What is that formula for?

and why does your method work. They don't look like they would give the same result.

It worked in all exercises and my teacher then admitted my way is better.

That formula is for machines that heat, like heaters and stuff. W is power/potency, Qq is heat from hot source and Qf is heat from cold source.

I remember last year for freshman Honors Geometry, to save time on any formula equations for polygons >4 sides in length, I created an equation that my teacher couldn't disprove.

Alas, I hated geometry and have forgotten the equation since.

Fuck yes!

I found it on my facebook:
Area of any regular polygon is (1/D)(P^2)√3
Where D is double the amount of sides, and P is the perimeter.

It only works for regular polygons.

Fuck yes!

I found it on my facebook:
Area of any regular polygon is (1/D)(P^2)√3
Where D is double the amount of sides, and P is the perimeter.

It only works for regular polygons.

I'd like to see a proof for that

Yeah, just imagine a circle which is a regular polygon with infinite sides (D = ∞). So 1/∞ = 0 and everything*0 = 0. So it doesn't make sense.

If y'all wanna find the area of a regula' polygon, y'all oughta use these.

S = sides
R = radius to a corner
L = length of one side

Area = 1/2 * R² * S * sin(360/S)

Area = 1/4 * L² * S * cot(180/S)

(cot(x) == 1/tan(x))


Edit:
   You'll have to use l'hopital's rule if you want to check circles.

Hmm, I was fooling around with my math notes when I came across an old proof. I saw that all numbers of the form 4n+2 could not be expressed as the difference of two squares, but all other numbers could. However, I pretty much left it at that. Let me tell you now that there will be a really cool hack with this, but there is more to the story. Anyway, I was looking through another notebook and some of my work on the Putnam two years ago and I saw a familiar proof where I showed that all integers >2 are in at least one pythagorean triple (I had broken down the problem to requiring only this to be true). Anyways, all I did was design a formula where, given a value A, you could get a B and C. Anyways, the process to designing the formula sparked an idea. Here is the awesome math hack:

I had shown that all integers not of the form 4n+2 could be expressed as the difference of two squares, so essentially, I have A²-B²=c. If we look at only odd integers for c, however, we get magic. If c is odd, there is a trivial solution for A and B:
A=(c+1)/2
B=(c-1)/2

But guess what? A²-B²=c can be factored to (A-B)(A+B)=c. DO you know what this means? Here:

• If c is prime, the trivial solution is the only solution
• If c is composite, there is more than one solution

I have been using this to write really fast factoring algorithms in assembly and I have been trying to figure out if there is a polynomial time method to finding if A²-B²=c has more than 1 solution (aside from using the AKS primality test on c). If you can find a way, then you can potentially run a very fast prime testing algorithm. Other than that, since I made this connection, I have been putting tons of energy toward following every path I can with this.

First, let me clarify that I am working solely with positive integers (natural numbers).
These proofs are not very formal, sorry. Typing them is a bit more tedious than writing them on paper

Theorem: All integers except those of the form 4n+2 are the difference of two squares.
Proof:
Here, I will simply show by cases.
Let A be odd, B be even. That is, A=2n+1, B=2m. Then:

• A²-B²=
  
• (2n+1)²-(2m)²=
  
• 4n²+4n+1-4m²=
  
• 4(n²+n-m²)+1=
  
• 4l+1

Now, let A be even, B be odd. That is, A=2n, B=2m+1. Then:

• A²-B²=
  
• (2n)²-(2m+1)²=
  
• 4n²-4m²-4m-1=
  
• 4(n²-m²-m)-1=
  
• 4k-1=
• 4(k-1)+3=
• 4l+3=

If both are even, you have:

• A²-B²=
  
• (2n)²-(2m)²=
  
• 4n²-4m²=
  
• 4(n²-m²)=
  
• 4l

If both are odd, you have:

• A²-B²=
  
• (2n+1)²-(2m+1)²=
  
• 4n²+4n+1-4m²-4m-1=
  
• 4(n²+n-m²-n)+1-1=
  
• 4l

[qed]

These are all the possible case, so you can have 4l,4l+1,and 4l+3 as the difference of two squares.

As a very simple proof that all odd integers can be the difference of two squares, just plug in the trivial solution (c+1)/2 and (c-1)/2 for A and B respectivels. Since c is odd, c+1 and c-1 are even, so you can divide by two. For a process to arrive to this conclusion, forst note that the difference of two consecutive squares is odd. That is:
(n+1)²-n²=n²+2n+1-n²=2n+1

So if you want to find, say, 23 as the difference of squares, 23=2n+1 means n=11. So 12²-11²=23. The powerful result of the math hack I presented says that because 23 is prime, this is the only solution.

Theorem: If c is composite and odd, then there is a non-trivial solution for A²-B²=c.
Proof:
Let c be composite. That is, let c=n*m where n and m are neither 1. Since A²-B²=(A-B)(A+B), let n=(A-B) and m=(A+B). Then:

n+2B=A+B
n+2B=m
2B=n-m
B=(n-m)/2      ;since c is odd, n and m are odd, so n-m is even.
A=(n+m)/2      ;This, too, is an integer.
[qed]

Theorem: If c is an odd prime, then the non-trivial solution for A²-B²=c is the only solution.
Proof:
Assume there is a non trivial solution for A and B, A=D, B=E. This implies that (D-E) and (D+E) are factors of c. Since c is prime, the only factors are 1 and c. Therefore, (D-E)=1, (D+E)=c. This means:

D+E=1+2E, D+E=c
1+2E=c
E=(c-1)/2
D=(c+1)/2

However, this is the trivial solution, so we have met a contradiction. Therefore, the trivial solution is the only solution
[qed]

EDIT: I opened up my notes and followed through some of my work and what I have found so far:
If there is a non-trivial solution for A²-B²=c, then it is of the form (A+2d)²-(B-4e)² where d and e are integers. This will let me speed up my code as I can forget about half or 3/4 of the potential values. I am still working on cutting that down even more :)