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 Trouble with 5.10 and what I am supposed to do. 
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Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi

Starting from Vasco's equation that $w^z=e^{z log w}$, which seems to be correct, because it is in Penroses book, and, as stated by Vasco, where we have to be careful that $log (w)$ is already taken to be the principal value (of the complex number w), and applying it to

$(e^{1+ 2 \pi i})^{1+ 2 \pi i}$ we have

$w=e^{1+ 2 \pi i}$, z=1+ 2 \pi i
then
$(e^{1+ 2 \pi i})^{1+ 2 \pi i}=e^{z log w}=e^{(1+2 \pi i) log e^{1+ 2 \pi i}}$,


but the logarithmic in the exponential has to be the principal value of the logarithmic of the complex number $w=e^{1+ 2 \pi i}$. This principal value is $log e^{1+ 2 \pi i}=log e+ 0$. This is obvious because the module of $e^{1+ 2 \pi i} is e and the argument between pi and -pi is 0.

Substituting you have

$(e^{1+ 2 \pi i})^{1+ 2 \pi i}=e^{z log w}=e^{(1+2 \pi i) log e^{1+ 2 \pi i}}=e^{(1+2 \pi i) log e}=e^{(1+2 \pi i) $, and for the same reason
e^{(1+2 \pi i)}=e

Therefore Eyedrop was right that

$e=e^{(1+2 \pi i)}= (e^{1+ 2 \pi i})^{1+ 2 \pi i}\ne e^{(1+ 2 \pi i)(1+ 2 \pi i)},

Thank you very much Vasco. Even if I think you were not right, It is ultimately thanks to your systematic explanations that the solution could be found.


10 Jun 2008, 09:53

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Sorry,

the solution is not really found, Now we have to understand why the last inequality of my previous post does not hold and as Eyedrop originally asked, which are the conditions under which it applies and the conditions under which it does not.


10 Jun 2008, 10:03

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
I guess that what really happens is that w^z as basco said is a multivalued function. If you choose the principal value you get (for our case) e, but a second value (for our case) is $e^{1- 4 \pi^2}$ and yet another third value $e^{1- 8 \pi^2}$ and another is $e^{1- 12 \pi^2}$ and so on...

What Eyedrop was doing is choosing the first one of the values on the LHS and the second one on the RHS, because in doing the actual multiplications of the exponents you automatically choose the second value. If we would have started with $(e^{1+4 \pi i})^{1+2 \pi i}=e^{(1+4 \pi i)(1+2 \pi i)}$ we wold have automaticall arrived at the third of the values $e^{1- 8 \pi^2}$ and so on. But $e^{1+ 2 \pi i}$ intrinsically has the potential to give rise to all these values. We will not choose one unless for instance we decide to choose the principal one, but then we have to consistently choose the principal value in both the LHS and RHS.


10 Jun 2008, 11:45

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Sorry to write another post about it. I just wanted to say that now that I am getting to understand the problem, the more I think about it, the more I realise that Vasco was right all the way long. When I said before that he was wrong it is only because at that time I did not properly understand what he meant. So forget about all my crap posts and read again his carefully.

It seems incredible to me that Penrose posted this problem with a smiley face anyway.


10 Jun 2008, 12:19

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi!

This is a question for ZZR Puig (or whoever else knows the answer):

have you found out whether the definition that you gave for the exponential

$f(z)^z=f(f(z)z)$ holds for any complex number z?


11 Jun 2008, 07:36

Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi damian.

A few things first. I don't have the book in front of me right now, but I think in my book this exercise is qualified as hard. No smiley face there if I remember well. This is also the reason that I felt a bit suspicious about the apparent simplicity of the solution.

As I said I think it's clear that e^{1+i2k\pi}\neq{e^{(1+i2\pi)^k}, but I see that this is not the real paradox presented here, but the fact that {\left(e^{(1+i2\pi)}\Right)^{1+i2\pi} could be equal to either:
a) ({e})^{1+i2\pi}=e
b) {e^{1-4\pi^2+i4\pi}}=e^{1-4\pi^2}

I think that the root of this problem is that \Left(f(z)\Right)^z is not equal to f(z\cdot z) for every z if we don't limit the imaginary part of z between -\pi and +\pi. [When defining f(z)=e^z].

In a more general case we would have f(w,z)=w^z, and the correspondent equalities to be checked for consistency for every value of zwould be:

\Left(w^z\Right)^z=f(w^z, z)=f\Left(f(w,z), z\Right)=f(w, z\cdot z)

Unfortunately I haven't had much time lately to try to give a good explanation to this issue. I'll try to work on it and see if I can report something useful.

By the way, as you may have noticed the expressionf(f(z)z) wasn't suitable for this problem, so it has been replaced by f(z\cdot z).


11 Jun 2008, 13:35
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Post Re: Trouble with 5.10 and what I am supposed to do.
vasco wrote:
Here is an observation which people may find useful.

In my last post I stated that

$w^z=exp\{z (\log w + 2k\pi i)\}.$ Formula 1

gives ALL possible values of $w^z$.

Let’s investigate $(w^z)^p$ where p is a complex number

Using formula 1 above, we can see that

$(w^z)^p=exp\{p(\log w^z+2k\pi i)\}= exp\{p(z\log w+2k\pi i)\}= exp\{pz\log w+2k\pi ip\} $
$=exp\{z(p\log w+2k\pi i\frac{p}{z})\}$
$=exp\{z(\log w^p+2k\pi i\frac{p}{z})\}$
$=exp\{z(\log w^p+2k\pi i-2k\pi i+2k\pi i\frac{p}{z})\}$
$=exp\{z(\log w^p+2k\pi i+2k\pi i(\frac{p}{z}-1))\}$
$=(w^p)^z$ only if p=z and/or k=0

So generally $(w^z)^p=(w^p)^z $ for all w and p ONLY if we use principal values


Edited on 12th May 2008 to correct an algebraic error which does not however affect the conclusion


12 Jun 2008, 06:27

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi ZZR Puig,

Thank you very much for your explanations!. If you find the time to think further about fthe problem, please post your conclusions!. I found your viewpoint very useful in understading the root of the paradox.

P.S. I am relieved to know that the problem does not have a smiley face :-))))


12 Jun 2008, 12:29
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Post Re: Trouble with 5.10 and what I am supposed to do.
This post has been edited as a result of some very useful comments from ZZR Puig - Vasco
I have now rejected these comments and re-posted below - Vasco
Damian
As you requested, I’ve been trying to find a simpler way to explain the paradox of 05.10. Here are some of my thoughts:

We know that generally exp(w)=exp(w(\log e+2k\pi i))

For w=1 this becomes exp(\log e+2k\pi i)=exp(1+2k\pi i)

Just as (in my earlier post) I used Log w to represent the multi-valued logarithm of w, and log w to represent the principal value, such that
Log w=\log |w|+i\arg w,
so I could use E to represent the various ways of expressing exp (1) and e the value when k=0:

E=exp(1+2k\pi i)=exp(1)exp(2k\pi i)=e\times e^{(2k\pi i)} , where k is any integer value, positive or negative.

if we take k in the range -2 to +2 then we get:

E= exp(1-4\pi i)=e\times e^{(-4\pi i)}  \ \ \ \ k= -2
E= exp(1-2\pi i) = e\times e^{(-2\pi i)}  \ \ \ \ k= -1
E= exp(1)=e\ \ \ \ k=0 (the e we all know and love!)
E= exp(1+2\pi i) = e\times e^{(2\pi i)}  \ \ \ \ k=1
E= exp(1+4\pi i)=e\times e^{(4\pi i) } \ \ \ \ k=2

These are all separate valid ways of expressing exp(1)

If we look at Ex 05.10 and take logarithms then we can see the fallacy immediately:

\log(E(k=0))=\log e=1 \ and\  \log(E(k=1))=1+2\pi i and it is clear that 1 \ne 1+2\pi i

A simpler, more obvious example of a similar fallacy would be for example sqrt(9). This is multi-valued and has the 2 possible distinct values -3 and +3.
However, writing

sqrt(9)=+3
sqrt(9)=-3
and then saying
Therefore +3=-3 is clearly fallacious.

Or, another example

sqrt(9)=+3
sqrt(9)/-3=+3/-3=-1

but we know sqrt(9)=-3 therefore

-3/-3=-1
1=-1 fallacious again.

When we say, for example above, that +3=-3 because they are both equal to sqrt(9), we are forgetting that sqrt(9) is a multi-valued function and therefore we are not allowed to equate the two values.

I hope this serves to clarify the fallacy and does not confuse you even more.


Last edited by vasco on 14 Jun 2008, 18:59, edited 4 times in total.

12 Jun 2008, 15:14

Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Trouble with 5.10 and what I am supposed to do.
That is a good point vasco, and gives a bit of understanding about this problem, but as we all know the function that is really multivalued is \log z, not e^z, and perhaps someone could still find some difficulties understanding the whole issue.

I'll try to show a solution that may help to clarify this.

We know that e^z=e^{z+ik2\pi}, so let's apply this to the original expression e=\Left(e^{1+i2\pi}\Right)^{1+i2\pi} to get:

e^{1+ik2\pi}=\Left(e^{1+i2\pi+iA2\pi}\Right)^{1+i2\pi+iB2\pi}, where k, A and B are integers.

If we now apply logarithms we are already taking into account the fact they are multivalued:

\log\Left(e^{1+ik2\pi}\Right)=\log\Left(\Left(e^{1+i2\pi+iA2\pi}\Right)^{1+i2\pi+iB2\pi}\Right), therefore:

1+ik2\pi=(1+i2\pi+iA2\pi}\Right)(1+i2\pi+iB2\pi)

Operating we get:

1+ik2\pi=1-4\pi^2-4B\pi^2-4A\pi^2-4AB\pi^2+i4\pi+i2B\pi+i2A\pi
1+ik2\pi=1-4\pi^2(1+A+B+AB)+i2\pi(2+A+B)

Separating the real and imaginary parts:

1=1-4\pi^2(1+A+B+AB) \qquad | \qquad ik2\pi=i2\pi(2+A+B)

0=1+A+B+AB \qquad \qquad | \qquad \qquad k=2+A+B


If now we choose k=0 so e^{1+ik2\pi}=e, the equation system

0=1+A+B+AB \qquad \qquad | \qquad \qquad 0=2+A+B

has only one solution, which is:

A=B=-1

Therefore the original equation e^{1+ik2\pi}=\Left(e^{1+i2\pi+iA2\pi}\Right)^{1+i2\pi+iB2\pi} becomes:

e^1=(e^1)^1, obviously.

The function may be multivalued, but there is only one possible set of parameters that give the right solution. We can't expect to chose A=B=0 and think we will find a valid one.

I hope you found this useful.


Note: Perhaps instead of beginning with e=e^{1+2k\pi} and e^{1+i2\pi}=e^{1+i2\pi+ik2\pi}, it would be better to say \log e=1+2k\pi and \log (e^{1+i2\pi})=1+i2\pi+ik2\pi, but it's all the same anyway.


12 Jun 2008, 18:27
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Post Re: Trouble with 5.10 and what I am supposed to do.
Hi ZZR
You are right that exp(z) is not multi-valued. It is the other way round of course - many different values of z giving the same value of exp(z). I will edit my post accordingly. Thanks
Vasco


13 Jun 2008, 06:19
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Post Re: Trouble with 5.10 and what I am supposed to do.
Damian and ZZR
I have thought about this again and I am now convinced that the post I edited as a result of ZZW’s comments was in fact correct and so I am reposting it here in a slightly modified form to make some things clearer I hope.

We know that generally exp(z)=exp(zLog e)=exp(z(\log e+2k\pi i)) (see my post on the general power w^z)

It’s important to make the distinction between

1)exp(z)=exp(zLoge)=exp(z(\log e+2k\pi i)) which is multi-valued and

2)$exp(z)=1+\frac{z}{1!}+\frac{z^2}{2!}+$ .... which is single valued

Looking at 1) above for z=1 we obtain

exp(\log e+2k\pi i)=exp(1+2k\pi i)

Just as (in one of my earlier posts) I used Log w to represent the multi-valued logarithm of w, and log w to represent the principal value, such that
Log w=\log |w|+i\arg w,
so I could use E to represent the multi-valued exp (z) and e the principal value:

E(1)=exp(1+2k\pi i)=exp(1)exp(2k\pi i)=e \times exp(2k\pi i) , where k is any integer value, positive or negative.

if we take k in the range -2 to +2 then we get:

E(1)= exp(1-4\pi i)=e\times exp(-4\pi i)  \ \ \ \ k= -2
E(1)= exp(1-2\pi i) = e\times exp(-2\pi i)  \ \ \ \ k= -1
E(1)= exp(1)=e    \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0 (the e we all know and love!)
E(1)= exp(1+2\pi i) = e\times exp(2\pi i)  \ \ \ \ k=1
E(1)= exp(1 + 4\pi i)=e\times exp(4\pi i)  \ \ \ \ k=2

These are all separate valid values for E(1)

Writing e=exp(1+2\pi i) as is done in Ex 05.10, is not a valid thing to do. It is the same as saying:

E(1)=exp(1)                  \ \ \ \ for \ k=0
E(1)=exp(1+2\pi i)        \ \ \ \ for \ k=1

And therefore exp(1)=exp(1+2\pi i).

A simpler, more obvious example of a similar fallacy would be for example sqrt(9). This is also multi-valued and has the 2 possible distinct values -3 and +3.
However, writing

sqrt(9)=+3
sqrt(9)=-3
and then saying
Therefore +3=-3 is clearly fallacious.

Or, another example

sqrt(9)=+3
sqrt(9)/-3=+3/-3=-1

but we know sqrt(9)=-3 therefore

-3/-3=-1
1=-1 fallacious again.

When we say, for example above, that +3=-3 because they are both equal to sqrt(9), we are forgetting that sqrt(9) is a multi-valued function and therefore we are not allowed to equate the two values.

This is exactly the same as saying E(1) \ for\  k=0 is equal to E(1) \ for \ k=1.
If we take logarithms then we can see the fallacy immediately:

\log(E(1)=\log e=1 \ and\  \log(E(1))=1+2\pi i and it is clear that
1 \ne 1+2\pi i


14 Jun 2008, 10:34

Joined: 06 Jun 2008, 08:59
Posts: 10
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi!!

Thanks to the comments of all of you, we have not only resolved the paradox, but more importantly we have learnt a lot about the logarithmic of a complex number. I do not think there are many profesional physisists or mathematicians that understand so much in depth about this topic as us now :-)))

I am so looking forward to continue further and learn more with the help of all of you. I cannot wait until I get to understand the topics from the actual frontiers of physics that the next chapters will bring us to (and for that grand purpose we have to be patient and first understand the complex numbers!!!). Step by step and slowly we will get there :-))


16 Jun 2008, 14:42

Joined: 11 Jul 2008, 05:14
Posts: 9
Post Re: Trouble with 5.10 and what I am supposed to do.
Hi newbie here, great forum, interesting discussion, but can one not just say for this 'paradox':

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

i.e. there is no paradox, only an apparent one, which is why Prof. Penrose put 'paradox' in inverted commas.

This ties in with the exercise in the book being attached to the statement: "This ambiguity leads us into all sorts of problems if we are not careful", i.e. the ramifications of e^{2{\pi}i}=1 whence e^{z+2{\pi}i}=e^z (p.95 hardback) can lead to all sorts of counter-intuitive formulae which have to parsed with care.


11 Jul 2008, 13:45
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Post Re: Trouble with 5.10 and what I am supposed to do.
dickdock wrote:
Hi newbie here, great forum, interesting discussion, but can one not just say for this 'paradox':

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


Nice try!

But e^{-4{\pi}^2}\ne (e^{2{\pi}i})^2 but is equal to e^{(2\pi i)^2}.

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

Remember (a^b)^2 =a^b \times a^b=a^{2b}


11 Jul 2008, 17:37
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