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[i2amroy]

As noted the math basics would be to sum the chances for you getting 2, 3, or 4 sixes respectively. Alternatively in this case you could also do 1 (100%) minus the chance for 1 six happening. Cumulative binomial probabilities like this can very quickly become a whole lot of math; as such if you just need a quick answer (instead of needing this for homework/etc. I find that the calculator at stattrek is pretty useful. Alternatively if you are looking for something aimed more specifically at dice you might check out Anydice.

[MrRoboto75]

Yeah, you can also work backwards by finding the inverse

Although in that case, remember to find p for 1 six and 0 sixes.

[Reelya]

I found a real-world use for the binomial stuff we were looking at - explaining how FPTP voting disadvantages third-parties as electorate sizes grow bigger. e.g. a third-party won 1/7 of seats in Canada's election. Canada's electorates are about 110000 people each, vs American Congressional Districts, which are about 750000 people each, so they're about 7 times as big. If you view a Congressional District as basically being 7 Canada-electorate sized areas, and assume that winning a majority of those will win you the Congressional seat, you can work out the odds using the binomial equations of winning 4,5,6 or 7 sub-areas out of 7, at odds of 1/7. I get a figure of 1% for that, which is much worse than the result in the smaller regions - winning 14% of them. This shows that high population by itself skews things towards a two-party system with FPTP voting, without needing any special trickery.

[hops]

How do I solve for the limit as x approaches 0 of cot(2x)cot(pi/2 - x)? I've tried everything, and even the computer can't solve it.

[TheBiggerFish]

Make sure the problem is right?  If Wolfram has no idea, that's generally a bad sign.

[hops]

Plugging it into Wolfram, it certainly does have a solution.

It has a solution from graphing the function. I'm trying to solve for it, because I'm a human being that is incapable of making more than ten complex calculations in my head per second.

[Culise]

Solve it by re-writing it into something considerably easier to solve. Such as something involving tangent.

That's what it sounds like to me as well.

Spoiler: Step 1: Useful Identities (click to show/hide)

For instance, cot(2x) can be simplified using its definition as the reciprocal of the tangent - that is, cot(x) = 1/tan(x).  It's much more pleasant to use the double-angle identities on the result of this rather than on cot(2x) directly. 

On the other side of the multiplication, the cot(pi/2 - x) is actually a co-function identity (or in other words, a reflection over the line x = pi/4 where x is, to be consistent with the original formulation, an angle rather than a cartesian coordinate - you'd more often see it as θ = π/4).  It's much simpler to reduce it by looking up and using that very reflection identity rather than rederiving it, in much the same way one can use the same reflection to shift between sine and cosine (the most common co-function identities).  If it's for classwork, though, certain teachers may dock points for not showing work if those identities haven't actually been taught, though.

EDIT: Apparently Wolfram Alpha is smart enough to simplify it as well, and it accords with my own simplification as well (albeit with the caveat that WA doesn't make note of particular cases where the original function was undefined, such as at 0 and π/2).  My, but this site is quite nifty; I wish it'd been around when I was taking calc, or even in college.

[Dozebôm Lolumzalìs]

Is this proof decent?

Spoiler: big (click to show/hide)

Let Z be a function whose domain is an open set (a, b), with possibly a number, an open or closed set, or a combination of the above removed from the domain. Let x be any number within that set. Z is considered to be "single-peaked" if at most one x exists such that Z(x) is "peaked". This means that functions Z without a number x within their domain such that Z(x) is peaked will still be called "single-peaked"; this is intentional.

Peakedness is defined thus: for any x, Z(x) is peaked if Z is left-peaked and right-peaked at x.

Z is left-peaked at x if:

1. If Z is leftwise continuous at x, then Z is left-peaked at x if all numbers y within a sufficiently small region (x-delta, x) have the property such that Z(y) is less than or equal to Z(x), and the second leftwise derivative of Z at x is not equal to zero.
2. If Z is not leftwise continuous at x:
   a. If the leftwise limit of Z at x exists but is not equal to x, then Z is left-peaked at x if the aforementioned limit is less than Z(x).
   b. If the leftwise limit of Z at x does not exist, because there do not exist any y in the domain of Z within a sufficiently small region (x-delta, x), then W is defined as the greatest number w such that w is within the domain of Z, and w is less than x. Z is left-peaked at x if Z(W) is strictly less than Z(x).

Right-peakedness is similarly defined.

If there exists a number x such that no number greater than x is within the domain of Z, then x is not peaked. If there exists a number x such that no number less than x is within the domain of Z, then x is not peaked.

Any function that is monotonic on a set is single-peaked on that set. Let us assume that the slope is always positive or zero within the domain. Then no number x can have the property than Z(x) is rightwise peaked, since every number f such that f is strictly greater than x must have the property that Z(f) is greater than or equal to Z(x).

Any function with a constant slope of 0 within a domain is single-peaked within that domain, since no number x within the domain can have the property that Z(x) is peaked - Z is continuous, and all numbers x have the property that Z(x) has a second derivative of zero.

Lemma A: Let us consider two domains, A (j, k) and B (l, m), such that Z is single-peaked over each. Let A be monotonic. Let every number within a sufficiently small region of l be greater than or equal to every number within a sufficiently small region of k. ((Let l be greater than or equal to k?)) Then Z is single-peaked over the union of A and B. No number x within A can have the property that Z(x) is peaked when considered within the domain of AUB.

Lemma B: Let us consider two domains, A (j, k) and B (l, m), such that Z is monotonic over each. Let the slope of Z over A be always positive or zero, and let the slope of Z of B be always negative or zero. Then Z is single-peaked over the union of A and B, since it cannot be the case that Z(k) is right-peaked and Z(l) is left-peaked - at most one can be strictly greater than the other. Every other Z(x) such that x is not equal to k or l will continue to not be peaked; we have at most one number such that Z(x) is peaked. QED.

=============

Let us consider a single-peaked function P, whose domain is the open set Q (a, b). There is only one number x such that P(x) is peaked. This x is X.

If we modify the domain by removing a number that is not X, then P will continue to be single-peaked, as we will have divided the domain into two sets, one of which is monotonic and thus single-peaked. Let us call the number which is to be removed from the domain R. Let us assume that R is less than x. Every number t such that t is strictly greater than R and strictly less than x will continue to have the property that P(t) is not peaked, as the right-peakedness has not been altered.

If we modify the domain by removing X, then P will have no number x such that P(x) is peaked. Let us consider the open set (a, X). Every number t within that set will continue to have the property that P(t) is not left-peaked. Then there can be no number t such that P(t) is peaked.

If we modify the domain by removing an open set not containing X, then P will continue to be single-peaked. The two resulting sets will satisfy the requirements for Lemma A. QED.

If we modify the domain by removing an open set containing X, then the two remaining sets will satisfy the requirements for Lemma B. QED.

Therefore, removing points or sets from the domain of a single-peaked function over that domain will result in the said function being single-peaked over the new domain. As "single-peaked" was defined to include functions with no peak on their domain, this does not exclude the possibility of there being no peak.

It's semi-rigorous. That's fine. More rigor is good, but not required. (It's not for an assignment, just a project.)

[MagmaMcFry]

Sorry, but I can't actually find the statement you're trying to prove. Please label your stuff with "Definition", "Lemma", "Claim", "Proof" and such.

[Dozebôm Lolumzalìs]

Oh, geez, I wrote that late at night. Gimme a sec, I'll get it tidied up.

[Helgoland]

Maybe just write straight up tech code. Might be easier to parse.

[Dozebôm Lolumzalìs]

Oops, meant interval instead of set. Thanks for catching that. I got confused because I started talking about the union of two intervals.

[Dozebôm Lolumzalìs]

Also, "within a sufficiently small region" can be simplified by using open balls.

That's what I was going for, but I didn't know the proper terminology. How would I say "within a sufficiently small region" more concisely?

[Dozebôm Lolumzalìs]

What domain are we working over here? R? If that's the case, label your "open set (a,b)" an open interval. Makes things a hell of a lot easier to read.

Also, you don't have to label your open intervals in the way you did. Consider subscripts.

No, R is the point that is removed from the domain. Clunky sentence there, I'll fix that.

[Dozebôm Lolumzalìs]

Ah, THE R. Okay. No, it's not R, it's a (closed? Bounded?) interval.

Wait, I might be misusing domain.

Edit: I meant open, not bounded. That's redundant with closed already included.