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[MagmaMcFry]

Try f(x) = 1 + floor(1/2 + sqrt(2*x)). Then f(1) = 2, f(2) = 3, etc...

[da_nang]

"W(z) is the Lambert W function"

...what is the Lambert W function?

Lambert W function is a multivalued, nonelementary function (curve is more correct if including all real-valued branches) that is the solution to the equation a = be^b where b = W(a).

http://en.wikipedia.org/wiki/Lambert_W

It's not something I'd expect one to come across at any education lower than university.

[ZetaX]

Is there a formula where I pop in n and it pops out the term I want.

From a mathematical perspective, this is a very unsatisfying definition. You need to specify the meaning of "formula", as shown by this valid (¿) solution:
a_0=2*sup_{zeta(s)=0} Re(s), a_{n+1} = a_n+[2n=x^2+x has a solution x \in \IZ]

[Skyrunner]

Well, from a linguistic perspective, it's a pretty clear request f(n) = something, substitute for n and get a result? o_O

In this case, n is the n'th term in the series.

[ZetaX]

If you really want to, you could also expand that recursion into a sum. So what makes it not satisfactory¿ 

[Owlbread]

I've got a problem that's been bugging me for months, I just haven't tried to work it out.

I have a cube. I stack 5 cubes on top of that cube in the shape of a cross. On top of those cubes I stack another 13 in the shape of a square with single cubes protruding above the points of the cross below. That pattern continues until I reach an indeterminate number (say 25 cubes) up in a straight line from the bottom cube. The pattern then reverses, creating a strange plumbob like shape made entirely of solid cubes.

At the centre of this plumbob will be enough space to create a large empty cube or "room". How do I calculate the length and height of one of the inner cube's walls, as if I was looking at them? The size would be cubes.

I've also got another question - how do I calculate the number of cubes I'd need to create the whole plumbob if I only know the length and height of one of the inner walls?

[MagmaMcFry]

I've got a problem that's been bugging me for months, I just haven't tried to work it out.

I have a cube. I stack 5 cubes on top of that cube in the shape of a cross. On top of those cubes I stack another 13 in the shape of a square with single cubes protruding above the points of the cross below. That pattern continues until I reach an indeterminate number (say 25 cubes) up in a straight line from the bottom cube. The pattern then reverses, creating a strange plumbob like shape made entirely of solid cubes.

At the centre of this plumbob will be enough space to create a large empty cube or "room". How do I calculate the length and height of one of the inner cube's walls, as if I was looking at them? The size would be cubes.

I've also got another question - how do I calculate the number of cubes I'd need to create the whole plumbob if I only know the length and height of one of the inner walls?

Let's say the small cubes have length 1, and the center cube is axis-aligned and placed at (0, 0, 0). Then there is a cube at any (x, y, z) exactly when |x| + |y| + |z| < 25. You basically get a voxelated octahedron.
Now let's imagine any axis-aligned cuboid with integer sidelengths (l, w, h) measured in small cubes. This cuboid fits into your shape exactly when |l| + |w| + |h| < 52, which means that the largest axis-aligned cube fitting into your plumbob has a side length of 17 cubelets.

About the other question: The volume of a plumbob with 2n-1 cubes diameter is (4n³-6n²+8n-3)/3.

[Skyrunner]

Can I do the question thing too?
I'm fairly sure this has already been asked, but what is the area of the largest equilateral triangle that can fit into a square which has sides of length n? And how would the triangle be placed? O_o

[MagmaMcFry]

Can I do the question thing too?
I'm fairly sure this has already been asked, but what is the area of the largest equilateral triangle that can fit into a square which has sides of length n? And how would the triangle be placed? O_o

Spoiler: Here it is (click to show/hide)

The really acute angle on the bottom left is 15°, therefore the side length s is n*sec(15°) = n*(sqrt(6)-sqrt(2)) and the area is s²(sqrt(3)/4) = n²(sqrt(12)-3).

[Virex]

Cool, but can you also prove that is the biggest triangle you can fit in there?


Edit: An answer of my own


Let the square be oriented so that one edge is along x and the other along y with x and y unit vectors.


Let the triangle consists of points A, B and C, and therefor of sides AB, AC and BC

Observe that if a triangle does not have all it's corners on the edges of the square, it is possible to translate and rotate the triangle so that only 1 corner is on an edge and then expand the edges of the triangle until one other point ends up on an edge of the square. Repeating this, one eventually ends up with all 3 corners on the square's edges (messy, this may need better proof)


We are then looking for a triangle with a projection along x equal to n and a projection along the y axis equal to n.

Observe that if there is no corner on the side of the square oriented along y, it is possible to rotate the square so that there is a corner on that side. Therefor we can start with point A on the side oriented along y.

We now define AB as starting on A and being at an angle t with x. This means that AC starts at A and is at an angle t+60 with x.

The projection of AB along x is |AB| cos t and it's projection along y is |AB| sin t. Likewise, the projections of AC are |AC| cos t+60 and |AC| sin t+60

Since the triangle is equilateral, |AB| = |AC| = s

Since all three corners of the triangle are on the square's edges, the projection of AB or AC on x must be n. The same holds for the projection on y. (remember we can always reorient and relable the triangle to turn BC[\i] int AB[\i] or AC[\i], so we do not consider it here)

s cos t > s cos t+60 and s sin t < s sin t+60, so

s cos t = n = s sin t+60

cos t = sin t+60

Since 0 <= t <= 90, t = 15 degrees

Now, since the projections of AB and AC are n, A must be at a corner of the square, but also at |x| = 0. Additionally, since we defined t to be positive, this must be the lower corner, meaning that A must be on the origin.

[MagmaMcFry]

Cool, but can you also prove that is the biggest triangle you can fit in there?

Sure I can, watch me.

Let there be a triangle inside a square. We can move this triangle around inside the square until one corner of the triangle is inside a corner of the square. Let A be this corner, and let ABCD be the square and AQR be the triangle, and let AQ'R' be the triangle in my previous post.

Spoiler: Diagram (click to show/hide)

Now either AQ is in ABQ' or AR is in AR'D, which means that either |AQ| <= |AQ'| or |AR| <= |AR'|, which means that AQR cannot be greater than AQ'R', and since AQR was chosen arbitrarily (save for translation), AQ'R' is maximal.

[Another]

A very informal take on this is that the maximal triangle should not be able to move inside the square and for it to not be able to move it has to touch all 4 sides of the square and that is only possible if one of the corners of the triangle is touching 2 sides i.e. is in a corner of the square and the other 2 sides are touching the rest sides which can only happen symmetrically.

[MagmaMcFry]

A very informal take on this is that the maximal triangle should not be able to move inside the square and for it to not be able to move it has to touch all 4 sides of the square and that is only possible if one of the corners of the triangle is touching 2 sides i.e. is in a corner of the square and the other 2 sides are touching the rest sides which can only happen symmetrically.

Actually the not-being-able-to-move is neither required nor sufficient for the maximality of a shape inside another. Examples: If you put an axis-aligned X pentomino into a square, you can't move it, but it would fit better diagonally; on the other hand, a maximal 3:4 rectangle has room to slide in the square.

[Another]

To slide, but not to rotate. And we are not talking about circles - they are a special case.

Is there a case when 1 non-circle figure inside another non-circle figure is maximal but can rotate?

[MagmaMcFry]

There are lots.