Let's discuss Mathematics

[dutchfire]

That answer fails at larger alpha. For example, at alpha > 0.6pi, x>1.

Re: 2) It looks hard to calculate it exactly, the integral is very ugly. It is much easier if you're doing this on a periodically continued square, because then there are only two variables effectively and the answer is 0.38. This is a strict lower bound for your question (although obviously not a very good one).

The question also looks much more tractable if you use another norm, for example the max-norm or the taxi-cab metric l_1 metric. Maybe that way it is possible to derive some other bounds on the answer.

 

[Atticus]

That's what I got too.

It's valid at least for cases alpha < pi/2, which the most brie pieces conform to. If alpha goes near pi, it's impossible to cut it in half from straight side to straight side, as the triangle becomes extremely small.

The problem #2 is from YouTube, which suggested a video posing this question. I haven't watched it yet (to not spoil the fun), but it could be something more accessible. If you use different norms, perhaps Hölder's inequality could give at least some estimate.

 

[LulThyme]

Problem 2 is ill-defined. The answer depends on the distribution you choose for the points. Even "natural" choices may lead to different answers. See

https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)

 

[Atticus]

I don't see how. Here's how I see the problem:

Points = real number pairs in [0,1]x[0,1].
Distance = the usual euclidic distance.
Average of an integrable function over set A is
1/|A| * \int_A f dm, where m = Lebesgue measure.

Thus, the average of the distance is
\int\int\int\int sqrt{ (x_1 - x_2)^2 + (x_3 - x_4)^2 } dx_1 dx_2 dx_3 dx_4.

I suppose you thought that it's the expected value of the distance of two random points?

 

[rhinokick]

Try this - For a project in math class, Anne is making a cover for a circular lounge chair 3 feet in diameter. The finished lounge chair needs to hang down 5 inches over the edge of the chair all the way around. To finish the edge of the cover, Anne will fold under and sew down 1 inch of the material all around the edge. Anne is going to use a single piece of rectangle fabric that is 60 inches wide. What is the shortest length of fabric, in inches, Anne could use to make the cover without needing to attached several pieces of fabric together?

Spoiler :

48 inches - Before doing anything else, make sure you convert all your measurements into the same scale. Because we are working mainly with inches, convert the table with a 3 foot diameter into a table with a (3)(12)=(36) inch diameter.

Now we know that the tablecloth must hang an additional 5+1 inches on EVERY side, so our full length of the tablecloth, in any straight line, will be:

1+5+36+5+1

48 inches.

 

[Timsup2nothin]

I just wanna know how putting this cover over it turned the lounge chair into a table.

 

[Lohrenswald]

feet and inches

I'll try to look at it later

 

[Lohrenswald]

Gotta be honest, I can't parse through the english in that problem

 

[uppi]

Before doing anything else, make sure you convert all your measurements into the same scale.

I thought this thread was supposed to be about mathematics and not conversions between silly units.

 

[Lohrenswald]

you convert by means of mathematic

 

[Robert FIN]

QUESTION: Show that number n^3 - 3n^2 + 2n is divisible by 3. n is integer.

I can't solve it. I tried sth like: the polynome = 3x, x is integer. But not sure how to show what I need by that... help?

 

[warpus]

Assume that it isn't divisible by 3 and try to prove by contradiction. Not sure if that'll work but that's what I would try first.

 

[Samson]

QUESTION: Show that number n^3 - 3n^2 + 2n is divisible by 3. n is integer.

I can't solve it. I tried sth like: the polynome = 3x, x is integer. But not sure how to show what I need by that... help?

Is it:

n^3 - 3n^2 + 2n = n (n - 1) (n - 2)

One of n (n - 1) (n - 2) will always be a multiple of 3.

Spoiler :

How to factorize n^3 - 3n^2 + 2n? Put it into Wolfram Alpha.

 

[Robert FIN]

Thank you. Awesome! The logic works now. Just didn't understand I had to factorize with the zero-points (in English??) method. I still got to think how to proof them that n(n-1)(n-2) is divisible by 3 but I think it will be an easier task, I'll ask if I have to. Thanks.

 

[warpus]

Actually instead of contradiction prove it using induction.

Base case is simple.

Assume true for n, prove for n+1 (induction step):

Also let x = (n+2)

We know that (n)(n-1)(n-2) is divisible by 3
Therefore (x+2)(x+1)(x) is also divisible by 3

Show that (x+3)(x+2)(x+1) is divisible by 3 to prove induction step

= (x)(x+2)(x+1) + (3)(x+2)(x+1)

We know that (x)(x+2)(x+1) is divisible by 3 as per our induction step assumption. (3)(x+2)(x+1) is also divisible by 3, obviously.

Add two numbers divisible by 3 together and the result will also be divisible by 3. (This would have to be proved in some way but I'm lazy)

Therefore (x+3)(x+2)(x+1) is divisible by 3
Therefore (n+1)(n)(n-1) is divisible by 3

Therefore the inductive step is proved.

Therefore n(n-1)(n-2) is divisible by 3

So of this notation is a bit off, you'll have to throw in k here and there and replace some of the n's I think, and expand on some of this, but you get the idea

 

[Kyriakos]

Finished reading a book account of the proof offered in the early 90s for a^x + b^x = c^x not true for x>2 and a,b,c integers (ie so-called 'Fermat's last theorem').
Book wasn't anything particularly good, (popularized account with general info, typically one chapter for each person of interest and his math used on this, and bio stuff). I liked the quest for unifying all math fields, cause i think it is obviously possible to do so (given all math is itself a subset in ways of human thought, and we use just that).
I do wonder what kind of proof Fermat had in mind, though. Maybe going with geometry (infinite collapse, or huge number of sides polygon leading to circle) and probability (since raised powers are directly tied to probability anyway; i suppose this is why x^0 = 1, cause there is only one case of empty set).

 

[Samson]

Finished reading a book account of the proof offered in the early 90s for a^x + b^x = c^x not true for x>2 and a,b,c integers (ie so-called 'Fermat's last theorem').
Book wasn't anything particularly good, (popularized account with general info, typically one chapter for each person of interest and his math used on this, and bio stuff). I liked the quest for unifying all math fields, cause i think it is obviously possible to do so (given all math is itself a subset in ways of human thought, and we use just that).
I do wonder what kind of proof Fermat had in mind, though. Maybe going with geometry (infinite collapse, or huge number of sides polygon leading to circle) and probability (since raised powers are directly tied to probability anyway).

I think I read that book. I have always assumed he was just wrong.

 

[Kyriakos]

I think I read that book. I have always assumed he was just wrong.

Me too. But i just assumed this for the book writer, not Fermat

 

[Ferocitus]

I wrote a memetic algorithm about 20 years ago that found this arrangement
of hulls that minimises the wave-making resistance at one particular speed
(technically, Froude number).
It was surprising at first, but then very obvious after a little thought
and some mathematical squiggling. I love that aspect of AI/nonlinear search.

 

[Kyriakos]

^Not commenting cause i don't know what that is.
But i do love use of computers in graphic representation of math. It is something which allows for a (seemingly) very different source of input/impression, and thus may lead to furthering ideas on math itself (and not just consciously imo).