You are absolutely right in saying this. However e^x is NOT the same as exp(x). exp is of course single valued I would not disagree.(Original post by DJMayes)
The exponential function is not multivalued; it is the logarithm that is multivalued. Your example involving the square root of e is nonsense, and using the power series definition you can quite easily see what you are saying is wrong:
This is an absolutely convergent series everywhere and is singlevalued. The square root by definition implies the positive root which is why is always written in the cases when both are used.
Is not technically true. It should read.
Which agrees with the principal value of e^x

Elie Bergman
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 05012014 16:09
Last edited by Elie Bergman; 05012014 at 16:15. 
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 05012014 17:04
(Original post by Elie Bergman)
You are absolutely right in saying this. However e^x is NOT the same as exp(x). exp is of course single valued I would not disagree.
snip 
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 05012014 17:15
(Original post by Elie Bergman)
You are absolutely right in saying this. However e^x is NOT the same as exp(x). exp is of course single valued I would not disagree.
Is not technically true. It should read.
Which agrees with the principal value of e^x 
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 05012014 17:23
e^x is not single valued though. For simplicity lets consider general powers; a^b is only single valued when b is an integer. For example, if a^(1/3) = x, then x^3 = a. This equation has 3 solutions in complex numbers. Thus a^(1/3) is ambiguous as it describes the 3^{rd} roots of a. Of course we can remove this ambiguity by defining a^(1/3) always denote one particular value of these roots but this does not get rid of the fact that there are many possible values which can all qualify as a^(1/3).
Substitute a=e and you'll see that e^(1/3) also suffers the same ambiguity.
The difference between DJMayes approach and mine is he is saying that e^x is defined as its power series. I don't think this is so. I think e^x is defined in the same way as a^x; By repeated multiplication; e^1 = e, e^2 = e*e etc...Last edited by Elie Bergman; 05012014 at 17:29. 
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 05012014 17:49
(Original post by Elie Bergman)
e^x is not single valued though. For simplicity lets consider general powers; a^b is only single valued when b is an integer. For example, if a^(1/3) = x, then x^3 = a. This equation has 3 solutions in complex numbers. Thus a^(1/3) is ambiguous as it describes the 3^{rd} roots of a. Of course we can remove this ambiguity by defining a^(1/3) always denote one particular value of these roots but this does not get rid of the fact that there are many possible values which can all qualify as a^(1/3).
Substitute a=e and you'll see that e^(1/3) also suffers the same ambiguity.
The difference between DJMayes approach and mine is he is saying that e^x is defined as its power series. I don't think this is so. I think e^x is defined in the same way as a^x; By repeated multiplication; e^1 = e, e^2 = e*e etc...
At no point is this multivalued, what you are talking about is finding roots to say x^3=5, where there is certainly more than 1 complex root, but this does not mean that 5^(1/3) is multivalued. f(x)=x^(1/3) is not the inverse function g(x)=x^3. 
Elie Bergman
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 05012014 18:05
(Original post by james22)
a^1/b is defined as the b'th root of a (the unique positive number, x, such that x^b=a). 
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 05012014 18:34
(Original post by Elie Bergman)
This definition certainly is consistent and implies the 'singlevaluedness' of powers in general. However it does seem to pose a problem. Namely is suggests that (1)^(1/2) has no value. This is because i is not 'positive'. Perhaps this can be circumvented? 
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 05012014 19:24
(Original post by james22)
This is because the extention of the definition for powers gives a^b=e^(b*log(a)) and log is multivalued
e^(1/2) = e^log(e^1/2) = e^(1/2*log(e)) = e^(1/2(1+2Pi*K*i)) "K is an integer"
= e^(1/2 + K*Pi*i)
As we let K vary, this expression will take two values, namely + sqrt(e).
Repeating the process for e^(1/3) gives three values. Repeating the process for e^(sqrt(2)) is more interesting. Since sqrt(2) is irrational, our method yields infinitely many values, all lying on a circle in the complex plane with radius of exp(sqrt(2)). From your own assumptions, we are led to the conclusion e^x is multivalued for certain x. In general it is not true that that a^b is well defined even for positive real a. (If we assume a^b = e^(b*log(a)). How to get around this problem? I would suggest that e^x is defined as the SET:
e^x = exp(log(e^x)) = exp(x*log(e)) = exp(x(1+2Pi*k*i)) = =exp(x)*exp(x*2*Pi*K*i) where K ranges over all integers.
Why believe in such a ludicrous definition? Because it poses no problems in the sense that it is perfectly well defined for all complex x. And also more importantly it agrees with what we would expect in simple cases.
Although it seems fantastical to suppose e^x is a set rather than a number, it does not lead to the problems the more conventional definition you gave does.Last edited by Elie Bergman; 05012014 at 21:28. 
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 05012014 20:33
(Original post by Elie Bergman)
e^(1/2) = e^(1/2*log(e)) = e^(1/2(1+2Pi*K*i)) "K is an integer"
= e^(1/2 + K*Pi*i) 
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 05012014 21:26
(Original post by james22)
This is wrong, the first equality is only true if you take log(e) to be 1, in which case the second equality is false unless k=0. It's a bit like me saying that 2*2=2*2*log(e)=2*2*(1+2*pi*i*k) so 2*2 is multivalued.
z = e^Log z . Where Log is the multivalued version. This is true for all z.
thus e^(1/2) = e^Log(e^1/2)= e^(1/2*Log(e)) = e^(1/2 + K*Pi*i)
Does this make more sense?Last edited by Elie Bergman; 05012014 at 21:33. 
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 05012014 22:32
(Original post by Elie Bergman)
I should apologize as I didn't insert the intermediary step which would make my derivation seem more logical.
z = e^Log z . Where Log is the multivalued version. This is true for all z.
thus e^(1/2) = e^Log(e^1/2)= e^(1/2*Log(e)) = e^(1/2 + K*Pi*i)
Does this make more sense? 
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 05012014 23:43
In complex arithmetic, either accept a^b is a multivalued function or to accept that the exponent law does not hold. Otherwise you can "prove" all sorts of nonsense like 1 = 0.00186744273
Likewise is also incorrect since it equates a multivalued function into a single valued function.Last edited by Llewellyn; 05012014 at 23:44. 
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 06012014 08:17
^ Lol, Mathematicians.

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 06012014 11:02
Well, let me just offer my 2cents to the whole complex logarithm discussion, from a marginally more daedalean perspective.
You people need to keep in mind that, the popular view is to consider the complex logarithm as defined on a Riemannian manifold, and well, say we have, complex manifolds, and a holomorphic vector bundle, and an open cover of X; then, if is a Stein open subset, then such an inclusion of the form induces an isomorphism of cohomology groups.
(See what's going on? I know its not immediately obvious, but, if you think about it for a little while, it should be clear.)
Yes, I know this is a vast generalization. But I find it interesting that even something as simplistic as the complex logarithm can have such profound extensions.
Also, for those unfamiliar with the notation, just note that is nothing more than the holomorphic Cech complex of its parent measure.Last edited by MW24595; 06012014 at 11:19. Reason: Latex typo. 
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 06012014 11:27
(Original post by Elie Bergman)
I would suggest that e^x is defined as the SET:
e^x = exp(log(e^x)) = exp(x*log(e)) = exp(x(1+2Pi*k*i)) = =exp(x)*exp(x*2*Pi*K*i) where K ranges over all integers.
Why believe in such a ludicrous definition?Last edited by MW24595; 06012014 at 11:27. Reason: Typo, again. 
Elie Bergman
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 07012014 17:18
Good to know I'm not that bonkers haha.

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 09012014 01:31
(Original post by Elie Bergman)
Good to know I'm not that bonkers haha. 
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 11012014 21:35
Problem 436**
Prove that if is a subgroup of , then .
Obviously it falls out from Lagrange's theorem directly, but is there another way? 
FireGarden
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 14012014 18:29
Well, 2 days have passed, and I have no idea what kind of proof you're looking for, but one proof certainly is trivial:
Last edited by FireGarden; 14012014 at 18:31. 
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 14012014 18:56
(Original post by FireGarden)
Well, 2 days have passed, and I have no idea what kind of proof you're looking for, but one proof certainly is trivial:
 The Proof is 'notso' Trivial  Physics Edition
 Matrices: detA=0 > there exists a nontrivial solution to Ax=0 ...
 Stuck on a proof!
 Slight ambiguity in STEP question
 Extremely difficult lower triangular matrices question proof ...
 Is there a bottom line to what should be proven?
 Proof by Induction ( Algebra Year 1 Uni)
 Elliptic functionsproof sum of residues =0, integral substitution ...
 Recursive unprovability
 Preparing for proofbased mathematics at university

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