Wikipedia:Reference desk/Mathematics

June 13

Regression

How would one perform a linear regression of Y on X and X² (simultaneously) using Excel?
Zain Ebrahim 11:06, 13 June 2007 (UTC)

Be very careful if you are using Excel for something like this. Excel has numerous errors in its statistical functions and Microsoft has a poor track record when it comes to fixing them. See, for example, [1], [2] and [3]. A short quote from the latter: "In its Excel XP release, Microsoft attempted to fix some statistical problems, but it did not do a good job (McCullough and Wilson, 2002). This failure presaged Microsoft’s attempt at a major overhaul with Excel 2003. While it fixed many functions, it failed to fix many others."

If you need accurate results, you would be much better off investing time and perhaps even some money in serious statistical software such as R (free) or the SAS System (not free). If you prefer to use a spreadsheet, Gnumeric has a better track record on accuracy than Excel.[4] If you must use Excel, don't use versions earlier than Excel 2003, and read the relevant parts of this site first: Errors, Faults and Fixes for Excel Statistical Functions and Routines. The LINEST function is Excel's regression workhorse. -- Avenue 13:05, 13 June 2007 (UTC)

Don't be too put off: for the simple regression you describe I'm sure Excel will do an adequate job. The papers referred to above say that no problems have been found with the linear regression routines in Excel 2003 and onwards, and in previous ones it was only some of the peripheral output that was questionable. You don't need a sledgehammer to crack a peanut. To actually answer your question:

Set out your Y, X and X² data in three successive columns. Then go to the menu item Tools|Data Analysis. If Data Analysis doesn't appear under Tools, you will first have to go to Tools|Add Ins and tick the Analysis ToolPak box. Under Tools|Data Analysis select Regression. In the following dialog box, identify your column of Y values in the Input Y Range box, and your double column of X and X² data in the Input X Range. The Help button in the dialog box will explain the other options. Nominate an output location for the results, and click OK. Good luck. --Prophys 10:42, 14 June 2007 (UTC)

abstract algebra

if G/Z(G)is cyclic then show that G is abelian

Perhaps reading Inner automorphism#The inner and outer automorphism groups will give you a hint.  --Lambiam^Talk 15:52, 13 June 2007 (UTC)

Just think about what it means for a group to be abelian; you need to show that for all $x,y\in G$, $xy=yx$. Since you know that G/Z(G) is cyclic, you can express any element in G/Z(G) as a power of a generator for G/Z(G). Think about what that means for arbitrary elements of G... –King Bee ^{(τ • γ)} 17:41, 13 June 2007 (UTC)

It's funny. A few hours ago I was working on an exercise in which I analyze the characters of a non-abelian group of order $p^{3}$, and had to effectively prove this statement in order to conclude that $|Z(G)|=p$. Perhaps the OP is a classmate :-) -- Meni Rosenfeld (talk) 20:01, 13 June 2007 (UTC)

abstract algebra

prove that Z/143Z is cyclic

Do your own homework. The reference desk won't give you answers for your homework, although we will try to help you out if there's a specific part of your homework you don't understand. Make an effort to show that you've tried solving it first. Dep. Garcia ( Talk + | Help Desk | Complaints ) 15:54, 13 June 2007 (UTC)

This is almost by definition. Take the time to read and understand cyclic groups (or any other topic that builds along the way) and you will find abstract algebra much easier. iames 15:56, 13 June 2007 (UTC)

(after edit conflict) What about: prove that Z/nZ is cyclic if n ≠ 0. See also Modular arithmetic#The ring of congruence classes.  --Lambiam^Talk 15:59, 13 June 2007 (UTC)

I suggest the following strategy: Prove that Z/3Z is cyclic. Prove that Z/4Z is cyclic. (Brute force may suffice.) Combine what you have have learned to attack Z/143Z. Question: How many groups have order 143? Have fun! :-) --KSmrq^T 16:26, 13 June 2007 (UTC)

I don't know much about group theory, and even less about Galois field theory, so maybe I am missing something here, but doesn't 11 * 13 = 143? Baccyak4H (Yak!) 16:35, 13 June 2007 (UTC)

Good catch; thanks. (That's what comes from using Factor instead of FactorInteger.) The fact that 1−4+3 = 0 shows that 143 is a multiple of 11. Fixed. --KSmrq^T 18:28, 13 June 2007 (UTC)

Composite in all bases

Assuming good faith on the part of KSmrq, I wondered if there exists a base > 5 for which 143 (expressed as 143 in that base) was prime, and at least up to base 32, the answer is no. Are there any results about this sort of thing? iames 16:47, 13 June 2007 (UTC)

I guess if I thought about it for 5 seconds I'd see how $n^{2}+4n+3=(n+3)(n+1)$ ... iames 16:59, 13 June 2007 (UTC)

I just recognized it as 12² − 1², which is of course (12+1)*(12-1). But ignore my comment if "prime" means something different in Galois theory. And even if it does not, I would definitely assume good faith; it wasn't like he said "60". Baccyak4H (Yak!) 17:09, 13 June 2007 (UTC)

One can assume good faith without assuming good judgment (a common Wikipedia confusion)! No sophisticated theory of bases or Galois fields is involved, I just made a mundane mistake. <:-O But the number of groups of order 143 remains of interest (for those interested in such things). --KSmrq^T 18:38, 13 June 2007 (UTC)

It turns out to be just 1[5] $Z_{11}\times Z_{13}$. I got side tracked into looking at groups whos order is the product of two distinct primes, p, q say, with p>q. A brief test on a few orders seems to suggest that if p % q == 1 then there will be two groups of order pq, but if p%q!=1 there is only a single group. Does any one know if this is true in general if so why? Also there seems to be two group of order 21 whats the other one look like? --Salix alba (talk) 00:06, 14 June 2007 (UTC)

See "example of groups of order pq". PlanetMath..  --Lambiam^Talk 00:23, 14 June 2007 (UTC)

That example generalises, of course: if p and q are primes, p < q, and p does not divide q-1, then there is a unique group of order pq. If p does divide q-1, then there is another group (a semidirect product of C_p and C_q). I'm pretty sure there's only one nontrivial semidirect product, but I can't remember if I've proved it or not and don't feel like doing it now. Exercise: for which n is there a unique group of order n? (this was set to me two years ago, and I have the answer, but my only proof uses the odd order theorem, so I'm still working on it) Algebraist 10:59, 15 June 2007 (UTC)

As for numbers that are composite in all bases, the simplest example would presumably be 100_n = 10_n × 10_n for any base n. —Ilmari Karonen (talk) 18:55, 16 June 2007 (UTC)

Analysis qualifying exam question

Hi all, I'm studying for a pesky PhD qualifying examination, and one of these complex analysis questions has me stumped. I was hoping to get some help. Here's the statement:

Let f be analytic and nonconstant in the unit disk and let 0 < r < 1. Suppose that on the circle {z : |z| = r}, |f| assumes its maximum at the point z₀. Prove that $z_{0}{\frac {f'(z_{0})}{f(z_{0})}}$ is real.

I would like to tell you what I've tried on this problem, but to be honest, I don't even know where to begin. A couple of my fellow graduate students are the same way. Can you help us out? Hints over solutions are preferred. Thanks! –King Bee ^{(τ • γ)} 17:56, 13 June 2007 (UTC)

I'll close my eyes and point to Maximum modulus principle and Borel–Carathéodory theorem (and then to sci.math when you find I wasn't helpful.) iames 19:22, 13 June 2007 (UTC)

I haven't tried a proof, but the first thing that occurs to me is that at a maximum the derivative (think "gradient") is perpendicular to the circle. Does that help? --KSmrq^T 19:24, 13 June 2007 (UTC)

A small correction, though - The derivative is not perpendicular to the circle. For example, if $z_{0}=1+i$ and $f(z_{0})=1$, then $f'(z_{0})\;\!$ is in the direction of $1-i$. There is no need for the Borel–Carathéodory theorem, but you do need to consider the directions and the Maximum modulus principle. Expressing $f(z)$ and $f'(z)\;\!$ in terms of real and imaginary parts may help clarify things. -- Meni Rosenfeld (talk) 19:50, 13 June 2007 (UTC)

I like these ideas, and I should admit that I knew that the Maximum Modulus Principle would be an important thing, but I'm having trouble trying to figure out how/where to use it.

Meni - I decomposed the expression $z_{0}{\frac {f'(z_{0})}{f(z_{0})}}$ into its real and imaginary parts, and it would suffice to show that the imaginary part is 0. For the imaginary part, I got (if z₀ = x₀ + iy₀)

${\frac {1}{u(z_{0})^{2}+v(z_{0})^{2}}}\cdot (y_{0}u(z_{0})u_{x}(z_{0})+y_{0}v(z_{0})v_{x}(z_{0})+x_{0}u(z_{0})v_{x}(z_{0})-x_{0}u_{x}(z_{0})v(z_{0}))$

Is this what you were thinking, or am I way off? –King Bee ^{(τ • γ)} 21:13, 13 June 2007 (UTC)

Maybe I've drunk too much, but if you take f = exp, isn't |exp z| = exp |z|, so z₀ can be any point on the circle |z| = r. But then z₀f'(z₀)/f(z₀) = z₀, which need not be real.  --Lambiam^Talk 21:35, 13 June 2007 (UTC)

Postscriptum. Simpler counterexample: f(z) = iz. Do I misunderstand the problem statement? 21:40, 13 June 2007 (UTC)

Lambiam : You have definitely drunk too much if you say |exp z| = exp |z|. Also, for $f(z)=iz$, the value in question is 1, which is real.

King Bee: I guess that should work. Now take a small real $\epsilon$ and see what the maximality of $f(z_{0})$ has to say about $f(z_{0}(1+i\epsilon ))\;\!$, and then use the Cauchy-Riemann equations. I think this will work even without the maximum modulus principle (you could use it to learn about $f(z_{0}(1-\epsilon ))$ for small $\epsilon >0$, but this doesn't seem necessary). -- Meni Rosenfeld (talk) 22:07, 13 June 2007 (UTC)

I don't think it makes the problem any easier, but it may simplify bookkeeping to assume z₀ = 1. (As far as I can tell, replacing f with g(z)=f(z₀ z) immediately reduces this to the case where 1 is a maximum. Of course g is then analytic in a disk strictly bigger than the unit disc, so there's no problem there.) Tesseran 01:22, 14 June 2007 (UTC)

Yes, this is certainly a good transformation. Building up on that, letting $h(z)={\frac {f(z_{0}z)}{f(z_{0})}}$ is even neater. -- Meni Rosenfeld (talk) 13:29, 14 June 2007 (UTC)

For an alternative (and in my opinion, more elegant) approach, read KSmrq's comment: the gradient of |f| is perpendicular to the circle. In other words, the derivative of |f| along the circle vanishes. You should now recognize f' / f as the logarithmic derivative and remember that the real part of the logarithm is the modulus, and a proof will follow. No need to use maximum modulus or Cauchy-Riemann or similarly advanced results. -- Jitse Niesen (talk) 00:26, 15 June 2007 (UTC)

• Thanks for all your help. There are some pretty clever folks here at the ref desk! –King Bee ^{(τ • γ)} 15:15, 16 June 2007 (UTC)

June 14

T-shirt

At [6], they have a t-shirt that poses the question, "If you consider the set of all sets that have never been considered, will it disappear?" Would it be correct to say that such a collection (for some reasonable definition of "considered") would form a proper class? Black Carrot 18:00, 14 June 2007 (UTC)

Seems to me the super-set (= the set containing the sets that have never been considered) is not a member of itself, because it has been considered. In the process of choosing the sets that form the super-set, is it possible not to consider them? I kinda doubt it. That suggests the super-set is inherently empty, but still existent. -- JackofOz 00:39, 15 June 2007 (UTC)

I'm not sure, but perhaps we end up in metamathematics (statements about statements, as in "This sentence is false.") or violating one of the set axioms (i.e., trying to define something that doesn't really qualify as a set). —Bromskloss 08:15, 15 June 2007 (UTC)

Since it seems reasonable to claim that we can only consider at most countably many sets (can you consider an object without being able to describe it?), the class of all unconsidered sets will presumably be a proper class. If we use a weaker definition of considered, it might even be the empty set (I considered all sets just now, while thinking about which ones have been considered). Algebraist 10:54, 15 June 2007 (UTC)

Which brings us neatly back to Black Carrot's "reasonable definition of considered". That seems to be the stumbling block. I can assert that I have just now "considered" all the individual atoms in the universe, but does it mean anything? Not to me, at least. -- JackofOz 11:04, 15 June 2007 (UTC)

I can assure you, Sir, that this, however, means a great deal to all those individual atoms out there. On behalf of the individual atoms in the universe, thank you for your kindness in considering us.  --Lambiam^Talk 13:22, 15 June 2007 (UTC)

Consider yourselves considered.  :) -- JackofOz 13:42, 15 June 2007 (UTC)

Puuhh! How did you do that so easily, without breaking sweat? It took me forever to get it done. I need a rest. Oh, and another thing. Were they uncountable? I lost count. —Bromskloss 20:04, 15 June 2007 (UTC)

I'm being told that they aren't even infinite! [7] I guess it didn't take me forever, then, but it sure felt like it! —Bromskloss 20:08, 15 June 2007 (UTC)

So saying that for each real number $x$, there's the set $\{x\}$ doesn't qualify? Perhaps I should really stop by at each set, hat in hand, and say it like I mean it? ;-) Like I did with the atoms (though I suspect JackofOz cheated). —Bromskloss 20:14, 15 June 2007 (UTC)

There seems to be some disagreement. Call the set (if such it be) of all sets that have never been "considered" A. As far as I can see, there are three options:

1. We've considered (in some sense) all sets already, perhaps by stating the axioms of set theory, perhaps merely by trying to find A, in which case A is empty and breaks no rules, hence is a well-defined set.

2. We have not considered all possible sets, where "considered" perhaps means "wrote down a rule individually defining". Then most sets have not been considered. As the collection of all sets forms a proper class, A would have to as well.

3. We have not considered A, where "considered" perhaps means "wrote down a rule individually defining each element of", and have neither considered nor individually defined some other set (making A nonempty), in which case A contains itself. Is a proper class allowed to contain itself? Black Carrot 17:44, 16 June 2007 (UTC)

By the way, the person above who complained about my "reasonable definition" had a fair point. As there are apparently several reasonable definitions available, but infinitely many trivial variations on each, let's rephrase it as "for each reasonable definition, up to isomorphism." :) Black Carrot 17:50, 16 June 2007 (UTC)

Regarding your question: No, classes are collections of sets, not of classes. A proper class is not a set, so it cannot be in a class, thus it cannot be in itself. -- Meni Rosenfeld (talk) 10:15, 17 June 2007 (UTC)

How hard is it to get a six

PUZZLE: How hard is it to get a 6? You have a single normal unbias 6 sided dice.

You can roll the six sided dice as many times as you want (including zero times) and add up all the results.

You can chose when to stop rolling.

The objective is to get a total result of 6.

You are NOT allowed to exclude the result of any roll.

What is the probability of obtaining a 6?

I tried a simpler game of getting a 1. The probability is 1/6

I tried a simpler game of getting a 2. The probability is 1/6 + 1/6 * 1/6

But six is hard. 202.168.50.40 22:00, 14 June 2007 (UTC)

Well, let's think about what the optimal strategy is, which it seems to me to be "roll until you are at, or over, 6" since if you've got less than 6 you've still got a chance of reaching it. Once you've decided that, let's see what happens. Either:

• You get a 6, and you stop.
• You get something less than 6, and you roll again.

If you roll again, then there is exactly one number on the die that will give you a total of 6, and up to 3 (why?) that will keep you under 6, allowing you to reroll.

Hopefully this should not be too bad a calculation, and I'll give it a shot now just to see. Confusing Manifestation 22:56, 14 June 2007 (UTC)

OK, I've tried it, and while my method may not have been the most efficient I brought it down to 18 separate probabilities in a form like "1 1 1-3 3-1 = 3/6^4", meaning that there's a 3/6^4 chance that you roll a 1, then a 1, then a 1 2 or 3 followed by a 3 2 or 1 respectively. I get a total probability of roughly 0.365, if anyone wants to check me. (It might have been easier to work with complementary probabilities - i.e., the probability that on any particular roll you have more than 6, especially since you could then probably write them all in terms of rolling six dice but only requiring the first n to add to 6). Confusing Manifestation 23:09, 14 June 2007 (UTC)

The way I'd approach this would be to find the partitions of 6:

6

5+1

4+2

4+1+1

…

1+1+1+1+1+1

The probability of getting any of these would be

${\frac {c}{6^{n}}}$

where $c$ is the number of ways to arrange the numbers (ignoring duplicates) and $n$ is the number of rolls in that particular sum.

For example, for the sum 2+2+1+1, we have ${\frac {4!}{2!2!}}=6$ so that particular probability would be 1/216. We add up all of these and get our overall probability. Donald Hosek 00:04, 15 June 2007 (UTC)

Oops, I just realized I did a different problem. I showed how to get exactly 6. Donald Hosek 00:05, 15 June 2007 (UTC)

Isn't that the problem? To get not exactly 6 is rather easy.  --Lambiam^Talk 00:15, 15 June 2007 (UTC)

So it is, the problem also, though is that the question isn't well-defined. You can find each of the individual probabilities, but you can't use them to get an overall probability since the total number of rolls is not defined. In fact, as the problem is currently stated, the probability of getting a 6 is (asymptotically) 1. (I say asymptotically in that there's no guarantee that you won't roll, e.g., 4-3-4-3 for an infinite number of rolls, but as the number of rolls approaches infinity, the probability gets closer and closer to 1. But you can use the method that I outlined above to at least see the different ways to roll a 6 and get the probabilities with $n$ die rolls. Donald Hosek 00:29, 15 June 2007 (UTC)

The way I interpret it: keep rolling until you feel like stopping. If then you have a total of exactly six, you score. What is the probability, under optimal play, of scoring? I conjecture that the probability of succeeding in getting (exactly) n, where 1 ≤ n ≤ 6, is equal to 7ⁿ⁻¹/6ⁿ. For n = 6, this is 16807/46656 = 0.36023... . I haven't proved this.  --Lambiam^Talk 00:33, 15 June 2007 (UTC)

I have now also proved this; there is an easy recurrence relation:

6p_n = 1 + Σ_{i < n} p_i.

 --Lambiam^Talk 00:49, 15 June 2007 (UTC)

The way I resolve the problem is to turn the problem into a similar problem.

Instead of getting to the score 6, I change the problem to starting with the number 6 (what I call the pot) and trying to get to the number 0 by subtracting the result of the dice roll(s) from the pot.

For example:

6 - "roll a 6" = 0 (with probability 1/6)

6 - "roll a 5" = 1 (with probability 1/6)

6 - "roll a 4" = 2 (with probability 1/6)

6 - "roll a 3" = 3 (with probability 1/6)

6 - "roll a 2" = 4 (with probability 1/6)

6 - "roll a 1" = 5 (with probability 1/6)

Let pr[n] be the probability of getting to a pot of 0 when currently you have a pot of n

Thus

pr[0]=1

pr[-1]=0 Once you go negative, you can never get to ZERO

pr[-2]=0

pr[negative number]=0

So

pr[6] = 1/6 * pr[0] + 1/6 * pr[1] + 1/6 * pr[2] + 1/6 * pr[3] + 1/6 * pr[4] + 1/6 * pr[5]

pr[1] = 1/6 * pr[0] + 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] + 1/6 * pr[-4] + 1/6 * pr[-5]

pr[1] = 1/6 * pr[0]

pr[2] = 1/6 * pr[1] + 1/6 * pr[0] + ( 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] + 1/6 * pr[-4] )

pr[2] = 1/6 * pr[1] + 1/6 * pr[0]

pr[3] = 1/6 * pr[2] + 1/6 * pr[1] + 1/6 * pr[0] + ( 1/6 * pr[-1] + 1/6 * pr[-2] + 1/6 * pr[-3] )

pr[3] = 1/6 * pr[2] + 1/6 * pr[1] + 1/6 * pr[0]

And so on

pr[6] = 16807/46656

er? I see a pattern here.

pr[0] = (1/6)^0

pr[1] = (1/6)^1

pr[2] = (1/6)^2 + (1/6)^1

pr[3] = (1/6)^3 + (1/6)^2 + (1/6)^1

pr[4] = (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

pr[5] = (1/6)^5 + (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

pr[6] = (1/6)^6 + (1/6)^5 + (1/6)^4 + (1/6)^3 + (1/6)^2 + (1/6)^1

opps! I think my pattern is wrong.

202.168.50.40 01:19, 15 June 2007 (UTC)

For some reason, I'm getting 0.171123 0.19876, which doesn't match anyone's. Let me look at it again... --Wirbelwindヴィルヴェルヴィント (talk) 01:41, 15 June 2007 (UTC)

Ok, assuming the problem isn't a trick question and that the ratio is 0 (cause you roll an infinite times every time, so you never get a 6), then I'm pretty sure that 0.171123 0.19876 is right. Have a look at my sandbox, where the Xs represent a roll that adds to 6, and the 0s represent no roll. I'd paste the results here, but that's way too long. --Wirbelwindヴィルヴェルヴィント (talk) 02:02, 15 June 2007 (UTC)

Oh, and to clarify, on the results I did, once you roll a sum of 6 or greater, you stop. And if you haven't gotten there, you roll again. Thus, I guess that's the optimal probabilities. And oh, fixed the ratios because I had runaway 7s in my rolls the first time. --Wirbelwindヴィルヴェルヴィント (talk) 02:05, 15 June 2007 (UTC)

The reason this returns the wrong result is that the 161 rolls in your list don't have equal probabilities. You're much more likely to roll 6-0-0-0-0-0 than 1-1-1-1-1-1, for example. -GTBacchus^(talk) 02:36, 15 June 2007 (UTC)

I'm getting ${\frac {1}{6}}+{\frac {5}{6^{2}}}+{\frac {10}{6^{3}}}+{\frac {10}{6^{4}}}+{\frac {5}{6^{5}}}+{\frac {1}{6^{6}}}={\frac {16801}{46656}}\approx 0.36010374$.

The first term is the probability of winning in one roll, the second term is the probability of winning in (exactly) two rolls, etc. I'm assuming the optimal strategy is to keep rolling until you either win or lose. I found this by simply drawing the complete tree of possibilities, writing down a probability for each branch, and adding up all the wins. I'm not sure why the binomial coefficients are popping up there.

Someone could program a computer to play the game a million or so times and see whether the empirical frequency agrees with any of our predictions. -GTBacchus^(talk) 02:29, 15 June 2007 (UTC)

As an example, the 10 ways of winning in exactly three rolls are: 4-1-1, 3-1-2, 3-2-1, 2-1-3, 2-2-2, 2-3-1, 1-1-4, 1-2-3, 1-3-2 and 1-4-1. Each of those occurs with a probability of 1/216. -GTBacchus^(talk) 02:33, 15 June 2007 (UTC)

Computationally, I'm getting ~0.3602. That's by rolling until the total equals or exceeds six. →Ollie (talk • contribs) 03:44, 15 June 2007 (UTC)

OK, I concur with the 0.3601 - I'd missed a term in the 1/6^5 lot and double-counted one in the 1/6^3, which gave me the 0.365 I stated above. Confusing Manifestation 04:43, 15 June 2007 (UTC)

I think that ${\frac {1}{6}}+{\frac {5}{6^{2}}}+{\frac {10}{6^{3}}}+{\frac {10}{6^{4}}}+{\frac {5}{6^{5}}}+{\frac {1}{6^{6}}}={\frac {16807}{46656}}$. That's 16807/46656 202.168.50.40 04:48, 15 June 2007 (UTC)

You're right, of course. I somehow missed that 6 in the ones place in the first term. Thanks. -GTBacchus^(talk) 09:27, 15 June 2007 (UTC)

... which is the answer I gave four hours earlier, together with a more general form and a proof.  --Lambiam^Talk 06:55, 15 June 2007 (UTC)

$\sum _{n=1}^{6}{}^{n}C_{r}/r^{n}$?

So you did. Isn't that interesting, that $\sum _{n=0}^{5}{^{5}C_{n}6^{5-n}}=7^{5}$, when you add it up properly? There must be some more general identity lurking behind that. -GTBacchus^(talk) 09:27, 15 June 2007 (UTC)

$\sum _{n=0}^{5}{^{5}C_{n}6^{5-n}}=\sum _{n=0}^{5}{^{5}C_{n}6^{n}}=(1+6)^{5}=7^{5}$

...binomial theorem (or did I miss the joke ?). Gandalf61 09:52, 15 June 2007 (UTC)

Huh... I guess the joke is that I was fairly drunk for my above post. I guess applying the binomial theorem to 1 and 6 just didn't occur to me. I still had fun working out the probability. -GTBacchus^(talk) 10:32, 15 June 2007 (UTC)

With a quick simulation, I've got an approximate of 0.360 as the probability. I'm lazy to compute an exact result now. – b_jonas 15:47, 16 June 2007 (UTC)

June 15

Comparing complex numbers

We all know that 2>1, and you can tell because 2 is farther to the right on the number line than 1. But since 1 and i can only be viewed together on a plane, how can they be compared? Would it even be accurate to say i is greater than or less than one? --65.31.80.100 22:06, 15 June 2007 (UTC)

Unlike the reals, we can't give the complex numbers a total ordering. But we can compare their magnitudes -- in which case 1 and i have the same magnitude. iames 22:12, 15 June 2007 (UTC)

It's not an Ordered field because squared elements can be negative. However, it should be noted that you could give the complex numbers an ordering (e.g. dictionary ordering), but that the ordering won't be compatible with the field axioms in the way we're accustomed. nadav (talk) 22:21, 15 June 2007 (UTC)

Is the lexical ordering a total ordering for this case? Twma 00:44, 16 June 2007 (UTC)

I was about to argue that certainly you could give them a total ordering but then I just realized nadav just argued the same thing. It's quite like sorting two fields in a Table (database); the two fields to sort by would be the real component and the imaginary component. However, you'd never want to sort by potential decimal representation (or any base for that matter) unless you could constrain numbers not to end with things like 0.999... Root⁴(one) 03:56, 16 June 2007 (UTC)

BTW a more direct answer is here: Total order#Orders on the Cartesian product of totally ordered sets.Root⁴(one) 03:59, 16 June 2007 (UTC)

June 16

Integral of the square root of tan(x)

We found a question in our maths text book asking us to find the integral of the square root of tan(x). Neither our teacher nor anybody in the class has been able to work it out. Can somebody give us some help? thanks! D3av 11:45, 16 June 2007 (UTC)

Unless I'm missing something, put u=tanx and it'll come straight out.81.159.77.229 12:29, 16 June 2007 (UTC)

I don't know the answer but I am sure it is more more complex than that, as I have heard this problem discussed before. Algebra man 13:02, 16 June 2007 (UTC)

ummm.. this integral is very complicated and a simple subtitution of u = tanx does not do the job..

my advice is to put = u² = tan(x), and it will reduce to:

= $2\int \ {\frac {u^{2}du}{u^{4}+1}}\$

= $2\int \ {\frac {u^{2}du}{(u^{2}+u{\sqrt {2}}\ +1)(u^{2}-u{\sqrt {2}}\ +1)}}\$

= $2(\int \ {\frac {(Au+B)du}{u^{2}+u{\sqrt {2}}\ +1}}\ +\int \ {\frac {(Cu+D)du}{u^{2}-u{\sqrt {2}}\ +1}}\ )$

now determine A, B, C and D ... then once you've done that, apply partial fractions again ... and you should get a function with some inverse tangents and natural logs.. if anybody wants to take it up from there or suggest a simpler method go for it..

or use the decomposition

$u^{4}+1=(\omega u+1)(\omega ^{3}u+1)(\omega ^{5}u+1)(\omega ^{7}u+1)\,,$

in which $\omega \,\!$ is a primitive 8th root of unity. If you make use of the properties of roots of unity, this should make the tedious exercise slightly less messy. Or you could cheat, and ask Mathematica.[8] It's always a good idea to check the answer, which I haven't done.  --Lambiam^Talk 14:47, 16 June 2007 (UTC)

Nasty result, not mine (can be found in Mathematics Today Vol. 41, No. 3, June 2005, Eric Watson CMath FIMA):

${\begin{array}{rl}\int {\sqrt {\tan x}}\,dx&=\int {\frac {\sin x}{\sqrt {\cos x\sin x}}}\,dx\\&=\int \left({\frac {1}{2}}{\frac {\sin x+\cos x}{\sqrt {\cos x\sin x}}}+{\frac {1}{2}}{\frac {\sin x-\cos x}{\sqrt {\cos x\sin x}}}\right)\,dx\\&=\int \left({\frac {1}{\sqrt {2}}}{\frac {\cos x+\sin x}{\sqrt {1-(\sin x-\cos x)^{2}}}}-{\frac {1}{\sqrt {2}}}{\frac {\cos x-\sin x}{\sqrt {(\sin x-\cos x)^{2}-1}}}\right)\,dx\\&={\frac {1}{\sqrt {2}}}\arcsin(\sin x-\cos x)-{\frac {1}{\sqrt {2}}}\operatorname {arccosh} (\sin x+\cos x)+c\end{array}}$ x42bn6 Talk Mess 15:35, 16 June 2007 (UTC)

Think that's nasty? Ha! We can directly ask Mathematica for a solution at The Integrator, where we get an answer with a more imposing appearance.

${\begin{aligned}\int {\sqrt {\tan(x)}}\,dx&{}={\frac {\arctan \left(1+{\sqrt {2\tan(x)}}\right)-\arctan \left(1-{\sqrt {2\tan(x)}}\right)}{\sqrt {2}}}+{}\\&\quad {\frac {\log \left({\sqrt {2\tan(x)}}-\tan(x)-1\right)-\log \left({\sqrt {2\tan(x)}}+\tan(x)+1\right)}{2{\sqrt {2}}}}\end{aligned}}$

Following Lambiam's wise advice, we can try algebraic manipulations to prove equality. That fails, and numerical tests confirm the answers are different. So is Mathematica wrong? No! Both results are antiderivatives of the square root of the tangent of x; they differ by the constant (1+i)π/(2√2). Most likely Mathematica converts the integrand to

${\sqrt {\frac {i(e^{-ix}-e^{ix})}{e^{-ix}+e^{ix}}}}$

so that it can apply advanced methods going back to Liouville. The transformation is clever and effective, but introduces complex numbers into a purely real calculation. --KSmrq^T 20:19, 16 June 2007 (UTC)

Nasty as in I wonder what motivated such a manipulation in the first place. x42bn6 Talk Mess 21:21, 16 June 2007 (UTC)

wow! thanks so much everyone. We got to the partial fraction point above and then just stopped because we knew it'd be a lot of work to do the partial fractions a few times and we weren't sure if we'd get a result. thanks! D3av 09:11, 17 June 2007 (UTC)

Can you find out if a list of numbers is random

What I mean is: I know that if you roll a die say, 300 times, and you count the number of times each number shows on top, then you can use a chi^2 test to see if the die is fair. But is there a way or a test to tell the difference between a computer outputting 50 1's then 50 2's then 50 3's and so on until 50 6's and then back to 1's again from a fair die?--71.175.128.187 16:15, 16 June 2007 (UTC)

There are several tests for statistical randomness, almost all of which would reject this sequence with long runs of the same value. No practical test, however, can distinguish with certainty between true randomness and good pseudorandomness.  --Lambiam^Talk 17:43, 16 June 2007 (UTC)

Or even bad pseudorandomness. —Bromskloss 20:17, 16 June 2007 (UTC)

Three points: (1) Certainty, no; confidence, yes. (2) Genuine random sequences includes long runs that make humans uncomfortable; "made up" sequences don't have enough long runs! (3) Practical applications like Monte Carlo integration suffer from runs, so use a different measure called discrepancy. --KSmrq^T 20:27, 16 June 2007 (UTC)

You might want to check out the program ENT - a Pseudorandom Number Sequence Test Program, written by John Walker. It runs several tests on sequences of bytes, and is useful for evaluating pseudorandom number generators. --NorwegianBlue ^talk 08:26, 17 June 2007 (UTC)

The Diehard tests are a well-known standard, but see also cryptographically secure pseudorandom number generator. --KSmrq^T 19:20, 17 June 2007 (UTC)

Well, of course a sequence of random numbers would, somewhere, contain the sequence 50 1's then 50 2's then 50 3's and so on until 50 6's and then back to 1's again. In fact, it would contain it infinitely many times. Or any other sequence however odd or long or short you may desire. So, the best you can do is come up with the odds, which is where statistics comes in. This would fall under the heading of nonparametric stats, which I like because it's smaller and more limited. There are several ways to analyze this kind of thing, that look at whether the next number goes up or down randomly enough; for instance your sequence 'same, same, same, same, up, same, same, same, same, up, same, same, same, same, up' etc. would immediately show up, but so would something like 'up, down, up, down, up, down' etc. They are named rank test, run test, etc. There's so much financial benefit in being able to determine whether something is going wrong in an industrial process or not, that it's all been studied very closely. Gzuckier 19:04, 20 June 2007 (UTC)

June 17

mechanics

(I'm assuming below rigid particles, rods, elastic collisions etc - the simplest case)

I'm stuck (totally) on a problem. There are two masses (A and B) (each mass m) connected by a rod of length L. For clarity I'll say A is at (0,0), and B at (L,0)

Another point mass C (also mass m) is at (0,-d) and is moving at velocity v in the direction (0,1) ie towards point A.

What happens when A and C collide - I'd expect some sort of rotation of A-B - but I can't get that to make any sense mathematically. It seems that what I've described is impossible (or maybe my stupidity is a factor) Any help appreciated. Thanks87.102.32.96 01:45, 17 June 2007 (UTC)

The animations in the Elastic collision article might help. I believe what would happen is this: C would move towards A at a constant velocity v until they collided. All the energy of the system would leave C and move into A, which would start A moving along the same line, at velocity v. This would send the bar rotating about its center point, without the center point moving, until B collided with C. The bar would stop moving, and C would start back on its merry way. Black Carrot 03:01, 17 June 2007 (UTC)

No, that's wrong. The first part is right, but I think the bar would resist rotation somewhat, and some of the system's energy would go into moving the bar forward. In other words, it would rotate about its center point a bit more slowly, and that centerpoint would move forward at a constant velocity. It would, then, miss C and keep moving forever. The article about Moment of inertia might help. Black Carrot 03:05, 17 June 2007 (UTC)

This one's not too bad if you take it step by step. At the instant that C collides with A, you can ignore the presence of B, because the collision is at right angles to the rod and so the rod can't exert any forces on A. So conservation of momentum and energy mean that C comes to rest and A heads off in the (0,1) direction with velocity v, as Black Carrot says. Now consider the pair of particles A and B. Their centre of mass (call it D) is halfway along the rod, initially at (L/2, 0), but since A is moving with velocity v, D is now moving at velocity v/2 in direction (0,1). By conservation of momentum, D will continue to move at this velocity, since there are no external forces on A or B. However, the rod joining A and B is now rotating with an instantaneous angular velocity w = v/L, and by conservation of angular momentum, will continue to do so (the rod between A and B is now under tension, and supplies the necessary centripetal force). So C has come to rest at the origin. Its original momentum mv is now taken up by the motion of D, with total mass 2m and velocity v/2. The original kinetic energy mv²/2 is now shared between the translational KE of the rod, which is 2m(v/2)²/2, and its rotational KE, Iw²/2, where its moment of inertia is I = 2m(L/2)². Add these together, and it all works out. C stays at rest, and A and B move off, spinning gently with an angular velocity of v/L (in radians/second) around their centre of mass, which itself moves steadily at speed v/2. In fact, A and B describe interlocking cycloids, each coming to rest instantaneously when they reach x coordinate L. --Prophys 10:03, 17 June 2007 (UTC)

Thanks - much clearer now.

1 + 1/2 + (1/2)/2...

Here's a bit of a math conundrum for you: say I took the number 1, and I added half of 1 (0.5), then I added half of half of 1 (0.25), and continued in this pattern an infinite number of times. Would I ever reach 2, or would you just end up with a number that has an infinite number of decimal places?

Here's the start of the math to explain what I mean a bit more clearly:
1 + 0.5 = 1.5
+ 0.25 = 1.75
+ 0.125 = 1.775
+ 0.0625 = 1.8375
continued...

Thanks for the help. I thought this up one day and I tried it with a calculator but wasn't able to get that far with my 7 digit screen. Taylor 08:13, 17 June 2007 (UTC)

This is just an example of an infinite series. Take a look at the article for a lot of interesting information. — Kieff | Talk 08:23, 17 June 2007 (UTC)

Thanks for that. Does this mean that the series would never reach 2, since there would always be a room at the end equivalent to the amount just added? Or does it count as 2 on the account of 0.999...=1?Taylor 09:11, 17 June 2007 (UTC)

If you summed the first 5 billion terms of that series, you'd end up with a number very close to 2, but not quite! If you took the sum of the first 5 thousand trillion terms, the sum would still be slightly smaller than 2. So, if the series has a finite number of terms, then it will never reach 2. BUT, if the series has an infinite number of terms, then by the same principle as the 0.9999... = 1 statement, it is equal to 2.

You can also consider how far you are from 2. For 1, that is 1. For 1.5, it is 1/2. For 1.75, 1/4. Then 1/8, 1/16, and so on. Each time you halve the distance. Half a positive number is again a positive number, so at each step you will still have a positive distance to 2. But you can get arbitrarily close: the limit of 1, 1/2, 1/4, ... is 0, since you can make the number as small as you want – provided you allow it to remain positive. To give an impression, here is where you would be after 100 steps if the screen size is not a factor:

1.9999999999999999999999999999992111390947789881945882714347172137703267935648909769952297210693359375

This falls short of 2 by the amount of

0.0000000000000000000000000000007888609052210118054117285652827862296732064351090230047702789306640625

See also Zeno's paradoxes.  --Lambiam^Talk 12:48, 17 June 2007 (UTC)

What you want to do is write the numbers in binary:

$1_{2}+0.1_{2}+0.01_{2}+0.001_{2}+\cdots =1.111\ldots _{2}\,\!$

This is, indeed, the binary equivalent of 0.999… in base 10. We must carefully distinguish two possibilities: (1) adding as often as we like, and (2) the infinite sum. For (1), we can get as close to two as we like, but always fall short; for (2), we get exactly two. And for those whose consider such mathematical conundrums silly, I offer the physical reality of quantum tunnelling: A particle approaches an insurmountable obstacle, but if the wall is thin enough we may find the particle on the other side because of quantum uncertainty in its position! What would Zeno make of that? :-) --KSmrq^T 19:36, 17 June 2007 (UTC)

0.999... = ${\frac {1+1/2+(1/2)/2+...}{2}}$. So just see the 0.999... article. 202.168.50.40 21:56, 17 June 2007 (UTC)

It's not a conundrum at all. If you limit yourself to adding up a finite number of terms of the series, then you will obviously never reach 2. Adding n terms will produce a sum of 2-2⁻ⁿ < 2 for all n ∈ N. If, however, you add up an infinite number of terms, the sum will be exactly 2. Confusion over this is simply confusion over the mathematical concept of infinity. Which is, admittedly, a subtle thing; it took mathematicians a long time to be sure they understood it.

Lagrange multipliers

Why does in this page [9] say the method won't catch a minimum if x,y,z can be negative? --Taraborn 09:09, 17 June 2007 (UTC)

The Lagrange multipliers method gives necessary conditions for the global minima and maxima. If the variables are assumed to be nonnegative, then adding the constraint results in a closed, bounded region, and since the target function is continuous, it must have a minimum and a maximum - and they will be found among the points which are "suspected" by the method. If, however, the variables are not assumed to be nonnegative, then the region is unbounded, and there is no guarantee that a minimum or a maximum exists. If, say, there is no minimum, then none of the points suspected by the method will be a minimum. However, the method does not give sufficient conditions, so by itself it cannot tell if any of the points found is actually a minimum. -- Meni Rosenfeld (talk) 10:09, 17 June 2007 (UTC)

Note also that the advice is specific to the example, not a general rule. 84.239.133.38 14:43, 17 June 2007 (UTC)

On closer inspection, the example given there is quite misleading. The final result is that the minimum is obtained at 3 points, when in reality it is obtained at infinitely many points (e.g., $(x,1-x,0)$ for any $0\leq x\leq 1$). More care needs to be taken when solving a system where the constraints involve both equalities and inequalities. -- Meni Rosenfeld (talk) 15:45, 17 June 2007 (UTC)

No. They are correct. The minimum is attained at only one point. --Spoon! 17:26, 17 June 2007 (UTC)

We are clearly not speaking about the same thing. I'm referring to "find the minimum and maximum of $xyz$ subject to $x+y+z=1$ and $x,y,z\geq 0$. -- Meni Rosenfeld (talk) 17:38, 17 June 2007 (UTC)

Consider the objective function 1+x². It attains a minimum at x = 0, but can grow arbitrarily large so has no maximum. Similarly, the objective function 1−x² attains a maximum at x = 0, but can grow arbitrarily small so has no minimum. Now multiply the objective function by y to give y(1+x²), but add the constraint that y = 1; clearly nothing should change. The first equations, now including Lagrange multiplier, become

${\begin{aligned}2xy&{}=0\lambda \\1+x^{2}&{}=1\lambda \\y&{}=1\end{aligned}}$

The only solution is x = 0, y = 1, λ = 1 — a minimum; but without bounds we again have no maximum. (Note that the positive value of λ tells us the constraint is pulling us away from a smaller minimum, at y = 0.)

Incidently, the recent question about a maximum on a circle invoked the same idea behind Langrange multipliers; if a component of the gradient is parallel to a direction of allowed movement, we must not be at an extremum. --KSmrq^T 20:49, 17 June 2007 (UTC)

Thanks for this incidental note -- pleasantly unexpected. Tesseran 09:13, 18 June 2007 (UTC)

Problem solving this equation

I started off with

${\frac {\text{dy}}{\text{dx}}}={\frac {w\cos(\beta )-v\sin(\theta )}{w\sin(\beta )-v\cos(\theta )}}$

Where sin theta = y/sqrt[x^2+y^2] and cos theta = x/sqrt[x^2+y^2]

And I got to this point (using u=y/x):

${\frac {w\sin(\beta )-{\frac {v}{\sqrt {u^{2}+1}}}}{w\cos(\beta )-w\sin(\beta )u}}{\text{du}}={\frac {\text{dx}}{x}}$

Basically I want to know if this can be solved?

${\frac {A-{\frac {B}{\sqrt {u^{2}+1}}}}{E-Fu}}{\text{du}}={\frac {\text{dx}}{x}}$

220.239.108.132 12:05, 17 June 2007 (UTC)

Are w, β and v all constants?  --Lambiam^Talk 13:17, 17 June 2007 (UTC)

Yes they are constants. 220.239.108.132 15:33, 17 June 2007 (UTC)

What I'd do is switch to polar coordinates, so that

$x=r\cos \theta \,,$

$y=r\sin \theta \,,$

in which r is considered a function of θ. Abusing the "prime" notation for the derivate with respect to θ, so that r' = dr/dθ, we have

${\frac {dy}{dx}}={\frac {dy/d\theta }{dx/d\theta }}={\frac {r'\sin \theta +r\cos \theta }{r'\cos \theta -r\sin \theta }}\,.$

Next, do a change of variable, in which r = exp z, so r' = z' exp z = z'r. Then we can rewrite the last fraction as:

${\frac {z'r\sin \theta +r\cos \theta }{z'r\cos \theta -r\sin \theta }}={\frac {z'\sin \theta +\cos \theta }{z'\cos \theta -\sin \theta }}\,.$

This is still the result of rewriting of dy/dx, which is equal to the r.h.s. of your first equation, and so, using A := w cos(β) / v, B := w sin(β) / v,

${\frac {z'\sin \theta +\cos \theta }{z'\cos \theta -\sin \theta }}={\frac {A-\sin \theta }{B-\cos \theta }}\,.$

You can solve this for z', leading to an equation of the form

$z'=f(\cos \theta ,\sin \theta )\,,$

in which f is a rational function. If we can find the antiderivative

$\int f(\cos \theta ,\sin \theta )\,d\theta \,,$

we're done. This is the point to call t := tan(¹/₂θ) to our aid:

$\cos \theta ={\frac {1-t^{2}}{1+t^{2}}}\,,$

$\sin \theta ={\frac {2t}{1+t^{2}}}\,,$

$d\theta ={\frac {2dt}{1+t^{2}}}\,,$

By this change of variable, the indefinite integral becomes

$\int {\frac {2f({\frac {1-t^{2}}{1+t^{2}}},{\frac {2t}{1+t^{2}}})}{1+t^{2}}}\,dt\,.$

The integrand is a rational function of t, and so the antiderivative can be found in the standard way by the method of partial fractions.  --Lambiam^Talk 00:56, 18 June 2007 (UTC)

I attacked this in a slightly different way, based on my own version of intuition (so it may not be the most elegant method).

• The RHS is, of course, easy to integrate so we ignore that.
• The LHS can be broken into two terms, the first one a simple fraction with a similarly simple integral, the second the bit containing the square root.
• When you have a ${\sqrt {a^{2}+u^{2}}}$, the traditional substitution is $u=a\tan \theta$, making use of the identity $\sec ^{2}\theta =1+\tan ^{2}\theta$.
• In this case it seems to simplify things significantly, leaving us with an integral of ${\frac {1}{E\cos \theta -F\sin \theta }}$, which can be solved via substitution of $t=\tan {\frac {\theta }{2}}$, and making use of various half-angle identities and partial fractions (although you need to watch out for special cases, such as when the denominator becomes a perfect square).

I haven't pulled out a solution myself, but it certainly looks doable without resulting to any particularly complicated integration techniques, thankfully. Confusing Manifestation 01:33, 18 June 2007 (UTC)

Help involving a Cauchy's integral formula problem

Hi all, it's me again. I was hoping to get some aid with another complex analysis problem. For reference, here's the statement I have to prove.

Let f be analytic in the the upper half plane and continuous in the closed upper half plane. Assume also that $|f(z)|\rightarrow 0$ as $|z|\rightarrow \infty$ in the upper half plane. If $f(z)=u(z)+iv(z)$, where u and v are real-valued, show that for $x\in \mathbb {R}$,

$u(x)=-{\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {v(y)}{x-y}}\ dy$ and $v(x)={\frac {1}{\pi }}\int _{-\infty }^{\infty }{\frac {u(y)}{x-y}}\ dy$.

I wanted to use the Cauchy integral formula using (as my countour) the semicircle centered at the origin with radius R, then send R to infinity and all that jazz. Playing free and loose (i.e., not checking whether or not this is legitmate), I was able to prove the statement I wanted, but each expression was divided by 2. Since f is not analytic ON my entire contour (although it is analytic within it), is there something I'm missing? Any help would be appreciated. –King Bee ^{(τ • γ)} 14:11, 17 June 2007 (UTC)

I think that's the correct approach. You might have missed that the semicircle contains the point x, while the formulation of Cauchy's integral formula in our article requires the contour to go around the point x. You should use a version of Cauchy's integral formula which allows the contour to go through x, or a version of the residue theorem which allows for poles on the contour. -- Jitse Niesen (talk) 21:04, 17 June 2007 (UTC)

Yeah, that's what I'm thinking. I wonder if the version I want says something like

$f(\omega )={\frac {1}{\pi i}}\int _{C}{\frac {f(z)}{z-\omega }}\ dz$,

since my function isn't necessarily analytic ON the contour. –King Bee ^{(τ • γ)} 21:10, 17 June 2007 (UTC)

• For those interested parties, I figured out the problem. I was using the wrong contour. What I want is an "archway" contour, that "jumps" over the given point x in the problem. Thanks for the help, however! –King Bee ^{(τ • γ)} 12:20, 19 June 2007 (UTC)

Calculating required power for lifts

How do calculate the power required to perform classic gym lifts like the deadlift in a certain time (1.5sec)? If it was only a matter of P = E/t it would be easy, but the fact that joints are working in different directions makes it difficult. I'm unable to make use of the information in the torque article, which I suppose should be of use. Could anyone give me a few guidelines how to calculate this kind of movements? Thanks in advance, Jack Daw 14:59, 17 June 2007 (UTC)

This is not really a maths question; it is more a question for the Science section of the Reference desk. If you post the question there, it may help if you could clarify what you mean by "effect". Do you mean power?  --Lambiam^Talk 18:06, 17 June 2007 (UTC)

I meant power, thanks. The reason I put my question here is that I need only the purely mathematical part to my answer - that is, the proper equations - rather than an explanation of the concepts of force, mass, power and so forth. Still think I should post it in the science section? Jack Daw 14:05, 19 June 2007 (UTC)

What are your assumptions? If you assume that there is no friction and that the mass of your own body is negligible, then by conservation of energy, the power your muscles output is exactly equal to the power received by the weight. Of course, if you do not assume these, then the power your muscles output is equal to the power received by the weight, plus the power received by the moving body parts, plus the power lost to friction. Do you aspire to calculate these other terms? Doing so requires quite a lot of data. -- Meni Rosenfeld (talk) 21:51, 19 June 2007 (UTC)

Inverse Function of x^2 + x

I have been trying to do this problem for a day or 2, and it is just not working. I'm supposed to find the inverse function of $x^{2}+x$(assuming $x=>-1/2$). I know the basics of inverse functions, but the 2nd x is screwing me up. I have no clue what to do with it. This is the only type of inverse function that I am currently having trouble with.69.49.70.112 15:35, 17 June 2007 (UTC)

Hint: It has something to do with a square root (I'm sure you know that from a basic parabola). See completing the square. x42bn6 Talk Mess 15:42, 17 June 2007 (UTC)

As far as I can tell, the function has no inverse, since it is not one to one. Both 0 and -1 get mapped to 0. I'm an idiot, and should really read the whole problem before responding. Follow the hint above, it should help you solve your problem. –King Bee ^{(τ • γ)} 15:59, 17 June 2007 (UTC)

Ok, I got through the problem I ended up with $x={\sqrt {y+(1/4)}}-(1/2)$ I checked it out on the graphing calculator and it looks pretty good. Thanks Guys!

It might be useful to note that the domain might change things. If one takes x<=-0.5, then you will have a negative square root. x42bn6 Talk Mess 16:17, 17 June 2007 (UTC)

abstract algebra

²²sir help me on this prolems as i am little bit confused.

• 1)find the two distinct right coset of R in (C+).

if i write a+(c+id) and b+(c+id) are'nt they distinct right coset of R in c for different value of a and b element R

• 2)sir tell me whether this statement is true or false give reason also to prove or disprove it ...........if every proper subgroups of group G is cyclic than group G is cyclic.

sir is'nt it true because even G is proper subgroup of itself and if it is cyclic than it is cyclic.

• 3)sir how should i start to show that M²(R×S)≈ M²(R)×M²(S) where M²(R)and M²(R) represent matrices of second order and R and S are rings.(≈) denotes isomorphism.
• 4)if H intersection K is a normal subgroup of G than H and K are normal subgroups of G where H and K are subgroups of G .............sir in my opinion it is false but what example i should give to disprove
• 5)let G and H be groups of orders P and q where p and q are distinct prime numbers show that any homomorphism from G to H must take any element of G to the identity element of H........... i am unable to think over it neither i am getting any example in book pls help

—Preceding unsigned comment added by 59.176.119.128 (talk • contribs) 16:41, 2007 June 17

You should sign your posts on talk pages, and remember that we don't do your homework for you here, but we can help you.

1) I think you're thinking about this one backwards. You need to find two elements $z_{1},z_{2}\in \mathbb {C}$ so that $z_{1}+\mathbb {R}$ and $z_{2}+\mathbb {R}$ have different elements in them. Concentrate on the imaginary parts of $z_{1}$ and $z_{2}$ and you should get it.

2) No group is a proper subgroup of itself. Also, the statement you are trying to prove is false. Now, go find a counterexample.

3) Smells like the Chinese Remainder Theorem to me.

4) I feel like this one is false. Pick one of your favorite pathological groups, and find two non-normal subgroups whose intersection is the trivial subgroup. That should suffice.

5) Smells like the fundamental homomorphism theorem to me. Think about what type of groups G and H are. Assume you have a homomorphism and look at its kernel.

Good luck. –King Bee ^{(τ • γ)} 17:44, 17 June 2007 (UTC)

How about this. Assume f homomorphism, f(e_G) ≠ e_H, e_G and e_H respective identities of G and H. Consider f(e_G)f(g), g ∈ G. To avoid spoiling all the fun, you can determine f(e_G)f(g) easily from the properties of homomorphisms. What are you forced to conclude? What does this imply? —Preceding unsigned comment added by 129.78.64.102 (talk • contribs)

He is trying to show that f maps all the elements to the identity of H, not just e_G. nadav (talk) 10:41, 18 June 2007 (UTC)

Ok, I misread. Never mind, you can still adapt the fact to get the necessary result. You need to know that any group of prime order is cyclic, so only one element has order 1. That should be all you need, the rest is piecing it together.

Well, it's a little bit more complicated than that (but not by much). The kernel of a homomorphism is a subgroup of the domain (of that homomorphism). Since H is of prime order, H has no proper nontrivial subgroups (and neither does G). Since $p\neq q$, what does this mean about the kernel of any homomorphism from G to H? (Think G/kerφ, which is isomorphic to Imφ....) –King Bee ^{(τ • γ)} 13:19, 18 June 2007 (UTC)

It can be more complicated, but it doesn't have to be. (This part is to King Bee, not the original poster.) Take G and H finite groups, with g in G. How might you prove (using additional assumptions on g and H if necessary) that this particular element g is sent to the identity by any homomorphism f: G → H? Tesseran 21:58, 18 June 2007 (UTC)

I suppose I would look at the orders of the elements. =)–King Bee ^{(τ • γ)} 22:43, 18 June 2007 (UTC)

June 18

Useless mathematics

After reading Von Neumann universe and some related articles, all horribly abstract, I came to think of the practical use of mathematics. Personally, I find the main reason to study mathematics being that it is fun, but other than that, are there areas of mathematics that aren't expected to be of any practical use in a long time (mabye hundreds of years), if ever, or are there areas that have been around for a long time but no one so far have found any use for? —Bromskloss 08:13, 18 June 2007 (UTC)

A lot of mathematics has no known practical use. For mathematics for fun, see recreational mathematics and Category:Recreational mathematics. PrimeHunter 12:07, 18 June 2007 (UTC)

This does not answer your question, but Number theory must be mentioned in this discussion. If I am not mistaken, for millennia it was thought to be completely useless, right until the advent of modern cryptography (perhaps some time before). This goes to show you that it is hard to predict what will or will not be useful in the future. -- Meni Rosenfeld (talk) 14:10, 18 June 2007 (UTC)

From Pure mathematics#Quotes: "There is no branch of mathematics, however abstract, which may not someday be applied to the phenomena of the real world." Nikolai Lobachevsky (1792-1856). PrimeHunter 14:19, 18 June 2007 (UTC)

I think that was true in his days, but had he heard of hyper-inaccessible cardinals, he would surely have changed his mind!  --Lambiam^Talk 14:56, 18 June 2007 (UTC)

Factorisation

Hi,
Is it possible to further factorise (x+1)^2 + 3? I vaguely have the impression you can be completing the square, but the article didn't shed any light. Thanks, --124.179.222.152 10:25, 18 June 2007 (UTC)

I think that's the best answer you'll find. When written out, the quadratic is $x^{2}+2x+4$. This has no real linear factors because its discriminant is negative. (However, you can factor it further via the complex roots) nadav (talk) 10:35, 18 June 2007 (UTC)

Yeah I had $x^{2}+2x+4$ as my original equation. I may as well tell you the full question. In the first part to the question they tell me to factorise $x^{2}+2x+4$, and then hence I have to anti-differentiate 1/($x^{2}+2x+4$). I think what I need to do is to use partial fractions to solve it. And so I'm really looking for something like (x+a)(x+b) as a factor for $x^{2}+2x+4$ --124.179.222.152 10:51, 18 June 2007 (UTC)

You're so close!! If you have to antidifferentiate, go back to your $(x+1)^{2}+3$ bit. Think about what the derivative of the inverse tangent function is... –King Bee ^{(τ • γ)} 10:55, 18 June 2007 (UTC)

Oh, right! Feel so silly :-) - thanks King Bee and Nadav! --124.179.222.152 11:10, 18 June 2007 (UTC)

Hooray! Glad to help. (Don't feel too silly. I find that my students often have troubles with questions like this. You did well!) –King Bee ^{(τ • γ)} 11:12, 18 June 2007 (UTC)

It has always annoyed me when teachers say "factorize [or factor] x² + 2x + 4" and want you to come up with (x + 1)² + 3. Where are the factors of the original polynomial? Tesseran 12:04, 18 June 2007 (UTC)

Yeah, I agree; the hint that the OP received is basically nonsense. To answer your question, the original polynomial has no real zeros, as mentioned above by Navdav (the dsicriminant is < 0). –King Bee ^{(τ • γ)} 13:15, 18 June 2007 (UTC)

What's the next real number after 0?

Because you can't have 0.an infinite number of zeros followed by a 1. Vitriol 12:14, 18 June 2007 (UTC)

There simply is no "next" number, just like there is no largest real number. We'll have to live with that. —Bromskloss 12:33, 18 June 2007 (UTC)

Yes. See also well-order which says: "The standard ordering ≤ of the positive real numbers is not a well-ordering, since, for example, the open interval (0, 1) does not contain a least element." PrimeHunter 12:49, 18 June 2007 (UTC)

True. Although, Zorn's lemma implies that there is a well-ordering of the real numbers (it's clearly not ≤). Under this ordering, there is a "next" number after zero. =) –King Bee ^{(τ • γ)} 13:13, 18 June 2007 (UTC)

Is there, by any chance, a way to find such a well-ordering? —Bromskloss 13:50, 18 June 2007 (UTC)

No way. You would effectively have proved the axiom of choice (which is not my axiom of choice). I expect that in intuitionistic mathematics the claim that such a well-ordering exists is actually contradictory.  --Lambiam^Talk 14:49, 18 June 2007 (UTC)

What do we then mean when we say that a well-ordering exists? —Bromskloss 15:16, 18 June 2007 (UTC)

Well, it can be proven that the Axiom of Choice, Zorn's Lemma, and the Well-Ordering principle are all equivalent. Hence, if you take one, you have to take them all. (Which leads to a funny joke: The axiom of choice is obviously true, the well-ordering principle is obviously false, and we're still undecided on Zorn's lemma...or something.)

Lambiam - I also hate the axiom of choice (in that you can prove wacky counterintuitive things), but you can prove wacky things in ZF without choice as well. Do you really reject the axiom (a la Banach, Tarski, and a host of other prestigious skeptics)? –King Bee ^{(τ • γ)} 15:20, 18 June 2007 (UTC)

I'm a closet intuitionist. Oops, now the cat is out of the bag. The main problem I have with AC is that it is false. It implies that there exists a real-valued function f from [0, 1] to itself such that |f(x) − x| > 1/3 for all x. However, as L. E. J. Brouwer already proved, all functions defined on [0, 1] are continuous, and so in this case f(x) = x for some x in [0, 1]. I don't understand why some people think ZF is a meaningful basis for doing maths.  --Lambiam^Talk 20:07, 18 June 2007 (UTC)

Yeah, I like that joke too. :-) Anyway, I don't understand why we can be so sure no one will ever find a well-ordering of the real numbers. —Bromskloss 15:52, 18 June 2007 (UTC)

Because, for any reasonable definition of 'find', it's provably impossible. For example, there is no formula of ZF which defines a well-oredering of the reals. Algebraist 16:40, 18 June 2007 (UTC)

I've seen an informal argument this before, but what about this statement in Well-order: "However it is consistent with ZFC that a definable well-ordering of the reals exists—for example, it is consistent with ZFC that V=L, and it follows from ZFC+V=L that a particular formula well-orders the reals, or indeed any set."? 70.21.2.80 03:42, 19 June 2007 (UTC)

Mean of circular quantities

Hi! I'm using the algorithm provided in Mean_of_circular_quantities. It works just fine for some sets but I have discovered that some sets gives a small error. For example: a mean of 90 deg, 90 deg and 45 deg I would expect to be (90+90+45)/3 = 75.00 deg, but the algorithm gives 75.36119... deg (using atan2). I just can't understand how this can be? Is it an incorrectness in atan2? some cumulative error? can it get bigger? can it be keept smaller? Or am I just wrong in assuming that it should be 75.00 deg? 82.182.115.156 12:30, 18 June 2007 (UTC)

You are wrong to assume that the answer should be 75 degrees. The algorithm decribed in mean of circular quantities does not calculate the arithmetic mean of a set of angles. It calculates the direction of the vector sum of a set of unit vectors. There is no reason, in general, for these two "means" to have the same value. For an even more extreme example of the difference between the two concepts consider the "circular quantities" mean of 0 degrees and 180 degrees - it is $\arctan({\frac {0}{0}})$, which is undefined. Yet the arithmetic mean of 0 and 180 is clearly 90. Gandalf61 12:56, 18 June 2007 (UTC)

Ok, thanks for that! 82.182.115.156 13:01, 18 June 2007 (UTC)

Just for the record, your computation is correct. I get 75.36119340482171369... degrees. PrimeHunter 13:03, 18 June 2007 (UTC)

Standard deviation

Why do you have to divide by n-1 instead of n?--71.185.137.94 15:41, 18 June 2007 (UTC)

Great question, this is explained (but not very clearly) at Standard deviation. The formula for standard deviation

${\frac {1}{n-1}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}$

is an unbiased estimator of $\sigma ^{2}$, meaning the result of the formula is as likely to be above the actual value of $\sigma ^{2}$ as below. Another formula

${\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}$

is biased, but it has other properties which some people prefer (it is the maximum liklihood estimate).

Some people also use the argument that since we know ${\bar {x}}$, we don't really have n points to calculate the standard deviation; If you know n-1 points and the mean, you can calculate the last value. So the "real" number of points is n-1. --TeaDrinker 16:12, 18 June 2007 (UTC)

The formulas above are estimators of variance. To get estimators of the standard deviation, you have to take the square root. The property of being unbiased for the first formula only holds for the variance; its square root is still a biased estimator of the standard deviation. For example, when n = 2, its expectation is only 80% of the population value. If you want a less biased estimator for standard deviation, you're better off using n – ³/₂ in the denominator.  --Lambiam^Talk 21:27, 18 June 2007 (UTC)

Surely the fact that an estimator is unbiased means that it has the right expectation (aka mean), not that it is as likely to be above the true value as below? Algebraist 16:37, 18 June 2007 (UTC)

• gasp* you're quite right, I stand corrected. --TeaDrinker 18:15, 18 June 2007 (UTC)

PayPal

As an example I charge $95 for a product plus $5 for shipping. If customers choose to use PayPal, however, I have to charge for that just like I have to charge for shipping. PayPal charges approximately 4% or $4 added for a total of $104. The problem should be obvious: the customer now pays $104 and 4% of $104 is 16 cents greater so the new required total payment is $104.16. But again PayPal will charge 4% so now I need to calculate on the basis of $104.16, ad infintium. My questions is what formula can I use to stop this maddness and get the final answer in the first shot? 71.100.3.132 16:31, 18 June 2007 (UTC)

What you want to do is say that you want a 4% discount of $x$ to be $100. That 4% discount can be thought of as $0.96x$ so now it's just a matter of solving the simple linear equation $0.96x=100$. Donald Hosek 16:44, 18 June 2007 (UTC)

Okay give me an eample: I have a product that is $188 plus $55.5 for shipping for a total cash payment of $243.5. PayPal charges 4% of $243.50 or $9.25 rounded which is added to $243.50 for a total payment of $253.25. But wait now PayPal is going to charge 4% of $253.25 when payment is made via PayPal or $10.13 insted of $9.25. Play this out and show me the linear result curve. I do not think that it will be linear at all. 71.100.3.132 16:59, 18 June 2007 (UTC)

Read Donald's reply again. PayPal charges 4% of what the customer pays. So if the customer pays x, paypal takes $0.04x$ and you get $0.96x$. If you are supposed to get 100$, then the customer should pay ${\frac {100}{0.96}}\approx 104.17$ dollars. 4% of that is 4.17$, and you get your 100$.

An alternative (yet hopelessly more complicated) way to look at it is along the line you suggest: If you start with 100$, then add 4%, then add 4% of what you added before, ad infinitum, you get an infinite geometric series, which also sums up to ${\frac {100}{0.96}}$.

On a completely unrelated note, I think your customers will appreciate it if you do not charge extra for using PayPal... -- Meni Rosenfeld (talk) 17:08, 18 June 2007 (UTC)

Okay, thanks. It is much clearer to me now. As for my PayPal policy... Many e-tailers do not charge directly for using PayPal but simply raise the price of their product to cover PayPal and then fail to give a discount when payment by check or money order is made. I think this is dishonest and deceptive and such business practices should be discouraged since the advantage of PayPal is speedy payment associated with securing a bid immediately in an auction situation such as eBay or the stock market.

Although I have long since raised my prices to cover PayPal at 4% I have also long since offered a discount for payment by check or money order. In this way my customer has the right to the option of saving by not using a service which he does not need. 71.100.3.132 18:35, 18 June 2007 (UTC)

Good point. I withdraw my comment, then. -- Meni Rosenfeld (talk) 18:55, 18 June 2007 (UTC)

A little off topic, but I believe credit card companies also charge a fee (and American Express and Diners Club are typically a bit higher than Visa and Mastercard), and different merchants handle this in different ways - some just raise the price across the board, some disallow credit cards - or just the more expensive ones, some don't allow them at all. I think that as long as the merchant is honest about how they charge their customers, it's not necessarily dishonest. Confusing Manifestation 22:53, 18 June 2007 (UTC)

::Credit cards certainly charge fees to the merchant. The base rate for Mastercard and Visa is 3.5%, I believe, and it's higher if the card has any associated rewards program. Other card companies are higher also. Tesseran 04:20, 19 June 2007 (UTC)

June 19

What is the probability that a vector came from a set?

Firstly, this isn't homework, this is for my job. I'm a programmer but most of the time I don't have to deal with this much math. I've been through my college stats book, and it has some hints but not the answer to this question:

There are N vectors (x₁, x₂, ... x_M) where N is usually around 10 and M is around 40, and another new vector with the same number of elements. All of the elements x_m are normally distributed and completely independent of each other. I need to know the probability that the new vector came from the same set as the N known vectors did. Note that I'm not trying to reject or accept the proposition at a given confidence level; I need to know the probability that the new vector is from the same population. Please help. 75.35.79.57 02:14, 19 June 2007 (UTC)

What do you mean, "came from": is a member of? is a linear combination of?

Is a member of: So, e.g., (1,2,3) is very likely to come from {(0.8,2.2,2.9),(1.1,2.1,3.2)} but (7,8,9) isn't. 75.35.79.57 03:07, 19 June 2007 (UTC)

That's not membership. Furthermore, you need to define some sort of tolerance value to get anywhere meaningful: (1,2,3) - (1.1,2.1,3.2) = (0.1, 0.1, 0.2) may be okay, but is (1,2,3) - (1.5, 1.9, 2.4) = (0.5, -0.1, 0.4)?

I mean in the statistical sense, just like a scalar has a certain probability of belonging to (having been taken from?) a normal distribution. In that case you would use a test to reject or accept membership at a certain confidence level. I need the analogous test for (1) a vector, and (2) providing the confidence level on the cusp of rejection. 75.35.79.57 04:08, 19 June 2007 (UTC)

So you're actually asking for the probability that vector x' comes from the same distribution that generated the other vectors? Unless you happen to know what the original distribution was (i.e. that it was N(0,1)^M or N(mu,sigma^2)^M or whatever), in which case you don't need the set of vectors, then you've got to assume that the set of vectors you have is a reasonable approximation of the distribution. In which case, then I guess you'd produce a statistic of the form $((x_{i}-\mu _{i})/\sigma _{i}^{2})$, where $\mu _{i}$ and $\sigma _{i}$ are the mean and sd of the ith components of the set of vectors. I suspect each component of the statistic vector would then be t-distributed, but I'm not 100% sure. Confusing Manifestation 04:14, 19 June 2007 (UTC)

Yes, that's exactly what I want and your stated assumptions are true. Those are standard scores for each element, right? So, any idea how to combine them? Pythagorean means? 75.35.79.57 04:38, 19 June 2007 (UTC)