r/numbertheory

*I know this is for number theory but I didnt find any other community. Please read itt....*
Okay, so I don't know if this is so worthy or not, but trust me, I was very happy to formulate it. I always wonder, since class 5th, if we have a formula for sine that gives output as sin(x) on an input x. Then I grew up and got to know about the Bhaskar approximation. I was amazed. I wanted to make one too upon realizing that no formula after him (yea, I didn't find any...) gives higher precision and is better for computing. Then I learned graphical transformations for my JEE prep and after realizing how I can tranform a degree two or degree four polynomial into a sine wave part, I opened desmos and worked on for next one and a half hours to formulate a graph that coincides almost perfectly with sine wave for x belonging to [0, π].

So I present:

click here for the graph and formula

It may look terrifying at first, but believe me it's not.
For a computer, it is the best possible sine approximation as:

Accuracy: Its mean error is just 0.00091 and it is astonishingly perfect 0, π, π/2 and closer to these poles.
Efficiency: Common! Taylor series may look elegant but it is very heavy for a computer hardware or a GPU. This formula makes it instant.

My previous formula was this:

click here to see my previous work

However it had a little more error than my final one, so I continues perfecting the coefficients. And ofcourse that thing 1.61803...., the golden ratio. Then I realized that this format was correct by what if I replaced phi with something, as whenever I didnt, and tries else, it bursted in waste. So I replaced phi with sqrt(8/π)

Even though many won't be surprised, won't be happy with this, I don't know if people will read this or not, but I just wanted to share this with real people who could understand this. You can also tell me what I can do with this thing now. Thank you for reading and please forgive me if I said or claimed anything wrong. I am a kid. I make mistakes. And my name is Mayank Kumar btw, but it doesn't matter anyway.

A^(n) + B^(n) ‡ C^(n) for all positive integers A,B,C and n, were n>2

A,B,C must be relatively prime for a non trivial solution to any such equation, if any integer solution exists. That requires that one and only one of those bases has at least one factor of 2.

For any two smaller objects (A & B) of order n to be equal in quantity to that of a larger n ordered object (C), the smaller objects when contained within the larger (on the longest line between 2 most distant vertices of the largest object and each of the smaller objects oriented in congruence with those opposite vertices of the largest object) must OVERLAP in a union, the n order quantity of that union (O^(n)) (this happens to be the minimum amount of union possible ) will equal the nth order quantity of the largest object that is disjoint from both of the smaller objects.

It becomes useful to express:

A=a+O

B=b+O

C=a+b+O O being the linear Overlap or C-(a+b)

When A^(n) shares a vertex and orientation with C^(n) and B^(n) is also located within C^(n) at the most distance vertex of C^(n) from that of A^(n) then it could be expressed that the overlap or union O^(n) would be required to equal the disjointed expressed by the binomial expansion of the(a + b)^(n) minus (a^(n)+b^(n).)

For n=2, O^(2)=2ab

For n=3, O^(3)= 3(a^(2)b) + 3a(b^(2)

For n=5, O^(5)=5(a4)b+10(a^(3)(b^2)+10(a^(2)(b^3)+5a(b^(4)

Because of the content of any such algebraic expansion of O^(n) (of particular interest were n is prime) O will be a even quantity and require either the factor a or b to be even (but not both).

That even integer would be the only source of factors of 2 for O^(n) and be required to contain at least n factors of 2. As a result O^(n) would contain 2n factors of 2 requiring that same even integer to also contain 2n factors of 2.

In short, A^(n) + B^(n) ‡ C^(n) under the conditions stated because the union of the two smaller objects when they are contained within the larger object can not have integer equivalence (the same number of factors of 2) to that which is disjoint

r/numbertheory

I am an 11th class student from Goa, India and I have made a sine approximation formula which could be super useful for hardware and GPUs.

Simple proof of Fermat’s last theorem