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 Exercise [05.10] 
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Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Exercise [05.10]
As a result of the discussion held on the exercise discussion board about this paradox I'm completely rewriting my previously posted solution.

Preface

Before facing the solution of the paradox itself I think it may be worth to make a brief clarification about the exercise wording or presentation. That is:

e=e^{1+i2\pi} therefore e=(e^{1+i2\pi})^{1+i2\pi}=e^{1+i4\pi-4\pi^2}=e^{1-4\pi^2}


As I stated in my previous paradox solution proposal, I think that this formulation can be misunderstood. In the way that the first equation don't necessarily involve the first equality of the second expression.

The initial equation can be understood as e=ee^{i2\pi}=ee^{i2\pi}e^{i2\pi}...=e^{1+ki2\pi} \Rightarrow e=e^{1+ki2\pi}

While the second one could be generalized as e=(e^{1+i2\pi})^{1+i2\pi}=e^{(1+i2\pi)^k} \Rightarrow e=e^{(1+i2\pi)^k}

I think it's quite obvious that e^{1+i2k\pi}\neq{e^{(1+i2\pi)^k} even if the are equal for k=1.


Main solution

Nonetheless the paradox continues to exist if we consider that {\left(e^{1+i2\pi}\Right)^{1+i2\pi} can have two different solutions depending on whether we solve the inner parenthesis first or we multiply the exponents.
a) \qquad ({e})^{1+i2\pi}=e
b) \qquad {e^{1-4\pi^2+i4\pi}}=e^{1-4\pi^2}

Let's resolve this paradox then.

We know that w^z=e^{w+ik2\pi} or that \log {(w^z)}=(z+ik2\pi) \cdot \log {w}. So let's apply logarithms to the original expression.

e=(e^{1+i2\pi})^{1+i2\pi}
\log {e}=\log {\left((e^{1+i2\pi})^{1+i2\pi}\right)}

Then,
1+ik2\pi=(1+i2\pi+iC2\pi)\cdot \log {(e^{1+i2\pi})}
1+ik2\pi=(1+i2\pi+iC2\pi)\cdot (1+i2\pi+iD2\pi), where k, C and D are integers.

For simplicity we redefine A=C+1 and B=D+1 to get:

1+ik2\pi=(1+iA2\pi)\cdot (1+iB2\pi)
1+ik2\pi=1+iA2\pi+iB2\pi-AB4\pi^2

Separating the real and imaginary parts:

1=1-AB4\pi^2 \qquad | \qquad ik2\pi=iA2\pi+iB2\pi

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


Therefore, the original equation is only valid if either A or B are zero (or both), and k is equal to its sum.

The function \log may be multivalued, and multiple values of z can give the same result for e^z, but there is only one possible set of parameters A, B and k (and consequently possible values of z) that give the right solution for the original expression. We can't expect to chose A=B\neq 0 and think we will find a valid one.

e^{1+ik2\pi}=\Left( e^{1+iA2\pi}\Right)^{1+iB2\pi}


I hope this helps.


17 Jun 2008, 12:12

Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise [05.10]
I've thought simply taking log on both sides of the first equal sign of the paradox would lead instantly to contradiction.


30 Jun 2008, 22:17

Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Exercise [05.10]
Of course there is a contradiction, that's why it's a paradox.

However we can't take logarithms on both sides so easily without taking into account it's a multivalued function. If we don't we may get to results like the following:

e=e^{1+i2\pi}
\log{e} = \log(e^{1+i2\pi})
1=1+i2\pi FALSE! \Rightarrow e\neq e^{1+i2\pi}
Obviously this conclusion is also false, so we are not doing it right this way.

Anyway I think that these comments should be included in the exercise discussion subforum to keep the solutions as clean as possible.


01 Jul 2008, 09:40

Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise [05.10]
Second eq is good, but the third one has gone too far. Sorry to ignore your opinion on "keeping the solutions as clean as possible".


01 Jul 2008, 10:32

Joined: 16 Jun 2011, 16:58
Posts: 1
Post Re: Exercise [05.10]
-4pi^2 can be written as 2piiX 2pi i

e = e^(1+2pi i X 2pi i) = e^1 X (e^2pii)^2pi i

and e^2pi i= 1

and 1^2pi i= 1

so e = e


16 Jun 2011, 17:15
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [05.10]
Hi
You obviously haven't looked at all the solutions to 5.10. If you sort the solutions you will see that there are three other solutions b,c,d which includes a lot of discussion.
Your solution is clearly fallacious since
e^(-4pi^2) is clearly not equal to 1.


16 Jun 2011, 22:01

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [05.10]
ZZR Puig wrote:

Let's resolve this paradox then.

We know that w^z=e^{w+ik2\pi} or that \log {(w^z)}=(z+ik2\pi) \cdot \log {w}. So let's apply logarithms to the original expression.


Is this a mistake?


22 Sep 2011, 14:12

Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Exercise [05.10]
Yes, of course. It should say:

w^z = w^{z+ik2·pi}

Unluckily Lattex is not working anymore in the forum, so I don't see any way to correct it properly. Anyway it's just a side comment, I guess it does not affect the understanding of the solution.


Last edited by ZZR Puig on 23 Sep 2011, 10:24, edited 1 time in total.

22 Sep 2011, 23:25

Joined: 22 Sep 2011, 10:17
Posts: 9
Post Re: Exercise [05.10]
ZZR Puig wrote:
Yes, of course. It should say:

w^z = w·e^{z+ik2·pi}

Unluckily Lattex is not working anymore in the forum, so I don't see any way to correct it properly. Anyway it's just a side comment, I guess it does not affect the understanding of the solution.


I still don't folllow. Shouldn't it say: w^z=e^{z(logw+ik2.pi)} ? sorry if I'm missing something obvious there.


23 Sep 2011, 08:07
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