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 Trouble with 5.10 and what I am supposed to do. 
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Joined: 11 Jul 2008, 05:14
Posts: 9
Post Re: Trouble with 5.10 and what I am supposed to do.
Thanks for the correction vasco, should have followed me own 'parse with care' guff!

However does this then not work:

e^{-4{\pi}^2}=e^{(2{\pi}i)^2}=e^{(2{\pi}i)(2{\pi}i)}=(e^{2{\pi}i})^{2{\pi}i}=(1)^{2{\pi}i}=1

which properly dissolves the paradox?


11 Jul 2008, 18:45
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Trouble with 5.10 and what I am supposed to do.
Ok but it seems to me that you have just expressed the paradox in a slightly different way because you are now saying

e^{-4\pi^2}=1 which is false!

I am almost sure that this paradox is connected with specifying a value for the logarithm and then not sticking to it on both sides of the identity as Penrose hints:

"This ambiguity leads us into all sorts of problems if we are not careful" and then he goes on to say
"The best way of avoiding these problems......adopt the notation w^z only when a particular choice of \log w has been specified....."

I am still working on this myself.

Have a look at Ex 5.15 and what Penrose says in the lead up to it.


11 Jul 2008, 19:22

Joined: 11 Jul 2008, 05:14
Posts: 9
Post Re: Trouble with 5.10 and what I am supposed to do.
Ah, yes, indeed, thanks again!

So after re-reading all the posts innumerable times and dredging the internet for hours and covering reams of A4 paper with e^{-4{\pi}^2}-sized pencil marks of unwavering dubiousness (all the while aggrievating a major case of crampo mentale), I think I've finally come to have some sort of a handle on this issue.

Basically of course yous had it sussed. (Your posts and susume's in particular deserve special mention.)

What continued to bug me though was the mechanical side of the issue. Surely it should be possible to specify a few unambiguous rules for complex exponentiation that if followed would allow one to mindlessly manipulate the symbols while pursuing one's higher goal - indeed, isn't this the main reason that algebra's so useful?

You pointed out Ex.5.15, and while coming up with a proposed solution for it (posted separately) I realized the following, which puts the whole business in a very clear light I think.

We have by definition w^z = e^{zlogw} = e^{z(w_0 + 2{\pi}ki)} = e^{zw_0}e^{z2{\pi}ki} where w_0 is a specific choice of logw.

But look what happens when we put w = e and naively follow the same manipulations:

e^z = e^{zloge} = e^{z(1 + 2{\pi}ki)} = e^ze^{z2{\pi}ki}

So 1 = e^{z2{\pi}ki} for any z! (Hence my revolutionary discovery that 1 = e^{-4{\pi}^2})

Posts here have shown that this is due to inconsistent choices of k. And having parsed the book many times I realized eventually that Prof. Penrose deals with it by stating the tacit convention of always taking loge = 1 for the special case of e^z, so e^z = e^{zloge} = e^{z1} = e^z always. (Though he states it after posing the paradox so it wasn't available to solve it.)

However, allow me to introduce the explicit or dickdockian convention, a variant of the tacit convention that I hope is clearer:

Define w^z = e^{zlogw} for all (nonzero) w except where w = e.

That's it. We don't need the definition (I think) for w = e so let's not include it in the definition.

How does this help? Looking at the original paradox:

e = e^{1+2{\pi}i} is fine, and (e^{1+2{\pi}i})^{1+2{\pi}i} = e^{(1+2{\pi}i)log(e^{1+2{\pi}i})} = e^{(1+2{\pi}i)(1+2{\pi}i)e_0} is fine also (where e_0 = the specific choice of loge in accordance with Ex.5.15). But we're not allowed to write

e^{1+2{\pi}i} = (e^{1+2{\pi}i})^{1+2{\pi}i} anymore than we're allowed to write w = w^{1+2{\pi}i} for arbitrary w.

Similarly, looking above, one can't write e^z= e^{zloge} anymore than one can write w^z = w^{zlogw}, though
e^{zloge} = e^{z(e_0 + 2{\pi}ki)} is fine, as of course is w^z = e^{zlog}.

Anyway as far as I can see (which isn't very) this, coupled with the convention outlined in Ex.5.15, allows us manipulate w^z and
e^z willy-nilly without having to use tacit assumptions and/or unerring choices and/or principal values et/aut al.

I don't know if any of this means anything to anyone or means anything full stop and I don't anticipate a big rush to adopt the dickdockian convention by the worldwide mathematics community but it's got rid of my crampo mentale (for the moment anyway).


16 Jul 2008, 17:06

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Trouble with 5.10 and what I am supposed to do.
The inverse operation of Anti-log(logx) requires a base counting system. Base 10 is useless in calculus. The number e = sum of 1/(n!), from n =0 to infinity, was invented because Newton's binomial expansion allowed f(x) = sum of (a(n)x**n)/n! from n= 0 to infinity.
e**1 is the value of sum of (a(n)x**n)/n! from n= 0 at x=1,,, 2.7....etc,,it can not be treated willy-nilly!
If you wish to choose your base e=sum of (1+2*PI)/n!, from n=o, to infinity that will work as a base for the inverse exponentiation operation (aka logarithm),,you fix it,, but you can't do e**1 =e**(1+2i*PI),,they are two different numbers,,to wit:
sum of 1/n! from n = 0 to infinity
and
sum of (1+2i*P)/n! from n=0 to infinity

The statement e**1 =e**(1+2i*PI) is wrong,,from the GITGO!!
Vasco talks this , ie Principle values,,but the heart of the matter is:
THE INVERSE OF EXPONENTIATION (AKA LOG) REQUIRES A BASE NUMBER!!
You need to know what number you multiplied by itself to find how many times you multiplied!!


31 Jul 2009, 23:31

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Trouble with 5.10 and what I am supposed to do.
Look, this whole mess starts with the mapping of z=x+iy to z=sqrt(x**2+y**2)*exp(iarctan(x/y)),,it's a one to multi mapping! (aka z= r*exp(iTHETA))! IT'S THE MAPPING THAT NEEDS TO BE DEALT WITH!!
You must define exponentiation and it's inverse ( ie log ),,,CHOOSE A PRINCIPLE VALUE!! The Exp and Log operations are inverses that are needed for the whole mess to be an ALGEBRA!!
PS:
I never quite understood how they went from z=x+iy = r*exp(itheta), and then started talking about the complex plane in terms of theta = 0, to infinity!!

Penrose has something to say in this regard late,,ie RIEMANIAN SURFACES! His treatment never explains the logic of the x+iy to r*exp(itheta) mapping.


04 Aug 2009, 04:06

Joined: 02 Oct 2010, 15:29
Posts: 2
Post Re: Trouble with 5.10 and what I am supposed to do.
Harried,
Your inital suspicion that the identity (b^n)^m=b^(n*m) does not hold in this problem was correct all along. While this identity holds for all real n and m, it does not hold in general when n and m are complex. This was first discovered in 1827 by Thomas Clausen. See this http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities.
The problem says e=e^(1+2*pi*i) which is true. Then, it goes on to say (e^(1+2*pi*i))^(1+2*pi*i)=e^(1+4*pi*i-4*pi^2) which could be expanded as (e^(1+2*pi*i))^(1+2*pi*i)=e^((1+2*pi*i)*(1+2*pi*i))=e^(1+4*pi*i-4*pi^2). This last equation is not correct because it makes use of the identity (b^n)^m=b^(n*m) which does not hold for complex exponents.
This was actually pointed out by Sameed Zahoor all the way back on 12 March 2008.
Sorry about the difficult to read equations. I couldn't get LaTex equations to show up. It seems like there is something wrong with this site's LaTex parser.


02 Oct 2010, 21:20
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