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        riddles >> easy >> e^pi = 1
        
(Message started by: Icarus on Aug 26^th, 2003, 5:03pm)

Title: e^pi = 1
Post by Icarus on Aug 26^th, 2003, 5:03pm

[smiley=e.gif][smiley=suppi.gif] = 1.

Proof:
Euler's formula tells us that

[smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif][smiley=supi.gif] = [smiley=c.gif][smiley=o.gif][smiley=s.gif] 2[pi] + [smiley=i.gif] [smiley=s.gif][smiley=i.gif][smiley=n.gif] 2[pi] = 1

so [smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif] = ([smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif][smiley=supi.gif])[smiley=supminus.gif][smiley=supi.gif] = 1[smiley=supminus.gif][smiley=supi.gif] = 1 (since 1 raised to any power is still 1).

[smiley=e.gif][smiley=suppi.gif] = [sqrt][smiley=e.gif][smiley=sup2.gif][smiley=suppi.gif] = [sqrt]1 = 1.
QED

Since the Gelfond-Schneider Theorem shows that [smiley=e.gif][smiley=suppi.gif] is a transcendental number, it follows that 1 is transcendental, and thus is not an integer as people have generally believed.

Title: Re: e^pi = 1
Post by Sir Col on Aug 26^th, 2003, 6:52pm

You're very naughty, Icarus.

Perhaps I could present one along similar lines...
e^[pi]i = cos[pi] + i sin[pi] = –1.

As e^2[pi]i = 1 (see previous post), we can raise both sides to the power 1/2, (e^2[pi]i)^1/2 = 1^1/2 [bigto] e^[pi]i = 1, but we have just shown that e^[pi]i = –1.

[therefore] 1 = –1, and multiplying both sides by x leads to the generalisation that x = –x; in other words, all positive numbers are exactly equal to their negative counterpart. Of course, we already knew that. ::)

It's an interesting general question that you pose, Icarus. I suspect that many problems are often misplaced, with regards to difficulty. I'm never quite sure where to post problems myself, as too many standards have been set with medium/hard problems being posted in the easy section. The boundaries are certainly not clearly defined – a little like the terms in the proof.

By the way, Medium got my vote, as in the grand scheme of things it's not Hard, but it certainly couldn't be described as Easy, either.

Title: Re: e^pi = 1
Post by towr on Aug 27^th, 2003, 2:26am

Part of those proofs are along the same lines as proving -1 = 1 by squaring both sides and noticing they're equal..
[sqrt]1 = 1 [vee] [sqrt]1=-1, you can't simply pick the one that suits you..
Of course [smiley=e.gif][smiley=suppi.gif] isn't -1 either.. :P

Title: Re: e^pi = 1
Post by Sameer on Aug 27^th, 2003, 7:01am

Aren't we supposed to use De Moivre's therom here to find roots ... this was like saying that if 1 and 3 are roots of equation f(x) = 0 then 1 = 3 ?

Title: Re: e^pi = 1
Post by TenaliRaman on Aug 27^th, 2003, 11:21am

i always knew e^pi = 1.But i voted for medium as this would have allowed me to post something in the medium section :D

Title: Re: e^pi = 1
Post by Icarus on Aug 27^th, 2003, 3:29pm

on 08/27/03 at 02:26:04, towr wrote:

Part of those proofs are along the same lines as proving -1 = 1 by squaring both sides and noticing they're equal..
[sqrt]1 = 1 [vee] [sqrt]1=-1, you can't simply pick the one that suits you..

This is the case for Sir Col's variation, but for mine, I am justified in picking the positive square root because [smiley=e.gif][smiley=suppi.gif] > 0.

Quote:

Of course [smiley=e.gif][smiley=suppi.gif] isn't -1 either.. :P

8)

on 08/26/03 at 18:52:41, Sir Col wrote:

It's an interesting general question that you pose, Icarus. I suspect that many problems are often misplaced, with regards to difficulty. I'm never quite sure where to post problems myself, as too many standards have been set with medium/hard problems being posted in the easy section. The boundaries are certainly not clearly defined – a little like the terms in the proof.

That's why I decided to add the poll. It's easy for me (and would be even if I wasn't the one to come up with it), but I have specialized background that makes it so. So I decided to put it where I thought it should go, but let others have a say. If significantly more people think it should be somewhere else, I'll move it.

Title: Re: e^pi = 1
Post by Sir Col on Aug 27^th, 2003, 7:09pm

Okay, I'll take the bite, Icarus...

on 08/26/03 at 17:03:38, Icarus wrote:

The Gelfond-Schneider Theorem states that [alpha]^[beta] is transcendental if [alpha][ne]0,1 and [beta] are algebraic. Neither e nor [pi] are algebraic, so the GS Theorem does not apply.

Quote:

(since 1 raised to any power is still 1).

Really? 1^1/3 = 1, true, but 1^1/3 = -(1[pm]i[sqrt]3)/2, also. ;)

Title: Re: e^pi = 1
Post by Icarus on Aug 27^th, 2003, 7:55pm

The Gelfond-Schneider theorem does apply. You just have to be canny about it. It was proved by Gelfond and Schneider independently for the express purpose of showing the [smiley=e.gif][smiley=suppi.gif] is transcendental. (This was one of Hilbert's problems.)

But that was not part of this puzzle. It was simply a side comment to show a curious consequence of this startling result.

(-1)² = 1, but 1^1/2 = 1, not -1. This is a straightforward consequence of the definition of exponentiation of Real numbers.

But to settle any doubts:

1 = [smiley=e.gif][sup0]. (Any base will do.) So,
1 [smiley=supminus.gif][supi] = ([smiley=e.gif][sup0])[smiley=supminus.gif][supi] = [smiley=e.gif][sup0][smiley=suplp.gif][smiley=supminus.gif][supi][smiley=suprp.gif] = [smiley=e.gif][sup0] = 1.

Title: Re: e^pi = 1
Post by towr on Aug 28^th, 2003, 2:33am

on 08/26/03 at 17:03:38, Icarus wrote:

[smiley=e.gif][smiley=suppi.gif] = 1.

[smiley=e.gif] > 1 and [smiley=pi.gif] > 1 thus [smiley=e.gif][smiley=suppi.gif] > 1
so we've got [smiley=e.gif][smiley=suppi.gif] = 1 and [smiley=e.gif][smiley=suppi.gif] > 1
clearly the universe should now explode :P

Title: Re: e^pi = 1
Post by Sir Col on Aug 28^th, 2003, 7:53am

Okay, I'm quite happy with, e^{2k[pi][smiley=i.gif]} = cos(2k[pi]) + [smiley=i.gif] sin(2k[pi]) = 1.

Of course, when k=0, everything is perfect: e⁰ = 1.

For k=1, I'll accept that, e^{2[pi][smiley=i.gif]} = 1.

However, I do not accept that, e^2[pi] = 1; clearly it is not true.

So it seems that raising both sides to the power of –[smiley=i.gif] causes a problem. I don't know why, but I suspect that is where the problem lies.

By the way, Icarus, do you know how we use the GS Theorem to show that e^[pi] is transcendental?

Title: Re: e^pi = 1
Post by Sameer on Aug 28^th, 2003, 11:52am

Assume the statement is correct:

i.e.

e^[pi] = 1

Taking Natural logarithm on both sides

ln e^[pi] = ln 1
[pi] ln e = 0
[pi] = 0

Hmm the egyptians definitely miscalculated the value of [pi] ;D

Title: Re: e^pi = 1
Post by James Fingas on Aug 28^th, 2003, 12:45pm

on 08/28/03 at 07:53:51, Sir Col wrote:

For the first line, we could say that:

e^{2k[pi][smiley=i.gif]} = e^{2m[pi][smiley=i.gif]} [forall] k,m [in] [smiley=bbi.gif]

but it is evident that

e^2k[pi] [ne] e^2m[pi] [forall] k,m [in] [smiley=bbi.gif] unless k=m

This encompasses the more specific case we're dealing with here.

Title: Re: e^pi = 1
Post by Sameer on Aug 28^th, 2003, 3:12pm

on 08/28/03 at 12:45:52, James Fingas wrote:

I guess k is a multiple of m would be better.

Title: Re: e^pi = 1
Post by Sir Col on Aug 28^th, 2003, 3:48pm

I think you've cracked it, James. The multi-value function, e^{2k[pi][smiley=i.gif]} = 1, only holds for complex numbers; we cannot apply the same identity to real numbers. Raising both sides to the power [smiley=i.gif], converts it to an equation that is exclusively real and, consequently, single valued.

Title: Re: e^pi = 1
Post by Icarus on Aug 28^th, 2003, 4:18pm

Close, but not quite correct.

Yes, I do know how to use GS to show that [smiley=e.gif][smiley=suppi.gif] is transcendental. But I am going to wait until someone has correctly stated what the problem is with the proof.

Hint 1: [hide]The nasty thing about the "proof" is that every statement by itself is true with the appropriate interpretation. [/hide]

Hint 2: [hide]My previous post was a deliberate obfuscation.[/hide]

Hint 3: [hide]Now why would I not want to show the GS thing until after the puzzle is solved?[/hide]

Title: Re: e^pi = 1
Post by James Fingas on Aug 29^th, 2003, 7:16am

The fundamental problem is that e^{2[pi][smiley=i.gif]} = 1 is true, but when you raise each side to the power of -[smiley=i.gif], you are implicitly doing the following:

a = b
e^ln(a) = e^ln(b)
e^{-[smiley=i.gif]ln(a)} = e^{-[smiley=i.gif]ln(b)}

This may break down if you use two different branches of the natural logarithm function. So the rule you've broken is: if you take the logarithm of a function, you must use the same branch of the logarithm everywhere.

Title: Re: e^pi = 1
Post by Icarus on Aug 29^th, 2003, 3:58pm

That isn't how I would of described it, but it amounts to the same thing.

Like the logarithm, exponentiation is not single-valued for complex exponents. Real exponentiation suffers from this as well, for rational exponents, but in that case there is a "natural choice" - one that is favored by the mathematics itself, not just the mathematicians: the positive real value.

The same is not true for complex exponents. With the exception of base [smiley=e.gif], where the multiple values collapse, complex exponents give infinitely many values, and none is "natural". So we get around this inconvenient fact by declaring complex exponentiation a "multi-valued" function. And such expressions can represent any of the values, depending on which "branch" of the exponential function you choose.

So 1[supminus][supi] = [smiley=e.gif][sup2][smiley=suppi.gif] is true for one branch of the exponential function.
1[supminus][supi] = 1 is also true for a branch. But they are true for different branches. Therefore it is incorrect to string them together as I did.

Sir Col: By one branch of the exponential function, (-1)[supminus][supi] = [smiley=e.gif][smiley=suppi.gif]. The expression on the left satisfies the requirements of the Gelfond-Schneider Theorem, so all of its possible values must be transcendental, including [smiley=e.gif][smiley=suppi.gif].

Last I knew, another of the numbers Hilbert asked about, [pi][smiley=supe.gif], is still unknown.

Title: Re: e^pi = 1
Post by SWF on Aug 30^th, 2003, 3:28pm

The Mathworld description of G-S Theorem does not mention that it applies when [beta] is non-rational and algebraic (allowing complex values)- only irrational and algebraic. I found some other references that say it permits non-rational values. Not that Mathworld's description is wrong, just that its description of the theorem is incomplete.