be said about how conformal self-maps of the unit disk affect the Euclidean norm of points in the disk? Is there some relationship to the characteristic constant? WTF, why can't I find anything on the entire interwebs about this?!
 09-27-2009, 12:48 PM #111 Pyromantha veteran   Join Date: Dec 2007 Posts: 2,227 Re: The Official Math/Physics/Whatever Homework questions thread What do you get if you write x = a+bi, y = c+di, z = e+fi and then expand the r.h.s of that inequality. (I have no idea, it's just what I would try if I was lacking any insight). Perhaps you can prove the triangle inequality by brute force like that for this map?
09-27-2009, 01:22 PM   #112
Subfallen
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Pyromantha What do you get if you write x = a+bi, y = c+di, z = e+fi and then expand the r.h.s of that inequality. (I have no idea, it's just what I would try if I was lacking any insight). Perhaps you can prove the triangle inequality by brute force like that for this map?
I don't think that's a feasible strategy; the expression is extremely unwieldy. Since I've become obsessed with this problem (extra credit in a class where I don't need the points), I've spent hours manipulating the equation without any progress.

I've dug up only one suggestive fact during my search. Under the Poincaire metric on the disk, one has...

dρ(z, w) = dρ(0, |ϕz(w)|)

...as shown on pages 47-48 of Complex Analysis: The Geometric Perspective. But my problem comes from a Real Analysis class, so I doubt that's even relevant here. :/

 09-27-2009, 01:40 PM #113 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread In other words, we know from complex analysis: --- dρ(x, z) ≤ dρ(x, y) + dρ(y, z) => dρ(0, |ϕz(x)|) ≤ dρ(0, |ϕy(x)|) + dρ(0, |ϕz(y)|) => ½log[(1+|ϕz(x)|) / (1-|ϕz(x)|)] ≤ ½log[(1+|ϕy(x)|) / (1-|ϕy(x)|)] + ½log[(1+|ϕz(y)|) / (1-|ϕz(y)|)] ...or written another way... tanh-1|ϕz(x)| ≤ tanh-1|ϕy(x)| + tanh-1|ϕz(y)| --- And we want to prove: --- |ϕz(x)| ≤ |ϕy(x)| + |ϕz(y)| ---
09-27-2009, 02:26 PM   #114
jason1990
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Subfallen Can someone tell me how tricky this is? ---- Prove that the function on the open unit disk... f(b, a) = ∥(b - a) / (1 - āb) ∥ ...is a metric. ---- I don't want to spend a lot of time on it, but I'd like to understand the basic idea. (Viz.---how to show f(b,c) ≤ f(b,a) + f(a,c). Inequalities own my soul.)
You also used the notation ϕa(z) = (z - a) / (1 - āz), so that f(b, a) = |ϕa(b)|. First show algebraically that
ϕab(z)) = wϕc(z),
where
$w = \frac{1+a\overline{b}}{1+\overline{a}b},$
and
$c = \frac{a + b}{1+a\overline{b}} = \phi_{-b}(a).$
Note that |w| = 1. Use this to show that
$f(\phi_c(a),\phi_c(b)) = |\phi_{\phi_c(b)}(\phi_c(a))|= |w\phi_b(a)| = f(a,b).$
Finally, given z1, z2, and z3, show that there exists a map of the form wϕa(z) sending z2 to 0 and z3 to some real number r > 0. Use this to show that it suffices to prove
f(z, r) ≤ f(z, 0) + f(0, r).
Then prove this last bit, which should be easy.

 09-27-2009, 02:42 PM #115 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread jason1990 - Thank you, thank you, thank you! Words fail, so I must turn to emoticons...!!!
09-27-2009, 09:18 PM   #116
Subfallen
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by jason1990 f(z, r) ≤ f(z, 0) + f(0, r). Then prove this last bit, which should be easy.
Quick question. By "should be easy", did you mean "easy for jason1990"? Or easy for a beginner like me? Because I've spent the last five hours on this line, with no apparent progress.

as;dlfkja;sdlfkja;sldkfj;aslkjdfasd;lkfjas;dlkjfa; sldkjfas;ldkjf

 09-27-2009, 09:50 PM #117 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,229 Re: The Official Math/Physics/Whatever Homework questions thread Heck, I can't even prove that for a, b ∈ (0, 1): |(a-b)/(1-ab)| ≤ a + bWhich surely begs the question of why I bother living. WTF is the point?
09-28-2009, 12:08 AM   #118
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Subfallen Heck, I can't even prove that for a, b ∈ (0, 1): |(a-b)/(1-ab)| ≤ a + bWhich surely begs the question of why I bother living. WTF is the point?
You want to make some reductions. First, you may as well assume that a>b. Then the inequality is just (a-b)/(1-ab) \leq a+b. Multiplying through by 1-ab (which is positive), you get the inequality a-b \leq (a+b)(1-ab) = a+b-a^2b-ab^2, which is equivalent to proving 0 \leq 2b-a^2b+b^2a. Can you see why that final inequality is true?

09-28-2009, 12:29 AM   #119
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Subfallen Quick question. By "should be easy", did you mean "easy for jason1990"? Or easy for a beginner like me? Because I've spent the last five hours on this line, with no apparent progress. as;dlfkja;sdlfkja;sldkfj;aslkjdfasd;lkfjas;dlkjfa; sldkjfas;ldkjf
I didn't even try it. I just figured it was a lot easier than the original. Now that I've tried it... it's a lot harder than I remembered it. Maybe there is a slicker way, but here's something. Be sure to check the details because it could be wrong. Let z = x + iy. Assume x > 0. (Hopefully, you can handle x < 0 on your own.) Then
$\bigg|\frac{z-r}{1-rz}\bigg|^2= \frac{(x^2+y^2+r^2)-2rx}{(1+r^2x^2+r^2y^2)-2rx}$
$f(r) = (1+r^2x^2+r^2y^2) - (x^2+y^2+r^2)$
$f'(r) = 2rx^2 + 2ry^2 -2r = 2r(|z|^2-1)<0$
Since f(1) = 0, this implies f(r) > 0 for all r in (0,1). Hence,
$(1+r^2x^2+r^2y^2) > (x^2+y^2+r^2).$
Now check that if 0 < A < B < C, then (B - A)/(C - A) < B/C. (You can check this with calculus like above.) Applying this to the above gives
$\frac{(x^2+y^2+r^2)-2rx}{(1+r^2x^2+r^2y^2)-2rx}\le \frac{x^2+y^2+r^2}{1+r^2x^2+r^2y^2}\le |z|^2 + r^2 \le (|z| + r)^2.$
Edit: Okay, calculus is obviously not necessary for (B - A)/(C - A) < B/C. Also,
$\frac{x^2+y^2+r^2}{1+r^2x^2+r^2y^2}= \phi_{-r^2}(|z|^2),$
which is in the unit disk, so the numerator must be smaller than the denominator, making calculus unnecessary even for that.

Last edited by jason1990; 09-28-2009 at 12:52 AM.

09-28-2009, 01:12 AM   #120
Subfallen
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by blah_blah You want to make some reductions. First, you may as well assume that a>b. Then the inequality is just (a-b)/(1-ab) \leq a+b. Multiplying through by 1-ab (which is positive), you get the inequality a-b \leq (a+b)(1-ab) = a+b-a^2b-ab^2, which is equivalent to proving 0 \leq 2b-a^2b+b^2a. Can you see why that final inequality is true?
(Red '+' should be '-', yes?) Is it just because 2b-a^2b-b^2a = b(1-a^2)+b(1-ba)? Meh...I simply MUST learn basic inequality skills. Are these good first books: Inequalities by Hardy, Littlewood, Polya; and The Cauchy-Schwarz Master Class by Steele? Would you recommend others?

jason1990 -

That explains a lot...I'm too frustrated to focus very well right now, but the basic theme seems clear. I will try to work through it on my own tomorrow. (I did recognize the x < 0 case is trivial, but algebraic manipulation didn't seem to help in the least when x > 0.)

If I ever learn this material, I will seek out people to help in your name.

Last edited by Subfallen; 09-28-2009 at 01:17 AM.

09-28-2009, 01:29 AM   #121
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Also, $\frac{x^2+y^2+r^2}{1+r^2x^2+r^2y^2}= \phi_{-r^2}(|z|^2),$which is in the unit disk, so the numerator must be smaller than the denominator, making calculus unnecessary even for that.
Or just...
|1-zr|^2 - |z-r|^2 = (1-|z|^2)(1-r^2) > 0
...yes? I think the (B - A)/(C - A) < B/C step is most opaque to me, since I completely failed to notice it after five hours.

09-28-2009, 08:33 AM   #122
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Subfallen Or just...|1-zr|^2 - |z-r|^2 = (1-|z|^2)(1-r^2) > 0...yes?
This looks right, and it shows that ϕr(z) is in the unit disk. A similar inequality holds when r is complex. But this should really be proven prior to this last step we are discussing. If nothing else, z in this last step is really wϕz2(z1), and we would like to know that |z| < 1.

I had a friend who worked on some very complicated research, but his papers were always much shorter than mine. I asked him why he thought that was. His answer was that much of his work involved long strings of equalities. According to him, most of these could be omitted and left to the reader, because there are only a limited number of ways to get from one end of a string of equalities to the other. My work, on the other hand, involves long strings of inequalities. And, again according to him, these need to be explicitly described to the reader, because the reader would not be able to easily reproduce them.

I am not sure how much truth is in my friend's comments. I do, however, feel there is a lot of art in doing inequalities. Whatever the case may be, there is no denying that inequalities are at the heart of analysis.

 09-28-2009, 08:30 PM #123 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread 3.4.15 Let Y be the random variable described in question 3.4.3. Define W = 3Y+2. Find f_W(w). For which values of w is f_W(w) does not equal 0. 3.4.3 let f_Y(y)=(3/2)y^2, -1<=y<=1 I guess my problem here is he part define W=3Y+2 am I substituting the equation in 3.4.3 for Y in 3.4.15?
 09-28-2009, 09:22 PM #124 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread IM confused on multiplying cycles 2 examples (1,4,5,6)(2,1,5) A={1,2,3,4,5,6} multiplying we get (1 2 3 4 5 6) (6 4 3 5 2 1) 2ND EXAMPLE (2,1,5)(1,4,5,6) multiplying we get (1 2 3 4 5 6) (4 1 3 2 6 5) same set A, If you could explain this to me it would be great, my book is a little vague. Last edited by smcdonn2; 09-28-2009 at 09:31 PM.
09-29-2009, 06:38 PM   #125
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 IM confused on multiplying cycles
(1,4,5,6)(2,1,5)

Always go right to left in terms of cycles. In each individual cycle go from left to right.

Start on the right hand side. 2 goes to 1, where does 1 go in the next cycle? It goes to 4.

1 goes to 5, then 5 goes to 6, and 6 doesn't go anywhere. So 1 goes to 6

5 goes to 2, and 2 doesn't go anywhere else. So 5 goes to 2.

3 goes to itself since it does not appear in any of the cycles.
4 goes to 5.
6 goes to 1.

(1 2 3 4 5 6)
(6 4 3 5 2 1)

You can always check by re-writing the cycles like
(1,4,5,6) =
1,2,3,4,5,6
4,2,3,5,6,1

(2,1,5)=
1,2,3,4,5,6
5,1,3,4,2,6

Then multiplying, but you will find with some practice (1,4,5,6)(2,1,5) is much faster.

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