Studying Mathematics Thread

[GrandizerII]

The area of the upper semicircle? I can do the definite integral from -1 to 1 intuitively, but indefinite integral without substitution I will have to think about for a while. But first, need to go to work now. So later.

Yes, the indefinite integral. If you can do it from -1 to 1, you have part of the basic idea.

Without substitution, I can't get to the expected answer. So show the way, master poly.

[polymath257]

Remember that the 'area function' is an anti-derivative.

So, the definite integral from 0 to x is an area under the circle. Draw a radius to the point on the circle, which divides the area into two pieces.

The lower piece is a triangle. It's area is (1/2) xy=(1/2) x sqrt(1-x^2)

The upper piece is an angular sector. The area is (1/2) rA, where A is the radian measure of the angle in question. But, r=1, and the picture shows that sin(A)=x. Hence,
A=sin^{-1}(x).

Putting the areas together gives an anti-derivative of sqrt(1-x^2) of

(1/2)x sqrt(1-x^2) + (1/2) sin^{-1} (x)

For the general anti-derivative, add +C.

[GrandizerII]

Remember that the 'area function' is an anti-derivative.

So, the definite integral from 0 to x is an area under the circle. Draw a radius to the point on the circle, which divides the area into two pieces.

The lower piece is a triangle. It's area is (1/2) xy=(1/2) x sqrt(1-x^2)

The upper piece is an angular sector. The area is (1/2) rA, where A is the radian measure of the angle in question. But, r=1, and the picture shows that sin(A)=x. Hence,
A=sin^{-1}(x).

Putting the areas together gives an anti-derivative of sqrt(1-x^2) of

(1/2)x sqrt(1-x^2) + (1/2) sin^{-1} (x)

For the general anti-derivative, add +C.

Interesting. Not sure I would've figured this out eventually on my own, but this is refreshing to have a new look on this, instead of always being exposed to the standard solutions available on YouTube and such.

[Fireball]

Here's a study I started on my own well over 25 years ago, but life got in the way. Essentially, I was interested in figuring out a mathematical description for how many ways cubical blocks could be attached to each other, where each block had at least one face attached to another. That is, to predict how many ways for each value of the number of cubes. So, no case where only an edge was the contact element between any two blocks. I was not interested in cases where there was symmetry (i.e., by rotation of the assembly or translation). For example, two blocks can be attached on any one of six faces, each. Boring- it is still just two blocks stuck together! I worked this most of the way through 6 blocks (it's a lot of shapes!) but life got in the way and it sat on a back burner. Then I lost the file containing the configurations. I recently found the file with all the pictures. I'll post page 1 of the set of drawings that I made to represent what I was interested in-




One block has only one configuration
Two blocks have only one configuration
Three blocks have two configurations
Four blocks have eight configurations. That's an additional two pages of pictures that I didn't want to splatter this thread with, unless there was some interest or insight. I know how to do this with geometry for the number of line segments that connect some set of points in a plane, for example, because I've done that (made a proof of the number of line segments) when teaching high school geometry. For all I know, someone has already figured it out. After all, it's been over 25 years since I started looking at it.

BTW, as near as I can tell, 5 blocks have 27 configurations. Maybe there is a cubic equation in there somewhere- the count is

1, 1, 2, 8, 27...  I don't think I've finished with the number of ways 6 blocks can be attached with the stated conditions. I made an earlier stab at using all rotational configurations, but it got out of hand pretty quickly, so I dropped that line of inquiry.

[GrandizerII]

You can post the further images in increments if you want. This is a go-all-out math thread, so why not. I also find it quite nice to think of really neat and "simple" but complex solutions to problems. It's rather peaceful when I personally come up with such neat solutions.

[Fireball]

You can post the further images in increments if you want. This is a go-all-out math thread, so why not. I also find it quite nice to think of really neat and "simple" but complex solutions to problems. It's rather peaceful when I personally come up with such neat solutions.

Yes, I get a great deal of satisfaction when I solve something for myself. Here is the case for four blocks. The 7th and 8th are mirror images; they are not rotationally symmetric.

[polymath257]

You can post the further images in increments if you want. This is a go-all-out math thread, so why not. I also find it quite nice to think of really neat and "simple" but complex solutions to problems. It's rather peaceful when I personally come up with such neat solutions.

Yes, I get a great deal of satisfaction when I solve something for myself. Here is the case for four blocks. The 7th and 8th are mirror images; they are not rotationally symmetric.

OK, there is a fair amount that can be said here.

First, this question makes sense in every dimension. For dimension 1, use line segments (this is a trivial case), for dimension 2, put together squares, dimension 3 uses cubes (this is your case), etc.

It is clear that the number of arrangements for n 'cubes' in dimension d+1 is at least as large as that for n 'lower dimensional cubes' in dimension d: just fatten up any arrangement in dimension d to a similar arrangement in dimension d+1.

So, while the dimension 1 case is trivial, it turns out that even the dimension 2 case is highly non-trivial. The figures in dimension 2 are known as polyominoes.

https://en.wikipedia.org/wiki/Polyomino

Even in the dimension 2 case, the growth rate is exponential. A similar proof works in dimension 3, where, if A_n is the number of arrangements of n cubes, we automatically get

A_n A_m <= A_{n+m}

We can see this by just attaching the 'upper right front' cube of an arrangement of n cubes to the 'lower left bottom' cube of an arrangement of m cubes.

This inequality is enough to prove exponential growth and gives a lower bound. So, based on the 2 dimensional results, the number in 3 dimensions grows at least as fast as (4^n)/n.

I also suspect that the use of 'twigs' can give a lower bound for any dimension, but have not yet worked through that.

I hope this helps!

[Kernel Sohcahtoa]

I've completed my self-study of real analysis, which IMO, is a beautiful and wonderful subject. That said, I'm hungry for more/different mathematical information, so I'm going to start self-studying elementary number theory; this topic seems fascinating to me.  One of my books, Mathematical Proofs by Gary Chartrand, has a brief chapter on number theory, so I'll read this first.  In addition, I've also ordered Elementary Number Theory by Jones and Elementary Number Theory by Dudley.

[Fireball]

Yes, I get a great deal of satisfaction when I solve something for myself. Here is the case for four blocks. The 7th and 8th are mirror images; they are not rotationally symmetric.

OK, there is a fair amount that can be said here.

First, this question makes sense in every dimension. For dimension 1, use line segments (this is a trivial case), for dimension 2, put together squares, dimension 3 uses cubes (this is your case), etc.

It is clear that the number of arrangements for n 'cubes' in dimension d+1 is at least as large as that for n 'lower dimensional cubes' in dimension d: just fatten up any arrangement in dimension d to a similar arrangement in dimension d+1.

So, while the dimension 1 case is trivial, it turns out that even the dimension 2 case is highly non-trivial. The figures in dimension 2 are known as polyominoes.

https://en.wikipedia.org/wiki/Polyomino

Even in the dimension 2 case, the growth rate is exponential. A similar proof works in dimension 3, where, if A_n is the number of arrangements of n cubes, we automatically get

A_n A_m <= A_{n+m}

We can see this by just attaching the 'upper right front' cube of an arrangement of n cubes to the 'lower left bottom' cube of an arrangement of m cubes.

This inequality is enough to prove exponential growth and gives a lower bound. So, based on the 2 dimensional results, the number in 3 dimensions grows at least as fast as (4^n)/n.

I also suspect that the use of 'twigs' can give a lower bound for any dimension, but have not yet worked through that.

I hope this helps!

Interesting. Wiki has an entry on polycubes that shows me that I missed two of the pentacube configurations. Oh, well. I hadn't looked at this in years, but the solution existed before I started working on it. I just didn't know where to look.

[polymath257]

I've completed my self-study of real analysis, which IMO, is a beautiful and wonderful subject. That said, I'm hungry for more/different mathematical information, so I'm going to start self-studying elementary number theory; this topic seems fascinating to me.  One of my books, Mathematical Proofs by Gary Chartrand, has a brief chapter on number theory, so I'll read this first.  In addition, I've also ordered Elementary Number Theory by Jones and Elementary Number Theory by Dudley.

If you have any questions, feel free to ask!