is a simple proportion one that is a one to one function of some sort. I dont understand this term.
To say A is poportional to B is to say that there is some constant k such that A=kB. It makes no other statement or assumptions on the behavior of A and B.
With that said I am going to bump this into the Math Forum
I don't want to create a new thread. This thread may be a start for my question:
I'm trying to determine whether there's a relationship between Phi=Golden Mean, and the natural log base=e. If there's a relationship, (other than by a simple constant - not what I'm looking for) it is almost certainly irrational.
I'm running into a brick wall; perhaps because there is no relationship. Can anyone help?
Of course, there's a relationship (there's a "relationship" between any two numbers!) but that's exactly what you say you are not looking for. It's pretty easy to calculate that phi= (1+ sqrt(5))/2 and don't think you'll find any simple relationship between an algebraic number like that and a transcendental number like e.
Some people might have once said that about Pi!
Although this number does not seem to be as important as Pi.
pi is transcendental like e.
You are not going to find a "simple" relationship between an algebraic number and a transcendental number.
Of course, it is possible to make up some formula that changes phi into e. Given any two numbers, it's possible to make up a formula that will change one into the other. That's clearly not what r637h meant.
I would like to avoid number theory (very uncomfortable):
Phi is an algebraic number, granted, and satifies the quadratic, x^2-x-1=0. (Although that contains the irrational square root of 5.)
But e is transcendental, and cannot satisy a quadratic or any other algebraic expression, as far as I know.
But Phi is also an irrational number, which can satisfy a Taylor Series or be expressed as a continued fraction.
I thought a number had to be either irrational or algebraic. Is that the flaw in my thinking?
Or is Phi "unique", in that it is both? Surely not.
Anyway, Pi and e can easily be related, and although the relationship may be complex, can be demonstrated in several series.
Am I running around in circles? Is the reasoning non-sequetur?
No, a number is either "rational" or "irrational".
Both rational and irrational numbers can be "algebraic".
An algebraic number is any number that can be found as a root of polynomial equation with integer coefficients. (If "n" is the lowest possible degree of such a polynomial, the roots are "algebraic of order n".)
Rational numbers are precisely those numbers that are algebraic of order 1: x= a/b if and only if x satifies bx- a= 0, a polynomial equation of order 1 with integer coefficients.
Square root of 2, on the other hand, is not rational but is algebraic of order 2.
Phi, since it satisfies the equation x^2- x- 1= 0, is also algebraic of order 2.
Transcendental numbers are those numbers that are not algebraic of any order. Pi and e are the best known of those, but, technically speaking, "almost all" numbers are transcendental.
To Halls of Ivy:
Thanks. I was confusing "rational" with "algebraic."
Right: Order 2.
Any thoughts on "constructing" a relationship of Phi with e.(other than a simple constant)?
If Pi with e, then why not Phi with e?
Am I meant to be aware of some result which states that there exists no 'simple' relationship between any transcendental number and a non transcental number.
Clearly e^1 = e,
Which gives a simple relationship between e and 1.
I think that we require a working definition of 'simple'.
As it turns out, relating phi to e is simpler than I thought:
phi=2cos(pi/5), etc. and the identity:
e^(i.pi)+1=0, (special case of Euler's Formula, with x=pi)
From there, a simple matter of substitution.
Now, the answer may be a complex number (I haven't figured it out, yet), but at least the relationship is expressed.
Now comes a tougher job: Trying to relate fundamental (or natural) # like i,pi,e,phi, to physical # such as Planck's constant, c, gravity acceleration, Avogadro's Number, etc.
Separate names with a comma.