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Picture 2: be[x]=ArcCos[Sin[x+Pi/2]]
Picture 1: x-floor{x}
How did I Came up with that?
I'll explain.
The function x - floor[x] = {x} is periodic with period 1. (picture 1)
I wanted to represent {x} with trigonometric functions.
First, ArcCos[Sin[x+Pi/2]]= |x| for x ∈ [-Pi,Pi] and this function is peridic with period 2Pi.
Let ArcCos[Sin[x+Pi/2]]=be[x]. (picture 2)
I reduced be[x] to have period 1:
ber[x]=be[Pi x].
Now you can notice that if x ∈(2n,2n+1) ber[x] ≡ {x} and if x∈(2m-1,2m) 1-ber[x]≡{x} ; n,m∈N. This is ecvivalent to:
Now i have to find function g[x]:
{|g[x]-ber[x]|}≡{x}, so
g[x]=
1. 0, if x ∈(2n,2n+1)
2. 1, if x ∈(2m-1,2m)
If I find function that changes its sign in period of 1, I'll be able to reduce it to g[x].
I can take a periodic function, for example sin[x]. (picture 3)
It has period Pi. We need period 1. So we'll use
sinr[x]=sin[Pi x] (picture 4)
The new function has period 1.
Now we construct function that gives us the sign of sinr[x]:
ss[x]=sinr[x]/(|sinr[x]|) (picture 5)
If x∈(2m-1,2m) ss[x]=-1
If x∈(2n,2n+1) ss[x]=1.
Now the only thing we need is a function q(x) for such
q[-1]=1
q[1]=0
This is linear function:
-a+b=1 && a+b=0 => 2a+1=0 => a=-1/2;b=1/2
so
q[x]=-1/2x+1/2
and q[ss[x]]=g[x] and
(|q[ss[x]]-ber[x]|)={x}
and
Floor[x]=x-(|q[ss[x]]|)
And when you simplify and substitute modulus with sqr(x^2) you should get the second example.
The final result depends on what periodic shall we take.
Here is another:
Here's better function:
I'll try to find function for all x.
Can someone help me?
I found that
Geometric proof
And we can form something like that for definite integrals:
Very good site, Keiichi.
Guess 4:2715
arctan(x) also has two horizontal asymptotes.
But I think that these functions cannot be adapted for this graph, because it is not symmetric.
I can't understand. Thes smallest solution is
{n^2}, n=1
or if n > 1 {i,n^2-i}, n=2.
Triangle MHO is simular to triangle OHB.
Let yr=h. then hr=yr²=x. But
x²=y²+h²
(yr²)²=y²+y²r²
Let f = r². Then
f²=1+f
f²-f-1=0
D=1+4=5
f= (1+sqrt(5))/2
r=x/h=2x/2h=AB/BC=sqrt((1+sqrt(5))/2)
here is a picture:
k+48:
And for k+47
Not I understood that the product of the numbers must be divisible by 10. When i wrote the my post i though that the SUM must be divisible by 10. My mistake.
k+28
My last proof wasn't correct at all.
In it i proved only that
LCM(1^n,2^n, ... ,n^n)/n^2!
But LCM(1^n,2^n, ...n^n) {!=} n!^n
Then I proved that
(n-k)^(n+k)/n^2!
but this wasn't enough.
I found a beautiful number theory proof with prime numbers, floors and an numberic theory theorem:
Let write n! as
Now we will use a theorem from the numeric theory:
The biggest power of the prime number p that divides n! is exactly
Now, at last, we will prove that
Yesterday I solved two of the unsolved problems. When i have time, i'll post the solutions.
x^4*3^x(5+x*ln 3)=x^4*3^x*x(5/x+ln3)=x^5*3^x*(5/x+ln3)
The third level is truely the worst. I've got 4 operations and very unlike logic for solving it. And first I wanted to solve it by mathematical way. You can form a matrc transformations and matric equations but it goes bery slow.
Hm... yes.