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Recursive:
Let use this:
a[i]=F[a[1],a[2],...,a[i-1]] depends of a[1..(i-1)]
We must fund iteractive approach.
What's wrong?
That is computable in a finite number of steps: )acc := 0
k:=1while i<n
{
acc:=acc+1/i^k
k:=k+1
}
in the end, return acc
You have forgotten one line:
while i<n
{
acc:=acc+1/i^k
k:=k+1
i:=i+1
}
First we must define what are the differences between the Iterative and Recursive approach.
Then R-I conjecture must be true.
Oh,so "iterative" meant this?
Then...
Thank you Siva, but it's not the full proof.
Just a joke.
As you see the coordinates of A in the second coordinate system A=={x',y'}II are equal to the coordinates of point A1 =={x',y'}I, which is A, rotated 45deg --clock.
So we must find what will be the coorinates of point A{x,y}, when we rotate it 45deg --clock.
Picture:
"thank us" ???
1. Coordinate transforms:
Let the point A has coordinates {a,b} in coordinate system xOy.
We must find the coordinates of a in coordinate system x'Oy', which is xOy rotated 45° ++clock:
For the first -you're right.
For the second-no.
I'm starting explaining everything.
Here is it:
(x-a1)(x-a2)(x-a3)=-a1 a2 a3 + (a1 a2 + a1 a3 + a2 a3) x - (a1 + a2 + a3) x² + x³.
The coeficients for the same powers of x must be same, so:
a1 a2 a3 = 15 AND
a1 a2 + a1 a3 + a2 a3 = 13 AND
a1 + a2 + a3 = -3
(
|equation1
|equation2
...
|equationm
means equation1 && equation2 && ... equationm
(system)
Happy new year!
Hi. Happy New Year!!!
Try to calculate
I'll find a counter-example
I don't think so.
I use CapturePlus. It's free and it works.
The things that cannot be proved are the axioms.
They are true without proving.
But something very very interesting:
Guedel proof that there doesn't exist a full set of axioms, e.a. there exist a statement that cannot be proved using finite number of axioms.
A factorization in complex numbers:
x^3+3x^2+13x-15=(x-a1)(x-a2)(x-a3)
So
|-a1a2a3=-15
|a1a2+a2a3+a3a1=13
|a1+a2+a3=-3
Yes. Other roots are immagineric.
And how is
∫(x^x)dx?
I think you are right. But in Minklovski's 2D space the length between two points (x1,y1) and (x2,y2) is
L=sqrt(|x1-x2|²+(i|y1-y2|)²)=sqrt(|x1-x2|²-|y1-y2|²)