Wikipedia:Reference desk/Archives/Mathematics/2016 July 6

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July 6

Just pretty...

If loving the bitruncated 5-cell made by cutting a 5-cube in half along the plane perpendicular to the long diagonal is wrong, I don't want to be right.Naraht (talk) 02:30, 6 July 2016 (UTC)

Is there a question? AllBestFaith (talk) 01:56, 8 July 2016 (UTC)

@AllBestFaith: Nope. Just an observation.Naraht (talk) 19:01, 8 July 2016 (UTC)

Winning the most money in a specific lottery drawing

As I'm aware of the miniscule odds of winning anything in a lottery, this isn't an advice request.

Like many other American lotteries, the jackpot for Mega Millions increases whenever a drawing yields no winners, but like many other American lotteries, if two or more tickets are sold with the winning number, the jackpot will be split among the winners. Tonight's news ran a story about people buying large numbers of tickets for the latest drawing, many excited by the unusually high jackpot. This leaves me wondering: for jackpots that increase for lack of winners (whether Mega Millions or other drawings), is it routinely possible to use publicly available information to calculate the jackpot value at which one is likely to win the largest amount of money? Obviously you'll win more money if the jackpot is bigger, but as the jackpot grows, people buy more tickets, so there's also an increase in the chance that a winning number will be chosen by two or more people, so your chances of winning-but-sharing also increase. Nyttend (talk) 03:15, 6 July 2016 (UTC)

That's really more of an application of game theory or decision theory than one of pure number-crunching statistics. Of course, there is an element of probability and statistics, but these are buried in the details of the n other players who also participate. For this type of game with many players, it's impossible to constrain a probabilistic model that could yield an analytically-optimal answer unless you have a fully-constrained model of other people's decision-making. There's a lot of mathematics and statistics that can be thrown at the problem - and it's an active area of applied research, because it has all sorts of practical applications to, say, financial markets - but at the end of the day, those methods all boil down to some kind of heuristic that describes how you expect yourself and the other players to behave. The question devolves more into a theory of crowd psychology than of mathematical probability.

If I may quote a famous text on mathematical economics:

“ Economic theorists, like French chefs in regard to food, have developed stylized models whose ingredients are limited by some unwritten rules. Just as traditional French cooking does not use seaweed or raw fish, so neoclassical models do not make assumptions derived from psychology, anthropology, or sociology. I disagree with any rules that limit the nature of the ingredients in economic models. ” — George Akerlof, An Economic Theorist's Book of Tales

Nimur (talk) 13:34, 6 July 2016 (UTC)

538.com took a shot at figuring this out recently: [1] shoy (reactions) 14:45, 6 July 2016 (UTC)

Thanks for that link. It's precisely what I was seeking: the rate at which purchases rise (thus reducing the chance of a winning number being picked by exactly one person) as jackpots also rise. Just the raw numbers from previous big jackpots, regardless of the crowd psychology that drives those purchases. Nyttend (talk) 16:04, 6 July 2016 (UTC)

Minimum Angle Formed By Incenter & Isogonal Conjugates

What is the minimum angle formed by the incenter with two isogonal conjugates ? (For instance, in the case of centroid-incenter-symmedian point, it is 160°. For orthocenter-incenter-circumcenter, we have 90°). Angles are considered positive and smaller than 180°. — 79.113.212.43 (talk) 03:50, 6 July 2016 (UTC)

How did you do those angle calculations? Loraof (talk) 17:48, 6 July 2016 (UTC)

I don't see how any triangle could have the orthocenter (H)–incenter (I)–circumcenter (O) angle be as low as 90°. H, the centroid G, and O are in that order on the Euler line. The incenter lies in the open orthocentroidal disk, whose diameter is HG. So even if I is arbitrarily close to the edge of the disk, angle HIG would, by Thales' theorem, be arbitrarily close to 90°; thus that angle must be greater than 90° since the incenter is inside the disk. And angle HIO is greater than HIG and hence greater than 90°. Loraof (talk) 18:10, 6 July 2016 (UTC)

Yes, that's what I mean. (Perhaps infimum would be a better word ?). I use GeoGebra. — 79.113.197.130 (talk) 04:00, 7 July 2016 (UTC)

Interesting and difficult question. I decided to check whether it is ever possible to have the angle actually be 90°. For a point P with trilinear coordinates p:q:r, the isogonal conjugate P' is at 1/p: 1/q: 1/r, and the incenter I is at 1:1:1. For a 90° angle at the incenter, we need the line PI perpendicular to the line IP'. By trilinear coordinates#Collinearities and concurrencies, three points P, I, and a variable point x:y:z are collinear if and only if the determinant of the matrix whose rows are the points' respective coordinates is zero. This gives the equation for the line through P and I as

(q–r)x + (r–p)y + (p–q)z =0.

Likewise, the line through I and P', with variable point x:y:z, is (after clearing denominators),

(pr–pq)x + (qp–qr)y + (rq–rp)z = 0.

Now Trilinear coordinates#Perpendicular lines gives the condition for two lines to be perpendicular. For the above two lines this condition becomes

(q–r)(pr–pq) + (r–p)(qp–qr) + (p–q)(rq–rp) = [(r–p)(rq–rp) + (qp–qr)(p–q)] cos A

+ [(p–q)(pr–pq) + (rq–rp)(q–r)] cos B + [(q–r)(qp–qr) + (pr–pq)(r–p)] cos C,

where A, B, and C are the angles of the triangle. Any triangle with angles A, B, C combined with a point p:q:r satisfying this gives a right angle from P to the incenter to the isogonal conjugate of P. But do any solutions of this, satisfying the constraint on the cosines of a triangle (and satisfying p, q, r ≠ 0 as required for isogonal conjugates to be defined), exist?

I decided to consider the special case of an equilateral triangle, in which all angles are 60° and thus all cosines are 1/2. The above equation simplifies to

6pqr – pq² – r²p – r²q –p²q – rp² – q²r = 0.

This does indeed have real solutions—for example, if p/r is very large or very small, the solution for q/r is a real number (though I don't know if there is a solution with all of p, q, r positive, as would be required for a point in the triangle's interior).

So I conclude that, under the strong assumption that I have made no algebra mistakes, there are some points P wrt an equilateral triangle that have PIP' = 90° Loraof (talk) 19:35, 7 July 2016 (UTC)

Apparently, it can get arbitrarily close to 0°, in certain situations where the point is exterior to the triangle, and arbitrarily close to one of its bisectors. So the initial question would have to be amended, by restricting it to the case when the point lies inside the triangle. — 79.118.183.27 (talk) 01:56, 8 July 2016 (UTC)

Why do you say it can get arbitrarily close to 0? Loraof (talk) 03:11, 8 July 2016 (UTC)

It can also be precisely 0°, if that's what you're asking. (Just take the point on the angle bisector, on the other half-plane than the one containing the triangle vertex through which the bisector passes, and it also has to be placed at a certain distance away from its corresponding side). — 86.125.207.132 (talk) 03:24, 8 July 2016 (UTC)

No. See the thread of July 1. If the point is on an angle bisector, the angle will be 180°, because the isogonal conjugate will be on the other side of the incenter. Take a look at isogonal conjugate for how it is constructed—all three lines to vertices are reflected around that vertex's angle bisector. Try drawing that for a point on an angle bisector and you'll see that. Loraof (talk) 03:41, 8 July 2016 (UTC)

What do you think that I've been doing ? The bisector has two halves. On one half, the angle is 180°, since the two conjugates are on opposing sides of the incenter. On the other half, however, the angle is 0°, since the two are on the same side of the incenter. — 86.125.207.132 (talk) 04:35, 8 July 2016 (UTC)

If the point is on an angle bisector, its isogonal conjugate is on the other side of the incenter, never on the same side. Consider a triangle with vertex A drawn at the top of the page, and B and C at the bottom. Suppose point P is on the bisector of A and is below the incenter. First we draw PA, which coincides with the bisector of A, and reflect it about the bisector of A, giving again the line PA. The conjugate P' is on this line. Next, draw PB, and reflect it about the bisector of B, which connects B to the incenter. The resulting line, which also contains the conjugate of P, intersects the bisector of A above the incenter. Likewise when you reflect PC: the reflected line crosses the bisector of A above the incenter, at the same intersection point. This intersection point, above the incenter, is the isogonal conjugate of P. Loraof (talk) 16:35, 8 July 2016 (UTC)

As I said, this is wrong. As you drag the point further and further away from the triangle-side opposite to the angle in question, after a certain while, the angle PIP' goes from 180° to 0°. (Could we please now focus on the case when the point P lies strictly inside the triangle ?) — 86.125.209.240 (talk) 06:34, 9 July 2016 (UTC)

Okay, now I see what you're saying—dragging beyond a certain point causes the reflected line to pass through parallel to the bisector and then beyond. (I would suggest that you create an account name so I'll know who I'm talking to—your original post and the next two were from 79.113... and 79.118..., but your later posts were from 86.125.... Until just now when I read that you mentioned dragging in your last post, I thought I was talking to someone different from the original poster, and I didn't realize I was still talking to someone who was using the graphing software. Your signature will be stable, avoiding confusion, if you get a sign-in name.) I'll think about the interior case some more. Loraof (talk) 14:46, 9 July 2016 (UTC)

It would appear that inside the triangle's circumscribed circle, the smallest angle cannot be lesser than 90°. — 86.125.214.167 (talk) 10:58, 17 July 2016 (UTC)

Resolved

Cubics, rationalizing radicals in denominator and more

1. Can someone tell me how did we add $z=-{\frac {1}{2}}+{\frac {{\sqrt {3}}i}{2}}$ into the equation

$t_{k}=z^{k}{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+z^{2k}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q}{2}}+{\frac {p^{3}}{27}}}}}},\quad k=0,1,2,$

all I know is the most simple solution

$t={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q}{2}}+{\frac {p^{3}}{27}}}}}}$ .

2. Does anyone know how to rationalize the denominator in ${\frac {1}{{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}+{\sqrt[{3}]{c}}}}$ , or even ${\frac {1}{{\sqrt[{n}]{a_{1}}}+\cdots +{\sqrt[{n}]{a_{k}}}}}$ ? יהודה שמחה ולדמן (talk) 07:43, 6 July 2016 (UTC)

1. The number $z=-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i$ satisfies the equation $z^{3}=1$. The equation $x^{3}=a$ has three solutions. If one of them is called ${\sqrt[{3}]{a}}$ then the other solutions are $z{\sqrt[{3}]{a}}$ and $z^{2}{\sqrt[{3}]{a}}$ . See root of unity. Bo Jacoby (talk) 11:07, 6 July 2016 (UTC).

Then why isn't it $z{\big (}{\sqrt[{3}]{a+b}}+{\sqrt[{3}]{a-b}}{\big )}$ and $z^{2}{\big (}{\sqrt[{3}]{a+b}}+{\sqrt[{3}]{a-b}}{\big )}$ ? יהודה שמחה ולדמן (talk) 11:16, 6 July 2016 (UTC)

Because those numbers don't satisfy the original equation $x^{3}+px+q=0$. Bo Jacoby (talk) 16:26, 6 July 2016 (UTC).

If the equation were as simple as $x^{3}=a$ that's what it would look like. But for the equation $x^{3}+px+q=0,$ which the indicated solution applies to, if you substitute your proposed solution into the original equation, you'll obtain a contradiction, rather than an identity. Loraof (talk) 16:23, 6 July 2016 (UTC)

This is discussed in Cubic function#Cardano's method, from right before equation (4) to the paragraph after equation (5). Loraof (talk) 16:35, 6 July 2016 (UTC)

2. ${\frac {1}{{\sqrt[{3}]{a}}+n}}={\frac {\prod \limits _{i=1}^{2}(z^{i}{\sqrt[{3}]{a}}+n)}{\prod \limits _{i=0}^{2}(z^{i}{\sqrt[{3}]{a}}+n)}}$ . The denominator does not contain ${\sqrt[{3}]{a}}$ . Bo Jacoby (talk) 16:26, 6 July 2016 (UTC).

The equations are

$t^{3}+pt+q=0$ and $z^{3}=1$ .

The 3 solutions are

$t=z{\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+z^{2}{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q}{2}}+{\frac {p^{3}}{27}}}}}}$

for

$z=1$ ,

$z=-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i$ ,

and

$z=-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i$

Bo Jacoby (talk) 16:46, 6 July 2016 (UTC).

2.

${\begin{aligned}{\frac {1}{{\sqrt[{n}]{a_{1}}}+\cdots +{\sqrt[{n}]{a_{k}}}}}&={\frac {1}{\sum \limits _{i=1}^{k}{\sqrt[{n}]{a_{i}}}}}\\&={\frac {\prod \limits _{j_{1}=0}^{n-1}\cdots \prod \limits _{j_{k}=0}^{n-1}(\sum \limits _{i=1}^{k}z^{j_{i}}{\sqrt[{n}]{a_{i}}})^{[0\neq \sum \limits _{i}j_{i}]}}{\prod \limits _{j_{1}=0}^{n-1}\cdots \prod \limits _{j_{k}=0}^{n-1}\sum \limits _{i=1}^{k}z^{j_{i}}{\sqrt[{n}]{a_{i}}}}}\end{aligned}}$

where [true]=1 and [false]=0 is the Iverson bracket, and where $z=e^{\frac {2\pi i}{n}}$ is a primitive n'th root of unity. Bo Jacoby (talk) 12:36, 7 July 2016 (UTC).

I think you're doing a bit a overkill in the products (in #2). The denominator is degree n^k in the ⁿ√a_i's which would make it degree n^k-1 in the a_i's. But if you leave out the first product the product is still a polynomial in the a_i's but of degree only n^k-2. For example ³√a+³√b+³√c | (a+b+c)³-27abc which is degree 3=3^3-2 in a, b, and c. --RDBury (talk) 07:24, 8 July 2016 (UTC)

I'd like to see a nice version of the above formula that did it right including for the case ${\frac {1}{\sqrt[{n}]{a_{1}}}}$ where there is only one term in the denominator. Dmcq (talk) 22:07, 8 July 2016 (UTC)

Just let k=1 in the above formula.

${\begin{aligned}{\frac {1}{\sqrt[{n}]{a}}}&={\frac {\prod \limits _{j=1}^{n-1}z^{j}{\sqrt[{n}]{a}}}{\prod \limits _{j=0}^{n-1}z^{j}{\sqrt[{n}]{a}}}}\end{aligned}}$

Bo Jacoby (talk) 20:01, 10 July 2016 (UTC).

Pretty sure a nice version, i.e. closed form, doesn't exist. For square roots, the numbers of terms in the rationalized denominator is: √a: 1, √a+√b: 2, √a+√b+√c: 6, √a+√b+√c+√d: 35, √a+√b+√c+√d+√e: 495 (unless there's some unexpected cancellation), etc. The formula for the number of terms is (2^k-2+k-1 choose k-1), again assuming there is no cancellation which seems to be the case. There might be a determinant formula but the size of the matrices would increase exponentially. Kind of proves that rationalizing the denominator is computationally impractical except for small n and k. --RDBury (talk) 08:11, 9 July 2016 (UTC)

Conformal geometry with the vector space model of geometric algebra

I am well aware of the conformal model of geometric algebra. But I was thinking, couldn't we do conformal geometry within the vector space model?

Euclidean transformations can be described by vector addition and rotor multiplication for translations and rotations respectively. Spherical inversion can be achieved by taking the multiplicative inverse of vectors. Does this work, and if so, are the any weaknesses with this approach?--Leon (talk) 11:49, 6 July 2016 (UTC)

I see no problem with the logic of this approach. However, it does have some drawbacks relative to conformal geometric algebra:

• It is algebraically more complicated: you have to track and algebraically manipulate the different types of transformation using different operations, whereas these all form part of the (obviously associative) composition of conjugation in CGA. Compare a rotation around an arbitrary point being an addition, rotation, addition sequence to a single conjugation in CGA. In CGA there is some added complexity hidden under the hood: two extra dimensions, translations to and from the vector model.
• Certain useful objects are naturally covered by the same operation (flats and k-spheres); these must each have special handling in the vector space model. Finding such objects given others (intersection, and finding the object given points on it) is simple in CGA.

So, basically it comes down to algebraic simplicity, as far as I can see. —Quondum 20:08, 6 July 2016 (UTC)