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April 10

LaTeX "compromise"

Is there a way of rendering \displaystyle in LaTeX such that it is slightly smaller, just as in a superscript or subscript?

e.g.

$e^{This\,\sum _{n}^{looks}asty\,with\,subscripts_{n}}$

$e^{This\,\displaystyle \sum _{b}^{looks}etter\,but\,is\,too\,large_{b}}$

x42bn6 Talk Mess 19:35, 10 April 2008 (UTC)

If you are concerned about the final subscript of a superscript:

$e^{This\,\sum _{better}^{looks}\,what\,do\,you\,think?_{\scriptstyle {n}}}$

$e^{This\,\sum _{nasty}^{looks}\,just\,like\,the\,original_{\scriptscriptstyle {n}}}$

If you are concerned about the later text of the superscript:

$e^{This\,\displaystyle \sum _{better}^{looks}\,\scriptstyle {what\,do\,you\,think?}_{b}}$

And don't forget \mathrm, \mathit, \mbox and their kin:

$e^{\mathrm {This} \,\displaystyle \sum _{\mathit {best}}^{\mathit {looks}}\,\scriptstyle {\mathit {what\,do\,you\,think\!?}}_{b}}$

Baccyak4H (Yak!) 02:22, 11 April 2008 (UTC)

Thanks, I'll take a look at those solutions. I was looking for a "shrunk" version of \displaystyle but couldn't find one. x42bn6 Talk Mess 23:15, 11 April 2008 (UTC)

Simplifying factorial equations

What software exists to help simplify equations involving large factorials? NeonMerlin 21:48, 10 April 2008 (UTC)

Any mathematical software should be able to handle factorials. Try Mathematica or Maple (or many others, but those are the two I know a little about). --Tango (talk) 22:39, 10 April 2008 (UTC)

Or SAGE, which is free, includes and unifies other pre-existing free mathematical software (among which GAP computer algebra system) and can use functions and packages from Mathematica, Maple etc., if you have them. Then again, perhaps this is a bit of an overkill just to simplify some factorials. Goochelaar (talk) 08:12, 11 April 2008 (UTC)

It will help if you give an example of what you have in mind. -- Meni Rosenfeld (talk) 09:43, 11 April 2008 (UTC)

Depending on what you're doing, you may be able to use the Gamma function or Stirling's approximation to simplify things (perhaps even reduce things to pencil & paper territory). -- 128.104.112.85 (talk) 18:01, 11 April 2008 (UTC)

Algebra

How do you solve

 4sinx0 +1 = -2

• ^^**^**^^*^*^*^**^22:12, 10 April 2008 (UTC)172.141.191.133 (talk)

DOVENSTONE

I assume you mean find x such that $4\sin x+1=-2$. Just rearrange the equation until you have $\sin x=a$ where a is some constant, and then take inverse sin of both sides. -mattbuck (Talk) 22:21, 10 April 2008 (UTC)

4sinx + 1 = -2 Isolate x or for now sinx so subtract 1 and divide by 4. Sinx = -3/4 then take the sin inverse of -3/4 I don't have a calculator with me but I'm sure you can do that so you should get x= #. I'm pretty sure that should be it

Draw a circle with center at B=(0,0) and radius 4. Let A=(4,0). Draw a horizontal line through the point (0,-3) meeting the circle at a point C on the left of y-axis. Then the angle ABC is your x. Using a compass, you can get an approximate value. I never use any powerful calculator in my life. twma 13:00, 16 April 2008 (UTC)

April 11

Polar to Cartesian

Given the formula ${\frac {r(1+e)}{1+ecosz}}\!$, where r is the radius from the focus to the ellipse border, e is the eccentricity, and z is the angle, how do you reformulate this into Cartesian coordinates?--十八 03:39, 11 April 2008 (UTC)

Write r and z as functions of x and y. Bo Jacoby (talk) 08:09, 11 April 2008 (UTC).

Start by defining the focus to be (x,y)=(0,0). The work out a formula in x and y for the distance from the origin (using Pythagoras). Then work out a formula in x and y for the angle, using basic trig. Then substitute it all in and simplify. --Tango (talk) 11:43, 11 April 2008 (UTC)

GMT-8

Odds

Is there a simple way to calculate the probabilities of an 11-year old boy in a developed country in the year 2008 living to a very old age, say 150? It was asked by him at the miscellaneous desk, but I'm curious about what the math answer might be if there is one. Any answer would help. Thanks, Julia Rossi (talk) 11:59, 11 April 2008 (UTC)

I would say probability=$\lim _{x\to 0}x\%$. C'mon, 150 is almost world record. Have you ever heard of anybody live over 120 in this 10 years? Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 12:51, 11 April 2008 (UTC)

Well, you have to consider that those people were born around 1900, healthcare was rather different then. I'd say it's entirely possible, but I don't know quite how you'd calculate it. In fact, I'd say it's impossible to calculate because it relies on knowledge of future conditions - whether global warming causes the spread of tropical diseases, how safe cars are in the future, how well we fend off the effects of ageing... You can't give odds because we can't say what will happen. 150 years is a LONG time. Consider that 150 years ago there were no such things as cars, planes, computers, fridges, plastics, widespread electricity, all the things we take for granted today. -mattbuck (Talk) 13:14, 11 April 2008 (UTC)

You can always calculate odds. If we don't know what's going to happen you work with the probabilities of what is going to happen. Probability is all about incomplete knowledge (unless you're working with Quantum Mechanics where things are actually random), so incomplete knowledge isn't going to stop you calculating it. --Tango (talk) 14:27, 11 April 2008 (UTC)

It's misleading to speak of "calculating" odds here. The subjectivist Bayesian_probability interpretation would be that any given individual can think about this situation and/or look at available evidence and come up with their own personal probability/odds (essentially, what their own betting odds would be on this). This is a complex situation and different people will likely come up with substantially different odds (as compared with rolling dice or flipping a coin, where almost everyone will come up with the same odds).

I'll also note that there are issues (in coming up with your own personal odds) with simply extrapolating from current data, as we do not know what the limits on human lifespan are, nor what medical advances may come in the next 100 years or so. kfgauss (talk) 07:35, 12 April 2008 (UTC)

You calculate the chance of someone living to a certain age using life tables, however no life tables are going to go up to 150 years since we have no evidence to base it on. You would have to decide on a way to extrapolate it, but with so little information to go on, it's going to be a pretty meaningless number. --Tango (talk) 14:27, 11 April 2008 (UTC)

Tango is right. Any mathematical model would be extrapolating so far beyond known data (the oldest person to have lived so far died at the age of 122) that it would be meaningless. There may be enough data to give a statistically meaningful answer up to an age of 110 or so, but beyond that - no chance ! Gandalf61 (talk) 14:46, 11 April 2008 (UTC)

Also note that for extremely old ages the chances of survival decrease probably at least exponentially because the body’s cells start running out of telomeres. At that point healthcare and modern technology become pretty much trivial factors. For a very long life span they’re certainly a limiting factor, and so you could make a model assuming a person lives an ideal life, figure out how many telomeres a person starts with (a random variable), the average rate of mitosis through the various stages of life (another few random variables), and figure out the probability of living to various ages of interest (again assuming an ideal life).

Not knowing the specifics, I don’t know what ages this model could be useful at: as it would only be applicable for people who have lived sufficiently old to get past the other hurdles of things that lead to an early death.

Another thing, although mostly unrelated, is to note is that a documented case of anyone living over 120 could potentially spark quite a religious debate because in the Bible, God is quoted as saying that man is limited to 120 years. GromXXVII (talk) 22:20, 11 April 2008 (UTC)

Um... Jeanne Calment... --Tango (talk) 22:40, 11 April 2008 (UTC)

Well, he did say that man is limited to 120 years. Black Carrot (talk) 23:18, 11 April 2008 (UTC)

Hi. I don't think this is a reliable source, but I think on a program called "Natural Cures" or something, it said that a man that is historicly documented lived to 150 years. Also, if I remember correctly, the Bible mentions Adam and Eve living to about 1000 years, and Noah living to about 700 years, but I don't remember exactly. Also, I don't really see how you can calculate such odds, if you don't know the exact circumstances of such future or the odds of such, or how you can calculate the possibility without a certain documentation of such an age, or the googolplexes of possible factors affecting this unlikely outcome. Thanks. ~AH1^(TCU) 23:55, 11 April 2008 (UTC)

Does this mean then, with things like telomeres to take into account, even given the history of life-expectancy increments, plus genetic intervention, changes such as ideal non-polluted environment, the body would have to stop around 120? Julia Rossi (talk) 01:12, 12 April 2008 (UTC)

My 2 cents, I would doubt that anyone born yet or in the next 20 years will reach 150, our environment is simply too polluted for that to happen I think. I'm not convinced that humans are limited to around 120 years maximum life span either, I'm no expert I doubt one can conclusively show that the rate of loss of Telomeres is necessary even approximately fixed either and I suspect is very much related to environmental conditions. I also doubt our environmental conditions will improve anytime soon either. As for the math, if you make a ton of assumptions (which removes it's relationship the real world which I assume you don't want) then yes it's calculable. A math-wiki (talk) 02:43, 12 April 2008 (UTC)

Well, leaving the real world aside, what would be the calculation process in math (and some english)? Would it involve a life mean (with projections taken into account), something like Ftbhrygvn's above with a couple of explanations? Nothing too involved, this is my first time here, thanks Julia Rossi (talk) 03:42, 12 April 2008 (UTC)

To clarify what I said a little , the telomeres comment and the age 120 have nothing to do with each other (I just thought the latter was interesting and noteworthy, while the former could actually be used to make a model). Every time a cell reproduces a telomere is expended, and so in a sense that starting value is determined at conception, although varies to some extent, but I don’t know the distribution.

As for when it becomes a factor that could lead to death, that also depends on a large part on how a person lives their life: because the bodies cells will reproduce when they need to, to account for losses for whatever reason. For instance, skin and bone marrow should be some of the first parts of the body to encounter this kind of problem because of it’s consistent reproduction throughout life.

What kind of lifestyle would be the needed ideal for this model? I’m not quite sure, but probably something fairly well balanced. For instance an athlete’s extreme exercise will cause greater cell reproduction, and a couch potato’s lethargy could cause perfectly good cells to go bad and cause others to reproduce to take its place.

That said, the ability to replace organs or large parts of the body could nullify this cause, especially because the brain is [I think] one of the lease active parts of the body in cell division, that by the time the brain runs out of telomeres, we’re talking about a timescale of probably tens of thousands of years if not more. However, there would be huge percentage of the body to replace to get around the decaying telomeres: it would probably be easier to figure out a way to graft telomerase (enzyme that adds telomeres) into the first somatic cells of a new organism.

As for an actual model for this, I’ll assume an ideal lifestyle, and a lucky enough person to avoid all disease or accidental death. Take one vital organ, say the skin, and some average starting number of telomeres S. While the body is capable, assume the skin has a roughly constant rate of use through adulthood, say R. That gives roughly S/R years until they run out and their skin can no longer function and death eventually follows. This has some obvious error because the rate will be greater during childhood growth, and less during old age when one starts to run out of telomeres. These errors have opposite signs, and so to some extent will cancel each other out: so perhaps with some luck S/R will be accurate to within a decade or two. There could be other errors though, like for instance I’m just guessing that skin has a constant rate of growth through adulthood.

Now if you want a real model, I’d say to find different models based on the leading causes of death at various age brackets, and weight them accordingly. For instance, childhood death is usually caused by something else, so take a model for that and weight to be the predominant influence during those years. Likewise the first few decades of old age will probably be more heavily affect by other things than telomeres. The error analysis of such a model will be quite ugly though. GromXXVII (talk) 12:01, 12 April 2008 (UTC)

The "natural limits" may no longer apply in another century. By that time we will be able to replace any body part that malfunctions, with the exception of the brain, because, of course, if your brain is replaced entirely it isn't really you any more. However, there may be ways we can replace it a few cells at a time and thus end up with the same "pattern", even though all the individual cells have been replaced. Now, as to how to calculate odds, I would take mortality tables from 1900, 1910, 1920, etc., and extrapolate how life expectancies increase with each decade. This wouldn't be all that accurate of a method, as it would be extrapolating well beyond our current data, and the more you extrapolate, the more the actual value tends to vary from the expected value. StuRat (talk) 04:00, 12 April 2008 (UTC)

For your fantastic help and consideration, my thanks to all, Julia Rossi (talk) 08:34, 12 April 2008 (UTC)

"it said that a man that is historicly documented lived to 150 years." Sounds to me like a biblical story. Imagine Reason (talk) 01:57, 13 April 2008 (UTC)

Rational Exponent solver

Is there a website that will solve any rational exponents? —Preceding unsigned comment added by 192.30.202.21 (talk) 19:56, 11 April 2008 (UTC)

What do you mean by a "rational exponent"? --Tango (talk) 21:04, 11 April 2008 (UTC)

Hi. I'm just guessing, and I'm not an expert on this, but perhaps an exponent by a rational number, like ^2.7 or ^7.9797..., and not an irrational number like ^π or ^$e$, and would a calculator work, or do you need an exact answer, as calculators can usually only hold finite number series? Thanks. ~AH1^(TCU) 00:02, 12 April 2008 (UTC)

Well, yes, that's what the words would normally mean, however the OP must mean something else, since you can't "solve" an operation. --Tango (talk) 00:05, 12 April 2008 (UTC)

Perhaps, in the idiolect of the questioner, "to solve" means "to compute the value of something involving".  --Lambiam 01:59, 12 April 2008 (UTC)

In which case, any scientific calculator will do the job. --Tango (talk) 13:24, 12 April 2008 (UTC)

Google will tell you that 3.14^2.72 = 22.4723579.  --Lambiam 02:01, 12 April 2008 (UTC)

Race

You have 25 horses, and a track on which you can race five of them at a time. You can determine in what order the horses in a race finished, but not how long they took, and so can not compare times from one race to another. A given horse runs at the same speed under all circumstances, and no two horses run at the same speed. How many races does it take to find the three fastest? I've found a way to do it in 7, and it can't be done in 5, so 6 is the big question mark. Black Carrot (talk) 23:14, 11 April 2008 (UTC)

I can do it in 25C5 races. I choose not to try and optimise my solution. -mattbuck (Talk) 23:21, 11 April 2008 (UTC)

Each horse would have to compete in 10626 races. I'm pretty sure there are laws against that. Black Carrot (talk) 23:32, 11 April 2008 (UTC)

He's my method, first off race 5 of them, and then take the 3rd horse and race it against 4 new ones, if anyone beats the 3rd horse, then the second horse will be in the next race as well, to determine if the new horse is 3rd, or 2nd or better. I'm also to lazy to check the efficiency of my method. A math-wiki (talk) 02:50, 12 April 2008 (UTC)

I don't think it fares too well if you pick slow horses near the beginning, since you'd focus a lot on getting rid of them one at a time. Black Carrot (talk) 03:20, 12 April 2008 (UTC)

What method do use to get 7 races ? StuRat (talk) 03:45, 12 April 2008 (UTC)

I think I've figured out the 7 race method. First you run each group of 5 in a race, for 5 races total. Then, in the 6th race, you race the champions, to get rankings for all 5 of those. Below are the results, with A1-E1 representing the champions, from fastest to slowest, and the remaining horses listed behind the champions in the order they finished in the eary races:

A1 A2 A3 A4 A5 <- early race order
B1 B2 B3 B4 B5 <- early race order
C1 C2 C3 C4 C5 <- early race order
D1 D2 D3 D4 D5 <- early race order
E1 E2 E3 E4 E5 <- early race order
^
|
Champions order

Now, we know the fastest horse is A1. The three fastest horses, in order, could be A1,A2,A3 or A1,A2,B1 or A1,B1,B2 or A1,B1,C1. Race A2,A3,B1,B2, and C1 to determine the 2nd and 3rd fastest horses. StuRat (talk) 04:27, 12 April 2008 (UTC)

That's exactly what I got. I wouldn't be surprised if it was unique, though I wouldn't know how to prove it. What do you think about the six-race case? I noticed that if you already know the three best, it's possible to prove you're right in only six races, so it's a close call either way. Black Carrot (talk) 07:14, 12 April 2008 (UTC)

I can't see any way to do it in 6 races. StuRat (talk) 13:58, 12 April 2008 (UTC)

I suppose so, but in many cases it could be quiet efficient. A math-wiki (talk) 19:56, 12 April 2008 (UTC)

I don't think it affects your answer, StuRat, but you left out a possible order: A1,B1,A2. --Tardis (talk) 15:15, 14 April 2008 (UTC)

Yes, good catch. StuRat (talk) 18:43, 14 April 2008 (UTC)

April 12

Statistics

This is a homework question but I've done the first bit of legwork for myself and I'm only asking for a hint.

There is a biased six sided diced such that the probability of getting a six is three times the probability of getting a 1, 2, 3, 4 or 5. This means that p(1)=0.125 and p(6)=0.375. Having worked out the expected value of X as 4.125, how do I calculate the variance? Thanks 92.0.233.217 (talk) 14:55, 12 April 2008 (UTC)

For a random variable X with expected value m, the variance of X is the expected value of (X-m)². You know that m is 33/8 or 4.125. So take each of the possible values of X, the number thrown on the dice; calculate (X-4.125)²; multiply this by the probability of throwing that number; and add these 6 values together. Gandalf61 (talk) 16:16, 12 April 2008 (UTC)

April 13

Not quite the Gamma function...

Hi everyone. Is there an elegant way of dealing with an integral that looks like this?

${\tilde {\Gamma }}(z)=\int _{0}^{\infty }t^{z-1}e^{-it}\,dt$

I'd like to take the real/imaginary parts of this in the end. Any ideas? --HappyCamper 06:22, 13 April 2008 (UTC)

I suggest trying a change of variables, such as $u=it$ (and thus $t=-iu$, $dt=-idu$ and $t^{z-1}=(-iu)^{z-1}=(-i)^{z-1}u^{z-1}$, etc.). --Prestidigitator (talk) 08:42, 13 April 2008 (UTC)

How many hours?

How many hours in 90 minutes? 60 minutes equals 1hr. If you cross multiply then 90 minutes equals 1.5 How do you make 1.5 an hour?71.142.208.226 (talk) 07:13, 13 April 2008 (UTC)Cardinal Raven

It's an hour and a half, that is, one hour plus half an hour. -- Meni Rosenfeld (talk) 07:20, 13 April 2008 (UTC)

Do you mean how to convert a number of hours into the relevant number of hours and minutes? The whole number part (here, the 1) is the number of hours. Multiply the rest by sixty (0.5 * 60 = 30) to give the number of minutes. Daniel (‽) 18:32, 13 April 2008 (UTC)

Sum

when (1/e)+(1/f)+(1/g)=0 and e+f+g=3, e^2+f^2+g^2=? the answer is 9, but why? note: ^ means the exponent. —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 08:03, 13 April 2008 (UTC)

Try multiplying the first equation by efg and squaring the second one. -- Meni Rosenfeld (talk) 08:27, 13 April 2008 (UTC)

Integration Problem

Hi, I'm wondering how to solve this integral: Given that E is the solid bounded by $y=2x^{2}+2z^{2}$ and the plane $y=8$, find $\displaystyle \int \displaystyle \int \displaystyle \int _{E}{\sqrt {3x^{2}+3z^{2}}}\,dV$ Thanks guys! —Preceding unsigned comment added by 70.111.95.226 (talk) 13:31, 13 April 2008 (UTC)

Take the projection of E on the x-z plane. For every point in the projection, the function is constant with respect to y. This will leave you with a 2-dimensional integral, which is best solved with polar coordinates. -- Meni Rosenfeld (talk) 14:14, 13 April 2008 (UTC)

Statistics problems

Hello! I ran into a couple of statistics problems, and wanted to check if my reasoning is correct.

1. Many public schools are implementing a "no pass, no play" rule for athletes. Under this system, a student who fails a course is disqualified from participating in extracurricular activities during the next grading period. Suppose the probability that an athlete who has not previously been disqualified will be disqualified is 0.15 and the probability that an athlete who has been disqualified will be disqualified again in the next time period is 0.5. If 30% of the athletes have been disqualified before, what is the unconditional probability that an athlete will be disqualified during the next grading period?

If we denote the event "an athlete is disqualified" with D, and the event "an athlete had been disqualified previously" with P, then we get that

$P(D|P^{c})=0.15$

$P(D|P)=0.5$

$P(P)=0.3$

$P(D)=?$

And I suppose that $P(D)=P(D|P^{c})\times P(P^{c})+P(D|P)\times P(P)=0.0561$

2. A research physician compared the effectiveness of two blood pressure drugs A and B by administering the two drugs to each of four pairs of identical twins. Drug A was given to one member of a pair; drug B to the other. If, in fact, there is no difference in the effects of the drugs, what is the probability that the drop in the blood pressure reading for drug A exceeds the corresponding drop in the reading for drug B for all four pairs of twins? Suppose drug B created a greater drop in blood pressure than drug A for each of the four pairs of twins. Do you think this provides sufficient evidence to indicate that drug B is more effectivfe in lowering blood pressure than drug A?

If I understand correctly, the probability that the reading for drug A exceeds the reading for drug B in a single pair of twins is 1/2. Therefore, the probability that the reading for drug A exceeds the reading for drug B should be $(1/2)^{4}=1/16$. Regarding the second question, I think that this is not enough evidence, as there is a 1/16 chance that this was just a coincidence.

Why I am wondering if my reasoning in these problems is correct is that these solutions seem far too easy, especially my solution to the second problem. So, if you find any errors, please point me towards them :)  ARTYOM  14:28, 13 April 2008 (UTC)

For problem 1, the formula is correct but the number is wrong - I got 0.255.

For problem 2, indeed assuming that the probability of an equal blood pressure drop is 0, symmetry guarantees that the probability is 1/2 each and 1/16 total. I agree that there is not enough evidence, but not for the reason you stated - the chance that this was a coincidence is not 1/16, in fact, we can't calculate it without knowing the prior probability of the treatments being different. -- Meni Rosenfeld (talk) 14:41, 13 April 2008 (UTC)

Thanks a lot! I got 0.255 for the first problem now too; no idea how I came up with 0.0561 the first time :D  ARTYOM  15:01, 13 April 2008 (UTC)

In future, it's good to "sanity check" your answers. What you're doing is taking a weighted average, so the correct answer must be somewhere between the two numbers you're averaging. Your answer was less than both, so couldn't possibly be right. There are often quick checks like that with maths problems - everyone makes mistakes, so it's good to check your answers are sane. --Tango (talk) 21:42, 13 April 2008 (UTC)

Perhaps the answer that the first part of problem 2 is "looking for" is 1/16, but this is incorrect. It can't be calculated using only the information given. Think about it: if the reading for drug "A" exceeds the reading for drug "B" on the first three sets of twins, your expectation that it will as well for the last set of twins should be greater than 1/2.

Only if you know beforehand that drugs "A" and "B" are equally effective (i.e. the probability of either one doing better in a given pair-of-twins trial is 1/2) is the answer 1/16. kfgauss (talk) 09:44, 14 April 2008 (UTC)

It is stated explicitly that we are assuming the treatments are equally effective. The question would obviously be meaningless otherwise. -- Meni Rosenfeld (talk) 10:02, 14 April 2008 (UTC)

Ah, reading the problem. How novel :). In any case, the concern I mention is relevant for the second half of the problem. As you say, having a prior probability distribution is key. If you live in a world in which two drugs are rarely of comparable effectiveness, then 4 trials may be fairly convincing. If you live in a world in which a two drugs are usually comparably effective, then you may need many many trials to determine with confidence which is better. A good reference for this stuff is Example 3 in Bayes'_theorem. kfgauss (talk) 18:50, 14 April 2008 (UTC)

Soon enough, we'll have probabilistically checkable problems, for which only 3 bits of the problem need to be read in order to know the solution :) -- Meni Rosenfeld (talk) 19:20, 14 April 2008 (UTC)

I didn't like the way question 1 was written. Do they mean the chances a particular athlete will be disqualified (that's how everyone here read it), or of any athlete in the class, school, or district being disqualified ? For the second interpretation you'd need to know the number of athletes in the group.

One other assumption for part 2 seems to be that the twins can't have identical BP readings. StuRat (talk) 19:02, 14 April 2008 (UTC)

Regarding the athletes problem - as far as I understand, your second interpretation is what the problem asks. However, despite the fact that I'm not "friends" with statistics (not at all! :D), I don't think that we have to know the total number of athletes in the group, since we are given the information that 30% had been disqualified before. And regarding the second problem--as far as I understand, again--they say that the drop in the BP is measured and compared, not the BP itself. The twins are taken because, I suppose, there is a great probability that the same medicine will have quite similar impact on both.  ARTYOM  21:12, 14 April 2008 (UTC)

No, the chances that at least one athlete would be disqualified would go up the more people you have in the group. As for measuring the drop in BP, twins could still have identical drops, at least to the accuracy measured by the apparatus. StuRat (talk) 02:34, 15 April 2008 (UTC)

And yet another question is giving me a hard time!

3. A survey of people in a given region showed that 20% were smokers. The probability of death due to lung cancer, given that a person smoked, was roughly 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006, what is the probability of death due to lung cancer given that a person is a smoker?

Initially this seemed like an easy problem for me. I denoted "smokers" event with S, and the "death due to lung cancer" with L. Then from the question we get that

$P(S)=0.2$

$P(L|S)=10\times P(L|S^{c})$

$P(L)=0.006$

$P(L|S)=?$

So I went on with

$P(L)=P(L|S)+P(L|S^{c})=11P(L|S^{c})=0.006$

Then, $P(L|S^{c})\approx 0.00055$, so $P(L|S)=0.0055$

But then I figured that perhaps the solution should be similar to the solution of problem #1 above (about athletes), and that $P(L)=0.006$ should in fact equal to $P(L|S)\times P(S)+P(L|S^{c})\times P(S^{c})$, which in the end gives me that $P(L|S)\approx 0.0214$. Which of these ways is the right way, as both of them make perfect sense to me, and why?  ARTYOM  21:32, 14 April 2008 (UTC)

I figure it this way:

P_n = Probability that a given Nonsmoker dies of lung cancer.

P_s = 10P_n = Probability that a given Smoker dies of lung cancer.

P_? = Probability that a given person of unknown smoking status dies of lung cancer.

Since we know 20% of the people are smokers, we get:

P_? = (0.2)P_s + (0.8)P_n

P_? = (0.2)10P_n + (0.8)P_n

P_? = 2P_n + (0.8)P_n

P_? = 2.8P_n

Now we toss in the actual value:

2.8P_n = 0.006

P_n = 0.0021428

This gives us:

P_s = 10P_n = 10(0.0021428) = 0.021428

Or, about a 2 1/7 % chance, which matches your second result. Your first result can be dismissed because you calculated the chances of a smoker dying of lung cancer at being less than those of the average person, and that's obviously wrong. You didn't even use the 20% smokers figure in the calculations, and that's obviously important. StuRat (talk) 03:09, 15 April 2008 (UTC)

Unknown Symbol

The other day I was at a science fair and saw a formula pertaining to gravity and general relativity. I can't remember the exact formula but it had the general format of ${\frac {a-b}{c^{n}+d}}{\Bigg |}_{x}^{y}$. What is the meaning of the | ? Thanks, Zrs 12 (talk) 18:16, 13 April 2008 (UTC)

It means "evaluated at", or in the case of a number at the top and bottom, "evaluated between". You evaluate it at the top value, and then subtract it's value at the bottom value. In the case of physics, the bottom value was probably the starting point, and the top value the end point. The energy required to lift something, for example, is the gravitational potential energy at the end minus the gravitational potential energy at the beginning - that could be written using such notation. --Tango (talk) 19:29, 13 April 2008 (UTC)

So $\int _{1}^{5}2x\ \operatorname {d} x=x^{2}|_{1}^{5}=F(5)-F(1)=24$? Zrs 12 (talk) 19:48, 13 April 2008 (UTC)

Yep. --Tango (talk) 19:59, 13 April 2008 (UTC)

Thanks, Tango Zrs 12 (talk) 23:02, 13 April 2008 (UTC)

Only that it should be $\int _{1}^{5}2x\ dx$. The dx is an integral part (pun intended) of the notation. -- Meni Rosenfeld (talk) 23:17, 13 April 2008 (UTC)

Haha. Nice one. Yep, I'm not in calculus yet so I tend to forget the dx quite often. Zrs 12 (talk) 00:32, 14 April 2008 (UTC)

Square Roots

Can anyone give me a way to find square roots by hand without the guess and check method? Zrs 12 (talk) 19:00, 13 April 2008 (UTC)

Take a look at Methods of computing square roots, in particular Methods of computing square roots#Digit by digit calculation. -- Meni Rosenfeld (talk) 19:30, 13 April 2008 (UTC)

Thanks, Meni Zrs 12 (talk) 23:04, 13 April 2008 (UTC)

Finding digits of Pi

Is it possible to calculate any digit of Pi (or other types of mathematical constants) without knowing the previous digits in the decimal expansion ? How/Why not ? -- Xedi (talk) 21:19, 13 April 2008 (UTC)

See Bailey-Borwein-Plouffe formula. PrimeHunter (talk) 21:29, 13 April 2008 (UTC)

Yes, for some constants and in some bases - this is called a spigot algorithm. As PrimeHunter says, the Bailey-Borwein-Plouffe formula is a spigot algorithm for the binary expansion of π. I don't know whether a spigot algorithm is known for the decimal expansion of π. Gandalf61 (talk) 21:36, 13 April 2008 (UTC)

See also, Science News Online, Ivars Peterson's MathTrek (2/28/98): Pick a Digit, Any Digit. – b_jonas 08:53, 14 April 2008 (UTC)

Thanks. -- Xedi (talk) 15:50, 14 April 2008 (UTC)

April 14

Minimization of three variables

Given that x, y, and z are all real numbers, find the minimum possible value of $x^{4}+y^{4}+z^{4}-4xyz$. I don't know how to tackle these kinds of problems. Thanks. —Preceding unsigned comment added by 70.111.95.226 (talk) 03:18, 14 April 2008 (UTC)

Trying to minimise a polynomial expression usually makes use of the fact that any real number raised to an even power must be positive. So, what you need to do is construct an expression that equals x^4+y^4+z^4-4xyz, but written in a form of (some stuff)^2 + a constant. Then you can say that the minimum value occurs when some stuff = 0 (you may have to prove that this is possible), so the minimum of the expression must be the constant. I would suggest looking at maybe expansions that look something like ((x - y)^2 + (y - z)^2 + (z - x)^2)^2, although that's just an educated guess. Confusing Manifestation(Say hi!) 03:32, 14 April 2008 (UTC)

Although, come to think of it, the fact that you've got a degree 3 term in there suggests that maybe something closer to (x + y + z - 1)^4 would be more helpful. Confusing Manifestation(Say hi!) 03:33, 14 April 2008 (UTC)

Since the function gets big for big values of x, y and z, the minimum value will be attained at a local minimum. Local minima can only occur at points where the gradient of the function is zero. By my calculations, that only gives you 8 points to check (or 2, if you do it cleverly). Is that any help? 134.173.93.127 (talk) 03:45, 14 April 2008 (UTC)

The function f(x,y,z)= x⁴+y⁴+z⁴−4xyz is symmetrical. (f(x,y,z)=f(z,y,x)=f(y,x,z)=&c). f(x,x,x)=3x⁴−4x³ has minimum f(1,1,1)=−1. Is that a local minimum for f(x,y,z)? Are there other local minima? Bo Jacoby (talk) 12:34, 14 April 2008 (UTC).

Yes, in addition to (1,1,1) there are also (-1,-1,1) & (-1,1,-1) & (1,-1,-1). StuRat (talk) 18:41, 14 April 2008 (UTC)

Check whether the Hessian matrix is positive definite at the singular point or not. The singular point is the one where the gradient vanishes.--Shahab (talk) 17:49, 14 April 2008 (UTC)

Also, you can use the AM-GM inequality applied to x^4, y^4, z^4 and 1^4.

(x^4 + y^4 + z^4 + 1^4)/4 ≥ xyz

therefore

x^4 + y^4 + z^4 + 1 ≥ 4xyz

x^4 + y^4 + z^4 - 4xyz ≥ -1

so you can get the minimum without finding the actual values of x,y and z. 91.143.188.103 (talk) 21:33, 14 April 2008 (UTC)

That gives you a bound on the minimum, it doesn't tell you if that bound is attained. It's a good place to start, but you do still need to find the values of x, y and z with give equality (which is trivial, of course). --Tango (talk) 22:41, 14 April 2008 (UTC)

We were actually only requested to find "find the minimum possible value of $x^{4}+y^{4}+z^{4}-4xyz$" which is −1, not "the values of x, y and z with give equality". Bo Jacoby (talk) 07:09, 15 April 2008 (UTC).

Yes, but if you've only found a lower bound, then you need to prove that the function attains that bound at some point, otherwise the actual minimum could easily be greater than that lower bound. Confusing Manifestation(Say hi!) 23:02, 15 April 2008 (UTC)

Yes, but the second part of the AM-GM inequality says that equality is attained if and only if x^4 = y^4 = z^4 = 1 (note that some care is needed over signs and roots to make sure you don't introduce spurious "solutions" - details left to student). Gandalf61 (talk) 11:06, 16 April 2008 (UTC)

aquestion about function

for quite some time the mathematicians have thought the mean value conditions is quite enough for afunction to be continouse at acertain point until RIEMANN gave his abs(x),where it satisfies the mean value at zero but not continouse at zero.my question is that i recall afunction but i cannot find it,it satisfies the mean value at every pints within it`s range but it is not continouse at all of those points.any one can tell me what is that function?thank you very much.Husseinshimaljasimdini (talk) 09:26, 14 April 2008 (UTC)

I think you are talking about the mean value theorem. The mean value theorem does indeed not apply to abs(x), because abs(x)=abs(-x) yet there is no point where the derivative of abs(x) is 0. Abs(x) fails to meet the conditions of the mean value theorem because it is not differentiable at 0. Note that this is not because abs(x) is not continous - it is continuous everywhere - differentiability is a stronger condition than continuity.

To construct a function that fails to meet the conditions of the mean value theorem for every interval in its range, you need a function that is nowhere differentiable. Since differentiability requires continuity you could use a function that is nowhere continuous, like the Dirichlet function, or a function that has a dense set of discontinuities, like Thomae's function. There are also functions that are everywhere continuous but nowhere differentiable, like the Koch curve. Gandalf61 (talk) 10:00, 14 April 2008 (UTC)

Thank you.I got it now.I realize that my informations were mixed up.Husseinshimaljasimdini (talk) 08:36, 15 April 2008 (UTC)

Scheduling problem

I've just about convinced myself that the following problem has no feasible solution - can someone either confirm this, or find one?

Four people who walk at the same speed are to complete a certain route. There is available one bicycle and one moped, each of which can carry only one person. Each person will ride the bicycle at the same speed and the moped at the same speed. Suppose that walk/bicycle/moped speeds are 5/10/20 mph. It is required that all four people start and end the journey together - this is to be achieved by each walking half the distance, riding the bicycle for one quarter of the distance and riding the moped for one quarter of the distance. These fractions can be made up of any number of smaller components.

Everything I've tried breaks down by the bicycle being required somewhere before it has arrived.—81.132.237.15 (talk) 18:29, 14 April 2008 (UTC)

Does it have be optimal? I have a solution where everyone waits an hour (call it an 80 mile track) out of 12 hours total -- maybe it's possible to do it in 11 still. 207.148.157.228 (talk) 19:31, 14 April 2008 (UTC)

I double-checked this, I think it works. Again, assume an 80 mile route.

A: 1hr moped, 8hr walk, 1hr rest, 2hr bike

B: 2hr bike, 1hr moped, 1hr rest, 8hr walk

C: 4hr walk, 2hr bike, 1hr moped, 1hr rest, 4hr walk

D: 8hr walk, 2hr bike, 1hr rest, 1hr moped

207.148.157.228 (talk) 19:59, 14 April 2008 (UTC)

I'd just break the bike apart, form it into two unicycles, sit a 2nd guy on the back of the moped and make him pull the unicyclists along. Lateral thinking ;) -mattbuck (Talk) 12:28, 15 April 2008 (UTC)

Better yet, hail a passing cab. --Tango (talk) 23:19, 15 April 2008 (UTC)

Euclidean prime numbers do not exist and euclid must should himself know it.

In 1996 Torsten Jensen was awarded the Millennium Leibniz Prize in logic, mathematics, physics, chemistry and medicine by providing a beautifully crafted very short proof that showed that all natural numbers can be divided by 3 and therefore Euclides theorem is false, invalid and worthless.

My question is this: 1. Do you know his ultra simple proof? 2. Why Euclides himself did not think of it? 3. Why did it take about 2300 years before a man saw the error in Euclid's theorem and destroyed about 12,000 rubbish theorems in number theory including Andrew Wiles's attempt at finding a proof for Fermat's last theorem?

Signed: T. Hansen, Lans, German Lutheran Church <email removed>

I do not mind at all if people read and steal my thoughts. —Preceding unsigned comment added by 81.152.51.207 (talk) 20:51, 14 April 2008 (UTC)

Strangely, our article on the Gottfried Wilhelm Leibniz Prize makes no mention of Herr Jensen and his incredible proof. Gandalf61 (talk) 21:02, 14 April 2008 (UTC)

Please do not write in all capitals, it's regarded as SHOUTING and is very irritating to the reader - Adrian Pingstone (talk) 21:08, 14 April 2008 (UTC)

Fixed that. — Kieff | Talk 23:31, 14 April 2008 (UTC)

Sure, all natural numbers can be divided by 3, but you don't get a natural number at the end of it 67% of the time. What's your point? --Tango (talk) 22:44, 14 April 2008 (UTC)

There is apparently no mention of +"Torsten Jensen" +Leibniz on the web, which means this whole thing is almost certainly made up and trolling. Not that any of us needed to look it up. — Kieff | Talk 23:31, 14 April 2008 (UTC)

I didn't know that 2 is divisible by 3... thanks for the post! --123.243.7.17 (talk) 02:03, 15 April 2008 (UTC)

Sure, you divide 2 by 3 and you get two thirds. --Tango (talk) 23:18, 15 April 2008 (UTC)

April 15

"Math Problems to be an Easy Task"

I'm struggling with my homework and by the policy of the reference desk I've painstakingly tried (with no avail) to solve the foremost on my homework, though found necessary data willing as a factor to solve the question. Generally, it informs the letter "e" is present about 44% of all words printed in books, magazines, and newspapers. Then it asks if a history books contain 275 worded text of page 219 it asks which is the best estimate of how many words will contain the letter "e".

I was unaware of how to establish a correct answer to answer the problem.

May someone of higher mathematical intelligence assist? --Writer Cartoonist (talk) 23:01, 15 April 2008 (UTC)

For each word, there are two options. Either it has an 'e', or it doesn't. That means you're looking at a binomial distribution. --Tango (talk) 23:17, 15 April 2008 (UTC)

In general, the fewer numerical values you are given, the easier the problem will be, as there is a limit on the number of things you can do with them. I can inform you, without violating the no-homework rule, that "219" has no bearing on the answer.—81.132.235.198 (talk) 09:22, 16 April 2008 (UTC)

Integrals

$\int _{0}^{\infty }{e}^{-x^{2}}\ \mathrm {d} x={\frac {\sqrt {\pi }}{2}}$ Therefore, $\int _{0}^{\infty }{e}^{-x^{2}}\ \mathrm {d} x=\int _{0}^{-\infty }e^{x^{2}}\ \mathrm {d} x$?

$\int _{-\infty }^{\infty }{e}^{-x^{2}}\ \mathrm {d} x={\sqrt {\pi }}$ So, $\int _{\infty }^{-\infty }e^{-x^{2}}\ \mathrm {d} x=-{\sqrt {\pi }}$ or possibly ${\sqrt {-\pi }}$? (By the way, I know an integral with an upper limit smaller than the lower limit is improprer.)

Futhermore, how does one go about evaluating integrals with infinite limits? I assume it can't be done with the method of $F(n_{u})-F(n_{l})$. Thanks in advance, Zrs 12 (talk) 23:23, 15 April 2008 (UTC)

I think you're getting confused with how minus signs and squares/square roots work. Your first "therefore" looks wrong to me - it looks like you've changed variables, x to -x, however (-x)²=x², not -x². --Tango (talk) 23:32, 15 April 2008 (UTC)

Oh, and $\mathrm {d} (-x)=-\mathrm {d} x$. --Tango (talk) 23:34, 15 April 2008 (UTC)

Oh, and to answer your last question. It can sort of by done using the F(b)-F(a) method, but you have to replace $F(\infty )$ with $\lim _{b\to \infty }F(b)$. See improper integral. --Tango (talk) 23:38, 15 April 2008 (UTC)

Yeah, I'm more confused about calculus. I'm not in the class yet and therefore just have to wonder and try to find out. Zrs 12 (talk) 23:47, 15 April 2008 (UTC)

Oh, and by the way, what does $\textstyle {:=}\,\!$ mean? Zrs 12 (talk) 23:52, 15 April 2008 (UTC)

I think it means "defined as". nneonneo ^talk 23:53, 15 April 2008 (UTC)

Yes, it does. (It's far from a universal notation, but I've never seen it used to mean anything else.) --Tango (talk) 23:56, 15 April 2008 (UTC)

April 16

Short exact sequences of abelian groups

I'm trying to determine which abelian groups A will fit into a short exact sequence of the form $\textstyle 0\rightarrow \mathbb {Z} \rightarrow A\rightarrow \mathbb {Z} _{n}\rightarrow 0$.

I know that I must have $\textstyle \mathbb {Z} _{n}\cong A/\mathbb {Z}$. And by Lagrange's lemma, $|A|=\left[A:\mathbb {Z} \right]\cdot |\mathbb {Z} |=|\mathbb {Z} _{n}|\cdot |\mathbb {Z} |$ but I haven't been able to make any further headway than this. I managed to use this technique to prove that any group B that fits into $\textstyle 0\rightarrow \mathbb {Z} _{p^{m}}\rightarrow B\rightarrow \mathbb {Z} _{p^{n}}\rightarrow 0$ is isomorphic to $\textstyle \mathbb {Z} _{p^{m+n}}$, but that worked because the orders were all finite, which fails in this case. Can anybody give me a suggestion on how to approach this? Thanks. Maelin (Talk | Contribs) 02:18, 16 April 2008 (UTC)

Okay, so you need to embed $\mathbb {Z}$ in A in such a way that its image is the kernel of a projection from A to $\mathbb {Z} _{n}$. I think one solution is to take A as $\mathbb {Z} \times \mathbb {Z} _{n}$, with embedding

$\mathbb {Z} \rightarrow \mathbb {Z} \times \mathbb {Z} _{n}:j\rightarrow (j,{\overline {0}})$

and projection

$\mathbb {Z} \times \mathbb {Z} _{n}\rightarrow \mathbb {Z} _{n}:(j,{\overline {k}})\rightarrow {\overline {k}}$

(where ${\overline {k}}$ is the congruence class k mod n). Not sure if this is the only solution. Gandalf61 (talk) 10:28, 16 April 2008 (UTC)

It isn't. There's also the solution A=Z, with the maps being multiplication by n and the natural projection. There may be other solutions, depending on the value of n (e.g. if n=ab with a,b coprime, we can take A=ZxZ_a, with Z mapping into A by multiplication by b into the first factor). I don't have time to solve this completely; the classification theorem may be useful, and possibly the matrixy techniques used to prove it. Algebraist 13:10, 16 April 2008 (UTC)

Calculus and Tensors

Will someone please give me a simple, easy to comprehend definition of vector space, tensor, vector feild, and tensor field? I have read the articles, but I still don't understand.

Also, what is the integral of a partial derivative? Thanks, Zrs 12 (talk) 02:55, 16 April 2008 (UTC)

For vectors, start with R², i.e. the Cartesian plane. Each point on the plane can be written as an ordered pair (x, y), and can be depicted either as just the point, or as an arrow from the origin to the point. You can then define addition on these arrows - to add two of them, join them head-to-tail and take the arrow that connects the tail of the first with the head of the second (equivalently, add the x and y coordinates separately). You can also define scalar multiplication - given the vector V and the scalar (i.e. real number) a, the multiplication aV is the vector heading in the direction of V, with length a times the length of V (if a is negative, then aV ends up pointing in the opposite direction to V). The plane, along with these two operations, forms a vector space over the real numbers - if you read the vector space article now, you can try and prove that these operations satisfy the vector space axioms given.

A tensor is a generalisation of vectors. Where a vector V can be written in terms of one dimension of components, say V_1 and V_2, a tensor can have any number of degrees in which to have said components - so, for example, T_11 T_12 T_13 T_21 T_22 T_23 would be a rank 2 tensor of dimensions (2, 3). However, it still has most of the linearity of a vector (i.e. you can multiply it by a scalar and add it to another tensor of the same rank), but with some trickier operations involved as well.

A vector or tensor field is just a region where each point in the region has an associated vector/tensor. You can think of it as a function V(x) where entering a position x gives you a vector/tensor V. For example, say you're in a wind tunnel. At every point in the tunnel, the wind pushes on you with a different force. If you mapped the force on you at various points in the tunnel, that map would show you the force vector field. Confusing Manifestation(Say hi!) 04:21, 16 April 2008 (UTC)

question

does always dy\dx=1\(dx\dy)?thank you.Husseinshimaljasimdini (talk) 14:04, 16 April 2008 (UTC)