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December 15

Is ther any site that calculates the expected return of some game?

Is ther any site that calculates the expected return of some game?
I want to calculate the expected return of my country lottery, but all I find on google are sites related to stockmarket.
201.79.67.114 (talk) 11:32, 15 December 2015 (UTC)

If you are referring to the Mega-Sena then our article says the the expected return is 32.2% after tax. Dbfirs 21:47, 15 December 2015 (UTC)

Limit of a sequence struggle

Can someone please explain how do we really prove that $\lim _{n\to \infty }\{a_{n}+b_{n}\}=\lim _{n\to \infty }\{a_{n}\}+\lim _{n\to \infty }\{b_{n}\}$ ?

I'll say now: I know this proof by heart, so please don't waste your breath repeating the same usual wording here - it won't help.

My question is:

How does ${\bigg |}a_{n}-L+b_{n}-M{\bigg |}\leq {\bigg |}a_{n}-L{\bigg |}+{\bigg |}b_{n}-M{\bigg |}<\varepsilon ,\ \forall n>k=\max\{k_{a_{n}},k_{b_{n}}\}$ prove anything?

I just don't understand where do I see in the triangle inequality the expression $\lim _{n\to \infty }\{a_{n}+b_{n}\}=\lim _{n\to \infty }\{a_{n}\}+\lim _{n\to \infty }\{b_{n}\}$, which we are trying to prove.

I can't see the formal logic here.

Is there another logic formal way to prove this expression? יהודה שמחה ולדמן (talk) 21:22, 15 December 2015 (UTC)

The triangle inequality is used to say that if $a_{n}$ is within ε of L and $b_{n}$ is within ε of M, then $a_{n}+b_{n}$ is within 2ε of $L+M$. Sławomir
Biały 21:31, 15 December 2015 (UTC)

So how does $2\varepsilon$ help us? And I don't get the point of defining those expressions above to be $<{\tfrac {\varepsilon }{2}}$ just for making the end look good.

Logically, how does $L+M<2\varepsilon$ give us any clue or indication to the equation we are proving? יהודה שמחה ולדמן (talk) 23:11, 15 December 2015 (UTC)

If $a_{n}$ is close to L and $b_{n}$ is close to M, then $a_{n}+b_{n}$ is close to $L+M$. Sławomir
Biały 23:57, 15 December 2015 (UTC)

You seem to be very confused.

Starting from the end, it's not $L+M<2\varepsilon$, it's $|(a_{n}+b_{n})-(L+M)|<2\varepsilon$.

Moving on, your qualms about "making the end look good" are misplaced. We want to prove something, so we do whatever it is that proves it.

What might be the real source of confusion is that the symbol $\varepsilon$ is used here in several different meanings. One time it's part of the definition of the limit of $a_{n}$, the other it's part of the definition of the limit of $a_{n}+b_{n}$ - and then it represents twice what the original epsilon did.

It might help to use different symbols for each time, or subscripts. To prove $\lim(a_{n}+b_{n})=L+M$ you need, by definition of limit, to prove that $\forall \varepsilon _{1}>0\ \exists k_{1}$ such that $\forall n>k_{1},|(a_{n}+b_{n})-(L+M)|<\varepsilon _{1}$. You know that $\lim a_{n}=L$ so by definition of limit, you know that $\forall \varepsilon _{2}>0\ \exists k_{2}$ such that $\forall n>k_{2},|a_{n}-L|<\varepsilon _{2}$. This is for all $\varepsilon _{2}$, meaning you can take any $\varepsilon _{2}$ you want - including $\varepsilon _{2}=\varepsilon _{1}/2$. Do the same for $b_{n}$, and the rest of the proof follows.

If it's still not clear, you should specify what exactly it is that you have trouble with. -- Meni Rosenfeld (talk) 00:05, 16 December 2015 (UTC)

Is there a way to graph all this with adding words? Or is it impossible for 2 limits? I'm confused with your $\varepsilon _{1},\varepsilon _{2}$ definition. How would you add them to the formal proof expression?

Is it possible to prove the sum law with the Squeeze theorem? This guy here claims he did, but something was metioned by others that it's not obvious or true. Would you please have a look at the link? יהודה שמחה ולדמן (talk) 16:02, 16 December 2015 (UTC)

In the linked SE question, there are steps that are not obviously true unless you already know that the limit of a sum is the sum of the limits. This makes that proof circular.

There's no avoiding using the foundational delta-epsilon until you have enough theorems to tackle anything at hand. And there's little point in trying to find alternative proofs - there is nothing wrong with the standard proof, and if it's not clear to you, you have to refine your mathematical understanding until it is - not to try and avoid it.

I'm not sure what you mean by "graph all this with adding words".

Can you repeat here a standard proof you've seen for the theorem, and specify where is the first step that you're having trouble with? Without this I don't think we can help. -- Meni Rosenfeld (talk) 16:45, 16 December 2015 (UTC)

The whole delta-epsilon thing was confusing when it was first sprung on me, but the difficulty is likely to be forgotten after a few semesters of real analysis and topology. It took mathematicians a couple thousand years to wrap their heads around these ideas (Euclid Book V Def. V to the late 19th century), so it's a bit much to expect college students to fully understand them in a day. Most of the problem is the double ∀∃ quantifier being thrown in suddenly, but there is also the fact that Greek letters are being used and most people (outside of Greece) are not familiar them them. To me, the best way of looking at these ideas is by way of measurement. All measurements involve error due to limitations in the accuracy of whatever you measuring with. If I measure one board as 96 in. to within 1/4 in., in a second board to be 48 in. also to within 1/4 in., then laid end to end they should measure 14 4 in. to with 1/2 in. The errors could cancel but in worst case they add together so if the possible error in each board length is 1/4 in. then the possible error for the sum is 1/2 in. Generally, if you've measured L and M to a given tolerance e, then L+M can only be said to be measured to tolerance 2e; this is where the triangle inequality comes in. In the proof you have to do this backwards, for example if I want two boards laid end to end so that the total error is less than 1/8 in., how accurately do I need to measure each board? In the proof, you're given the tolerance for the sum is ε, so in order to get this you need the tolerance for each piece to be ε/2. There is another feature of mathematical proof here that can be confusing when you first see it, namely that sometimes you need to start at end and work backwards to set up what you need to get there. Then you write it all down in the proper order as if you knew what you would need from the start. For the limit of a sequence you make the tolerance for L smaller my increasing n, and in another case of working backwards you ask, how big to I need to make n so that a_n and b_n are both within the tolerance ε/2 of their respective limits L and M?

The epsilon-delta type argument is unavoidable, at least if you want to progress into higher mathematics, and it is to some extent intrinsically difficult, but there are ways of dealing with it to make it easier to understand for those who have never seen it before. It would help as well if there was more instruction on how a proof should be constructed rather than showing the finished product and saying "Make it look like this." Despite tradition, Euclidean geometry is a bad way of doing this in a lot of ways besides the fact that it doesn't really prepare people for epsilons and deltas. --RDBury (talk) 19:35, 16 December 2015 (UTC)

Wait, how do we prove that just like $\lim _{n\to \infty }\{a_{n}\}$ and $\lim _{n\to \infty }\{b_{n}\}$ are convergent, so is $\lim _{n\to \infty }\{a_{n}+b_{n}\}$ ?

And, is the idea of $2\varepsilon$ making not much of a difference than $\varepsilon$ because of the expression $\forall \varepsilon >0$ ? יהודה שמחה ולדמן (talk) 00:25, 17 December 2015 (UTC)

What do you mean, "how do we prove that $\lim _{n\to \infty }\{a_{n}+b_{n}\}$ is convergent"? That's exactly the theorem we're proving - that the sequence converges and that its limit is L+M. We do this by showing that it satisfies the definition of limit. You can see a proof at https://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules (Proof of the Sum Rule for Limits).

Yes, the idea is that the definition uses "for all epsilon", so we can choose any epsilon we want - including, half the epsilon we started with. It's confusing because the symbol "epsilon" is overloaded, which is why I suggested using different symbols for each variable. -- Meni Rosenfeld (talk) 10:41, 17 December 2015 (UTC)

The equation

$\lim _{n\to \infty }\{a_{n}+b_{n}\}=\lim _{n\to \infty }\{a_{n}\}+\lim _{n\to \infty }\{b_{n}\}$

is not necessarily true. Let

$a_{n}=n$

and

$b_{n}=-n$.

Then the left hand side

$\lim _{n\to \infty }\{a_{n}+b_{n}\}=\lim _{n\to \infty }\{n+(-n)\}=\lim _{n\to \infty }\{0\}=0$

is defined while the right hand side

$\lim _{n\to \infty }\{a_{n}\}+\lim _{n\to \infty }\{b_{n}\}=\lim _{n\to \infty }\{n\}+\lim _{n\to \infty }\{-n\}$

is undefined.

Bo Jacoby (talk) 12:56, 17 December 2015 (UTC).

Yes. The questioner loosely implied convergence of the individual sequences via the use of L and M in the snippet of the proof they questioned, but that is an important point. In our Limit of a sequence#Properties, the convergence of the individual sequences is given in the "If $a_{n}\to a$ and $b_{n}\to b$,".-- ToE 15:40, 17 December 2015 (UTC)

OK, why isn't $\lim _{n\to \infty }\{a_{n}+b_{n}\}\leq \lim _{n\to \infty }\{a_{n}\}+\lim _{n\to \infty }\{b_{n}\}$ true? it makes more sence. יהודה שמחה ולדמן (talk) 20:20, 19 December 2015 (UTC)

Do you agree that if $\lim _{n\to \infty }\{a_{n}\}=a,$ then $\lim _{n\to \infty }\{a_{n}-a\}=0$ and vice-versa?

or that together with $\lim _{n\to \infty }\{b_{n}-b\}=0$

implies that $\lim _{n\to \infty }\{(a_{n}-a)+(b_{n}-b)\}=0=\lim _{n\to \infty }\{(a_{n}+b_{n})-(a+b)\}$

this is basically what that inequality you had in the original question is about. With the vice-versa bit you can eliminate the constant parts on both sides.

$\lim _{n\to \infty }\{a_{n}-a\}+\lim _{n\to \infty }\{b_{n}-b\}=\lim _{n\to \infty }\{a_{n}\}-a+\lim _{n\to \infty }\{b_{n}\}-b$

$\lim _{n\to \infty }\{(a_{n}+b_{n})-(a+b)\}=\lim _{n\to \infty }\{a_{n}+b_{n}\}-(a+b)$

These are both zero and you can equate them and remove the $a+b$ on both sides.

All the epsilons are just formalizing this. Dmcq (talk) 21:13, 19 December 2015 (UTC)

December 16

Whether to add to Linear Equation how to find the general form of a linear equation given two points

Hi everyone.

I constantly come to Wikipedia to find how to solve already solved problems, but the Linear Equation page failed me. I wanted to know how to find the general form of a two-dimensional linear equation given any two points (already solved; see later). The page only provides a solution for a point lying on the x-axis and another on the y-axis. I read several links from the first few pages of a Google search, but the only solution was to find the slope–intercept form and then transform to the general form. The problem is that this solution doesn't work for vertical lines. So, the solution, according to the internet, is to first test if it's a vertical line, and then choose the appropriate method. My limited mathematical skills tells me there should be a more generic approach, and I found it (still not the point of this question, read on):

• To find the equation for a 2D line in the form $Ax+By+C=0$ that passes through points $P=\left[{\begin{matrix}p_{x}\\p_{y}\end{matrix}}\right]$ and $Q=\left[{\begin{matrix}q_{x}\\q_{y}\end{matrix}}\right]$.

• $A={\begin{cases}q_{y}-p_{y}&\quad p_{x}>q_{x}\\p_{y}-q_{y}&\quad {\text{otherwise}}\end{cases}}$

• $B=\left|p_{x}-q_{x}\right|$

• $C=-\left(A\cdot p_{x}\right)-\left(B\cdot p_{y}\right)=-\left(A\cdot q_{x}\right)-\left(B\cdot q_{y}\right)$

By replacing x and y by the coordinates of a point, the value on the left hand side will be 0 if the point is on the line, a positive number if the point is on one side of the line, and a negative number if the point is on the other side of the line. One problem is that this number doesn't give the distance to the line, but I don't know if that's expected from the general form.

Now, the actual point of this question. I wanted to add this to the Linear Equation article. First thing I did was to consult the Talk:Linear_equation page. However, the first thing I found was "This is not a forum for general discussion of the article's subject.", and I suppose this issue would require a discussion. The second thing I found was "No original research", which is exactly what I did by comparing some points to the general form equation obtained from the slope–intercept form.

So, I want two things by leaving this here: for people to read it and decide whether it requires further action, and to leave this as a reference for if someone searches the same thing that I did. — Preceding unsigned comment added by GuiARitter (talk • contribs) 18:35, 16 December 2015 (UTC)

Edit: forgot to say I tested this with pairs of points with randomly generated coordinates between -7.0 and 7.0.

What you're looking for is already in the article, look for mentions of "determinant form". -- Meni Rosenfeld (talk) 19:08, 16 December 2015 (UTC)

Specifically, the article says that the line passing through $(p_{x},p_{y}),(q_{x},q_{y})$ is given by

${\begin{vmatrix}x&y&1\\p_{x}&p_{y}&1\\q_{x}&q_{y}&1\end{vmatrix}}=0\,.$

In other words, $(p_{y}-q_{y})x+(q_{x}-p_{x})y+(p_{x}q_{y}-p_{y}q_{x})=0$, which is equivalent to what you wrote, and simpler.

Also, to clarify, this question could fit perfectly in Talk:Linear equation. The first warning simply says the discussions must be aimed at improving the article, and by your account this is indeed your aim. The second warning means that the article shouldn't include things that you are the first to discover and that were not published anywhere else - your proposed addition is clearly already well known. (The problem, again, is that the article already includes this, so there is nothing to add. Good job deriving it independently, though). -- Meni Rosenfeld (talk) 19:16, 16 December 2015 (UTC)

Actually, the expansion of the determinant also appears in the article already - in the very row above the first one mentioning the determinant form. -- Meni Rosenfeld (talk) 19:25, 16 December 2015 (UTC)

Thanks for your response. Yeah, I see it now. As I said, my math skills are limited, so I didn't tried to find the determinant to see what would happen, and I skimmed the Two-point form sub(sub?)section because it involved the slope. So, maybe the article could be improved by adding $(p_{y}-q_{y})x+(q_{x}-p_{x})y+(p_{x}q_{y}-p_{y}q_{x})=0$ to the General (or standard) form sub(sub?)section, but it doesn't seem that necessary anymore. Also, I saw your response accidentally because I came here to add something I forgot. So, I've just seen that the "Watch this page" feature doesn't work as I expected... GuiARitter (talk) 19:40, 16 December 2015 (UTC)

@GuiARitter:: To see changes in pages you've watched, you need to go to your "watchlist" - the link is found at the top of the Wikipedia interface. I'm not sure how to get more active notifications for a page. Also, people can tag you so you get an active notification, as I've done now (but it's not customary in the RefDesk to tag people when answering). -- Meni Rosenfeld (talk) 21:12, 16 December 2015 (UTC)

December 20

apeirogon and zerogon

Yesterday's Wiktionary "Word of the day" was "apeirogon" - a polygon with an infinite number of sides and vertices. Wikipedia also has an article about that: apeirogon.

The Wiktionary definition for apeirogon says that "zerogon" is a synonym...yet the definition it provides for zerogon defines it as a polygon with zero edges and vertices.

1. Seems to me that wiktionary is incorrect - zero is not infinity (well, DUH!)
2. Seems to me that the reason someone there thought these were synonyms is that (allegedly) a circle could be described as either a zerogon (it has no vertices) or as an apeirogon (a regular apeirogon is arguably a circle). I was under the impression that polygons had to have straight edges - so I don't think a circle is a zerogon...but then IANAM.
3. If wiktionary is correct, or even if the terms are loosely related, why no mention of zerogon in our apeirogon article?
4. If zerogon is a real term, shouldn't we have an article about it and an entry in Template:Polygons - which already has entries for weird shit like the monogon and digon?

Either way, something needs fixing - I'm just not quite sure which. SteveBaker (talk) 14:13, 20 December 2015 (UTC)

In the case of a regular polygon, they seem synonymous to me, as the sides on the regular apeirogon are all infinitely short, which really makes them points, not sides. But for the general case, you could have an apeirogon with a finite number of measurable sides and an infinite number of (infinitely short) sides, elsewhere. That's not a zerogon. StuRat (talk) 18:16, 20 December 2015 (UTC)

"Zerogon"? I guess I can accept "0-gon", but this looks really weird. What's it supposed to be in Ancient Greek? "Medenagon"? Anyway, digons are way above monogons and 0-gons in legitimacy, being proper abstract polytopes. A regular apeirogon may also be construed as a line subdivided equally into an infinite number of congruent segments. Each interior angle is thus pi, as expected.

A 0-gon would have to have a boundary and no sides, so drawing it on a surface would have to give a single point with the interior being the rest of the surface. This very OR-ish. I have not seen references to 0-gons in respectable sources at all. (Coxeter at least sort of mentions monogons.) Further, it has an unbounded area unless the surface you draw it on is closed, like a sphere.

To me, though, the biggest problem with polygons with 0 or 1 sides is that thry have no 1-polytopes as elements. Double sharp (talk) 18:30, 20 December 2015 (UTC)

Sounds like a joke. A null polytope or nullitope is an abstract polytope with no vertices, so maybe its a null polygon or nulligon? Tom Ruen (talk) 18:39, 20 December 2015 (UTC)

Is there any concept besides Division by zero that does not exist, but gets analysed throughly?

Is there any concept besides division by zero that does not exist, yet has a Wikipedia article?--Jubilujj 2015 (talk) 19:11, 20 December 2015 (UTC)

Bottom type, Russell's paradox, Least interesting number, Penrose triangle, Cantor's diagonal argument, Fermat's Last Theorem, an enormous amount of mathematics is about things that don't exist. Dmcq (talk) 20:47, 20 December 2015 (UTC)

Division by zero is in fact possible in some natural and useful structures, which the article is at pains to explain. --Trovatore (talk) 20:57, 20 December 2015 (UTC)

Once upon a time the idea that a number might exist, but not be the ratio of two integers, seemed totally irrational. Later, a number that was the square root of a negative number was a similarly imaginary concept. In both cases, analyzing these concepts led to important extensions of mathematics by giving rise to the understanding that they were possible after all. --76.69.45.64 (talk) 00:03, 21 December 2015 (UTC)

I am really confused by your question. Are there any concept besides "Division by zero" that does not exists? Dude, there are no concepts that does not exists because by definition, if it is a concept then that concept exists! Concepts exists. You are like asking. Are there any imaginations that are not imagined by any humans? 175.45.116.66 (talk) 05:13, 21 December 2015 (UTC)

Cubic function formula proof

Anyone know How do we solve the cubic $ax^{3}+bx^{2}+cx+d=0$ step-by-step, without skipping any logical steps?

Most proofs I've seen so far always do this, and I end up not believing a word. I just wonder if it's easy to solve just like the quadratic function $ax^{2}+bx+c=0$.

Please find me honest answers because I'm slow. For example, no-one explains why it's allowed to erase $bx^{2}$ somehow out of the equation. יהודה שמחה ולדמן (talk) 19:56, 20 December 2015 (UTC)

See cubic equation#Reduction to a depressed cubic Sławomir
Biały 20:10, 20 December 2015 (UTC)

There is nothing new under the sun. You want a formula for solving the cubic equation like a formula for solving the quadratic equation? Here it is. [Formula for cubic equation]. 175.45.116.66 (talk) 00:55, 21 December 2015 (UTC)

Multiplicatively perfect numbers

A number n is multiplicatively perfect if the product of the divisors of n is equal to n². Is it true that the multiplicatively perfect numbers are just 1, the cubes of primes, and the product of two distinct primes? 172.56.30.117 (talk) 23:54, 20 December 2015 (UTC)

This seems like a homework problem. Hint: if p and q are primes, what are all the divisors of pq? And what are all the divisors of p³? --76.69.45.64 (talk) 00:06, 21 December 2015 (UTC)

More generally, suppose you want to find numbers n so that the product of the divisors is n^k. If n = p^aq^b... then the exponent of p in the product of divisors is (1/2)a(a+1)(b+1)..., the exponent of b is (a+1)(1/2)a(b+1)..., ans so on. So the problem amounts to solving the simultaneous equations (1/2)a(a+1)(b+1)... = ka, (a+1)(1/2)b(b+1)... = kb, etc. But with cancellation these reduce to the same equation (a+1)(b+1)... = 2k. In other words look at every way of factoring 2k and subtract 1 from the exponents to get the possible values for a, b, .... So for k=3 you get numbers of the form p⁵ and p²q; for k=18 you get numbers of the form p³⁵, p¹⁷q, p¹¹q², p⁸q³, p⁵q⁵, p⁸qr, p⁵q²r, p³q²r², and p²q²rs. A corollary is that the product of the factors of n is always a power of n unless unless n is a perfect square. Also he product of the factors of n² is always an odd power of n. --RDBury (talk) 06:38, 21 December 2015 (UTC)

December 21

Why is there no Roman numeral to represent zero?

Why is there no Roman numeral to represent zero? 2602:252:D13:6D70:6533:6D2D:ACBE:8031 (talk) 05:22, 21 December 2015 (UTC)

I don't know why, but see Roman_numerals#Zero. They had the concept of zero, but zero really aids in decimal calculations, so it isn't really applicable to Roman Numerals. Bubba73 ^{You talkin' to me?} 05:33, 21 December 2015 (UTC)

Thanks. Yes, I had read that (which is why a placed the Wikipedia link -- Roman numeral -- in my original question). But what do you mean it is not "applicable" to Roman numerals? How -- for example -- is the number "7" (or VII) applicable? And how is zero not applicable? I don't follow what you mean. Thanks. 2602:252:D13:6D70:F5CC:81BD:3808:6FD5 (talk) 05:42, 21 December 2015 (UTC)

There is the number zero and the numeral for zero. The numeral zero is essential in a Positional number system but Roman Numberals aren't a positional number system. I don't think having a Roman numeral for zero would really help. Bubba73 ^{You talkin' to me?} 06:45, 21 December 2015 (UTC)

Help for what? The OP didn't mention any application of zero; he/she just asked how you represent zero (or more precisely, why you can't). The answer seems to be, as 76.69.45.64 explains below, because the Romans didn't know about zero and therefore didn't feel the need to represent it. The same reason, in other words, that you can't represent −1 or i or the smallest infinite ordinal. --Trovatore (talk) 07:07, 21 December 2015 (UTC)

Bubba is saying, I think, that in our system of numerals you might have need a 0 in any position so you can tell the number 23 from 203 or 2300. In Roman numerals they did not have the concept of positional notation where the same symbol might mean 2 or 20 or 200 or 2000 depending on where it occurs. They wrote 2 as II, 20 as XX, 200 as CC, and so on. So their system did not need zero within a numeral the way ours does.

This is a separate issue from having the concept that zero is a number. In our time someone writing up a report on a train accident might say there were "0 killed, 2 seriously injured, 4 slightly injured, 214 uninjured". The 0 here is just as much a number as the 2, the 4, or the 214. The ancient Romans did not have this concept; it developed later. They would do the equivalent of writing "none were killed" in words. Note that the linked article says what word was used in medieval times, not in ancient times when Roman numerals were invented. Medieval writers might have invented a symbol for zero as a number to be used with Roman numerals, but they didn't, or if some of them did, it didn't catch on. --76.69.45.64 (talk) 06:08, 21 December 2015 (UTC)

OK. That makes some sense. But, I think of two examples that might need further clarification. One: What would they do in a case such as this? In our way, we would write 7 - 7 = 0. They would write VII - VII = _____ (what)? And Two: What would they do for a decimal number that requires a "zero" position? For example, 5.907 (where the zero is positioned between the 9 and 7). Thanks. 2602:252:D13:6D70:B04B:668:A3D0:7138 (talk) 07:10, 21 December 2015 (UTC)