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Exercise [05.10]
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Joined: 22 May 2008, 19:08
Posts: 18
Re: Exercise [05.10]

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:

therefore

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

While the second one could be generalized as

I think it's quite obvious that even if the are equal for .

Main solution

Nonetheless the paradox continues to exist if we consider that can have two different solutions depending on whether we solve the inner parenthesis first or we multiply the exponents.

We know that or that . So let's apply logarithms to the original expression.

Then,

, where , and are integers.

For simplicity we redefine and to get:

Separating the real and imaginary parts:

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

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

I hope this helps.

17 Jun 2008, 12:12

Joined: 30 Jun 2008, 22:14
Posts: 25
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
Re: Exercise [05.10]

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:

FALSE!
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
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
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
Supporter

Joined: 07 Jun 2008, 08:21
Posts: 235
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
Re: Exercise [05.10]
ZZR Puig wrote:

We know that or that . 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
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
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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