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October 30

Relationship between $IJ$ and $I\cap J$

Let $I$ and $J$ be ideals of a ring. Is there a relationship between $IJ$ and $I\cap J$? Other than the obvious one $IJ\subset I\cap J$?--AnalysisAlgebra (talk) 01:09, 30 October 2012 (UTC)

If the ring is commutative, they have the same radical: ${\sqrt {IJ}}={\sqrt {I\cap J}}$. This also means that if P is a prime ideal, then P contains IJ if and only if it contains I∩J. Rckrone (talk) 05:07, 30 October 2012 (UTC)

Canonical map

What is the canonical map from a ring $R$ to $R[x]/(ax-1)$ where $a\in R$?--AnalysisAlgebra (talk) 05:26, 30 October 2012 (UTC)

(p.s. and showing that its kernel is $\{b\in R|\exists n\in \mathbb {N} :a^{n}b=0\}$ would be nice too...)--AnalysisAlgebra (talk) 05:29, 30 October 2012 (UTC)

There is a natural inclusion of R in R[x]. R[x] is the ring of polynomials in x with coefficients in R. The elements of R can also be considered polynomials (with degree 0). Then there is a natural surjection from R[x] to R[x]/(ax-1) mapping a polynomial to its equivalence class mod (ax-1). The composition of these two is the canonical map from R to R[x]/(ax-1). Rckrone (talk) 05:54, 30 October 2012 (UTC)

Generalization of the Laplacian and partial derivative

Let $u(x,y)$ be the solution of the steady-state heat equation ${\frac {\partial ^{2}u}{\partial y^{2}}}+\sum _{j=1}^{d}{\frac {\partial ^{2}u}{\partial x_{j}^{2}}}=0$ with $u(x,0)=f(x)$
A generalization of repeatedly applying the Laplacian is $(-\Delta )^{a}f(x)=\int _{\mathbb {R} ^{d}}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$ and it is quite easy to show that this coincides with the regular Laplacian when a is a positive integer, if f is in the Schwartz space (${\widehat {f}}$ is the Fourier transform of $f$).
Show that $(-\Delta )^{k/2}f(x)=(-1)^{k}\lim _{y\rightarrow 0}{\frac {\partial ^{k}u}{\partial y^{k}}}(x,y)$ for positive integer $k$.Widener (talk) 11:07, 30 October 2012 (UTC)

Take the Fourier transform of u wrt to the x variables and solve the steady state heat equation: ${\hat {u}}=e^{-2\pi y|\xi |}{\hat {f}}$ (this has the correct initial conditions in y=0 and boundary conditions at infinity). Apply an inverse Fourier transform and again differentiating enough times under the integral gives $(-\Delta )^{k/2}f(x)=(-1)^{k}\lim _{y\rightarrow 0}{\frac {\partial ^{k}u}{\partial y^{k}}}(x,y)$. Sławomir Biały (talk) 00:36, 31 October 2012 (UTC)

Heisenberg Principle Implies Theorem

Show that the Heisenberg uncertainty principle implies $\int _{-\infty }^{\infty }(-{\frac {d^{2}f}{dx^{2}}}(x)+x^{2}f(x)){\overline {f(x)}}dx\geq \int _{-\infty }^{\infty }f(x){\overline {f(x)}}dx$ for every $f$ in the Schwartz space.
I have a version of the Heisenberg uncertainty principle which states that $\left(\int _{-\infty }^{\infty }x^{2}|f(x)|^{2}dx\right)\left(\int _{-\infty }^{\infty }\left|{\frac {df}{dx}}\right|^{2}dx\right)\geq 1/4$. I can get the integral on the left hand side of the inequality to $\int _{-\infty }^{\infty }{\frac {df}{dx}}(x){\overline {{\frac {df}{dx}}(x)}}+x^{2}f(x){\overline {f(x)}}dx$ but this is a sum instead of a product, and I don't see how you can get the inequality anyway. Widener (talk) 14:57, 30 October 2012 (UTC)

Your version of Heisenberg is wrong. It should be

$\left(\int _{-\infty }^{\infty }x^{2}|f(x)|^{2}dx\right)\left(\int _{-\infty }^{\infty }\left|{\frac {df}{dx}}\right|^{2}dx\right)\geq {\frac {1}{4}}\left(\int _{-\infty }^{\infty }|f(x)|^{2}\,dx\right)^{2}.$

So, this is an inequality of the form $AB\geq C^{2}/4$ and you want to show an inequality of the form $A+B\geq C$. You can use the AM-GM inequality. Sławomir Biały (talk) 00:40, 31 October 2012 (UTC)

Properties of the generalized Laplacian

How do you verify smoothness of a vector valued function? In particular, I want to show that the function $g(x)=\int _{\mathbb {R} ^{d}}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$, where $f$ is in the Schwartz class, and real $a>0$, is smooth.
Also, unlike the standard Laplacian, the function $\int _{\mathbb {R} ^{d}}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$ is not necessarily in the Schwartz class if $a$ is not an integer. How do you prove this?Widener (talk) 18:56, 30 October 2012 (UTC)

Differentiation under the integral sign proves smoothness. Non-Schwartzness follows since the frequency domain representation, $(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )$, is non-smooth. Sławomir Biały (talk) 22:21, 30 October 2012 (UTC)

Okay, but remember, this is a vector valued function. What exactly do you mean by "differentiation" in this context?

That is to say, x is a vector. Widener (talk) 23:23, 30 October 2012 (UTC)

Partial derivative Sławomir Biały (talk) 23:40, 30 October 2012 (UTC)

Okay. ${\frac {\partial }{\partial x_{i}}}(-\Delta )^{a}f(x)=\int _{\mathbb {R} ^{d}}2\pi i\xi _{i}(2\pi |\xi |)^{2a}{\widehat {f}}(\xi )e^{2\pi i\xi \cdot x}d\xi$. Hmm. The partial derivatives commute essentially because multiplication is commutative. Does this prove that these partial derivatives are $C^{1}$? I know the converse is true at least. But do you even know that the integral still converges?Widener (talk) 23:47, 30 October 2012 (UTC)

Integral converges because ${\hat {f}}$ is rapidly decreasing. Sławomir Biały (talk) 00:25, 31 October 2012 (UTC)

Yes, of course. Is my logic about the partial derivatives being commutative therefore $C^{1}$ correct? I know that partial derivatives that are $C^{1}$ commute, but I don't know if the converse is true. And does that indeed imply that the whole function is smooth? Widener (talk) 00:32, 31 October 2012 (UTC)

BY the way, how do you justify interchanging the partial derivative and the Integral?Widener (talk) 00:54, 31 October 2012 (UTC)

That's an argument that requires using dominated convergence theorem. Sławomir Biały (talk) 01:28, 31 October 2012 (UTC)

Also, could you explain the non-Schwartzness in more elementary terms?

Thanks Widener (talk) 23:21, 30 October 2012 (UTC)

A function is Schwartz iff it's smooth and its Fourier transform is smooth. Sławomir Biały (talk)|

So, for instance, choosing $a=1/4$, $(2\pi |\xi |)^{1/2}{\widehat {f}}(\xi )$ is not differentiable at $\xi =0$? Widener (talk) 23:54, 30 October 2012 (UTC)

October 31

limit percentage of no duplicates drawing half...

Let P(2n) be the probability that drawing from a bag of 2n distinct items n times with replacement will not draw the same item twice... Is the limit of P(2n) as n goes to infinity zero or something positive? I think it should be zero, since if n is a 500, then the probability is 501/1000 times 502/1000 times ... 999/1000 of which there are 166 entries less than 2/3 and none greater than 1 so P(2*500) < 2^166/3^166 which is pretty small... But I'm not sure what a formal proof would look like, pick an epsilon, get k so that (2/3)^k < epsilon and then use n=3k?Naraht (talk) 03:07, 31 October 2012 (UTC)

This is basically the birthday problem. You are correct that the probability of no repeats goes to zero if the number of draws is linear in the number of objects, and the strategy you suggest for proving it will work. You have a product $\prod _{i=1}^{n}(n+i)/2n$, and over 1/4 of the terms are less than 2/3, so the product is less than (2/3)^n/4, which goes to zero. In fact, if there are k objects you only need on the order of √k draws to have high probability of a repeat. Rckrone (talk) 04:13, 31 October 2012 (UTC)

• You can get $n^{2n}$ different sequences as a result, each equally probable, ${(2n)}^{\underline {n}}={\frac {(2n)!}{(2n-n)!}}={\frac {(2n)!}{n!}}$ of which are sequences without repetition (see Permutation#Counting sequences without repetition). The probability of getting such a sequence is a ratio of the two values $P(2n)={\frac {\ \ {\frac {(2n)!}{n!}}\ \ }{n^{2n}}}={\frac {(2n)!}{n!\ n^{2n}}}$. The Stirling's approximation will be probably helpful in showing that P converges to zero as n grows indefinitely. --CiaPan (talk) 09:46, 31 October 2012 (UTC)
  Or one can easily show that ${\frac {P(2(n+1))}{P(2n)}}={\frac {2(2n+1)}{{\left(1+{\frac {1}{n}}\right)}^{2n}\,(n+1)^{2}}}\approx {\frac {4}{e^{2}\,n}}$ for large n, so P converges to zero faster than any geometric sequence. --CiaPan (talk) 11:09, 31 October 2012 (UTC)

Thank you for the information, I wasn't looking for rate to zero, but I'm glad that information is available as well...

Related Problem

OK, new related problem. Let f be defined as a function of n. Let P(k,n) be defined as the probability of drawing k items from a bag with n items in it with replacement without drawing a duplicate. For example, if f(n)=b*n for a constant b less than 1, then the limit of P(f(n),n) is zero, if on the other hand f(n)= c, then the limit of P(f(n),n) is one. So, if the limit of P(f(n),n) as n goes to infinity is a constant m where 0<m<1, what do we know about f(n)? Is it log, a square root or something like n^(1+epsilon)?)

Apply same reasoning and you'll get ${\frac {n!}{(n-k)!\,\cdot \,k^{n}}}=m.$ I'm afraid it is pretty hard to solve for k... But if you do, k(n) with m as a parameter will be your f. --CiaPan (talk) 12:01, 2 November 2012 (UTC)

Yeah. I can see that it is relatively ugly. Thanx.Naraht (talk) 17:32, 2 November 2012 (UTC)

f(n) is on the order of square root of n, as the birthday problem article mentions. More specifically they claim $f(n)\approx {\sqrt {-2n\ln(m)}}$. Also I think the expression above should be ${\frac {n!}{(n-k)!\,\cdot \,n^{k}}}=m.$ Rckrone (talk) 06:04, 3 November 2012 (UTC)

Can Obama win?

The United States is the only current example of an indirectly elected executive president, with an electoral college comprising electors representing the 50 states and the federal district.

Suppose you are GOD. Suppose you know the true probability of Obama winning in all states and the federal district. Please note that you only know the true probability for each state (and federal district) and not if he will definitely win.

How do you calculate the chances of Obama retaining his office. For example: State A is 40%. State B is 60%, etc etc 220.239.37.244 (talk) 09:13, 31 October 2012 (UTC)

See Electoral College (India) for another instance.

If you know the probabilities for each vote and they are independent then it is quite easy to get an overall probability though a trick is needed. I'll illustrate with just three voters A B C who have probabilities .1 .6 and .7 of voting yes. The total probability of 2 or more yes votes is the sum of each individual possible overall vote that has 2 or more yes votes. For instance A ye, B yes, C no would give 2 votes and the probability of that combination is .1×.6×(1-.7)=0.018. For this problem the overall chance of 2 or 3 ves votes is (.1×.6×(1-.7)) + (.1×(1-.6)×.7) + ((1-.1)×.6×.7) + (.1×.6×.7) = 0.466

For the american president doing the sums that way would take an awful long time and even the fastest computers nowadays couldn't do it before the sun engulfed the earth. However there is a nice simple way of doing the job. We calculate the probability of each total yes votes adding one elector at a time. Suppose we have calculated the probability of 0, 1 or 2 yes votes for A and B above then when we add C the probability of 1 total yes vote is the sum of the probability of 1 so far times a no by C plus the probability of 0 yes votes so far time a yes by C. Doing the sums this way the whole business can be done by a computer before your finger has raised from the enter key.

There is a problem with the original supposition however, i.e. if I were God. That is a false assumption therefore I can prove anything from it. Dmcq (talk) 10:13, 31 October 2012 (UTC)

You can also write down a generating function f(x) that has as its coefficient of x^n the probability of winning n electoral college votes

$f(x)=\prod _{j=1}^{50}\left[1-p_{j}+p_{j}x^{n_{j}}\right]$

where the product over the j is over all States, $p_{j}$ is the probability of winning state j and $n_{j}$ is the number of electoral college votes you win if you win state j. Count Iblis (talk) 18:33, 31 October 2012 (UTC)

Thank you. I think I got it. for example: Ohanian (talk) 08:53, 1 November 2012 (UTC)

It is important here to realize where the simplification comes from. Mathematica is now doing the heavy lifting, so you may think that computing f(x) is done by expanding out all the factors but that is then equivalent to the direct method, so it may look like there is no real gain in computing f(x). However, this is not the case. The efficient way to compute f(x) is to compute the series expansion of Log[f(x)], this is the sum of the series expansion of the logarithms of the individual factors, which you only need to compute to the required order. Then one can compute the exponential function of that series expansion. This only requires a small number of squarings and multiplications. But Mathematica uses an even more efficient algorithm to do this.

Functions of series expansions can usually be most efficiently computed by making use of the following algorithm for division. For given y, computing x = 1/y can be considered as solving the equation

y - 1/x = 0, if you do that using Newton-Raphson, you get the successive approximants as x_{n+1} = 2 x_{n} - y x_{n}^2. The point of this is that there is no division involved in this, only multiplication. In each step the number of significant digits are squared. Now, if y is a polynomial, then this algorithm yields the series expansion if 1/y, and in each step you are doubling the number of correct expansion coefficients.

To compute the series expansion of Log[y(x)], you can use the above algorithm to compute y'(x)/y(x) and integrate term by term. The most efficient way to compute s(x) = exp[y(x)] is to solve the equation

log[s(x)] = y(x) using Newton-Raphson, so this is then a nested Newton-Raphson as computing the logarithm also requires using Newton-Raphson. But this is still much more efficient than computing exp[y(x)] directly. This then allows you to compute the series expansion of any elementary function as you can write them as compositions of logarithms and exponential functions. Count Iblis (talk) 16:24, 1 November 2012 (UTC)

Another shortcut you can take is to consider "solid Obama states" and "solid Romney states" to be out of play, since the chance of an upset in those is virtually 0. This reduces the number of swing states (and swing electoral votes, for states lacking the winner-take-all rule) to a small enough number that every possible combo can be considered. For example, if you end up with 20 swings states and electors, that's 2²⁰, or a bit over a million, calculations, which is simple for a computer to do. Just plug in the probability of a Romney or Obama win for each, and run the numbers to get the total probability. For example, this map shows only 16 states which aren't strongly for one candidate or the other: [1]. StuRat (talk) 18:44, 31 October 2012 (UTC)

• It's easy to calculate the expected value of Obama's total, and its variance. Those should give a decent approximation of his probability of winning. If that isn't good enough, I would probably do this using a Monte Carlo method -- run a million or so simulated elections on a computer, and count how many of them Obama wins. Looie496 (talk) 19:03, 31 October 2012 (UTC)

Based on info easily found on the internet. Obama seems to have a 62% chance of wining the election. Ohanian (talk) 00:37, 4 November 2012 (UTC) File:Obama wining 20121104.png

I don't think the hypothesis that all those probabilities are independent is realistic. Polling results have ups and downs that are highly correlated between states. Fivethirtyeight.com is the best known probability forecast site, that has been in the news a lot recently, and it currently puts Obama at over 80%. 67.119.3.105 (talk) 08:41, 4 November 2012 (UTC)

November 1

if a^2 =1 in a ring

In a Ring with identity if a^2 = 1 ,then a =1 or -1 ,i want a proof of it — Preceding unsigned comment added by 182.187.75.148 (talk) 00:14, 1 November 2012 (UTC)

What makes you so sure this is even true, if you have no proof of it? Sławomir Biały (talk) 02:01, 1 November 2012 (UTC)

sorry, there is one more condition, R has no zero divisor — Preceding unsigned comment added by 182.187.75.148 (talk) 02:20, 1 November 2012 (UTC)

If you had zero divisors, this would be false (consider, for instance, $\mathbb {Z} _{9}$; this is a ring with identity, and 5 times 2 is 1). I have a strange feeling this might just be a homework question, so I don't want to give it away, but the proof would be the same as the proof you'd see if you had a field (or even if you were just dealing with real numbers). Hope that helps. Wgunther (talk) 05:56, 1 November 2012 (UTC)

I'm not even aware that -1 makes sense for any ring with identity? In, say, $\mathbb {Z} _{n}$ you might call the number n-1 "-1" if you want to, but is this possible for any ring with identity? --KnightMove (talk) 08:45, 1 November 2012 (UTC)

Yes. -1 means the additive inverse of 1.--80.109.106.49 (talk) 10:16, 1 November 2012 (UTC)

Ok, you're right. --KnightMove (talk) 11:35, 1 November 2012 (UTC)

Well if there are no zero divisors and $(a-1)(a+1)=0$, what do you conclude about a? Sławomir Biały (talk) 23:25, 1 November 2012 (UTC)

Is there a name for the set of primes?

Is the set of primes just "the set of primes", or are there any other, preferably shorter names in use? --KnightMove (talk) 08:38, 1 November 2012 (UTC)

"Prime numbers", or possibly just "primes". — Preceding unsigned comment added by 83.70.170.48 (talk) 08:55, 1 November 2012 (UTC)

The letter "P" in blackboard bold (ℙ) is sometimes used, but this is not unambiguous. Gabbe (talk) 20:09, 1 November 2012 (UTC)

In a commutative ring, the set of prime ideals is called the "spectrum" but this is probably not what you want (even though the elements of Spec(Z) are exactly the ideals generated by prime numbers, plus the zero ideal). Rckrone (talk) 02:11, 2 November 2012 (UTC)

Indeed I mean the classic primes within the natural numbers. It seems there is no other term, thanks to all. --KnightMove (talk) 12:23, 2 November 2012 (UTC)

Einstein would have approved of this: "Things should be as simple as possible, but no simpler". -- Jack of Oz ^[Talk] 08:56, 3 November 2012 (UTC)

November 2

Generalizing the birthday problem

Can anyone find a good way to calculate the following: you have 15 people, and each person has 5 distinct birthdays, chosen uniformly from a 50 day year. What is the probability that there is a day on which at least 5 people have a birthday?--149.148.254.159 (talk) 14:28, 2 November 2012 (UTC)

Forgot to add: each person's birthdays are chosen independently of other people's birthdays.--149.148.254.159 (talk) 14:29, 2 November 2012 (UTC)

You can roughly approximate this as follows. Each day has on average $\lambda =1.5$ birthdays. If we approximate the number of birthdays per day as a Poisson process, then the probablity that a day has less than 5 birthdays is:

$p=\left(1+\lambda +{\frac {\lambda ^{2}}{2}}+{\frac {\lambda ^{3}}{6}}+{\frac {\lambda ^{4}}{24}}\right)\exp(-\lambda )\approx 0.981424$

The probability that all days have less than 5 birthdays is p^50, the probability that one or more days have 5 or more birthdays is thus 1 - p^50 = 0.6084. Count Iblis (talk) 17:12, 2 November 2012 (UTC)

OP here. It's not hard to calculate p exactly using binomial distributions, but when you then calculate p^50, you seem to be treating days as independent. Are you sure that's justified?--80.109.106.49 (talk) 18:17, 2 November 2012 (UTC)

can a person have all their birthdays on the same day or do they have to be distinct? 122.60.245.210 (talk) 08:46, 3 November 2012 (UTC)

All must be distinct.--80.109.106.49 (talk) 17:39, 3 November 2012 (UTC)

OK then, we can use direct numerical simulation, here with the R programming language:

> table(replicate(1e6,any(rowSums(replicate(15,sample(c(rep(1,5),rep(0,45)))))==5)))

FALSE   TRUE 
576416 423584 

So about 42.3% of the time. HTH, Robinh (talk) 08:11, 4 November 2012 (UTC)

This answer is surprising to me, since assuming days are independent, and calculating p exactly, I get 47.3%. Of course they're not independent, but rather knowing one day had 5 birthdays decreases the probability that another day did; equivalently, knowing one day did not have 5 birthdays increases the probability that another day did. So the actual answer should be greater than 47.3%, yet your number is smaller.--80.109.106.49 (talk) 10:06, 4 November 2012 (UTC)

Apologies, I was checking for *exactly* five birthdays and you asked for *at least* five:

R> table(replicate(1e6,any(rowSums(replicate(15,sample(c(rep(1,5),rep(0,45))))) >= 5)))

 FALSE   TRUE 
504936 495064 

So I get 49.5%, more or less. Let me know if you want a more accurate value. Sorry about that. Robinh (talk) 21:47, 4 November 2012 (UTC)

Ah, makes sense. That's plenty accurate, this was just for settling a little debate (which I've reframed; it wasn't actually about people with multiple birthdays). Thanks.--80.109.106.49 (talk) 00:18, 5 November 2012 (UTC)

Integration

I was asked solve the following:

$\int d(zx)+\int d(xy)=0$

At first, I thought the integration can be evaluated directly as,

$\int d(zx)+\int d(xy)=zx+xy+g(y,z)=0$

Then I had to look for the value of $g(y,z)$, but found things complicated. When I turn to basics, evaluating this integral,

$\int dx=0$,

I assume, it should be $x+c=0$, however, wolframalpha ignores the constant. Am I wrong somewhere?--Almuhammedi (talk) 14:53, 2 November 2012 (UTC)

This problem is ill-defined as stated. The meaning of "solve" needs to be clarified, and we need to know what is a function of what. Looie496 (talk) 17:27, 2 November 2012 (UTC)

I'm not sure of the proper descriptive term (you may say, evaluate).--Almuhammedi (talk) 18:20, 2 November 2012 (UTC)

Urgh. Wolframalpha appears to interpret this as the definite integral from x=0 to x=x₀, and that it is to find x₀. More obvious if you try this. Ignore its result. — Quondum 16:17, 5 November 2012 (UTC)

History of knowledge of various geometric shapes

Our articles circle, hyperbola, and parabola have history sections. But ellipse, triangle, Reuleaux triangle, curve of constant width, quadrilateral, parallelogram, and others do not. Can someone give references on when each of these was first studied, and what was known at that time? Also, I invite any interested editors to contribute history sections to these articles. Duoduoduo (talk) 16:50, 2 November 2012 (UTC)

Franz Reuleaux lived from 1829 to 1905, so that is some indication for his triangle. Euclid's Elements was one of the first works on geometry and contains some basic facts on triangles. Rojomoke (talk) 18:03, 2 November 2012 (UTC)

Apollonius of Perga did some quite remarkable work on conics, including ellipses, he studied their normals and centers of curvature. See [2] Aristaeus the Elder did some earlier work as did Euclid.--Salix (talk): 18:15, 2 November 2012 (UTC)

Some of those were likely studied "in antiquity". That is, before writing. Therefore, we don't know the specifics. For example, the hypotenuse of a right triangle being longer than either leg but less than the total of the two legs was probably known way back, although the Pythagorean theorem wasn't discovered until later (when they knew how to find irrational square roots). StuRat (talk) 05:06, 4 November 2012 (UTC)

November 3

More or less?

Can we say that there exists more irrationals than rationals or vice-versa, more even that odd or vice versa among other things. I think the answer is no, but if so why we say more transcendental exists than algebric and so on? 117.227.3.241 (talk) 16:40, 3 November 2012 (UTC)

I'd think you could say there are more irrationals than rationals, but should be the same number of odds and evens. For the rationals, consider just those which can be created by integers between 1 and 10 in both the numerator and denominator. That gives 100 possibilities, some of which are multiples of each other. If you then allow the numerators and denominators to be reals, we then get an infinite number of irrationals using the same ranges. StuRat (talk) 16:46, 3 November 2012 (UTC)

Infinite sets may be countable or uncountable. Rationals are countable but irrationals are not, so we usually say there are more irrationals with an implicit definition of "more". Algebraic numbers are countable but transcendentals are not, so it's the same here. See cardinality for a definition which also allows us to compare two uncountable sets. Integers and all subsets of integers are countable but other definitions of "more" can be used for integers, such as natural density. With this definition we can for example say there are more composites than primes. A more advanced definition will allow comparison of sets with natural density 0 and for example say there are more squares than cubes. It all depends on the used definition. Odds and evens will be the same in any meaningful definition where they can be compared. PrimeHunter (talk) 17:34, 3 November 2012 (UTC)

There are definitely more irrationals than rationals. PrimeHunter talks about using an "implicit" definition, which is true, but it really is the correct definition; there really, Platonistically, are indeed more of them, in the natural sense of the term. --Trovatore (talk) 08:53, 4 November 2012 (UTC)

I don't like heating the atmosphere to produce bitcoins

hi, i don't like heating the atmosphere to produce bitcoins. Please propose an alternative scheme with the same properties that does not involve proof-of-work "mining" but has the correct incentive scheme for a distributed, anonymous, peer-to-peer currency with the same properties (cryptographic and social) as bitcoin. thanks. --82.131.132.190 (talk) 22:34, 3 November 2012 (UTC)

I have just the thing, and will give you a 10% discount off of my standard $20,000 (US) consulting fee. (I do accept bitcoins.) Best, Sławomir Biały (talk) 23:07, 3 November 2012 (UTC)

I think this is an incredibly rude demand at the volunteer reference desk, and you should be ashamed. if you don't want to volunteer, then don't, but to offer consulting services to those who can't afford it is just rude. would you "volunteer" at a children's library, ask the children what they want to read, and then not only not give it to them, but try to sell it to their parents for a huge markup? shameless. 86.101.32.82 (talk) 08:57, 4 November 2012 (UTC)

I think you have overestimated the sincerity of Sławomir's offer. --Trovatore (talk) 09:35, 4 November 2012 (UTC)

Ok, fine, 25% off. But that's my final offer! Sławomir Biały (talk) 20:35, 4 November 2012 (UTC)

Before I blindly hand over money for something, I would like to have some assurance that the thing works. Since the only requirement (besides not heating the atmosphere with proof-of-works) is that it have "the same properties (cryptographic and social) as bitcoin", and you write, "(I do accept bitcoins.)", will you accept payment in the resulting currency? --91.120.48.242 (talk) 13:32, 5 November 2012 (UTC)

See Trovatore's post. I had assumed that you were trolling, but in the light of this post I am not so sure. Sławomir Biały (talk) 14:13, 5 November 2012 (UTC)

And I had assumed that where you wrote, "I have just the thing" it meant that you can produce it. Let's be serious for a moment. What would it take for you to produce an alternative to the BitCoin protocol that does not depend on "proof-of-work" 'mining' which exists literally to no other purpose than to heat the atmosphere. This is the definition of "work" in this sense. This takes one of the worse possible solutions for a monetary system - wasting resources to mine gold just to use it instead of paper - and brings it into a cryptographic digital solution where it has no place whatsoever. I would like all the benefits of bitcoin without the silly mining portion. For example, wouldn't it be possible to give coins out in exchange for "availability" - thereby incentivizing participation in the network - which can be checked statistically from time to time with simple pings, instead of for number-crunching? Other solutions are also possible. What would it take? --91.120.48.242 (talk) 14:59, 5 November 2012 (UTC)

The CPU cycles could be spent doing something useful for humanity, rather than just heating the atmosphere. Sławomir Biały (talk) 15:11, 5 November 2012 (UTC)

That would be acceptable to me as long as you don't produce a "nonce" need (i.e. that you just happen to discover). Also I hold a level of incredulousness toward some distributed (e.g. Seti@home or finding twin primes or something) projects, I think they are a waste of resources. So you would have to make a pretty good case for the work. Do you have any other solutions or suggestions? (To be sure, what you've just said is a good, acceptable solution.) The idea of *making* someone do work to produce the scarcity is one that I feel is cryptologically unsound. There should be something better. --91.120.48.242 (talk) 15:37, 5 November 2012 (UTC)

Instead of backing your bitcoins with precious metals, you can back them with property. Since land values are low now, this would be a good time to buy it up and parcel it out, no mining required. StuRat (talk) 23:21, 3 November 2012 (UTC)

The proof-of-work system in Bitcoin serves two purposes:

1. Determine the initial distribution of the bitcoins.
2. Synchronize transactions to prevent double-spending.

The reason Bitcoin didn't exist before 2009 is that prior to Satoshi Nakamoto's invention of the Blockchain, there was no known way to do #2 in a decentralized way (and the known ways to do #1 without being prone to Sybil attacks also involved proof of work). Thus there is no known way to operate a decentralized digital currency without proof of work, and developing one, even if such a thing is possible, is way beyond the scope of this reference desk.

By your own admission, having a cryptographic peer-to-peer currency is extremely important, and the energy expenditure required to maintain it is a small price to pay (arguably, smaller than the resource waste of maintaining the current monetary system).

Purpose #1 is only temporary, the long-term purpose is #2. And there are hypothetical proposals for how to prevent double-spending while minimizing energy consumption, using some form of proof of stake. There is also an alternative digital currency called PPCoin which uses related ideas.

See also https://en.bitcoin.it/wiki/Myths#Bitcoin_mining_is_a_waste_of_energy_and_harmful_for_ecology.

FWIW, I'll be away for a week and unable to follow up on this discussion. -- Meni Rosenfeld (talk) 04:53, 4 November 2012 (UTC)

November 4

"Random article" chances of landing on a football-related page.

What are the chances of clicking Wikipedia "Random article" and recieving an article related to football? Nicholasprado (talk) 06:23, 4 November 2012 (UTC)

I guess one could use a Category-searching tool to estimate what fraction of articles are related to football. But not all articles have an equal chance of being chosen ... —Tamfang (talk) 06:39, 4 November 2012 (UTC)

Random article is not really random in the sense of having uniform probability. There are tweaks in it, to increase the likelihood of showing the reader something interesting. I suggest just clicking it 20 times or so, counting how many times you click it and how many football articles you get. 67.119.3.105 (talk) 08:43, 4 November 2012 (UTC)

Can you provide a link or any more information about those "tweaks"? —Steve Summit (talk) 13:46, 4 November 2012 (UTC)

The explanation at Wikipedia:TFAQ#random suggests that the selection is not truly random due to limits in the methodology. But beyond that, it does not mention any tweaks, and looking at other related pages, I can find no evidence of the existence of any such tweaks. A rough estimate of the proportion of football articles can be gained from the table at Wikipedia:Version 1.0 Editorial Team/Football articles by quality statistics, which lists all pages tagged as falling under the scope of WikiProject Football. While not every football-related article is tagged, in my experience the proportion is over 90%. After stripping out non-article categories, at the time of writing Wikipedia:Version 1.0 Editorial Team/Football articles by quality statistics lists 141,850 tagged football articles, and Special:Statistics gives the total number of articles as 4,090,691. This gives a lower bound for the proportion of football articles of 3.47%. Oldelpaso (talk) 22:42, 4 November 2012 (UTC)

more odd than even natural numbers?

with the formula "for all even natural numbers n, n<->n+1" (goes with), we have drawn a 1:1 relationship between every even number and an odd number. e.g. 2 goes with 3, 4 goes with 5, 6 goes with 7 and so on. Every even number is thus matched. There are an equal number of them as odd numbers, since you can bring them into 1:1 correspondence.

But wait, I trickily left out one odd number, namely "1". So after we have matched every even number with an odd number (without exception - the formula gives the match for every even number) we still have an odd number left over that no even number is matched with! So, in this pairing, there isn't it fair to say there is one more odd natural number than even natural number? Is this reasoning correct? Thanks. 87.97.96.159 (talk) 16:50, 4 November 2012 (UTC)

No, because you could just as well do it the other way around, and leave out an even. StuRat (talk) 17:07, 4 November 2012 (UTC)

That doesn't address the "flaw" in the proof. All you've shown is that counting one way, there is one more odd number than even number, with all the even numbers being paired 1:1 with an odd number, but one odd number left out; and counting another way, there is one more even number than odd number, with all odd numbers being paired 1:1 with an even number, but one even number left out. So, are both these statements true? 87.97.96.159 (talk) 17:16, 4 November 2012 (UTC)

The reason is that one number, or indeed any finite set of numbers, is infinitely small when compared to infinity. In other words, the limit of 1/x, or c/x, goes to zero, as x approaches infinity. StuRat (talk) 17:53, 4 November 2012 (UTC)

But that does not really address my argument. Look, if every boy is matched with a girl, and no girl is unmatched but there is one boy unmatched, then there is 1 more boy than girl, is that not fair to say? So, with infinite boys and girls, perhaps it is possible to match every girl to a boy but leave one boy alone, and in a different arrangemnt it's possible to match every girl to a boy but leave one girl alone? 87.97.96.159 (talk) 18:07, 4 November 2012 (UTC)

With infinite numbers (cardinalities), there is always the possibility of mismatching exactly matched sets by pretty much any number you choose, e.g., one could say that the set of natural numbers is ten elements larger than itself. The sensible (and standard) way of matching set sizes, then, is to say they have the same cardinality if a one-to-one mapping between them exists, or one is larger than the other if it has unmatched elements with every possible matching. — Quondum 18:31, 4 November 2012 (UTC)

Thank you, this makes a lot of sense and resolves my question. 87.97.96.159 (talk) 20:02, 4 November 2012 (UTC)

Read the definition of equipollence carefully — it says 'there exists a bijection from the set A to the set B', but not 'any function from the set A to the set B is a bijection'. You can present many functions that leave some items not paired in either set, but that doesn't prove or disprove anything. On the the other hand if you can show a single function which matches two sets with a one-to-one correspondence, this implies immediately the two sets have equal cardinality.
We can math even and odd integers in multiple ways, one of which is 2n ↔ 2n+1 — and that's enough to say they are equinumeros. --CiaPan (talk) 12:06, 5 November 2012 (UTC)

November 5

Teach integration

Hi. I realize that mathematical integration contains dozens of human variants of the How-wise, including the Quadrature methods, Archimedes' quadruplets and likewise. For example, I may need to integrate a piecewise function, and require to explanify the various modulus on f(x) to derive the antiderivative in two, three, n and n-1.5 dimensions. My brain works best in acute visual calculus, and so to predict a subspace or manifold I must first visualize, and then derive, otherwise my dopamine level approaches 0 as t approches 1. Since I will need to use integrative methods for processing the universe within the next three weeks, I would great-infinitely appreciate some explanatory guidance unto this matter. As it stands, I have absolutely no background in integration, and only limited edusophical background in calculus. My understanding of complex erudite logic is far superior to my thought-attention span in numerical and differential mathematics. Some history of integration across the lands of the English, the Arabs and the Chinese would also serve interest. You may recall that I asked a question in Science many years ago on the flow of water into a geo-sedimentary basin of lower height, yet this is likely a common application of integral calculus. I find the quadrature of the Parabola highly interesting, and the proposition of self-same continuity across x as the scale rotates about the x-y plane is also. Please understand the meaning of this post, and give any resources based on my understanding or lack thereof of the subject that which I would highly very like to learn about regarding. Thank you! Verily. ~AH1 ^(discuss!) 19:54, 5 November 2012 (UTC)

The phrasing of this question is so clever that I can't figure out what the question is. I get that it has something to do with integration, but what exactly? Looie496 (talk) 22:29, 5 November 2012 (UTC)

localization

in the article about localization it reads:

The ring Z/nZ where n is composite is not an integral domain. When n is a prime power it is a finite local ring, and its elements are either units or nilpotent. This implies it can be localized only to a zero ring. But when n can be factorised as ab with a and b coprime and greater than 1, then Z/nZ is by the Chinese remainder theorem isomorphic to Z/aZ × Z/bZ. If we take S to consist only of (1,0) and 1 = (1,1), then the corresponding localization is Z/aZ.

I don't agree with the statement that for n = prime power, the localization must be a zero ring. For example:

Z/pZ where p is prime is a field. So any element a!=1 in Z/pZ* generates Z/pZ*.

Take for example: Z/5Z and a = 2, then S:=<2> = {2,4,8=3,6=1} = Z/5Z* but then S^-1 Z/5Z ~= Z/5Z (because the field of fraction of a field is the same field)

--helohe (talk) 21:36, 5 November 2012 (UTC)

algebra inequality

If

${\sqrt {N-1}}+(N-1){\sqrt {K-1}}<(N-1){\sqrt {K}}$

then does this mean that $K<{\frac {N^{2}}{4(N-1)}}$ ?

I am reading that this is the case but I am having trouble getting from a to b. Is there some inequality trick involved? — Preceding unsigned comment added by Thorstein90 (talk • contribs) 22:28, 5 November 2012 (UTC)

Do we know anything about the possible values for N and K ? Specifically, can they be 1 or less ? StuRat (talk) 02:51, 6 November 2012 (UTC)

Derivative word problems

Hello! I have three problems that I need help on:

1) A rectangle has its base on the x-axis and its upper two vertices on the parabola y = 12-x^2. What is the largest area the rectangle can have, and what are its dimensions?

2) A 500ft^3 tank is to be made by welding thin stainless steel plates together using as little weight as possible. How do I use derivatives to solve this?

3) A 1125-ft^3 open top rectangular tank with a square base x ft on a side and y feet deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy:

If the total cost is C = 5(x^2 + 4xy), what values of x and y will minimize it?

If you could offer any assistance, I would greatly appreciate it. --Colonel House (talk) 01:20, 6 November 2012 (UTC)

I fixed your numbering (auto-numbering doesn't work with blank lines). StuRat (talk) 01:55, 6 November 2012 (UTC)

1) Start by coming up with a formula for the area. Let's define one vertex on the X-axis as (j,0). Since it's going to be symmetric about the Y -axis, that gives us (-j,0) for the other vertex on the X-axis. The width, therefore, is 2j. Plugging either coord in to the parabola equation, we get (±j,12-j²) for the upper coords, with a height for the rectangle of 12-j². Multiplying by the width, we get an area of 24j-2j³. Now that you have an area formula, you can maximize for that by varying j. Specifically, find the derivative of the area formula and set it equal to zero, then solve for j. Next plug that value of j back into the area formula to find the answer. As a check I suggest trying values for j which are .01 more or less than the optimal value. When you plug those into the area formula, you should get a slightly smaller area. Show your work here and we will help with any problems. StuRat (talk) 02:07, 6 November 2012 (UTC)

(2) doesn't tell us enough about the constraints. —Tamfang (talk) 02:29, 6 November 2012 (UTC)

Yea, I'm thinking it will be a cube, and they want us to prove it, but it is confusing, as stated. StuRat (talk) 02:45, 6 November 2012 (UTC)

3) Find the equation for the volume: x²y = 1125. Solve for y: y = 1125/x². Now plug that into the cost equation: C = 5(x² + 4xy) = 5(x² + 4x(1125/x²)) = 5(x² + 4500/x) = 5x² + 22500/x = 5x² + 22500x^-1. As before, find the derivative, set it equal to zero, and solve for x. From this you can use y = 1125/x² to find y, then plug x and y into the original cost equation. Check, as before, by trying values of x which are .01 higher or lower. Find y for each case, then plug that x and y into the original cost equation. The cost should be slightly higher, in each case. StuRat (talk) 02:42, 6 November 2012 (UTC)

Thank you so much everyone! I am working on another assignment now, but will post any work if I need more help. Thanks again!--Colonel House (talk) 02:51, 6 November 2012 (UTC)