Wikipedia:Reference desk/Archives/Mathematics/2015 December 15

= 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.    D b f i r s   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 $$\bigg|a_n-L+b_n-M\bigg| \le \bigg|a_n-L\bigg|+\bigg|b_n-M\bigg| < \varepsilon, \ \forall n > k = \max\{k_{a_n},k_{b_n}\}$$ 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 &epsilon; of L and $$b_n$$ is within &epsilon; of M, then $$a_n+b_n$$ is within 2&epsilon; of $$L+M$$.  S ławomir 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 $$L+M$$.  S ławomir  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 an and bn 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, 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 $$\lim_{n\to\infty}\{a_n+b_n\} \le \lim_{n\to\infty}\{a_n\}+\lim_{n\to\infty}\{b_n\}$$ 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)