I don't know what benefits I think people will have by watching this youtube? It took me little time to find. It "proofs" MW's 0^0 = 1 theorem. Enjoy and discover the flaw. http://www.youtube.com/watch?v=b9q24AS2mR0.
if y * x = y then necessarily x = 1, isn't it ? example 0 * 7 = 0 so 7 = 1 !
It is the oldest hamete in math and it is exactly the same reasoning the smart lady in the video is using.
Shouldn't that be "-1/0" on the second line I quoted? I don't see how you get from one of these to the other. Unless you're trying to assign some sort of literal value to x/0, in which case it should probably be ([-inf,inf]) and the solution from that point forward will look like a resolution for a quantum mechanics equation or something.
Re: Math puzzles
Posted: Tue Jan 25, 2011 5:44 am
by cyclops
@ethanb
If it is secret I'll hide too. Trust Wikileaks! If I understand you well you question my 5e line. At least its derivation. So I assume you buy my fourth line. _ 4_ : 0 * ( -inf ) = 0 . ( it comes from 0^0 = 1, taking the log from 1, and applying the powerrule for logs )
If a * b = c then 1/a * 1/b = 1/c ( even in some sense if c equals zer0 : both sides of the second equation here are inf then.) so from _ 4_ we get 1/0 * 1/(-inf) = 1/0 q.e.d.
About quantummechanic equations: Schöderinger's and Dirac's equations allow healthy solutions in physical cases. In the computing proces maybe distributions are handy but even these are respectable mathematical objects. Only with renormalisation it gets tricky, only actually to deal with infinities from the recursive aspects that comes along with general relativity. Disclaimer: long time ago I studied these things.
Re: Math puzzles
Posted: Tue Jan 25, 2011 10:45 am
by ethanb
cyclops wrote:@ethanb
If it is secret I'll hide too. Trust Wikileaks! If I understand you well you question my 5e line. At least its derivation. So I assume you buy my fourth line. _ 4_ : 0 * ( -inf ) = 0 . ( it comes from 0^0 = 1, taking the log from 1, and applying the powerrule for logs )
If a * b = c then 1/a * 1/b = 1/c ( even in some sense if c equals zer0 : both sides of the second equation here are inf then.) so from _ 4_ we get 1/0 * 1/(-inf) = 1/0 q.e.d.
About quantummechanic equations: Schöderinger's and Dirac's equations allow healthy solutions in physical cases. In the computing proces maybe distributions are handy but even these are respectable mathematical objects. Only with renormalisation it gets tricky, only actually to deal with infinities from the recursive aspects that comes along with general relativity. Disclaimer: long time ago I studied these things.
Well, if it's a puzzle, I didn't want to spoil anything.
It's not the derivation of the 5e line I'm questioning... I'm pretty comfortable with distribution and commutation!
It's the derivation of the 6e line. How do you find this: -1 * (1/0) = inf ?
Re: Math puzzles
Posted: Tue Jan 25, 2011 2:45 pm
by cyclops
@ethanb
One of us has a blind spot. Only one-eyed I'm the suspect. Better to discuss string theory. For the ease of reading I included some brackets. LHS: push the minus sign to the front. RHS: substitute inf for 1/0. Can't make it easier
Actually I'm not too proud of this whole "derivation". I think there is a stronger case. For example that for nice operators "op" we expect lim ( f(x) op g(x) ) = ( lim f(x)) op ( lim g(x) ) if RHS exists. MW's definition would make ^ unnice.
Re: Math puzzles
Posted: Tue Jan 25, 2011 9:38 pm
by ethanb
cyclops wrote:@ethanb
One of us has a blind spot. Only one-eyed I'm the suspect. Better to discuss string theory. For the ease of reading I included some brackets. LHS: push the minus sign to the front. RHS: substitute inf for 1/0. Can't make it easier
Actually I'm not too proud of this whole "derivation". I think there is a stronger case. For example that for nice operators "op" we expect lim ( f(x) op g(x) ) = ( lim f(x)) op ( lim g(x) ) if RHS exists. MW's definition would make ^ unnice.
Er, I understand *what* you did... I'm just saying it doesn't make sense to me, so I'm asking why you think it's possible. Could be I'm wrong.
Usually laws of equality and commutation work together to help solve equations so that you do something like this (same thing on each side of the equality symbol): LHS: multiply every term by -1. RHS: multiply every term by -1
You seem to have done... well, what you say above, which doesn't make sense mathematically (unless I'm being very foolish somehow.) It looks like this to me: LHS: multiply every term by -1. RHS: 1/0 ??? infinity ??? profit!
Shouldn't that be "-1/0" on the second line I quoted? I don't see how you get from one of these to the other. Unless you're trying to assign some sort of literal value to x/0, in which case it should probably be ([-inf,inf]) and the solution from that point forward will look like a resolution for a quantum mechanics equation or something.
I'm not sure I see your problem, ethanb - what cyclops appears to have done is to simply rewrite the left-hand side, using that (-a)*b=a*(-b), and then write "inf" for 1/0 on the right, which at least seems intuitively plausible.
Of course, this is all horribly unrigorous, and makes especially little sense if you care about signs and are going to make a distinction between plus and minus infinity (as appears to be the case here). Saying 1/0 is infinity seems reasonable when we consider that 1/x tends to infinity as x tends to 0 from above, but if x tends to 0 from below then 1/x tends to minus infinity!
D - the series of sums which contain the first n terms of the sequence doesn't converge, so the sum of the entire sequence is undefined. Specifically, the radius never gets smaller than 1 - if s_n =1, s_n+1 =0, and vice versa.
