View Single Post 01-24-2010, 03:12 PM   #903
abraxasinas
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Join Date: Dec 2009
Location: Queanbeyan/Canberra; NSW, Australia
Posts: 635 Re: The Occult Reptilian Agenda and the Council of Thuban

Quote:
 Originally Posted by berathebrain In this thread http://projectavalon.net/forum/showthread.php?t=18834 you mention that all perfect numbers are the sums of the ODD NUMBERS CUBED. That is true only if a number is perfect, but the other way around is not necessarily true. Of course; A implies B does not necessarily imply B implies A. All bachelors are Men is true, but does not mean that all Men are bachelors. For example, take first 16 odd numbers and cube them, you will get 130816, which is the next number after 8128. This number is equal to 511*512/2, where 511 is not a Mersenne prime, so this can not be a perfect number. It is true that this number can be expressed as the sum of first 511 numbers, but that doesnt make it a perfect number. Here is the definition of a Perfect number: http://mathworld.wolfram.com/PerfectNumber.html. You have just reiterated what I have stated in the post you mentioned. Please make sure you understand the presented data corrctly, before critisizing it. Interesting formula thou. I have been always fascinated by numbers, and so I did notice the fabulous connection of number 2. Namely, number 2 satisfies the equation x+x=x*x=x^x. I have also noticed that 2^4=4^2. Yes but this follows by definition of 4=2^2; so 4^2=(2^2)^2=2^4 by definition of the exponential functions. So I have wondered If any other numbers satisfies the equation x^y=y^x. So I found a formula to find infinite many numbers that satisfies the above condition. Here it is: x=n*n^(1/(n-1)) y=n^(1/(n-1)) Here n*n^(1/(n-1))=n^[1+1/(n-1)]=n^[n/(n-1)] Then x/y=n^{n/(n-1) -1/(n-1)}=n^{1}=n and so x=ny Taking natural logs lnx=ln(ny) for x=e^[n.lny] or y=e^[lnx/n] This reduces to your initial statement x^y = y^x in taking natural logarithms and using the definition for x=ny ylnx=xlny=y.ln(ny)=ny.lny or ln(ny)=n.lny But ln(ny)=ln(n)+ln(y) by definition of exponential/logarithmic function and so ln(n)+lny=n.lny for ln(n)=(n-1)lny=ln(y)^[n-1] and n=y^[n-1] Now eliminate n=x/y for x/y=y^[(x-y)/y] and x=y^[(x-y+y)/y]=y^[x/y] Raise both sides by the power of y for: x^y = y^x This equation so is INDEPENDENT of n in the definition n=x/y. where n is some real number. If you take n=2 you will get 2^4=4^2. If you take n=3, you will get (3*(3^(1/2)))^(3^(1/2))=(3^(1/2))^(3*(3^(1/2))) and so on. Could you elaborate on this equation. And Notice that when n goes towards number 1, x and y is equal to e=2.7182818284590452353602874713527. Of course, because n=x/y=1 implies x=y for y=e^[lnx]=x=y So now you are using the form n/(n-1)=(n+1-1)/(n-1)=1+1/(n-1), which normalises to the mathematical definition of the exponential function f(n)=(1+1/n)^n in the limit ofr asymptotic approach of n. This also engaes the function f(x)=e^x being its own derivative f'(x)=f(x). Below are some details Dejan. You do make a competent mathematician. I always thought this equation could be used to factorize numbers through some fractal type algorithm Dejan
Consider the following function:

f(n)=(1+1/n)n.

Evaluate for increasing counts n:

f(1)=(1+1/1)^1=2

f(2)=(1+1/2)^2=9/4=2.25

f(3)=(1+1/3)^3=64/27=2.370370...
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f(10)=(1.1)^10=2.59374246........
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f(100)=(1.01)^100=2.704813829...
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f(Infinity)=℮=2.718281828.... This function is limited by a number 'e' for increasing n.

Definition: ℮=lim{1+1/n}n for n->Infinity

But this function approches infinity as n approaches -1 from -1- (from -2 say).
This function is limited by 1 as n approaches 0 from 0+ from above (>1) is UNDEFINED for n=0 and is limited by 1 from below (<1) as n approaches 0 from 0-.
The function is again UNDEFINED for n=-1 and behaves 'complexified irrationally' in the interval from n=-1 to n=0. Say the function assumes the complex value 1/i=i/ii=-i for n=-1/2.
As n decreases from -1 to negative infinity, it decreases from Infinity at n=-1 to e as n approaches large negative values.

f(1/100)=(101)^1/100=1.04723.
f(1/2)=(1+2)^1/2=√3
f(-100)=(1-1/100)^-100=2.732.
f(-10)=(1-1/10)^-10=2.867972.
f(-3)=(1-1/3)^-3=27/8=3.375
f(-2)=(1-1/2)^-2=4
f(-1)=(1-1)^-1=1/0
f(-1/2)=(1-2)^-1/2=1/√-1=1/i=-i

f(0)=(1+1/0)^0=(1+Infinity)^0=Undefined These basic considerations in regard to this (transcendental) number e lead to an infinite series expansion for this number and where factorial function n!=n(n+1)..2.1, relates to a recurrence relation, known as Gamma Function: Gamma(x+1)=x.Gamma(x)=n! as a special case. The Gamma Function is similarly undefined for n=0 and n=-1.

{The Gamma Function is defined in a Euler Integral below and has the value 1 for n=0. The integral evaluates then as a limit for -1/e^x for the interval from Infinity to 0, which is (0+1)=1.

But now we have Gamma(x+1)=0!=x.Gamma(x) and by the recurrence of the integral as shown below, for n=1, we also have Gamma(x+1)=1!=1=x.Gamma(x).
Without evaluating the integral, we find the integrals as relating to each other in the Factorial Function.

I1=1.I0 this is 1.0!=1!
I2=2.I1=2.1.I0 this is 2.1!=2!
I3=3.I2=3.2.1.I0 this is 3.2!=3!
I4=4.I3=4.3.2.1.I0 this is 4.3!=4!
..
In=n.In-1=n! generalising as n.(n-1)!=n!

For the Integration by parts simply use d(uv)/dx =(udv/dx+vdu/dx).
Let u=x^n and v=-e^-x. Then du/dx=n.x^n-1 and dv/dx=e^-x.
Then since Integral (d(uv)/dx)dx=uv; the procedure below follows in writing: uv=Integral{d(uv)/dx}dx=Integral{udv/dx}dx + Integral{vdu/dx}dx.

The integral equation is then: Integral{-udv}=In=uv-Integral{vdu}=uv-Integral{-e^-x.n.x^n-1}dx.

This is an interesting application of integration by parts. Let's have a look at the integral
In = xne-x dx

where n is some non-negative whole number.

This seems to be a candidate for integration by parts. Differentiate the xn and integrate the e-x. This will give us
In = - xne-x + n xn - 1e-x dx

It is an important property of the exponential function that, whatever the value of n, xne-x 0 as x  (ex grows far faster than any power of x). So, if n > 0 the first term on the RHS vanishes at both ends. The other term on the RHS involves an integral which is just the same as our original one with n replaced by n - 1. So we have

In = n.In - 1 for n 1

You grumble that this has not evaluated the integral--it has just turned it into another similar integral. True, but that does actually make it easy to evaluate the original integral. The point is that the above formula is valid for any positive integer n. So, climbing up the ladder, we have

I1= 1.I0I2= 2.I1 = 2.1.I0I3= 3.I2 = 3.2.1.I0I4= 4.I3 = 4.3.2.1.I0

--and so on. You will probably believe me if I say that the overall result is that In = n!.I0.

This has reduced everything to the single problem of evaluating
I0 = x0e-x dx = 1

(as you should check).

So we have obtained the rather nice result that xne-x dx = n!

Now for a generalisation. We had a positive integer n in the integrand as a power. There was actually nothing in our integration procedure (the integration by parts) that relied on the fact that n was a whole number. We could just as well look at the integral
f (a) = xae-x dx

where a is any positive value. Doing integration by parts on this will give

f (a) = a.f (a - 1) if a > 1

as before. If a is a whole number then f (a) = a!, but f (a) makes perfectly good sense for any positive value a. So we have produced a kind of generalisation of the factorial function. We can now talk about the factorial of or . This may sound a rather silly thing to do, but it does actually have wide application.

Traditionally, we do not work with the function f (a) but with the function (a) = f (a - 1). This is known as the Gamma Function, (a) = xa - 1e-x dx, (a + 1) = a. (a)

for a > 0. Of course, we have (1) = (2) = 1.
To satisfy idle curiosity I can tell you that ( ) = } An important consequence then follows for a similar expansion for the exponential function. AA 