Definition of the imaginary unit
matt
Recently, while contemplating the meaning of ... the meaning of ...
... erm ... oh yes ... the meaning of exponentiation in the domain of
complex numbers, a thought suddenly struck me.
I have found nothing through google, nor in any of the mathematical
books I have propping up my wobbly table, so maybe someone can
enlighten me.
It's usually said that i is defined as the square root of -1. But we
know that the equation
x^2 = -1
has TWO solutions, which we would write as i and -i.
My question therefore is:
1. How do we know which value to designate i, and which to designate
-i?
2. If we DON'T know, then almost any expression involving i would seem
to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is
there any easy way to understand why this never matters?
Virgil
matt271...@yahoo.co.uk (matt) wrote:
The mapping x + i*y -> x - i*y is a field isomorphism, so that, as far
as the field of complex numbers is concerned, there is no difference
that makes any difference.
Arturo Magidin
matt <matt271...@yahoo.co.uk> wrote:
>Hello again
>
>Recently, while contemplating the meaning of ... the meaning of ...
>... erm ... oh yes ... the meaning of exponentiation in the domain of
>complex numbers, a thought suddenly struck me.
>
>I have found nothing through google, nor in any of the mathematical
>books I have propping up my wobbly table, so maybe someone can
>enlighten me.
>
>It's usually said that i is defined as the square root of -1. But we
>know that the equation
>
> x^2 = -1
>
>has TWO solutions, which we would write as i and -i.
>
>My question therefore is:
>
>1. How do we know which value to designate i, and which to designate
>-i?
We don't. But in a very real sense "it doesn't matter."
Suppose martians are all left handed, and so when they came up with
complex numbers, what they designated as "i" is what ->we<- would
designate as "-i". There is then a perfect translation between what
the martians will do and what we do. It is called "complex
conjugation."
Namely, complex conjugation is a bijection function from C to C which
has the following properties:
(i) f(x+y) = f(x) + f(y) for all complex numbers x and y;
(ii) f(xy) = f(x)f(y) for all complex numbers x and y;
(iii) f(r) = r for all real numbers r;
(iv) f(i) = -i.
This automorphism of the complex numbers tells you that, regardless of
which of the two roots of x^2+1 you choose to designate by "i" and
which you choose to designate by "-i", everything you can do with one
choice you can also do with the other choice. The two choices yield
algebraically indistinguishable constructions.
>2. If we DON'T know, then almost any expression involving i would seem
>to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is
>there any easy way to understand why this never matters?
No, it is not ambiguous. Choose your favorite one to be i; the only
other choice will be algebraically indistinguishable, and therefore
there is no problem with the arbitrariness of this choice. Just
reflect everything about the X axis if you made the other choice.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Ken Pledger
> ....
> It's usually said that i is defined as the square root of -1. But we
> know that the equation
>
> x^2 = -1
>
> has TWO solutions, which we would write as i and -i.
>
> My question therefore is:
>
> 1. How do we know which value to designate i, and which to designate
> -i?
>
> 2. If we DON'T know, then almost any expression involving i would seem
> to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is
> there any easy way to understand why this never matters?
That's a good question, which has come up before. Here's what I
wrote about it in the sci.math news group in july 2003.
> At 11.28 AM +1200 21-7-03, Ken Pledger wrote:
>
> In article <bfeqqi$pam$1...@green.tninet.se>, "Mattias Wikstr–m"
> <mattias...@hotmail.com> wrote:
>
> > There is an oddity in the way complex numbers are introduced....
> > -1 has two square roots, which are conjugates of each other, but
> > only one of them is given its own symbol. Why pretend there is asymmetry
> > when there is none?
> >
> > If we accept that there is complete symmetry, then we are faced with a
> > strange consequence: we cannot name any single non-real complex number....
> > with complex numbers we can only name pairs of numbers, such as the pair
> > that satisfies the equation x^2+1=0....
>
> Some respondents have seized upon details of what you wrote, and
> perhaps evaded the real issue. I agree with your main concern, and I
> think the answer lies in the connection between mathematics and our
> physical diagrams.
>
> As you suggest, complex conjugation is an automorphism of the complex
> field, leaving invariant every real number, but swapping i with -i.
> That means that most algebraic statements we make about complex numbers
> remain true if i and -i are transposed, so how do we know which is
> which? The mathematics assures us that the equation z^2 + 1 = 0 has two
> solutions, but it treats them even-handedly, giving us no way to stick a
> special label on one of them rather than the other.
>
> The down-to-earth everyday answer is that our diagrams represent the
> real line horizontally with positive numbers to the right, and the
> imaginary axis perpendicular to it with i above and -i below. That
> depends upon our notions of right and left (or clockwise and
> anti-clockwise), which we acquired as small children by being shown
> something *physical*. You can't define right or left mathematically. A
> mathematical argument (perhaps about real vector spaces) can tell you that
> there are two sides, but it treats them even-handedly, giving us no way to
> stick a special label on one of them rather than the other. I've repeated
> those words, to suggest that your question about i versus -i is pretty
> close to the more basic question of right versus left.
>
> If you're doing really pure mathematics, then calling certain complex
> numbers i and -i isn't going to cause any trouble. You never *need*
> to decide which is which. But if you're a normal human being who uses a
> physical plane diagram to help in thinking about complex numbers, then you
> put 1 on the right, -1 on the left, i above and -i below. It's
> as humdrum as that.
>
> Ken Pledger.
>
William Elliot
> matt271...@yahoo.co.uk (matt) wrote:
>
> > It's usually said that i is defined as the square root of -1. But we
> > know that the equation
> > x^2 = -1
> >
> > has TWO solutions, which we would write as i and -i.
> >
> > My question therefore is:
> > 1. How do we know which value to designate i, and which to designate
> > -i?
> >
i = x + (x^2 + 1)
-i = -x + (x^2 + 1)
Naturally, that's how.
Otherwise you'd get cross eyed looking at
i = -x + (x^2 + 1)
-i = x + (x^2 + 1)
matt
I think I get what you're saying. Let me just explain it back in a
slightly different way to make sure:
"Any true (false) mathematical statement involving i will be equally
true (false) if i and -i are interchanged."
Is that correct?
The only potential counterexample I can think of so far is
arg(i) = pi/2 (1)
Swapping i and -i, you get
arg(-i) = pi/2 (2)
which is wrong.
Although this looks to be a counterexample, I THINK there must be a
subtle reason why it isn't. Maybe the DEFINITION of the arg function
depends on the choice of i vs -i, so the arg function in (1) is not
the same as the arg function in (2)?
[After writing this, have just read Ken Pledger's response, which
includes:
"... most algebraic statements we make about complex numbers remain
true if i and -i are transposed..."
Almost the same as mine, except that I have "all" and KP has "most"!!]
Arturo Magidin
No. Any ->algebraic<- statement. The complex numbers and the complex
numbers after conjugation are ALGEBRAICALLY indistinguishable. I said
that many times. What made you think that the word "algebraically" was
superfluous?
>The only potential counterexample I can think of so far is
>
> arg(i) = pi/2 (1)
>
>Swapping i and -i, you get
>
> arg(-i) = pi/2 (2)
>
>which is wrong.
Because "arg" is a function which is not defined algebraically.
>Although this looks to be a counterexample, I THINK there must be a
>subtle reason why it isn't. Maybe the DEFINITION of the arg function
>depends on the choice of i vs -i, so the arg function in (1) is not
>the same as the arg function in (2)?
>
>[After writing this, have just read Ken Pledger's response, which
>includes:
>
>"... most algebraic statements we make about complex numbers remain
>true if i and -i are transposed..."
>
>Almost the same as mine, except that I have "all" and KP has "most"!!]
You are doing the same thing you did with the exponentials thread: you
read a response, drop all sort of assumptions, and then wonder why
your conclusions are false after you dropped the bunch of
conditions.
mathedman
wrote:
>Hello again
>
>Recently, while contemplating the meaning of ... the meaning of ...
>... erm ... oh yes ... the meaning of exponentiation in the domain of
>complex numbers, a thought suddenly struck me.
>
>I have found nothing through google, nor in any of the mathematical
>books I have propping up my wobbly table, so maybe someone can
>enlighten me.
>
>It's usually said that i is defined as the square root of -1. But we
>know that the equation
>
> x^2 = -1
>
>has TWO solutions, which we would write as i and -i.
HUH???
The complex plane is the ordinary x-y plane with a definition of a
multiplication and addition operations which make it an algebraic
field.
The Number "i" (Europeans largely use "j" so as to mimic vectors
in the x-y plane) is the complex number (0,1). "-i" is the complex
number (0, -1) . When we casually say the "square root of -1", as we
do for folks who have never known about the complex plane,
we mean the square root of the complex number (-1, 0). It has those
two complex-number roots named above. And surely one can tell which is
which!
Once logarithms are defined in this plane [and ln(z) is NOT a
function but an infinitely many valued relation], exponentiation is
defined by:
z^w = e^(w*ln(z)) . This MAY have infinitely many values or a
finite number or a single value depending upon the values of z and w.
Arturo Magidin
mathedman <math...@hotmail.CUT.com> wrote:
>On 21 Nov 2004 14:49:31 -0800, matt271...@yahoo.co.uk (matt)
>wrote:
>
>>Hello again
>>
>>Recently, while contemplating the meaning of ... the meaning of ...
>>... erm ... oh yes ... the meaning of exponentiation in the domain of
>>complex numbers, a thought suddenly struck me.
>>
>>I have found nothing through google, nor in any of the mathematical
>>books I have propping up my wobbly table, so maybe someone can
>>enlighten me.
>>
>>It's usually said that i is defined as the square root of -1. But we
>>know that the equation
>>
>> x^2 = -1
>>
>>has TWO solutions, which we would write as i and -i.
>
>
>HUH???
>The complex plane is the ordinary x-y plane with a definition of a
>multiplication and addition operations which make it an algebraic
>field.
This is not the only way to define the complex numbers. One can also
define them to be equal to R[x]/(x^2+1), for example. In that case,
the question not only makes perfect sense, but understanding the
answer is the first step along the way to Galois Theory.
> The Number "i" (Europeans largely use "j" so as to mimic vectors
>in the x-y plane)
I didn't know this was a "European vs. non-European thing". Are you
sure it's not just some field vs another? I think some engineering
schools prefer to use j (since the use i for indices)...
> is the complex number (0,1). "-i" is the complex
>number (0, -1) . When we casually say the "square root of -1", as we
>do for folks who have never known about the complex plane,
>we mean the square root of the complex number (-1, 0). It has those
>two complex-number roots named above. And surely one can tell which is
>which!
And, if you start from R[x]/(x^2+1), can you tell me please whether
x(x^2+1) corresponds to i or to -i? And why?
The World Wide Wade
> It's usually said that i is defined as the square root of -1.
That is never said in any well grounded introduction to complex numbers;
furthermore it doesn't make mathematical sense. As mathedman said, the easy
way out (which goes back to Gauss I believe) is to define a multiplication
on pairs of real numbers (x,y), identify the reals with the pairs (x,0),
and then define i = (0,1). It is then a theorem that i^2 = -1.
Ladnor Geissinger
any particular naming of a "complex square root of -1" if we point out
the many different possibilities for identifying such roots in
concrete situations.
For example, in the real algebra R^(2X2) of 2X2 matrices, we might
choose J=[0,1; -1,0], the standard matrix of a rotation thru -pi/2,
and then notice that J^2=-Id (I write Id for the identity matrix) so
that the subalgebra generated by Id and J, the set of all c*Id+d*J for
real c,d is a field which is just like the complex numbers. Here J
and -J are the complex square roots of -Id. But of course there are
many more such matrices K such that trace(K)=0 and det(K)=1, i.e. so
that K^2=-Id. In fact, any such K is gotten as [a,b; -(1+a*a)/b, -a]
for any a and any non-zero b. The subalgebra generated by Id and any
such K is a copy of the complex numbers and has as its "square roots
of negative numbers" the "pure imaginaries of form d*K", and in
particular the square roots of -1 are K and -K. In each of these
fields we have complex conjugation taking K to -K and we have the
usual notion of absolute value |c*Id+d*K|^2=(c*c+d*d)=det(c*Id+d*K),
etc. However, the only cases where conjugation actually coincides
with the familiar transpose operation on matrices is when K=J or K=-J,
so this is a specially nice choice.
Consider another setting. We know the geometry and algebra of
addition and multiplication on a Cartesian line, which we identify
with the real numbers R. Then we extend this to the Cartesian plane
and still think of R as sitting inside R^2 as those points with
coordinates (a,0) - the "horizontal axis". For geometric purposes we
find it useful to extend addition in R to the usual vector addition in
R^2. And multiplication on the line R we extend to sclar
multiplication a*(c,d)=(a*c,a*d) which we could also write as
(a,0)*(c,d)=(a*c,a*d). And we know that it is easy to extend this
multiplication so as to make the plane into (a copy of) the complex
numbers, just by defining (0,1)*(0,1)=(-1,0) and then making
multiplication bilinear (i.e. distributive properties). Then the
square roots of -1 are (0,1) and (0,-1). But we could extend scalar
multiplication in many other ways to get the plane as a copy of the
complex numbers, by choosing any vector (a,b) with b non-zero and make
its square be (-1,0). Then the two square roots of -1 are (a,b) and
(-a,-b) and all the points (t*a,t*b) are the pure imaginaries in that
copy of the complex numbers. Note however, that the absolute value in
such a complex field will be the same as the ordinary Euclidean
distance in the plane if and only if (a,b) is (0,1) or (0,-1). So
this special choice has this desirable geometric property as well as
the simplicity of the formula for computing products.
Ken Pledger
mag...@math.berkeley.edu (Arturo Magidin) wrote:
> In article <41a209d1...@netnews.worldnet.att.net>,
> mathedman <math...@hotmail.CUT.com> wrote:
> ....
> > The Number "i" (Europeans largely use "j" so as to mimic vectors
> >in the x-y plane)
>
> I didn't know this was a "European vs non-European thing". Are you
> sure it's not just some field vs another? I think some engineering
> schools prefer to use j (since they use i for indices)...
For instantaneous current, I think. That's why "j" is often
preferred when complex numbers are used in A.C. circuit calculations.
Ken Pledger.
George Cox
>
> ...
> The Number "i" (Europeans largely use "j" so as to mimic vectors
I don't think so. Electrical engineers use "j" so that it doesn't get
confused with a common name for current.
matt
> In article <ed8acaa4.04112...@posting.google.com>,
> matt <matt271...@yahoo.co.uk> wrote:
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<cnr85g$4ld$1...@agate.berkeley.edu>...
> >> In article <ed8acaa4.04112...@posting.google.com>,
> >> matt <matt271...@yahoo.co.uk> wrote:
...
> >
> >"Any true (false) mathematical statement involving i will be equally
> >true (false) if i and -i are interchanged."
> >
> >Is that correct?
>
> No. Any ->algebraic<- statement. The complex numbers and the complex
> numbers after conjugation are ALGEBRAICALLY indistinguishable. I said
> that many times. What made you think that the word "algebraically" was
> superfluous?
>
The answer is simple, my friend. To you, the word "algebraically"
obviously has a precise technical meaning which is essential to the
point you are making. To me, the word "algebraically" merely conveyed
the idea of "some collection of symbols with a mathematical meaning"
and did not register as being particularly important.
matt
> On 21 Nov 2004 14:49:31 -0800, matt271...@yahoo.co.uk (matt)
> wrote:
>
> >Hello again
> >
> >Recently, while contemplating the meaning of ... the meaning of ...
> >... erm ... oh yes ... the meaning of exponentiation in the domain of
> >complex numbers, a thought suddenly struck me.
> >
> >I have found nothing through google, nor in any of the mathematical
> >books I have propping up my wobbly table, so maybe someone can
> >enlighten me.
> >
> >It's usually said that i is defined as the square root of -1. But we
> >know that the equation
> >
> > x^2 = -1
> >
> >has TWO solutions, which we would write as i and -i.
>
>
> HUH???
> The complex plane is the ordinary x-y plane with a definition of a
> multiplication and addition operations which make it an algebraic
> field.
> The Number "i" (Europeans largely use "j" so as to mimic vectors
> in the x-y plane) is the complex number (0,1).
In the UK at least I think that "i" is pretty much universal in maths.
I think that "j" might be used in physics possibly, but I know even
less about physics than I do about maths so that's not saying much...
> "-i" is the complex
> number (0, -1) . When we casually say the "square root of -1", as we
> do for folks who have never known about the complex plane,
> we mean the square root of the complex number (-1, 0). It has those
> two complex-number roots named above. And surely one can tell which is
> which!
How, exactly?
Mattias Wikstr?m
> It's usually said that i is defined as the square root of -1. But we
> know that the equation
>
> x^2 = -1
>
> has TWO solutions, which we would write as i and -i.
>
> My question therefore is:
>
> 1. How do we know which value to designate i, and which to designate
> -i?
>
> 2. If we DON'T know, then almost any expression involving i would seem
> to be ambiguous (except, obviously, things like i^2, i^4 etc.) Is
> there any easy way to understand why this never matters?
The symmetry between i and -i is essentially the symmetry between
clockwise and anti-clockwise. You have this symmetry in any plane.
Here is a way to see what this symmetry means. Suppose I turn a bike
upside down, rotate one of its wheels, and ask if the wheel is
rotating clockwise or anti-clockwise. The answer depends on which side
we choose to look at the bike from, so the wheel in itself is rotating
neither clockwise nor anti-clockwise, it is simply rotating. Suppose
now that I rotate both wheels of the bike and ask if they are rotating
in the same direction. The answer is either yes or no. Suppose finally
that in addition to rotating a wheel on this bike, I rotate a wheel on
another bike. I then ask if the wheels are rotating in the same
direction. Unless the bikes are aligned somehow, this question will be
meaningless.
In my opinion, one should stop speaking about /the/ complex plane and
/the/ imaginary unit and instead speak about complex planes (in
plural), each having two imaginary units. Mathematical objects are
only determined up to isomorphism, and one should not pretend
otherwise.
Some things to note:
* Although the square roots of -1 have identical properties in C, they
have different properties in Z_(4k+1), where k is any positive
integer. For example, in Z_5 we have (2)(2)=4=-1 as well as
(-2)(-2)=4=-1, but 2 and -2 have different properties in Z_5.
* Although sqrt(2) and -sqrt(2) have different properties (sqrt(2)>0
while -sqrt(2)<0), if we forget about the ordering properties of the
real numbers and just consider the algebraic properties (i. e.,
properties that can be expressed in terms of addition and
multiplication), the two square roots of 2 have identical properties.
For example, both are roots of the equation x^2-2=0, and neither is a
root of the equation x^2+1=0.
* Some expressions, like (exp(ix)-exp(-ix))/(2i) at first appear to
contain a reference to an imaginary unit i of a complex plane, but can
be rewritten into a form where no imaginary unit appears (the
expression above can be simplified into sin(x)). The imaginary unit
works like a bound (a.k.a. apparent or dummy) variable in this case,
and if we take into account that in (Integral from x=a1 to x=a2 of
f(x))+(Integral from x=b1 to x=b2 of g(x)), x in the first integral
has nothing to do with x in the second integral, it seems we should
say that in (i/i)+(i/i), i in the first parenthesis has nothing to do
with i in the second parenthesis.
Mattias
matt
> In article <41a209d1...@netnews.worldnet.att.net>,
> mathedman <math...@hotmail.CUT.com> wrote:
> >On 21 Nov 2004 14:49:31 -0800, matt271...@yahoo.co.uk (matt)
> >wrote:
> >
> >>Hello again
> >>
> >>Recently, while contemplating the meaning of ... the meaning of ...
> >>... erm ... oh yes ... the meaning of exponentiation in the domain of
> >>complex numbers, a thought suddenly struck me.
> >>
> >>I have found nothing through google, nor in any of the mathematical
> >>books I have propping up my wobbly table, so maybe someone can
> >>enlighten me.
> >>
> >>It's usually said that i is defined as the square root of -1. But we
> >>know that the equation
> >>
> >> x^2 = -1
> >>
> >>has TWO solutions, which we would write as i and -i.
> >
> >
> >HUH???
> >The complex plane is the ordinary x-y plane with a definition of a
> >multiplication and addition operations which make it an algebraic
> >field.
>
> This is not the only way to define the complex numbers. One can also
> define them to be equal to R[x]/(x^2+1), for example.
I'm trying hard to keep up with this, but much of the maths is over my
head. This "R[x]/(x^2+1)" thing also appeared in an earlier response.
Could someone explain what this means? What is R? ("R" stands for
"Real"?) What is x? What is R[x]? How does R[x]/(x^2+1) define the
complex numbers?
matt
Ignore that question. Having read some of the further postings I now
understand what you're saying.
matt
That's interesting.
Would defining i = (0,-1) make any difference to anything, or is it
just an arbitrary choice?
William Elliot
> > >matt wrote:
> > >
> > >>It's usually said that i is defined as the square root of -1. But we
> > >>know that the equation
> > >> x^2 = -1
> > >>
> > >>has TWO solutions, which we would write as i and -i.
> > >
> > >multiplication and addition operations which make it an algebraic
> > >field.
> >
> > This is not the only way to define the complex numbers. One can also
> > define them to be equal to R[x]/(x^2+1), for example.
>
> I'm trying hard to keep up with this, but much of the maths is over my
> head. This "R[x]/(x^2+1)" thing also appeared in an earlier response.
> Could someone explain what this means? What is R? ("R" stands for
> "Real"?) What is x? What is R[x]? How does R[x]/(x^2+1) define the
> complex numbers?
>
coefficients, ax^2 + bc + c for example in R[x]. (x^2 + 1) is
all multiplies of the polynomial x^2 + 1
R[x]/(x^2 + 1) is the polynomials modulus x^2 + 1. Just like integers
modulus n, you divide away factors of n to keep the remainder.
For example,
x^3 + x^2 + 2x + 1 = x(x^2 + 1) + x^2 + x + 1
= (x + 1)(x^2 + 1) + x
would have remainder x.
Auturo means not x(x^2 + 1), but instead x + (x^2 + 1)
or the polynomial x modulus x^2 + 1
Thus x^2 = x^2 + 1 - 1 = -1 modulus x^2 + 1
Also (-x)^2 = x^2 + 1 - 1 = -1 modulus x^2 + 1
Tho _algebraically_ it makes no difference which is i,
the natural way to define i would be
i = x + (x^2 + 1) in R/(x^2 + 1) or the polynomial x modulus x^2 + 1
-i = -x + (x^2 + 1) or the polynomial -x modulus x^2 + 1
Otherwise you'll be crossed i'd with
i = -x in R/(x^2 + 1)
-i = x in R/(x^2 + 1)
Arturo Magidin
Look at calculus textbooks. Some of them have the words "Early
Transcendentals" on them. This refers to the early introduction of
functions such as the logarithms, the exponential functions, and most
importantly, the trigonometric and inverse trigonometric
functions. They are transcendealt functions. Algebraic functions are
those that involve only the use of addition, multiplication, division,
exponentiation, and fractional powers. The arg function in complex
analysis is defined, usually, via a branch of the inverse tangent, a
transcendental function.
matt
> In article <ed8acaa4.04112...@posting.google.com>,
> matt <matt271...@yahoo.co.uk> wrote:
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<cnstr2$mok$1...@agate.berkeley.edu>...
> >> In article <ed8acaa4.04112...@posting.google.com>,
> >> matt <matt271...@yahoo.co.uk> wrote:
> >> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<cnr85g$4ld$1...@agate.berkeley.edu>...
> >> >> In article <ed8acaa4.04112...@posting.google.com>,
> >> >> matt <matt271...@yahoo.co.uk> wrote:
> >
> Look at calculus textbooks. Some of them have the words "Early
> Transcendentals" on them. This refers to the early introduction of
> functions such as the logarithms, the exponential functions, and most
> importantly, the trigonometric and inverse trigonometric
> functions. They are transcendealt functions. Algebraic functions are
> those that involve only the use of addition, multiplication, division,
> exponentiation, and fractional powers. The arg function in complex
> analysis is defined, usually, via a branch of the inverse tangent, a
> transcendental function.
>
Copy that. I'm actually pretty familiar with all the ... erm ....
familiar transcendental functions, and the basic calculus associated
with them. Just didn't make the contrast "transcendental" vs
"algebraic" here. I think of everything like arctan(x), sin(x), exp(x)
etc. as just "algebra". (Comes from having no formal training you
know!)
Let me coin the term "parity violation" for the situation when
interchanging i and -i turns a true statement into a false one, or
vice versa. (I don't know the correct term, or even if there is one.)
Most familiar transcendental functions can be (or, if you prefer, ARE)
defined as the sum of an infinite series of algebraic terms. I can't
currently see how an infinite sum can violate parity when none of its
individual (algebraic) terms do. Though I DO know that some very weird
and counterintuitive things can happen with infinite sums. Maybe this
one of them?
Right now, I actually think that the arg function causes a parity
violation NOT because it's transcendental, but because its definition
depends on the handedness of the complex plane (which, debatably, is
arbitrary). For example, I could equally well define the very simple
"quadrant" function, which takes the values 1, 2, 3, 4 (ignoring
boundary cases here). Then,
quadrant(1+i) = 1
and yet
quadrant(1-i) =/= 1
Thus violating parity.
In fact, I'm currently thinking that IF a statement (algebraic,
transcendental, anything) violates parity, then some part of its
DEFINITION must depend on the (debatably arbitrary) assignation of i
(aka handedness of the complex plane) - like arg and quadrant.
None of that was terribly precisely explained, but hopefully you will
get the gist of what I am trying to say.
Can you think of any parity-violating statements involving plain old
arctan (NOT the "handed" version used by arg), or plain old sine,
cosine, exp, log etc.? I'd be very interested to see an example, as
that would blow my current theory on this matter out of the water.
Arturo Magidin
Fair enough.
>Most familiar transcendental functions can be (or, if you prefer, ARE)
>defined as the sum of an infinite series of algebraic terms. I can't
>currently see how an infinite sum can violate parity when none of its
>individual (algebraic) terms do.
Again, you seem to be confusing things. When I said that i and -i are
algebraically indistinguishable, we are talking about certain kinds of
statements; infinite sums involve limits, and limits are not algebraic
statements in this sense.
When you say "an infinite sum" vs "the individual terms", the
statement is really meaningless. If you are defining the value of,
say, arctan, via an infinite series, then you are saying that
arctan(x) is the limit of the values of the partial sums.
You can certainly talk about the partial sums. But what statement are
you making about the partial sum? The only statement you could make
about the partial sums without involving the value of the limit would
be, perhaps, that the partial sums form a Cauchy-sequence. Even that
statement is already non-algebraic, since it involves the notion of
distance and proximity in the complex plane, which is an analytic (not
algebraic) concept. So what statement are you making about partial
sums which is algebraic and "does not violate parity"?
> Though I DO know that some very weird
>and counterintuitive things can happen with infinite sums. Maybe this
>one of them?
Before we can talk about whether something is weird and
counterintuitive, we must have a clear idea of just what it is that we
are talking about. What is it that you are talking about? What
algebraic statement are you making about the partial sums, and what
statement about the infinite series? And how are you defining
convergence of the infinite series without first fixing the complex
plane in some way?
>Right now, I actually think that the arg function causes a parity
>violation NOT because it's transcendental, but because its definition
>depends on the handedness of the complex plane (which, debatably, is
>arbitrary).
I did not say that the transcendence of arctan (or of arg) is the
->cause<- of the violation. The point was that the violation by arg
does NOT invalidate the statement that i and -i are algebraically
indistinguishable because arg is not an algebraic function. The
transcendental nature of arg is not what ->causes<- the violation; it
is what ->allows<- the violation to happen. Do you see the difference
there?
matt
Google groups is as buggy as ... well ... a thing that is very buggy.
mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<co09pt$1ood$1...@agate.berkeley.edu>...
Let me try to clarify. Consider the general statement:
Finite algebraic expression 1 = Finite algebraic expression 2
(1)
(By "finite" I mean an expression consisting of a finite number of
algebraic operations. Throughout the remainder of this I will also
assume that EVERYWHERE we are considering complex-valued expressions
and functions. Obvious from the context, but some other things I've
thought were "obvious from the context" haven't been!)
From what's been said earlier, (1) cannot be parity-violating because
it involves only algebraic operations. Correct?
Now introduce one or more transcendental functions into the mix -
maybe the familiar ones, maybe any others you choose - PROVIDED that
the function(s) can be defined as the limit of an infinite sequence of
algebraic operations. (Don't know how big a "provided" that might be.)
This gives, in general
Infinite algebraic expression 1 = Infinite algebraic expression 2
(2)
(By "infinite" I mean the limit of an infinite sequence of algebraic
operations.)
If (2) can be parity-violating but (1) can't, then this is the
possible "weird and wonderful" thing about infinite sums I was talking
about. My current view is that it is impossible for (2) to violate
parity, but proving it is well beyond me. If you know otherwise then
I'd be interested to know!
In particular, if you can show me a parity-violating statement of the
form
Transcendental expression #1 = Transcendental expression #2
using only transcendental functions defined (or definable) as the
limit of an infinite sum of algebraic expressions, then my theory is
demolished.
> >Right now, I actually think that the arg function causes a parity
> >violation NOT because it's transcendental, but because its definition
> >depends on the handedness of the complex plane (which, debatably, is
> >arbitrary).
>
> I did not say that the transcendence of arctan (or of arg) is the
> ->cause<- of the violation. The point was that the violation by arg
> does NOT invalidate the statement that i and -i are algebraically
> indistinguishable because arg is not an algebraic function.
Agreed.
> The
> transcendental nature of arg is not what ->causes<- the violation; it
> is what ->allows<- the violation to happen. Do you see the difference
> there?
That depends on the definition of "transcendental". If
"transcendental" literally just means "non-algebraic" then my
"quadrant" function is transcendental (because quadrant(z) cannot be
defined in terms of algebraic operations on z), and I agree. (I note
that this definition would also make such functions as "abs", "floor"
etc. "transcendental". I've never considered them to be so, but you
may be able to correct me on this one. Could be just another
terminological blunder on my part.)
If, however, my quadrant function isn't transcendental then no, I
don't understand what the property of transcendence has to do with it.
Whatever the case, I think that the transcendental nature of the
(vanilla) ARCTAN function is irrelevant to the parity-violation of
arg. As I say, my quadrant function has nothing to do with arctan and
yet exhibits an identical parity-violation.
Let me introduce another example which may help illustrate my point
further. Define the function f of some complex number by
f(a + ib) = b (3)
This statement is TRUE (by definition). Swap i with -i and you get
f(a - ib) = b
which is FALSE. Parity is violated. My contention is that this is
because (like arg and quadrant) the choice between i and -i is
implicit in the DEFINITION of f.
(Question: is f transcendental?)
To help understand this, I'll extend your Martian metaphor. Suppose
that somewhere "out in space" the complex numbers exist and are
uniquely defined. That is, the complex numbers a + ib and a - ib are
distinct and distinguishable.
(Actually, I am not currently entirely convinced that they ARE
distinguishable, either by us or the Martians. Ref KP's response: "The
mathematics assures us that the equation z^2 + 1 = 0 has two
solutions, but it treats them even-handedly, giving us no way to stick
a special label on one of them rather than the other.")
But anyway, suppose that here on earth we DO know which of this pair
we call a + ib and which we call a - ib. The left-handed Martians do
the opposite to us. Now pull a unique complex number, z, out of space.
We call it a + ib, and the left-handed Martians call it a - ib. To us,
f(z) = b
but to the Martians
f(z) = -b
Even though the complex number z is the SAME. I.e. the DEFINITION of f
depends on the choice of i vs -i. It depends on whether we call the
complex number that we picked out of space a + ib or a - ib, and THAT
is why (3) is parity-violating.
I am ASSUMING that the definitions of the familiar transcendental
functions (exp, log, trig, inverse trig, hyperbolic etc.) do NOT
depend on the choice of i vs -i, and are therefore NOT
parity-violating.
matt
Google groups is as buggy as ... well ... a thing that is very buggy.
mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<co09pt$1ood$1...@agate.berkeley.edu>...
Let me try to clarify. Consider the general statement:
> >Right now, I actually think that the arg function causes a parity
> >violation NOT because it's transcendental, but because its definition
> >depends on the handedness of the complex plane (which, debatably, is
> >arbitrary).
>
> I did not say that the transcendence of arctan (or of arg) is the
> ->cause<- of the violation. The point was that the violation by arg
> does NOT invalidate the statement that i and -i are algebraically
> indistinguishable because arg is not an algebraic function.
Agreed.
> The
> transcendental nature of arg is not what ->causes<- the violation; it
> is what ->allows<- the violation to happen. Do you see the difference
> there?
That depends on the definition of "transcendental". If
matt
Google groups is as buggy as ... well ... a thing that is very buggy.
mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<co09pt$1ood$1...@agate.berkeley.edu>...
Let me try to clarify. Consider the general statement:
> >Right now, I actually think that the arg function causes a parity
> >violation NOT because it's transcendental, but because its definition
> >depends on the handedness of the complex plane (which, debatably, is
> >arbitrary).
>
> I did not say that the transcendence of arctan (or of arg) is the
> ->cause<- of the violation. The point was that the violation by arg
> does NOT invalidate the statement that i and -i are algebraically
> indistinguishable because arg is not an algebraic function.
Agreed.
> The
> transcendental nature of arg is not what ->causes<- the violation; it
> is what ->allows<- the violation to happen. Do you see the difference
> there?
That depends on the definition of "transcendental". If
Arturo Magidin
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<co09pt$1ood$1...@agate.berkeley.edu>...
>> In article <ed8acaa4.04112...@posting.google.com>,
>> matt <matt271...@yahoo.co.uk> wrote:
[.snip.]
You are already introducing horrible notation that is bound to cause
confusion. A limit of an infinite sequence is ->not<- an algebraic
expression; it is an ->analytic<- expression. The use of the adjective
"algebraic" here is ->bound<- to cause all sorts of headaches when you
try to square it with the statement that "i and -i are algebraically
indistinguishable".
>If (2) can be parity-violating but (1) can't, then this is the
>possible "weird and wonderful" thing about infinite sums I was talking
>about.
You are talking about ENTIRELY DIFFERENT ANIMALS. Just because you
throw in the word "algebraically", it doesn't mean that there is such
a close relation between them.
>My current view is that it is impossible for (2) to violate
>parity,
How can that possibly be your "current view", if you already have that
the arctan function gives you different values for i and -i, and is an
instance of (2)?
>In particular, if you can show me a parity-violating statement of the
>form
>
> Transcendental expression #1 = Transcendental expression #2
>
>using only transcendental functions defined (or definable) as the
>limit of an infinite sum of algebraic expressions, then my theory is
>demolished.
You already did!
>> The
>> transcendental nature of arg is not what ->causes<- the violation; it
>> is what ->allows<- the violation to happen. Do you see the difference
>> there?
>
>That depends on the definition of "transcendental".
No. It depends on the meaning of the word "cause". Which you
improperly used.
Is English your native language? It isn't mine, and it is obvious we
are having some serious problems communicating. I have the feeling
that I say things clearly, and they wash off you like water off a
duck's back. You pick a few drops, misinterpret them, misrepresent
them, misstate them, and then wonder where "things" went wrong.
matt
> ... it is obvious we
> are having some serious problems communicating.
You're spot on there! From your reply it's clear that none of the
points in my last posting got across at all. It obviously just didn't
make any sense to you.
I'm still interested in your views, so if you get time read it again
and see if you can figure out what I'm trying to say. If that proves
impossible then no worries ... let's call it a day on this one!
rgds
matt
Mattias Wikstr?m
> Let me coin the term "parity violation" for the situation when
> interchanging i and -i turns a true statement into a false one, or
> vice versa. (I don't know the correct term, or even if there is one.)
To make this precise: You have a statement S(z) which asserts
something about the complex number z=a+bi. S is then a parity violator
if there exist a, b such that S(a+bi) is true but S(a-bi) is false.
>
> Most familiar transcendental functions can be (or, if you prefer, ARE)
> defined as the sum of an infinite series of algebraic terms. I can't
> currently see how an infinite sum can violate parity when none of its
> individual (algebraic) terms do. Though I DO know that some very weird
> and counterintuitive things can happen with infinite sums. Maybe this
> one of them?
No, it is not. The algebraic VS. transcendental distinction has
nothing, or almost nothing, to do with the symmetry between i and -i.
>
> Right now, I actually think that the arg function causes a parity
> violation NOT because it's transcendental, but because its definition
> depends on the handedness of the complex plane (which, debatably, is
> arbitrary).
You are absolutely right. The function arg causes "parity violation",
as does the function Im. However, if we define f(z)=i*Im(z) and
g(z)=i*arg(z), then neither f nor g will cause parity violation.
Let me use z' to denote the conjugate of z. Then the functions Re, Im,
abs, and arg can be defined by the following symmetrical-looking
expressions:
(z+z')*(1/2)=Re z
(z-z')*(1/2)=i*Im z
(z*z')^(1/2)=abs(z)
(z/z')^(1/2)=exp(i*arg z)
(To use the last expression as a definition of arg, you must of course
first define the exponential function for complex numbers without
using the arg function, but this can be done using power series.)
Mattias
matt
> matt271...@yahoo.co.uk (matt) wrote in message news:<ed8acaa4.04112...@posting.google.com>...
> > Let me coin the term "parity violation" for the situation when
> > interchanging i and -i turns a true statement into a false one, or
> > vice versa. (I don't know the correct term, or even if there is one.)
>
> To make this precise: You have a statement S(z) which asserts
> something about the complex number z=a+bi. S is then a parity violator
> if there exist a, b such that S(a+bi) is true but S(a-bi) is false.
>
Absolutely, except that the statement needn't be concerned with just
one complex number. It could be, for example, z_1 * z_2 = z_3, or
arctan(z_1)/z_2 + z_3 = 0, or whatever. In this case i and -i must be
interchanged wherever they occur.
> >
> > Most familiar transcendental functions can be (or, if you prefer, ARE)
> > defined as the sum of an infinite series of algebraic terms. I can't
> > currently see how an infinite sum can violate parity when none of its
> > individual (algebraic) terms do. Though I DO know that some very weird
> > and counterintuitive things can happen with infinite sums. Maybe this
> > one of them?
>
> No, it is not.
Thanks. That's immensely reassuring. I wondered if I was (i) going
bonkers or (ii) making a meaningless statement.
> The algebraic VS. transcendental distinction has
> nothing, or almost nothing, to do with the symmetry between i and -i.
>
Right. That's interesting. We're getting to the heart of something I'm
pretty confused about.
Here's three statements that I currently think are true, but people
can agree or disagree with:
1. No algebraic parity-violating statements can be constructed (In
other words, algebraic operations cannot break the symmetry.) All
parity-violating statements are therefore non-algebraic. (Ref. AM)
2. No partity-violating statements consisting of algebraic operations
and transcendental functions definable as the limit of an infinite
sequence of algebraic operations can be constructed. In other words,
contrasting with (1), the algebraic vs transcendental distinction IS
irrelevant when the transcendental function(s) in question are of this
nature.
3. Every parity-violating statement contains an operator, function,
whatever you want to call it, whose definition depends on the
(debatably arbitrary) assignation of the symbol i to one of the roots
of x^2 + 1 = 0, or, if you prefer, the assignation of i to the point
(0,1) or (0,-1) on the complex plane.
Now, IF every non-algebraic statement is transcendental, then, from 1,
all parity-violating statements are transcendental.
BUT ... I'm not clear if this is the standard definition of the term
"transcendental". Many sources say that "transcendental functions are
non-algebraic", but I haven't been able to find an explicit statement
that ALL non-algebraic functions are transcendental. For example, is
the parity-violating Im() function transcendental? Is my
parity-violating "quadrant" function transcendental?
Not sure how much sense that made to anyone, but any views are, as
always, very welcome...
Mattias Wikstr?m
> mattias...@hotmail.com (Mattias Wikstr?m) wrote in message news:<e47110b3.04112...@posting.google.com>...
> > something about the complex number z=a+bi. S is then a parity violator
> > if there exist a, b such that S(a+bi) is true but S(a-bi) is false.
> >
>
> Absolutely, except that the statement needn't be concerned with just
> one complex number. It could be, for example, z_1 * z_2 = z_3, or
> arctan(z_1)/z_2 + z_3 = 0, or whatever. In this case i and -i must be
> interchanged wherever they occur.
That is, a statement S(z1, z2, ..., zn) about complex numbers z1, z2,
..., zn is a parity violator if there exist complex numbers z1, z2,
..., zn such that S(z1, z2, ..., zn) is true, but S(z1', z2', ...,
zn') is false, where ' denotes conjugation.
> > The algebraic VS. transcendental distinction has
> > nothing, or almost nothing, to do with the symmetry between i and -i.
> >
>
> Right. That's interesting. We're getting to the heart of something I'm
> pretty confused about.
>
> Here's three statements that I currently think are true, but people
> can agree or disagree with:
>
> 1. No algebraic parity-violating statements can be constructed (In
> other words, algebraic operations cannot break the symmetry.) All
> parity-violating statements are therefore non-algebraic. (Ref. AM)
Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all
z, and (II) sqrt(-1)=i) is algebraic "in spirit" but can be used to
construct the parity-violating statement S(z) <-> z=sqrt(-1). I would
need an exact definition of "algebraic" to tell whether statement 1.
is correct.
> 3. Every parity-violating statement contains an operator, function,
> whatever you want to call it, whose definition depends on the
> (debatably arbitrary) assignation of the symbol i to one of the roots
> of x^2 + 1 = 0, or, if you prefer, the assignation of i to the point
> (0,1) or (0,-1) on the complex plane.
Your basic intuition here is correct: If you combine ingredients, none
of which can lead to parity-violation, the result cannot either.
[...]
> BUT ... I'm not clear if this is the standard definition of the term
> "transcendental". Many sources say that "transcendental functions are
> non-algebraic", but I haven't been able to find an explicit statement
> that ALL non-algebraic functions are transcendental. For example, is
> the parity-violating Im() function transcendental? Is my
> parity-violating "quadrant" function transcendental?
"Transcendental" is synonymous with "non-algebraic", but the
algebraic/non-algebraic distinction does not apply to anything you can
think of. Just as it would be strange to call the number i
non-negative, so it would be strange to call your quadrant function,
which is not even continuous, non-algebraic.
Mattias
matt
> matt271...@yahoo.co.uk (matt) wrote in message news:<b21cacc3.04112...@posting.google.com>...
> > mattias...@hotmail.com (Mattias Wikstr?m) wrote in message news:<e47110b3.04112...@posting.google.com>...
> > other words, algebraic operations cannot break the symmetry.) All
> > parity-violating statements are therefore non-algebraic. (Ref. AM)
>
> Well, the function sqrt (which satisfies (I) sqrt(z)*sqrt(z)=z for all
> z, and (II) sqrt(-1)=i) is algebraic "in spirit" but can be used to
> construct the parity-violating statement S(z) <-> z=sqrt(-1). I would
> need an exact definition of "algebraic" to tell whether statement 1.
> is correct.
The definition of "algebraic" that I'm working with is "Algebraic
functions are
those that involve only the use of addition, multiplication, division,
exponentiation, and fractional powers." That would seem to include
Sqrt.
The statement Sqrt(-1) = i is an interesting one. Here's how I look at
it.
If one accepts that Sqrt has two values then the function is not
parity-violating in this instance because
Sqrt(-1) = {i, -i} is TRUE
Sqrt(-1) = {-i, i} is TRUE
If one considers Sqrt as a single-valued function then Sqrt(-1) has
the single value i (presumably), and
Sqrt(-1) = i is TRUE
Sqrt(-1) = -i is FALSE
which is parity-violating. This is because the assignation of i to one
of the roots is then implicit in the DEFINITION of Sqrt (as per my
earlier statement 3).
Where this leaves us with the algebraic vs non-algebraic vs
transcendental question I do not know.
Any ideas?
David W. Cantrell
I haven't been following this thread, but I may now be able to shed some
light (or, failing that, to create more confusion).
"Algebaic" can be used in two different ways. In the link below, I mention
a text describing the ways. See "The function Sqrt(x) is transcendental!" at
<http://groups.google.com/groups?selm=20030113105343.853%24kq%40newsreader.com>
As you have already noted, symmetry between i and -i is broken by the
principal-valued square root function Sqrt, while symmetry is preserved
under the multivalued relation sqrt. Using the sense of the term
"algebraic" which is appropriate to this thread, sqrt is algebraic, while
Sqrt is not. Perhaps that resolves your problem.
David Cantrell
matt
Yes, that would certainly resolve it. I guess the only downside is
that this solution makes "the square root function" (i.e. the
single-valued function) non-algebraic, which contradicts quite a few
sources I've found.
Thanks for the link too. I found a few other bits and pieces about
"algebraic" vs "transcendental", and it this is definitely a
terminological minefield for the unwary!