Spacial Geometry (I think)

[miniGOINGS]

If there was a pyramid made out of blocks, and it was a perfect pyramid;

The 1st layer would be 1 block wide and 1 block long.
The 2nd layer would be 2 blocks wide and 2 blocks long.
The 3rd layer would be 3 blocks wide and 3 blocks long.
etc.

Therefor;

The 1st layer would have 1*1 = 1 block.
The 2nd layer would have 2*2 = 4 blocks.
The 3rd layer would have 3*3 = 9 blocks.
etc.

Therefor;

A pyramid 1 layer high would have 1 block.
A pyramid 2 layers high would have 1 + 4 = 5 blocks.
A pyramid 3 layers high would have 1 + 4 + 9 = 14 blocks.
etc.

This means that;
Layers|Blocks
....1...|....1...
....2...|....5...
....3...|...14..
etc.

...(after a few hours and 2 pieces of paper later)...

The formula I found for connecting the number of layers and the number of blocks is;

{[(a)(2) + 1]^2 + [(a)(2) + 1]}
{----------------------------} * (a/2) = x
{.................6.................... }

Were a = the number of layers and x = the number of blocks.

After a few minutes of simple algebra I was able to shorten it to;

......2a^3 + 3a^2 + a..
x = -------------------
................6..............

My questions are;

1. Does this formula have any practical use?
2. Has this formula already been found?
3. Is it possible to condense the formula even more?

Thanks,

Matt

 

[Johannes91]

2. Has this formula already been found?

Sure, it's well known and the second one here: http://en.wikipedia.org/wiki/List_of_mathematical_series#Sums_of_powers. But good job finding it on your own.

 

[miniGOINGS]

Effffff...

 

[nitrocan]

Yeah, I've known this for a while.

x*(x+1)*(2x+1) / 6

 

[miniGOINGS]

Yeah, I've known this for a while.

x*(x+1)*(2x+1) / 6

So apparently I wasted 4 hours of my life for nothing.

 

[robertpauljr]

No you didn't! Figuring stuff out on your own rather than looking it up is a good thing. We don't always have time to do it. It isn't always practical. But when you have the time and interest to work through something, it isn't a waste of time.

Consider people who figure out how to solve twisty puzzles on their own as opposed to looking up and learning someone else's solution. Is it a waste of time? I think not.

 

[miniGOINGS]

Ohhhhh....

 

[Lucas Garron]

1. Does this formula have any practical use?
2. Has this formula already been found?
3. Is it possible to condense the formula even more?

1. How about orders of pyraminx?
2. Yes.
3. Depends on your density metric.

 

[miniGOINGS]

1. How about orders of pyraminx?
2. Yes.
3. Depends on your density metric.

1. They are not made out of blocks
2. ...ihatemylife...
3. Oh.

EDIT: This isn't spacial geometry, is it?

 

[cmhardw]

There is a very similar, although slightly more complicated due to evenness and oddness, formula for puzzles like these. I took about 3 hours to derive the formula I use, which is pretty well condensed.

I don't understand why you're upset that the formula you discovered was discovered before you. The formula I "discovered" for the puzzle linked to above was discovered before me. Yet I had an absolute blast coming up with it without any help.

If you like deriving formulas on your own, try the puzzle I just linked to. That one was very fun to analyze/look at.

Chris

 

[miniGOINGS]

There is a very similar, although slightly more complicated due to evenness and oddness, formula for puzzles like these. I took about 3 hours to derive the formula I use, which is pretty well condensed.

I don't understand why you're upset that the formula you discovered was discovered before you. The formula I "discovered" for the puzzle linked to above was discovered before me. Yet I had an absolute blast coming up with it without any help.

If you like deriving formulas on your own, try the puzzle I just linked to. That one was very fun to analyze/look at.

Chris

Yea, I guess it was fun. I was like "...just for the heck of it lets saaaaaay 2a-1*a/2..." and that ended up being part of my initial solution. I love analyzing things! It's so much fun!!

 

[nitrocan]

Yeah, I've known this for a while.

x*(x+1)*(2x+1) / 6

So apparently I wasted 4 hours of my life for nothing.

Not at all. One day, you might find something really useful

It's fun even just figuring it out anyway.

 

[miniGOINGS]

So, decided to revive this thread in hope of a response. Recently I've been having a lot of these, like a lot. And just now, I think I figured something out, but I'm not sure if it's right.

So basically, there is a zoo. In the zoo, there are some exhibits with different types of animals. The number of exhibits we will call n. In each exhibit there are the same number of animals z. Each exhibit has a certain number of male animals, x, and a certain number of female animals, y. This assumes that all of the animals can only be one of two genders and not both.

The question is, if one animal was chosen out of each of the exhibits, what is the probablity that at least one is male?

The formula that I found out to solve this is 1 - [y^n / z^n].

Please post if you hava anything to say about this.

 

[qqwref]

The formula that I found out to solve this is 1 - [y^n / z^n].

Seems right to me. It's just one minus the probability that all chosen animals are female.

 

[miniGOINGS]

The formula that I found out to solve this is 1 - [y^n / z^n].

Seems right to me. It's just one minus the probability that all chosen animals are female.

Exactly. It just popped into my head and I have no idea if it has any relevance whatsoever. Now I'm intrigued by how the probability increases according to the number of exhibits. It's kind of like walking half the distance to a destination, and then walking half that distance and so on.