Is there one? What is it?
'tis Ergot's question, but he's lazy, so eh?
i is an imaginary unit, a complex number whose square equals -1.
i, of course, then leads us into entirely new fields of mathematics altogether...
Quote from: Explicit[nK] on January 29, 2006, 05:33 PM
i is an imaginary unit, a complex number whose square equals -1.
Then
i is basically a logic bomb, because there is no way to get x*x to equal -1. =/
Yes there is: i * i
Didn't you pay attention to what we just said?
Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p
Quote from: Joe on January 30, 2006, 07:49 PM
Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p
It shouldn't hurt your head seeing as to how it's pretty straight forward.
Quote from: Joe on January 30, 2006, 07:49 PM
Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p
You don't stick your tongue out when your head hurts. You squint your eyes and clench your jaw. You are obviously just trying to toy with us.
i = sqrt -1
i2 = -1
Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.
With the introduction of the imaginary number i, it can be proven that every polynomial of degree n can be factored as a product of n polynomials of degree 1. It's called the Fundamental Theorem of Algebra.
http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html
http://mathworld.wolfram.com/PolynomialFactorization.html
This is a very important result. It means that, by extending numbers onto 2 axes, algebra is "complete" (purely algebraic equations, such as x^2 + 1 = 0, are solvable).
Quote from: Shout on January 30, 2006, 11:40 PM
i = sqrt -1
i2 = -1
Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.
Assuming that x is being limited to real numbers, this is true. However, if complex numbers are allowed, then your statement goes to shit. Other than that, way to tell Joe off :0
Quote from: Shout on January 30, 2006, 11:40 PM
i = sqrt -1
i2 = -1
Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics
Obviously you have no idea how useful imaginary numbers can be,
especially in applied mathematics
and Real Analysis.
Also, here's something to think about:
(-1)^(1/2) * (-1)^(1/2) = (-1*-1)^1/2 = 1^(1/2) = 1
As far as "imaginary" numbers not existing, and being hard to think about, they "exist" just as much as negative numbers do. You just haven't thought about them that much.
Now prove or disprove that R^n can be related to R^(n-1) in the following fashion:
There exists x in R^(n-1) and 1-1 f(x) s.t. f(x)=r is not a member of R^(n-1) (and is not infinite!).
Furthermore f^(-1)(r)=x
Example:
Find a 1-1 f(x) that could generate an imaginary number relative to Z (call it j).
Hm.
e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0
Somehow I find that wrong... o well.
Quote from: rabbit on February 07, 2006, 07:07 PM
Hm.
e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0
Somehow I find that wrong... o well.
Um.. ln(-(0!)) != -ln(0!). ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi
... + 2pi*k*i ;)
Quote from: dxoigmn on February 07, 2006, 07:56 PM
Quote from: rabbit on February 07, 2006, 07:07 PM
Hm.
e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0
Somehow I find that wrong... o well.
Um.. ln(-(0!)) != -ln(0!). ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi
HAH! I forgot logs don't like negatives...which explains why I confused myself...
Quote from: rabbit on February 08, 2006, 09:05 PM
Quote from: dxoigmn on February 07, 2006, 07:56 PM
Quote from: rabbit on February 07, 2006, 07:07 PM
Hm.
e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0
Somehow I find that wrong... o well.
Um.. ln(-(0!)) != -ln(0!). ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi
HAH! I forgot logs don't like negatives...which explains why I confused myself...
Don't like negatives?
Quote from: rabbit on February 07, 2006, 07:07 PM
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
This transformation is the wrong one.
1) ln(e^pi X) != pi * X, more like pi + ln(X).
2) ln(-X) != -ln(X), more like ln(X) + ln(-1).
And the correct value of ln(-1) is pi*i, of course.
Quote from: Rule on February 08, 2006, 09:49 PM
Don't like negatives?
I've always been taught that ln(t) is undefined for negative values of t, is this wrong?
z = e^(w)
w = u + iv
z = e^(u)e^(iv)
x + iy = e^(u)e^(iv)
re^(itheta) = e^(u)e^(iv)
r = e^(u) ---> u = log r
Also, v = arg (z)
Therefore log (x+iy) = log(z) = log (r) + iarg(z) = log(|z|) + iarg(z)
Therefore log(-1) = log(1) + ipi + 2*pi*k*i, where k is any integer.
The principal branch of log (-1) = log(1) + ipi
Although my comment about the "principal branch" may seem trivial, "branch chasing" of multi-valued functions becomes a very important topic in analysis.
Although, we like to define logarithms unambiguously by imposing the restriction that 0 <= Im[log(z)] < 2*pi.
Edit: Im[], not arg() ;)
Quote from: Yoni on February 11, 2006, 11:11 AM
Although, we like to define logarithms unambiguously by imposing the restriction that 0 <= Im[log(z)] < 2*pi.
Edit: Im[], not arg() ;)
I think the most often used branch of log(z) is the "Principal Branch,"
Log(z) = log(|z|) + iArg(z), where Arg(z) (capital 'a') takes the argument of z
from the set (-pi, pi]. I believe this is also a standard in computer science. One of the reasons we like the argument in (-pi, pi] is because it allows us to define 1-1 analytic inverse trigonometric functions, that take the values we're used to in real analysis.
But there are problems with just using this branch of log(z); this principal branch has a branch cut along the negative real axis (it isn't analytic there). If we want to find the derivative of a complex function at z = x < 0, we have to choose another branch of log.
Also, for various applications, we may want a branch of a function like
(z^2-1)^1/2 analytic outside of the circle |z-i/2| = 1, etc. An immediate application of this sort of thing is "
Keyhole Contour Integration": if you want to find say "Integral(0, 1) x^(alpha-1) * (1-x)^(-alpha) dx" 0< alpha < 1, you'd start by finding a branch of z^(alpha-1)(z-1)^(-alpha) with a cut on the real axis from 0 to 1.
Here are some fairly well written articles on these topics:
Multivalued Functions (http://www.math.ubc.ca/~ward/teaching/m301/m301mval.pdf)
Complex Integration (http://www.math.ubc.ca/~ward/teaching/m301/m301integ.pdf)
Integration of Multi-valued Functions (http://www.math.ubc.ca/~ward/teaching/m301/m301imval.pdf)
Keyhole integration ("residue at infinity") (http://www.math.ubc.ca/~ward/teaching/m301/m301res.pdf)
Not really quite as related, but you might find this interesting after reading the notes titled "Complex Integration":
Tricks for summing series (http://www.math.ubc.ca/~ward/teaching/m301/m301sum.pdf)