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riddles >> hard >> Picard's Theorem Proof that 0 = 1
(Message started by: Eric Yeh on Aug 2nd, 2002, 9:34pm)

Title: Picard's Theorem Proof that 0 = 1
Post by Eric Yeh on Aug 2nd, 2002, 9:34pm
Ok, one last new problem and then I'll go to sleep.  I was motivated by seeing the 2 = 1 problem in the Medium section; I know a whole family of these sorts of things, but this is my best one.  Sorry, it perhaps requires a tiny bit of math.  ;)

In complex analysis, an entire function is defined as a function which is
infinitely differentiable at every point in C (for example:  constants,
polynomials, e^x, etc.).  Picard's Theorem says that every nonconstant
entire function f misses at most one point (i.e. f(C) = C or C-{x0}).  
For example, every nonconstant polynomial hits every point, and e^x misses
only 0.

Now consider the function f(x) = e^(e^x).  Since e^x is entire, f is also
entire by the chain rule.  But it misses 0 since the base e^y misses 0,
and it misses 1 since the top e^x misses 0 so that e^(e^x) misses e^0 = 1.
But by Picard's Theorem there can be only one missing point, so the two
missing points must be the same.  Therefore, 0 = 1.

Where's the flaw in the argument?

Happy Puzzling,
Eric

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by oliver on Aug 3rd, 2002, 3:54am
...and it misses 1 since the top e^x misses 0.




Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by william wu on Aug 3rd, 2002, 4:31am
Hiya Eric; thanks for the cool riddles!

With regards to this one, I'm confused. While it's true that e^x "misses" 0, we can also say that it "misses" all the negative numbers. And that's a lot more than just one point.

Perhaps some clarification on Picard's Theorem is in order?


Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by oliver on Aug 3rd, 2002, 5:17am
william,
we have to thank you for the cool website. Thank you very much.

With regard to this riddle, I'm quite sure that my post (yes, I have a login now) is right. The riddle was to show the error in the reasoning, and I cited the part which has the error.

To your question about the negative numbers, we are talking about the complex plane, where we have the identity
ez = eRe(z)*ei*Im(z)
                  = eRe(z)*(cos(Im(z))+i*sin(Im(z)))
(hope I got that right, it's been a while ....)
Therefore, ez indeed hits all complex number besides 0.
What is wrong is the assumption that exp: C -> C is injectiv, i.e. that the only solution exp(z) = 1 is z=0.
It isn't, as one can easily see from the identity above.





Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by Eric Yeh on Aug 3rd, 2002, 6:10am
Will,

Ollie is right -- to be explicit, e^(pi i) = -1.  e^(pi i / 2) even = i.  You can figure out the rest...

Nice job Ollie.  Wow, it's so tough to keep answers secret in these forums, huh??  I guess that's the only problem with posting them on these msg boards instead of on the main page.  By the time people read the new problem, there's already an answer sitting there!!!

And people don't post much unless it's an answer or a clarification, so if the problem is stated well it'll never get to "hot topic" status until there's a soln sitting there...

Oh well.  At least a couple of mine are still waiting for solns!

Best,
Eric

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by Paul Hsieh on Aug 5th, 2002, 6:46pm
Yeah, this one isn't hard at all.  Unfortunately, the if one has not had at least some complex analysis, then people are going to be turned off from even attempting it.  Shouldn't the difficulty of the problem be related its true difficulty rather than its inaccessibility due to context?

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by Eric Yeh on Aug 5th, 2002, 8:22pm
Yes, I suppose that this one isn't really meant to be a "puzzle" in some sense, more an amusement.  I just dropped it in as a parallel to the algebraic proof of 0 = 1.

I guess I think it's interesting because it's not as easy as some of the other 0 = 1 pfs, which are usu immediately obvious (there's a div by zero, there's a limit you can't take, etc.).  In my experience, it can take a minute or two even for very bright people to see.

Best,
Eric

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by Eric Yeh on Aug 5th, 2002, 9:54pm
Actually Paul, another thought:  The soln is quite elementary, so there is no barrier of entry regarding complex analysis proper, just some knowledge of high-school level complex arithmetic (i.e. algebra II).  I've asked people who had never taken a complex analysis course in their lives, and had never heard of a holomorphic function (day 1 cmplx anal), but could still answer the question.  So while there is certainly some mathematical barrier, that barrier isn't one that requires formal training per se...

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by Paul Hsieh on Aug 6th, 2002, 4:56pm
Yes, I agree you can solve it with elementary complex number mathematics.  But in high school, things like *onto* versus *into* functions are not typically taught with any great depth.  You are asking the participant to quickly understand the core nature of the ex function in complex numbers.

A high schooler might also be very skeptical of Picards theorem -- how it is that a function can have complete continuity everywhere on the complex plane except that its missing one single point.  Heck, at an intuitive level, *I* have trouble with this, and it might overshadow the otherwise simple nature of this problem.

Title: Re: NEW PROBLEM: Picard's Theorem Proof that 0 =
Post by andi L. on Jul 22nd, 2005, 9:51am
To find a flaw in the argument it is enough to show that there is one z element C, such that exp(exp(z))=1. The following z will do the job:
z = ln(2 pi) + i pi/2.



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