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# The Proof is Trivial! watch

1. (Original post by DJMayes)

The exponential function is not multi-valued; it is the logarithm that is multi-valued. 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 single-valued. The square root by definition implies the positive root which is why is always written in the cases when both are used.
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
2. (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-
That isn't the point DJMayes was trying to make. e^x is actually a single-valued, many-to-one function. This is why the inverse (logarithm) is multi-valued, giving infinitely many solutions for a given input.
3. (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
Aren't e^x and exp(x) interchangable? If e^x is not defined to be that sum or one of the other equivalent definitions then what is it defined as?
4. 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 3rd 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...
5. (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 3rd 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...
When defining powers for rational a and b, a^b is defined as expected when b is an integer, a^1/b is defined as the b'th root of a (the unique positive number, x, such that x^b=a). Then a^(b/c) is defined as (a^(1/b))^c. If b is irrational then it is defined as the limit of the sequence a^(b_n) where b_n converges to a, and it can be shown that this gives teh same value whatever sequence you take. If a is irrational then it is defined by a^b=e^(a*log(b)) where e^x is defined by the power series. It can be shown that this is consistent with the earlier definitions.

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.
6. (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).
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?
7. (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?
You are sort of right, things get more complicated when you are raising a non-positive number to some power. If a is not a positive real, then a^b is, in general, multivalued. This is because the extention of the definition for powers gives a^b=e^(b*log(a)) and log is multivalued. Like logs, powers do not general have a well defined value when working with complex numbers. However things still work out very nicely when a is positive, so e^x is still single valued.
8. (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
Starting with this assertion lets see what trouble we can stir:

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 multi-valued 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.
9. (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)
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.
10. (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.
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 multi-valued 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?
11. (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 multi-valued 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?
Yes, but it has another flaw. When you take logs on the complex plane you lose some of it's properties. For example it is not true that log(a^b)=b*log(a) which is what you used here.
12. 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.
13. ^ Lol, Mathematicians.
14. Well, let me just offer my 2-cents 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.
15. (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?
Not really that ludicrous. All you're doing is considering cosets in the quotient ring () rather than individual elements, (which yes, is a good idea). No single-valued function can cover all the possible properties we demand of the complex logarithm, and so, rather than look at the logarithm as mapping the plane to itself, one can look at it as mapping the plane to its quotient ring with being the associated ideal. Here, in this modified arena, we find that the complex logarithm truly comes into its own, and satisfies almost all the properties one needs it to.
16. Good to know I'm not that bonkers haha.
17. (Original post by Elie Bergman)
Good to know I'm not that bonkers haha.
"not that bonkers"
18. Problem 436**

Prove that if is a subgroup of , then .

Obviously it falls out from Lagrange's theorem directly, but is there another way?
19. Well, 2 days have passed, and I have no idea what kind of proof you're looking for, but one proof certainly is trivial:

Spoiler:
Show
Solution 436 If , then of course
20. (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:

Spoiler:
Show
Solution 436 If , then of course
Sn has size n! not n.

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