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[edit] June 28

[edit] Divisible by 126

What numbers are divisable by 126? Thanks --WontYouPoloMarco (talk) 17:59, 28 June 2009 (UTC)

There's not really any answer much better than 'those which are divisible by 126'. You can rephrase that as 'those n such that there exists m such that n=126m' but that's hardly an improvement. What sort of answer did you have in mind? Algebraist 18:06, 28 June 2009 (UTC)

0 126 252 378 504 630 756 882 1008 1134 1260 1386 1512 1638 1764 1890 2016 2142 2268 2394 and so on. Also their negatives. Bo Jacoby (talk) 18:48, 28 June 2009 (UTC).

Multiples of 126? --84.221.69.148 (talk) 19:19, 28 June 2009 (UTC)

The OP probably has Divisibility rules in mind. Since $126=2\cdot3^2\cdot7$ , you need to verify that the number is divisible by 2, 9 and 7. For that you can use the rules in the linked article. -- Meni Rosenfeld (talk) 20:02, 28 June 2009 (UTC)

Meni Rosenfeld's answer is too complicated. One need not go into prime factorizations. Here's the answer:

0 × 126,

±1 × 126,

±2 × 126,

±3 × 126,

±4 × 126,

±5 × 126,

±6 × 126,

.....

Michael Hardy (talk) 22:55, 28 June 2009 (UTC)

Meni's answer provides an efficient procedure for deciding if any arbitrary number, given in base 10, is divisible by 126. Sure, it's 'complicated', but that could be exactly what the OP needs (assuming the OP wasn't just doodling on the internet). --COVIZAPIBETEFOKY (talk) 00:38, 29 June 2009 (UTC)

I have just enough faith in humanity to assume that the OP was not looking for a list of numbers divisible by 126. That means he was looking for something else. Both divisibility rules and factors are good guesses, and being "complicated" is meaningless if that is the simplest way to answer what the OP meant to ask. -- Meni Rosenfeld (talk) 11:48, 29 June 2009 (UTC)

The OP didn't ask for an algorithm for ascertaining divisibility by 126; but rather: which numbers are divisible by 126? It is altogether possible that that was the intended meaning of the question. Michael Hardy (talk) 01:08, 30 June 2009 (UTC)

Just in case the question was supposed to be "What numbers is 126 divisible by?" and it got mangled, the answer would be:

1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126.

Michael Hardy (talk) 23:00, 28 June 2009 (UTC)

[edit] calculate pi with hot dogs

How does this work? Mac Davis (talk) 19:49, 28 June 2009 (UTC)

See Buffon's needle. -- Meni Rosenfeld (talk) 19:56, 28 June 2009 (UTC)

And Buffon's noodle, for another gastronomic version. --84.221.69.148 (talk) 05:24, 29 June 2009 (UTC)

[edit] June 29

[edit] dependency with a fractional exponent

If y is dependent on x through a proportional constant, we say that's a linear dependency. If y is proportional to x², it's quadratic; for xⁿ we'd say polynomial. What about something like x^1.57238? "Polynomial" might still be accurate in the computation complexity-theoretic sense, but in other contexts (this is about statistics, if it matters) it seems to me that it connotes an integer exponent. Is there some other word for this sort of relationship? Thanks. 208.70.31.206 (talk) 01:49, 29 June 2009 (UTC)

If y=x^α, we may simply say that the logarithms of x and y are proportional, or that x and y are proportional in logarithmic scale. The natural expression would be that they are "logarithmically proportional", but I see this is already used, with the meaning "y is proportional to log(x)". (Not very satisfying use, for in this acceptance, "logarithmic proportionality" is not an equivalence). --84.221.69.148 (talk) 06:24, 29 June 2009 (UTC)

A relationship of the form y = ax^k where k is a constant but not necessarily an integer is generally called a power law. Gandalf61 (talk) 08:18, 29 June 2009 (UTC)

[edit] June 30

[edit] Linear functionals

Another qual problem:

Let $1 < p < \infty$ , and suppose $φ$ is a continuous linear functional on $L^p(\mathbb{R})$ such that for every $f \in L^p(\mathbb{R})$ , $φ(f (x)) = φ(f (x - 1))$ . Show that $φ$ is the zero functional. Show that this is false if $p = 1$ .

I'm not sure how to do this, obviously, or I wouldn't ask. But I have gotten somewhere, whether it is helpful or not, I do not know. First, by the Riesz Representation Theorem, there exists $g \in L^q(\mathbb{R})$ such that

$\phi(f) = \int_\mathbb{R} fg$

for all $f \in L^p(\mathbb{R}$ . In particular, let $f = χ [a, a + 1]$ for any real a. Then, using the condition on $φ$ and using a change of variables gives

$\int_{a}^{a+1} g = \int_{a+1}^{a+2} g$

for any real a. Here I am stuck. My thought was perhaps I could show this implies g is periodic... I could use $χ a, a + ε$ instead maybe. I don't know. Then, if I get that g is periodic, perhaps I could pick another choice of f that gives a contradiction. Any suggestions? Thanks. StatisticsMan (talk) 02:20, 30 June 2009 (UTC)

This is basically fine. L^p(R) for finite p does not contain any nonzero periodic functions. For every e>0, there is some interval outside of which |g| is less than epsilon. On the other hand, its integral is constant on same-size intervals, a big no-no for functions with a finite integral. This breaks down when p is infinity, and you can have periodic g. In terms of f, this is when p=1. JackSchmidt (talk) 04:21, 30 June 2009 (UTC)

That's substantially correct, but notice that the integral of g a priori could be constant and zero on same-size intervals; and also that a g in L^q(R) for 1<q needs not be integrable (so |g| may have infinite integral on R). So it's safer and more direct to use |g(x)|^q in your argument, when showing that there are no periodic nonzero functions in L^q(R) . --pma (talk) 06:05, 30 June 2009 (UTC)

Notice that g 1-periodic also follows from $\scriptstyle0=\int_\R g(x)\{f(x)-f(x-1)\}dx=\int_\R\{g(x)-g(x+1)\}f(x)dx$ for all f, whence $\scriptstyle g(x)-g(x+1)=0$ a.e. --pma (talk) 06:16, 30 June 2009 (UTC)

By the way, notice that the fact mentioned by JackSchmidt (there is no nonzero periodic function in L^p for finite p) also answers negatively to the question: can a nonzero function in L^p for a finite p have zero integral over all unit intervals? extending the analog result in L¹ mentioned around your preceding post. The reason is that the integral function of such a function g (i.e. $\scriptstyle f(x);=\int_0^x g(t)dt$ ) would be 1-periodic, so g(x) itself would be 1-periodic, being the derivative a.e. of f(x), therefore necessarily g=0 as we know. pma. --131.114.72.186 (talk) 11:24, 30 June 2009 (UTC)

Thanks a lot. I read through what you all said and it made sense to me after some thought and I was able to complete the proof. I have not come up with an example to show it is not true for p=1, but I understand what you are saying and I will think about that some. StatisticsMan (talk) 19:35, 30 June 2009 (UTC)

Hint: the p=1 case is really easy. Algebraist 19:42, 30 June 2009 (UTC)

So easy, the answer has already been given. Let g(x) = sin(2pi x). Then g(x) is in L^\infty(R). By a Proposition in Royden,

$\phi(f) = \int_{\mathbb{R}} f(x) \sin(2\pi x) \,dx$

defines a bounded linear functional on L^1(R). And, then I can use

f (x) = χ [0,1 / 2]

which is in L^1 and the integral is not 0 so phi is not the 0 functional. Thanks! StatisticsMan (talk) 21:04, 30 June 2009 (UTC)

I was thinking just set g(x)=1. Algebraist 23:49, 30 June 2009 (UTC)

Good point, thanks! StatisticsMan (talk) 13:42, 1 July 2009 (UTC)

[edit] Unemployment vs. application rate

What is the theoretical relationship between the unemployment rate and the number of applicants per job opening? Neon Merlin 03:17, 30 June 2009 (UTC)

In basic theory, unemployment is positively correlated with the number of applicants per job opening. If you look specifically at a single sector such as accounting jobs, it's easier to think of the basic principles. If there are 900 accounting jobs for 1000 accountants, then I would imagine if I were an unemployed accountant, I would broaden my job search since I have more hungry accountants pursuing limited jobs. The more unemployed accountants relative to me, the more broadly I would apply to accounting jobs. I think that there are two positive correlations. In addition to the part mentioned about every unemployed accountant expanding his job search and applying more aggressively and exercising less discretion, you also have more unemployed accountants who all think the same way. I think it is squared relationship (quadratic relationship?). When the unemployment rate was 5%, you never really panicked like facing 10%. Also, humans are a species driven by exuberance, fear, and emotional reasoning. Lots of theoretical relationships only apply to rational decision makers acting in their best interest. I think a theoretical relationship exists, but it's probably not an economics question--perhaps a consumer psychology model would best explain it by using "unemployed people" as the "consumers" of scarce new jobs. See also Bigger fool theory which shows that a stable equilibrium may not exist for building a reasonable model for your question. If you would like more help, try the article on Financial modeling and Econometrics 74.5.237.2 (talk) 09:06, 30 June 2009 (UTC)

[edit] "Scoring" a product, correcting for incomplete scores.

Please could you check that my thinking on this problem is mathematically sound. The company that I work for tenders for various products and services. As part of the evaluation process we come up with a "scoring" matrix, which scores each company against a number of categories. The points in each category are set to give a weighting. A simplified example might be:

	COMPANY 1	COMPANY 2	COMPANY 3
PRICE (0-10)
FUNCTIONALITY (0-20)
SUPPORT (0-10)
COMPANY STABILITY (0-20)

In practice there would be many functional areas and criteria, i.e. many rows. Each scorer would potentially put a score in the range given against each company. If everyone filled in the table entirely then scoring would be easy - just totalling the scores for each company. In practice two things happen:

Some people only score certain areas This is expected, technical people might only be able to answer questions about functionality and not (for example) company stability. The thing is that if we just add up the scores some areas would not get the weighting they need; for example only three people might score company stability but ten functionality. I figure that the way to cope with this is to give the people who have not scored in an area a score equal to the average of the scores that have been given.

Some people do not score all companies Ideally this would not happen, but due to ongoing work, unexpected calls, sickness, etc. some people may miss some of the presentations. Obviously it would be wrong to mark a company down because fewer people attended their presentation, so I figure that a solution to this is to give companies they missed an average of the scores they gave to ones they attended. I thought that this is better than giving an average of the scores given by other people because some people seem to mark high and some low.

Is this mathematically sound? I can see no reason why I should apply the correction for missing rows before I apply the correction for missing columns, but it feels right. In practice we have used these corrections and they do tend to come out with results that match the "general feelings" that people had. Thanks in advance . -- Q Chris (talk) 09:36, 30 June 2009 (UTC)

It's a complicated topic. We have an article, imputation (statistics), but I think it really just scratches the surface. --Trovatore (talk) 09:45, 30 June 2009 (UTC)

I think about it this way. The strength of a category of a company should be a number between 0 and 1, like a probability, P, of success. This number P is unknown, but some knowledge of P is gained by scoring. The number of times it is scored is n, and the number of successes in scoring is i. So the score i is an integer between 0 and n, while the strength P is a real between 0 and 1. The likelihood function of P is the beta distribution having mean value m = (i+1)/(n+2) and variance s² = m(1-m)/(n+3). So if a category is not scored, simply put n = i = 0 in the above formula and get m = 1/2 and s² = 1/12. Bo Jacoby (talk) 06:27, 1 July 2009 (UTC).

[edit] Expected value of the reciprocal of the gcd

Our greatest common divisor article says that the expected value E(k) of the gcd of k integers is E(k) = ζ(k-1)/ζ(k) for k > 2. Does anyone have a reference for this? (While you're at it, if you're knowledgeable, you can try to fix up the sentences following this statement in the article which currently don't make any sense to me.) My real question: Is a similar formula known for the expected value of the reciprocal of the gcd? Staecker (talk) 14:46, 30 June 2009 (UTC)

Assuming the stuff in the article is correct, then the same argument shows that the expectation of the reciprocal of the GCD is ζ(k+1)/ζ(k). Algebraist 14:51, 30 June 2009 (UTC)

OK- I agree. I should've read the derivation more closely. Thanks- Staecker (talk) 23:19, 30 June 2009 (UTC)

[edit] Calculating residues

Hi. I just made an edit to the section of Residue (complex analysis) on calculating residues. I wonder if someone could have a look at it to double-check that I didn't say anything false, or otherwise break the article. Thanks in advance. -GTBacchus^(talk) 15:27, 30 June 2009 (UTC)

I think it's a good idea to have the special case of a simple pole mentioned explicitly. This should probably be discussed at Wikipedia_talk:WikiProject Mathematics, though. --Tango (talk) 18:19, 30 June 2009 (UTC)

Ah yes, that would be a better forum. I'll head there now. -GTBacchus^(talk) 18:22, 30 June 2009 (UTC)

[edit] Double integral = 0 over all rectangles

Suppose f(x, y) is a bounded measurable function on $[0, 1] \times [0, 1]$ . Show that if for every a < b and c < d

$\int_{[a, b]} \int_{[c, d]} f(x, y) \,dxdy = 0$

then f = 0 a.e.

In our study group, we looked at a similar problem we have yet to figure out, which is that the integral over each open disk in R^2 is 0 and we are supposed show f = 0 a.e. [as far as the Lebesgues measure on R^2].

Can any one help us figure these out? Thanks! StatisticsMan (talk) 20:47, 30 June 2009 (UTC)

A naive attempt but did you try proof by contradiction? Assume that the function f is nonzero on a set of measure nonzero. Then you show that f is strictly positive (or negative) in a open set so its integral over that open set will be also nonzero. Notice that being nonzero is not enough. Because nonzero functions on a set of measure nonzero CAN have zero integral (for example, when volume above the plane is the same as volume below the plane, they cancel). So you need a set where the function is always positive or always negative. You fill in the details using all the assumptions you assume. I hope this will be enough.-Looking for Wisdom and Insight! (talk) 21:13, 30 June 2009 (UTC)

Alternatively, just check the answers to the last question of this type you asked. Some of them are straight-up applicable, others require only minimal amounts of adaptation. Ray^Talk 21:15, 30 June 2009 (UTC)

Well, the two hints I suggested for the 1-dimensional version to your question here easily generalize to any dimension. Alternatively, you can reduce the problem to 1 variable using Fubini's theorem. For fixed a and b consider the function $\scriptstyle f_{a,b}(y):=\int_{[a, b]} f(x, y) \,dx$ . It has vanishing integral over all intervals [b,c]. Therefore it is identically zero, according to the 1 dimensional case. This means that for a.e. y, the function $\scriptstyle x\mapsto f(x,y)$ has vanishing integral over all intervals [a,b], and you conclude. PS: can you see how to prove in an elementary way that if a function f in L¹(R²) has zero integral over all unit squares, then it is identically zero? Note that this also works for the analog with unit cubes in dimension three, &c. However, if you replace unit squares with unit disks, the thing is still true but (as far as I know) no elementary proof is available (you can do it via Fourier transform as I mentioned) --pma (talk) 22:08, 30 June 2009 (UTC)

I haven't thought very long about it, but can't you just use Lebesgue's density theorem, which works either for squares or balls? If a function f is positive on a set of positive measure then there is a set P of positive measure on which f is uniformly bounded above 0. Also f is bounded overall. So if you find a ball or square in which P has sufficiently high density, then integral of f over that ball or square will be positive. — Carl (CBM · talk) 22:37, 2 July 2009 (UTC)

well, if you are talking of f(x,y) having vanishing integral on disks, or squares, of all size, then yes; but then you can do in even more elementary ways. I was talking of unit disks and unit squares.--pma (talk) 06:42, 3 July 2009 (UTC)

[edit] July 1

[edit] Triangle with all angles less than 180º

I know that, in elliptic geometry, it's possible to have a triangle with three right angles — thus adding up to 270º rather than the Euclidian 180º. Is there any system in which it's possible to have a triangle with less than 180º between its angles? I've looked through Hyperbolic geometry, but it's been a few years since I last took any maths, so I can't understand whether this is such a system. Nyttend (talk) 04:00, 1 July 2009 (UTC)

Yes, triangles having angles totalling less than 180 degrees characterises hyperbolic geometry. I'm surprised our article doesn't make that clear - I'll take a look at it. --Tango (talk) 04:07, 1 July 2009 (UTC)

I've added a bit about hyperbolic triangles to the lede of hyperbolic geometry. Somebody may want to review it. It would also be good if someone found a reference, I haven't had a chance to look for one but any geometry textbook that touches on non-Euclidean geometry will do. --Tango (talk) 04:15, 1 July 2009 (UTC)

[ec] Actually, it does, in the Escher section; I missed it. However, your assurance made me look more carefully and confidently, so thanks :-) Nyttend (talk) 04:16, 1 July 2009 (UTC)

I missed it too! I think it is good to mention it early on, it's a pretty key characteristic. --Tango (talk) 04:20, 1 July 2009 (UTC)

[edit] MathPsy and QuantPsy: Difference

What is the difference between Mathematical Psychology and Quantitative Psychology, if any?```` —Preceding unsigned comment added by Ultraluna (talk • contribs) 18:01, 1 July 2009 (UTC)

I can't speak to Psychology, but the distinction also exists in Economics. In Mathematical Economics, one builds mathematical models to describe economic phenomena. What results from the models are (generally) qualitative solutions (e.g., an increase in X accompanies a decrease in Y). The focus is on mathematically representing the dynamics of a process so as to better understand the process rather than predicting what outcome the process will yield. In Quantitative Economics (we call it Econometrics), one fits data to mathematical models in an attempt to obtain quantitative solutions (e.g., an increase in X of one unit accompanies a decrease in Y of 10 units). The focus here is more on measuring the outcome of a process. Wikiant (talk) 18:28, 1 July 2009 (UTC)

Thanks for your response. That helps, albeit, somewhat. Without demeaning Econometricians in any way, from what I understand then is that Mathematical Economics is more important than Econometrics. The former builds the model on which the latter works on: like an Architecht and a Mason+Civil Engineer.```` —Preceding unsigned comment added by Ultraluna (talk • contribs) 04:51, 2 July 2009 (UTC)

[edit] Solve for z

Hi, I'd like guidance on solving the following equation for z,

$xy+yz+xz-z-z^2 = 0 \,$

The presence of the $z 2$ term makes this trickier; normally you might be able to extra the z term through factorisation and seclude it that way. I know I probably need to factorise the left hand expression in some way, but I don't really know any factoring techniques that apply here. Thanks. 94.171.225.236 (talk) 20:51, 1 July 2009 (UTC)

It's just a quadratic equation

$az^2 + bz + c = 0 \,$

where

$\begin{align} a & = -1 \\ b & = y + x - 1 \\ c & = xy. \end{align}$

Michael Hardy (talk) 21:10, 1 July 2009 (UTC)

Understood, thankyou! Can I just ask why c isn't equal to xy? Wouldn't all terms lacking z imply that they constitute the constant term? 94.171.225.236 (talk) 21:21, 1 July 2009 (UTC)

Because Michael Hardy made a typo. Algebraist 21:25, 1 July 2009 (UTC)

Sorry. Typo. xy is correct. Michael Hardy (talk) 01:41, 2 July 2009 (UTC)

Is there a better way than using the formula

a (x - r 1)(x - r 2)

(where

r 1, r 2

are roots of the polynomial) in order to achieve a factorisation? The roots in quadratic formula form are, if my calculations are correct, something like this:

$r_1, r_2 = \frac{(y+x-1) \pm \sqrt{y^2+6xy-2y+x^2+2x+1}}{2xy}$

Which seems unecessarily complicated, considering this is an exercise. I haven't tested to see wether the whole expression for

a (x - r 1)(x - r 2)

for the above polynomial expands to its original form, but I'm fairly certain there's a more elegant approach to this. 94.171.225.236 (talk) 08:07, 2 July 2009 (UTC)

The equation

$z^2-(y+x-1)z-xy = 0 \,$

has the solutions

$z_1, z_2 = \frac{y+x-1 \pm \sqrt{y^2+6xy-2y+x^2+2x+1}}{2}$

Your calculation was almost correct. Bo Jacoby (talk) 09:15, 2 July 2009 (UTC).

[edit] July 2

[edit] The empty function

Suppose f:A->0, where 0 is the empty set and A is non-empty. Using vacuous truth, we can prove f is a bijection.

Surjectivity: "For all y that belongs to 0, there exist a x in A such that f(x)=y"

Since there aren't any y in 0, this proposition is automatically satisfied by vacuous truth

Injectivity: "For all x1 and x1 in A that aren't equal, f(x1) isn't equal to f(x2)"

Since for all x in A, f(x) isn't even defined, again this proposition is satisfied.

I'm not an expert in anything, but this "thing" shows every set has "0 elements". Is there something wrong with my reasoning? Or is it just the strangeness of vacuous truth? —Preceding unsigned comment added by Standard Oil (talk • contribs) 07:47, 2 July 2009 (UTC)

When we say f:X->Y we require that f have a value for every element of X. That means if f:A->emptyset, then A must also be empty. Because if A has an element x, that would give f(x) is an element of the empty set. So there are no functions like you describe (i.e. with A non-empty). 208.70.31.206 (talk) 08:39, 2 July 2009 (UTC)

You are misusing the phrase "vacuous" which if we have a predicate φ(x), where x ranges over X, then ∀x.φ(x) is true when X is empty, even if φ(x) is not true for any x. Instead, if one supposes f:A->0, where 0 is the empty set and A is non-empty, then one can derive 0=1 because the premises are contradictory. This does not establish that 0=1 by vacuous truth, although it does establish that 0=1 for every f that satisfies the premises. — Charles Stewart (talk) 09:13, 2 July 2009 (UTC)

[edit] Safety Stock Formula

Why does safety stock vary with SQRT(Lead Time) & not Lead Time? —Preceding unsigned comment added by Bharat Kantharia (talk • contribs) 09:01, 2 July 2009 (UTC)

Because the safety stock level is determined by the standard deviation of total demand during the lead time. If we model the demand per unit time period as a normally distributed random variable with volatility σ² (and volatility is independent of time) then the total demand over a lead time of n periods is a normally distributed random variable with volatility nσ² (see sum of normally distributed random variables). So the standard deviation of total demand over the lead time is sqrt(n)σ, which is proportional to sqrt(n). Gandalf61 (talk) 09:17, 2 July 2009 (UTC)

[edit] I don't understand why the sign changes when grouping polynomials for factoring.

Hi, I am doing some self study of precalculus.

When factoring polynomials by grouping, you are supposed to change the sign of each term in the if there is a minus sign before the group. This seems wierd to me because I don't see how the act of adding parenthises changes the expression.

To see what I am talking about look at Factorization#Factoring_by_grouping, and see how the signs change in the second group.

I can't find any resources that explain why the signs change, what am I missing here? Thanks for any insight or pointers to more info. -- 99.129.153.2 (talk) 14:10, 2 July 2009 (UTC)

Take a second term of the second group for example. Adding 234x² is the same operation as subtracting minus 234x², so if you include the term into a group which is itself subtracted (has a minus before its left bracket) you need to change that term's sign. This is because parentheses cause applying the preceding plus or minus to all terms included:

–2 – 7 = (–2) + (–7) = –(2 + 7)

so

30 – 2 – 7 = 30 – (2 + 7)

and similary

30x – 70y + 13z = 30x – 70y – (–13z) = 30x – (70y – 13z)

HTH --CiaPan (talk) 14:44, 2 July 2009 (UTC)

If you work backwards (expand out the brackets) you'll see that the minus sign gets distributed over the bracket and changes all the signs. When you are grouping the terms together you are doing the inverse of expanding brackets so you have to do the inverse of changing all the signs, which is just changing all the signs again. --Tango (talk) 15:15, 2 July 2009 (UTC)

10 - 3 + 2 = 9, but 10 - (3 + 2) = 5, so you get an idea how parentheses can change the meaning of an expression when there is subtraction involved. To group the terms in the first expression without affecting the value, we have to consider how the negative implied by subtraction will distribute over the whole term in parentheses, so 10 - 3 + 2 = 10 + (-3 + 2) = 10 - (3 - 2). Notice how the sign of every term inside the parentheses flips when we stick a (-) outside the parentheses. In other words adding -3 becomes subtracting +3 and adding +2 becomes subtracting -2. Rckrone (talk) 17:23, 2 July 2009 (UTC)

...and here's an example with a minus sign on the inside becoming a plus sign on the outside:

1000 − (200 − 10) = 1000 − 200 + 10.

Say you have $1000 in your checking acount. Something that normally costs $200 is on sale for $200 − $10. After you pay for it with your debit card, your checking account balance is $1000 minus the price, and the price is $200 − $10, so your account balance is now $10 MORE than it would be if you'd paid the regular price. Hence plus rather than minus. Michael Hardy (talk) 18:02, 2 July 2009 (UTC)

Thanks everyone I think I've got my head wrapped around this now. Great examples! -- 99.129.153.2 (talk) 19:08, 2 July 2009 (UTC)

[edit] July 3

[edit] Development of calculus

Why did calculus develop as late as it did? I would have thought that, given how easy it is to derive the derivatives of most functions and how elementary questions like "what's the rate of change of this quantity?" are, calculus should have been "discovered" thousands of years ago. I certainly pondered what were basically calculus problems long before I knew about calculus; I'm surprised that the ancient Greek mathematicians didn't do the same. --99.237.234.104 (talk) 07:58, 3 July 2009 (UTC)

As can be seen here, some of the basic concepts of differential and integral calculus floated around for millennia; it's just only in the past few hundred years that rigorous axioms and a solid branch of mathematics (calculus) have emerged. —Anonymous Dissident^Talk 10:49, 3 July 2009 (UTC)

(e/c) They did (Archimedes for instance). They did not develop the general theory, which is probably related to the fact that they did not think about functions in the algebraic way we are accustomed to. — Emil J. 10:55, 3 July 2009 (UTC)

I knew that certain aspects of calculus were around for a long time, but I'm surprised that they didn't develop further, considering the utility of calculus in solving mathematical problems. The complexity of what Archimedes and other mathematicians of the time managed to derive--like the surface area and volume of a sphere--is amazing; I'd never understand how they came upon such complicated proofs. That's why I'm surprised that nobody developed something as simple and intuitive as the basic concepts of calculus. --99.237.234.104 (talk) 11:19, 3 July 2009 (UTC)

A systematic development of calculus can't get very far without modern algebraic notation and the synthesis of geometry and algebra in analytic geometry. Diophantus started the development of algebraic notation, but even apparently simple advances like plus, minus and equals signs didn't emerge until the 16th century. Descartes didn't develop the key insight that an equation describes a curve and vice versa until the early 17th century. Major paradigm shifts like calculus affect our perspective so much that it can be difficult to appreciate how difficult they were to develop in the first instance. Gandalf61 (talk) 13:44, 3 July 2009 (UTC)

Discovering calculus is a harder problem than it appears to be because when you learn calculus, you can see only the easy part. The easy part is to answer all the questions after you know the questions. Michael Hardy (talk) 23:47, 3 July 2009 (UTC)

However, recall that the Hellenistic science has been abruptly stopped by the Roman expansion on the Mediterranean. For the story of the Hellenistic science I suggest Lucio Russo's wonderful book "The forgotten revolution".--pma (talk) 06:36, 4 July 2009 (UTC)

[edit] Method to solve 3-variable equation

x=(y/(z^y))

If I know x and z, what's the best way to obtain a value for y? I assume it can't be solved algebraically because I end up with y and log(y) in any rearrangement which can't be combined in a single term to make it y=f(x,z). If Newton's method is to be used then I'm not sure how to differentiate it or apply the method in this case. 86.163.186.102 (talk) 08:55, 3 July 2009 (UTC)

You can express the solution as $y=\frac{W(-x\,\log z)}{-\log z}$ , where W is the Lambert W function, if that's any help. — Emil J. 11:02, 3 July 2009 (UTC)

(ec) I don't have the time to explain this with much depth right now, but for any integer n, $y = - \frac{W_{n}(-x \log(z))}{\log(z)}$ for your case. This is assuming a non-zero log(z). W_n(x) is the Lambert W function. Hope this helps. I expect someone will be able to come along shortly and give something more elaborate and explained. —Anonymous Dissident^Talk 11:05, 3 July 2009 (UTC)

If you want to solve it numerically using Newton's method, you need to express the equation as

f (y) = 0

. In your case $f(y) = x - \frac{y}{z^y}$ is one way, but

f (y) = x z y - y

is simpler. To differentiate it, you need to treat x and z as constants and use the normal rules of differentiation. The result is

f'(y) = x z y log z - 1

, and the Newton iteration is $y_{n+1} = y_n - \frac{xz^{y_n}-y_n}{x z^{y_n}\log z -1}$ . -- Meni Rosenfeld (talk) 11:37, 3 July 2009 (UTC)

Thanks everyone. Since using W is iterative too but is more complicated I just used Newton's method with your f(y) and f'(y) and it's working great. 86.163.186.102 (talk) 11:56, 3 July 2009 (UTC)

[edit] Mininmum Sample size for correlation study

I know of a formular for determining the minimum sample size for a prevalence study, which most research studies use. Is there any formular to determine minimum sample size for a correlation study, like a correlation between birth weight and fetal cord leptin concentration-a research?Tunmisadej (talk) 19:43, 3 July 2009 (UTC)

If you're doing homework or trying to replicate established procedures, look at the testbook/published literature. Otherwise, I can't think of a formula off-hand, but in general the best way to approach these sorts of problems is to work backwards. Figure out what sort of confidence level you want to have (0.05, 0.01, 0.001, etc.) in your results, and then using worst-case estimates for the various other parameters apply the equation you're going to use to analyze the completed study. You can then predict how the confidence level varies for various sample sizes. The sample size you want is the one that will give you at least the desired confidence level. A little extra work will get you information on how sensitive your confidence interval is to fluctuations in the various parameters. -- 76.201.158.47 (talk) 21:00, 4 July 2009 (UTC)

[edit] Connected component

How does one know the existence of a connected component (i.e., maximal connected subset) in a topological space? Put in another way, does one need Zorn's lemma to show the existence? (Our connected space doesn't address this question, maybe it should unless this is a trivial matter.) Somehow related question: the article says: "The components in general need not be open"; true, but this pathology almost never happens in practice, I believe. What guarantees that connected components are open? -- Taku (talk) 23:38, 3 July 2009 (UTC)

The connected component of the point x is simply the union of all connected subsets containing x. It's not hard to show that this set is connected. Alternatively, the connected components are the equivalence classes of the relation x~y iff there is a connected set containing both x and y. Neither of these characterizations require heavy machinery, and they certainly don't require choice.

In the rationals, the connected components are not open. Surely Q doesn't count as a pathology? Anyway, the most obvious condition that forces components to be open is that there be only finitely many of them. Algebraist 00:14, 4 July 2009 (UTC)

Yeah, taking union. Why didn't I think of that. Anyway, thank you. You answered a lot more than I asked :) -- Taku (talk) 01:34, 4 July 2009 (UTC)

A kind of a followup-question. This is probably trivial too, but how do we know that there is a connected set containing given a point x at all? (Obvious, say, in a metric space, but in general) -- Taku (talk) 01:50, 4 July 2009 (UTC)

{x} Algebraist 02:02, 4 July 2009 (UTC)

Right :) -- Taku (talk) 02:25, 4 July 2009 (UTC)

Probably the most marvellously succinct answer I've ever seen given on a refdesk, Algebraist :D Maelin (Talk | Contribs) 03:22, 4 July 2009 (UTC)

Let me expand slightly on Algebraist's (sufficient) response. In a topological space, the components are always closed since the closure of a connected set is connected. Thus Algebraist's remark is justified. Secondly, the perhaps more general requirement for the components of a space to be open, is that the space be locally connected. Local connectedness is both a necessary and sufficient requirement for the components of any open subspace of the space in question to be open.

Similarly, there may be only one connected set containing a point. For instance, in the space of rationals (as mentioned above), there exists no connected subspace having more than one point. Thus each point is contained in precisely one connected subspace.

With regards to the necessity of the axiom of choice/Zorn's lemma to justify the existence of connected components, I think that Taku was comparing the situation with that in ring theory. More specifically, the truth of the idea that any proper ideal in a ring is contained in at least one (proper) maximal ideal requires the application of Zorn's lemma.

To conclude, I think that the word "pathology" has no specific meaning and depends on the context in which it is used. For instance, I do not consider any space to be pathological (because I am a point-set topologist :)), but on the other hand, people may consider the Bug-eyed line to be a pathological example of a manifold. Some people may even consider the Vitali set to be pathological. To take a completely different view point, pathological examples are often the most interesting. In particular, if a theorem does not hold unless one rules out specific pathologies, it is probably less natural than a theorem which holds for a class, including every pathology. --PS T 04:40, 4 July 2009 (UTC)

Ah, that's very nice (and complete solution): locally connected-ness is necessary and sufficient. Thanks a lot. (It's just so much quicker to simply ask than going through topology books myself. Not to mention that the library is closed for the July 4-th). As for Zorn's lemma, actually, I was thinking of cases of irreducible components, since I thought I read somewhere that one uses Zorn's lemma to construct a maximal irreducible subset (i.e., irreducible component]). As for "pathology", I used the term because to me connected components that are not open are counterintuitive. I guess one could say a topological space that is not locally connected is not intuitive: In fact, locally connected has this: "In fact the openness of components is so natural that one must be sure to keep in mind that it is not true in general". -- Taku (talk) 12:51, 4 July 2009 (UTC)

Careful: local connectedness is not necessary for openness of components. It's necessary for openness of components of every open set. Algebraist 15:03, 4 July 2009 (UTC)

Right :) so not exactly the complete solution, then. -- Taku (talk) 02:00, 5 July 2009 (UTC)

Actually, if you read my response carefully, I noted this - "Local connectedness is both a necessary and sufficient requirement for the components of any open subspace of the space in question to be open." Perhaps Taku did not notice the wording because it is easy to miss.

Yes, I know. Where do you think I cribbed it from? Algebraist 02:44, 5 July 2009 (UTC)

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