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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)[reply]

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)[reply]

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[reply]

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

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. D aniel (‽) 18:32, 13 April 2008 (UTC)[reply]

I believe the question concerns dimensional analysis. Here's why your intuition that the units should work out is indeed correct:

90 min × 1/60 h/min = 90/60 h = 1.5 h.

The general algorithm is to start with the quantity you're given and repeatedly multiply by conversion factors. So long as each new multiplier equals one, you know your final product equals your initial quantity. As another example, consider the problem of expressing the speed 45 ft/s in miles per hour:

45 ft/s × 60 s/min × 60 min/h × 1/5280 mi/ft ≈ 30.7 mi/h.

PaulTanenbaum (talk) 00:33, 17 April 2008 (UTC)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)=0.255

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 :) AR TYOM 14:28, 13 April 2008 (UTC)[reply]

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)[reply]

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 AR TYOM 15:01, 13 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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. AR TYOM 21:12, 14 April 2008 (UTC)[reply]

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)[reply]

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? AR TYOM 21:32, 14 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

Precisely what would you do after that? Most cabs won't let you to put the bicycle in. – b_jonas 09:26, 18 April 2008 (UTC)[reply]

Leave it behind - who needs a bicycle once the cab gets there? --Tango (talk) 14:30, 19 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

If I understand the question, you're told about what percentage of words tend to have a property, and then asked about what percentage of words probably have that property. The answer is the same - about 44% of the total. See Percentage. Black Carrot (talk) 07:02, 17 April 2008 (UTC)[reply]

Just multiply 275 words by 44%, or, in other words, find 275 × 0.44. The only trick here is that they tossed in an extraneous value, the page number, which you must ignore. StuRat (talk) 18:00, 17 April 2008 (UTC)[reply]

Unless it's asking how many words with e there are in the entire book, and telling you that there are 219 pages. In that case, you can assume that there are roughly 275 words on each page, hence 275*219=60225 words in total. 60225*0.44=26499, which is an estimate for the whole book. D aniel (‽) 12:47, 20 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

Oh, and

\mathrm {d} (-x)=-\mathrm {d} x

. --Tango (talk) 23:34, 15 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

Oh, and by the way, what does

\textstyle {:=}\,\!

mean? Zrs 12 (talk) 23:52, 15 April 2008 (UTC)[reply]

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

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)[reply]

I've seen some variation: a:=b can either mean 'a is defined to be b' or 'b is defined to be a'. I believe the former is more common, but both are used. Algebraist 16:20, 16 April 2008 (UTC)[reply]

The first is the only one I've seen, I expect that is the more common. --Tango (talk) 16:39, 16 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

From the theory of group extensions, any such group A must have a presentation

\langle x,y\mid y^{n}=x^{s},y^{-1}xy=x^{t}\rangle

for some integers s and t. Since we want A to be abelian we must have t = 1 and then it is not too hard to see that A is isomorphic to

\mathbb {Z} \times \mathbb {Z} _{(n,s)}

. Hence A is isomorphic to

\mathbb {Z} \times \mathbb {Z} _{d}

for some divisor d of n.

Also, for any integer s the homomorphisms

\iota :\mathbb {Z} \rightarrow A

and

\phi :A\rightarrow \mathbb {Z} _{n}

such that

\iota (1)=x

,

\phi (x)=0

,

\phi (y)=1

do make the sequence exact, so d can be any divisor of n.

I think that you might have made a mistake in the finite case. There is an exact sequence

\textstyle 0\rightarrow \mathbb {Z} _{p^{m}}\rightarrow \mathbb {Z} _{p^{m}}\times \mathbb {Z} _{p^{n}}\rightarrow \mathbb {Z} _{p^{n}}\rightarrow 0

but

\mathbb {Z} _{p^{m}}\times \mathbb {Z} _{p^{n}}

is not isomorphic to

\mathbb {Z} _{p^{m+n}}

when both m and n are positive.

Matthew Auger (talk) 03:30, 17 April 2008 (UTC)[reply]

Matthew Auger's answer is quite clear, systematic, and elementary, so there is nothing really to improve that solution. Variety is nice, however, so I wanted to mention a couple of other ideas. A different approach that interested me was given earlier this year at sci.math. It introduces a number of more advanced concepts through example in a nice setting. Another approach that might be helpful to some would be to consider Ext(Z,Z/nZ)=Z/nZ is cyclic, so the generator Z -n-> Z -> Z/nZ can be Baer summed repeatedly to get the other extensions. These (including MA's solution) also work in the "finite case" above. JackSchmidt (talk) 00:47, 18 April 2008 (UTC)[reply]

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)[reply]

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)[reply]

question

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

As long as both dy/dx and dx/dy are defined and non-zero, I think so, yes. It's important to note that

{\frac {\partial y}{\partial x}}\neq {\frac {1}{\frac {\partial x}{\partial y}}}

in general, it only works for total derivatives. --Tango (talk) 14:15, 16 April 2008 (UTC)[reply]

Perhaps I am being dense here, but I can't think of a case where

{\frac {\partial y}{\partial x}}\neq {\frac {1}{\frac {\partial x}{\partial y}}}

, as long as both partial derivatives are defined and non-zero. Do you have an example ? Gandalf61 (talk) 14:30, 16 April 2008 (UTC)[reply]

Now that you mention it, neither can I, but I swear I remember being taught that... I see three possibilities: 1) I was taught wrong, 2) I'm remembering wrong or 3) We're both being dense. Anyone want to help out? --Tango (talk) 15:25, 16 April 2008 (UTC)[reply]

I believe either (1) or (2) holds. Algebraist 16:16, 16 April 2008 (UTC)[reply]

Yeah... perhaps it was

{\frac {\mathrm {d} y}{\mathrm {d} x}}\neq {\frac {1}{\frac {\partial x}{\partial y}}}

, which is true. --Tango (talk) 16:38, 16 April 2008 (UTC)[reply]

Maybe you're thinking of the fact that

{\frac {\partial x}{\partial y}}{\frac {\partial y}{\partial z}}{\frac {\partial z}{\partial x}}=-1

(provided all three derivatives are defined). The minus sign shows up when the number of variables in the chain is odd. Partial derivatives are weird. -- BenRG (talk) 10:34, 17 April 2008 (UTC)[reply]

Mathematics

Does anybody here know how to do the rational expopnents? 25 3/2, —Preceding unsigned comment added by 192.30.202.29 (talk) 14:32, 16 April 2008 (UTC)[reply]

Did you mean

25^{3/2}

? There are two rules for exponentiation, which, when combined, allow you to calculate rational exponents. First,

a^{1/b}={\sqrt[{b}]{a}}

, and second,

a^{b\cdot c}=(a^{b})^{c}

. See Exponentiation with real numbers for more information. nneonneo ^talk 14:55, 16 April 2008 (UTC)[reply]

Also see this simple example at Wikiversity: [1]. Note that since there is an even root implied (a square root, in this case), there are two roots, a postivie and a negative. StuRat (talk) 17:30, 17 April 2008 (UTC)[reply]

Paradox

This brings up an interesting paradox. What's wrong with the following logic:

 $1=1^{1}=1^{2/2}={\sqrt {1^{2}}}={\sqrt {1}}=-1$

StuRat (talk) 17:53, 17 April 2008 (UTC)[reply]

Well, first a notational point -

{\sqrt {1}}=1

, that symbol means "positive square root". However, that doesn't really matter - the key point is that there are two square roots to any number (except 0), but you don't always have a choice of which one to take. One of them will always work, but not necessarily both. See Extraneous and missing solutions for more information. --Tango (talk) 19:30, 17 April 2008 (UTC)[reply]

Hmm, it's news to me that

{\sqrt {1}}

means only the positive root while

1^{1/2}

means both roots. Are you sure this is a universal convention ? StuRat (talk) 04:12, 18 April 2008 (UTC)[reply]

It's not an isolated one, at least. In my experience, especially when working in the complex numbers, notation such as

z^{1/2}

tends to refer to the dual-valued square root, whereas the square root sign tends only to be used as an operator on real numbers and returns a single value. Confusing Manifestation(Say hi!) 05:20, 18 April 2008 (UTC)[reply]

That's my experience. It's also what it says in the square root article, if memory serves. --Tango (talk) 18:08, 18 April 2008 (UTC)[reply]

Reflection matrix about any line rotated theta degrees from the x-axis

I am trying to find a transformation matrix that will reflect any vector (compatible of course) rotated theta degrees from the x-axis (counterclockwise). I remember seeing a matrix for this somewhere, but when I attempted to look this up I only got reflections across the x-axis, y-axis and the line x=y. John Riemann Soong (talk) 19:33, 16 April 2008 (UTC)[reply]

If I understood your question correctly, you are looking for

\left({\begin{array}{cc}\cos 2\theta &\sin 2\theta \\\sin 2\theta &-\cos 2\theta \end{array}}\right)

. -- Meni Rosenfeld (talk) 20:08, 16 April 2008 (UTC)[reply]

Quotient map which is neither open nor closed

This is from Munkres's Topology, section 22, problem 3.

Let $\pi _{1}:\mathbb {R} \times \mathbb {R} \to \mathbb {R}$ be the projection on the first coordinate. Let A be the subspace of $\mathbb {R}$ consisting of all points $x\times y$ for wich either $x\geq 0$ or $y=0$ (or both). Let $q:A\to \mathbb {R}$ be formed by restricting $\pi _{1}$ . Show that q is a quotient map which is neither open nor closed.

I think that I can show that q is not open by considering $q([0,1)\times (1,2)\mapsto [0,1)$ . Am I correct? I'm stumped on showing that it's not closed, however. Donald Hosek (talk) 23:30, 16 April 2008 (UTC)[reply]

Hint: the fancy subspace isn't needed for the closed part, just the projection from R² to R. Try hitting {1/n|n in N} with a closed set. Algebraist 00:17, 17 April 2008 (UTC)[reply]

So the line on y=0 shooting into the left half-plane is a mathematical McGuffin then? Donald Hosek (talk) 00:32, 17 April 2008 (UTC)[reply]

Without it, the map would not be a quotient map (it wouldn't even be surjective). Algebraist 10:08, 17 April 2008 (UTC)[reply]

April 17

Calculus

What is the integral of a partial derivative? (The integral of the full derivative of $f(x)$ is $f(x)$ .) Also, how would this be written? $\textstyle {\int {\frac {\partial }{\partial {y}}}f(x)\ \mathrm {d} x}$ ? Also, can a partial integral be taken by replacing $\textstyle \operatorname {d} x\,\!$ with $\textstyle \partial {x}$ ? Thanks, Zrs 12 (talk) 02:15, 17 April 2008 (UTC)[reply]

In a general sense, all integrals are "partial integrals", in the sense that you hold one variable constant when integrating with respect to another. So

\int {f_{x}(x,y)\ \mathrm {d} x}=f(x,y)+C

. As far as I know there is no notation that replaces the total dx with a partial symbol in an integral. Confusing Manifestation(Say hi!) 06:56, 17 April 2008 (UTC)[reply]

The antiderivative on the other hand would be the integral of

f_{x}(x,y)

plus some function purely in terms of y. -mattbuck (Talk) 11:21, 17 April 2008 (UTC)[reply]

Ah, of course. Very important when solving differential equations, particularly separable ones. Confusing Manifestation(Say hi!) 05:17, 18 April 2008 (UTC)[reply]

Applications of Cauchy's Integral Theorem and Formula

HOW TO SOLVE THIS PROBLEM?

INTEGRATE (Z²-Z+1)/Z-1 WITH RESPECT TO Z OVER THE UNIT CIRCLE —Preceding unsigned comment added by 59.93.24.178 (talk) 10:38, 17 April 2008 (UTC)[reply]

Hello, this question looks like homework, and as a rule the reference desk will not answer homework questions unless you have actually tried to answer them yourself. I suggest looking at the page Cauchy's integral formula. Also, please turn off your Caps lock button, as typing in capitals is generally construed as shouting. -mattbuck (Talk) 11:30, 17 April 2008 (UTC)[reply]

Assuming that your function is actually (z²-z+1)/(z-1), then you have a bit of a conundrum, as the pole at z=1 is actually on the unit circle. Now you might "tweak" the path to "wiggle" around the pole. But then you will get a different answer depending on whether you wiggle to one side of the pole or the other - no matter how small your make your wiggle - can you see why ? Gandalf61 (talk) 11:53, 17 April 2008 (UTC)[reply]

Understanding bases, and expanding them

Hello, all. I've read up on 1089 (number), which filled me with joy. I've been attempting to find the pattern between +1 digit answers, and it so far evades me. I'd be happy if anyone could lend a hand with that, perhaps some sumsign expression for all digit values 2 (99) to 10 (1,099,989,000)?

Anyway, this had me thinking about something else. Without thinking (contrary to my initial statement, I know, but the thinking follows from what ensued), I put the digit listings up for order. 1, 2, 3-... 10, 11, 12. Wait, this will not work beyond the 10th digit, because one is not allowed to repeat any digits! That made me think I might better be able to plot a pattern if I had more numbers. Hexadecimal would allow me this, but I'm not familiar with that, and it'd take me a long time. But wait! My computer should do these calculations in any photo editor, should it not? What's the base that computer graphics use? 81.93.102.185 (talk) 15:09, 17 April 2008 (UTC)[reply]

Computers use binary, or base 2, so I expect graphics processing is done in binary too. When you see computer stuff done in hex, that's just to make it easier for humans - since 16 is a power of 2, you can covert from binary to hex (and back) very easily. Each block of four bits (binary digits) corresponds to precisely one hexadecimal digit (so you can convert 4 bits at a time and don't need to look at the whole number at once as you would when converting to decimal). The computer uses binary, but converts to hex when displaying it for humans, since the numbers are shorter than way. --Tango (talk) 16:07, 17 April 2008 (UTC)[reply]

Also, stuff tends to be converted to hex (rather than base 32 or something) because 4 divides 8, and most things are organized into groups of 8 bits (=1 byte). This makes it a clear conversion of two hex digits into 8 binary digits. You can also take a look at wikipedia's article on bases. Someletters<Talk> 00:38, 18 April 2008 (UTC)[reply]

I would say the main reason for not using base 32 is because that's too many digits to remember and would confuse people, but the fact that it wouldn't work well to express a byte is also significant (but we could just use 10-bit bytes - I'm not sure having a power of two bits is essential) 16 is quite close to 10, so you only have a handful of extra digits. It would be great if you could express a byte as a single digit, but it would be more confusing that expressing it as two. You can't use octal because there wouldn't be an integer number of digits per byte, and using base 4 would be pretty pointless, it's not much better than binary. Base 16 is the smallest useful base. --Tango (talk) 15:50, 18 April 2008 (UTC)[reply]

On a related note, you mention "this will not work beyond the 10th digit, because one is not allowed to repeat any digits!" What's important to keep in mind here is what, precisely, is meant by "digit" in this context. It's simply a unique symbol agreed to be representative of something. For hexadecimal, the general agreement is 0123456789abcdef, with the understanding that "a" means base-10 "10". However, you could define a-f to instead be α-ζ if you preferred the Greek alphabet, and so long as the mapping is understood, those are perfectly valid digits. On the other hand, when you discuss binary, the symbol "2" is no longer a digit -- it's just a squiggle that can't be interpreted by the rules in question. So it's not a case of running out of digits but just understanding what the symbols represent. — Lomn 15:40, 18 April 2008 (UTC)[reply]

Calculus

Given the maclaurin series of f(x)

1/2! - x²/4! + x^4/6! - x^6/8! + ... + [(-1)^n x^2n]/(2n+2)!

How do I write this in terms of a familiar function, without series?

Thanks Thermophylae (talk) 17:37, 17 April 2008 (UTC)Thermopylae[reply]

Well, the standard series you are likely to know are for the exponential function, sine and cosine. Have a look at some of their series, and see which seems most like yours, and what the relation is. -mattbuck (Talk) 18:28, 17 April 2008 (UTC)[reply]

1+x²·f(x)=?. Bo Jacoby (talk) 18:05, 18 April 2008 (UTC).[reply]

Looks like the integral of some function... --wj32 ^t/c 03:55, 19 April 2008 (UTC)[reply]

OK, it isn't. But heres a clue: it's (1 - something) / (something else). I used Mathematica. --wj32 ^t/c 03:59, 19 April 2008 (UTC)[reply]

April 18

Really confusing probability question

I read this question in some blog that talked about the Monty Hall 3-door problem: "1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?" Apparently the answer is 1/3, because "Three equally likely combinations are BG, GB and BB, and only one of them has the (rather ill-defined) “other” child being a boy." But it makes no sense to me. Consider that the second child is as of yet not conceived when the question is asked. Its chance of coming into the world as a boy should be entirely independent of its sibling's sex. And yet this question suggests that somehow one is dependent on the other. Is this really right? Can somebody explain this to me in a convincing way that does not use those permutation possibilities (cause I just can't buy them.)? Thanks, 140.247.237.63 (talk) 04:47, 18 April 2008 (UTC)[reply]

We have an article on this actually, Boy_or_Girl. What you've pointed out is in fact a subtle distinction which actually affects the problem. Counter-intuitiveness aside, the explanation on that page is quite solid. Someletters<Talk> 05:01, 18 April 2008 (UTC)[reply]

The issue is the conditional probability involved. In particular, it's about how much you know about the situation. If you only know that at least one child is a boy, but don't know whether it's the eldest or youngest, then that's when you get the 1/3 case. If you know that the eldest is a boy, then of course the sex of the youngest is independent of that. Imagine 100 random families, each with two children. Now of those 100, roughly 25 will have two girls, 25 will have two boys, and 50 will have one of each. Hence, 75 families have at least one boy, and of those, 25 have two, giving you 25/75 = 1/3 probability that a family has two boys, given that it is known it has at least one. Confusing Manifestation(Say hi!) 05:08, 18 April 2008 (UTC)[reply]

Thanks! 140.247.237.63 (talk) 06:52, 18 April 2008 (UTC)[reply]

Reversing Normal Distribution Function

I have the normal distribution function as follows:

$\phi \left(Z\right)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{Z}{e^{-{\frac {1}{2}}t^{2}}dt}$

I'm trying to rearrange it so that Z is the subject. However, I cannot work out how to integrate the above (which I assume I need to) in order to reverse the equation. The nearest I've seen is the integral of $e^{x^{2}}$ , which has a long result, though it's still not the one I want. So I was wondering if anyone knew how to integrate the above? Robert (talk) 13:07, 18 April 2008 (UTC)[reply]

If by "integrate" you mean "find an antiderivative for", you're out of luck.

e^{-{\frac {1}{2}}t^{2}}

has no elementary antiderivative. –King Bee ^{(τ • γ)} 13:09, 18 April 2008 (UTC)[reply]

Well, you can always write it in terms of

{\mbox{erf}}(x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}dt

(though that might not be of great help, because it's not an elementary function as said King Bee)

Letting u² = ½ x²,

\int e^{-{\frac {1}{2}}x^{2}}dx=\int e^{-u^{2}}\cdot {\sqrt {2}}du={\frac {\sqrt {\pi }}{2}}\cdot {\sqrt {2}}\cdot {\mbox{erf}}(u)={\frac {\sqrt {\pi }}{2}}\cdot {\sqrt {2}}\cdot {\mbox{erf}}\left({\frac {{\sqrt {2}}x}{2}}\right)

-- Xedi (talk) 14:30, 18 April 2008 (UTC)[reply]

The rephrase what the others have said - you can't integrate it algebraically, you have to do it numerically. That's why you have look up tables for the normal distribution. --Tango (talk) 15:45, 18 April 2008 (UTC)[reply]

In algebra real numbers the only completely ordered field?

The article Number says:

In abstract algebra, the real numbers are uniquely characterized by being the only completely ordered field. They are not, however, an algebraically closed field.

The article Total order says:

if X is totally ordered under ≤, then the following statements hold for all a, b and c in X:

If a ≤ b and b ≤ a then a = b (antisymmetry);

If a ≤ b and b ≤ c then a ≤ c (transitivity);

a ≤ b or b ≤ a (totality or completeness).

Would this not also apply to Integers and rational numbers? -- Q Chris (talk) 14:49, 18 April 2008 (UTC)[reply]

Well, the integers aren't a field (there are no multiplicative inverses except for 1 and -1). I'm not sure why the rational numbers don't count - by my understanding, they are a totally ordered field. They're not complete, in the analytical sense, but I don't think that's what it's talking about. --Tango (talk) 15:41, 18 April 2008 (UTC)[reply]

There is indeed an error in the article Number. The Reals are the only (up to isomorphy) complete ordered field (not the missing -ly). This changes the meaning of this phrase. "Ordered field" means "totally ordered field", i.e. a field being totally ordered as explained at Total order. The "complete" does note modify "ordered" but adds the property of Completeness (order theory) to the phrase. This is where the rationals fail. The article Real numbers has it correct. —Preceding unsigned comment added by 80.130.141.245 (talk) 15:57, 18 April 2008 (UTC)[reply]

An ordered field is a more narrow concept than that of a field with a total order on it. The order has to respect the field operations. The complex numbers with the lexicographical ordering is Cauchy complete and has a total order, but it is not an ordered field. 134.173.93.127 (talk) 05:33, 19 April 2008 (UTC)[reply]

Now fixed in Number 80.130.141.245 (talk) 15:59, 18 April 2008 (UTC)[reply]

Actually, the reals aren't complete in the sense of order theory. They're what our article calls bounded complete, and is sometimes called Dedekind complete. This is distinct from the property of being Cauchy complete (which the reals also have); there are lots of nonisomorphic Cauchy complete ordered fields. Algebraist 00:13, 19 April 2008 (UTC)[reply]

number line

Which of the following statements about the number line is true?

A. Number lines end with the number 10.
B. Numbers get smaller as you move to the left on the number line.
C. Number lines can't help you compare numerals.
D. Numbers get larger as you move to the left on the number line
— Preceding unsigned comment added by 67.55.21.71 (talk • contribs) 18:51, 18 April 2008 (UTC)[reply]

E. People who post homework questions should be ignored. --LarryMac | Talk 18:57, 18 April 2008 (UTC)[reply]

We are glad to help with homework, but only if you do your part. Tell us what answer you think is correct, or at least which answers you can eliminate, and we will tell you if we agree. One hint is that B and D appear to say the opposite of each other, so, if one is true then the other is false. StuRat (talk)

Please see our Number line article. --hydnjo talk 22:50, 18 April 2008 (UTC)[reply]

April 19

Chain rule in multi-variable calculus

From Wiki, I found the chain rule of composite functions of several variables with notations in partitions and blocks from The electronic journal of combinatorics 13 (2006), # R1. Since this formula has been used extensively in generalized functions, does Wiki provide any information of this rule in terms of multi-indices? More precisely, let y=f(x) where x,y are vector variables, z=g(y) is a real variable and m=(m_1, ... , m_k) is a multi-index where m_j are integers >=0. What is an explicit formula to express \partial^{|m|} z/\partial x^m in terms of partial derivatives of f,g? Thank you for your help in advance. twma 07:11, 19 April 2008 (UTC)[reply]

female mathematicians

in the world of mathematics,is there any famous female mathematicians?i mean aname related to afamous theory?Husseinshimaljasimdini (talk) 08:26, 19 April 2008 (UTC)[reply]