unreality Posted January 5, 2009 Report Share Posted January 5, 2009 This is a good one eiπ = -1 [square both sides] e2iπ = 1 [take the natural logarithm of both sides] 2iπ = ln(1) = 0 But something is wrong! 2*pi*i does not equal 0, it equals 2pi*i, ie, ~6.28i. What went wrong? Post answers or your own mathematical fallacies Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 5, 2009 Report Share Posted January 5, 2009 this had me going for a bit the final step is where it looses. cant take ln of i. would have to evaluate in expanded form ln[cos(2*pi) + i*sin(2*pi)] = ln[1] = 0 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 5, 2009 Report Share Posted January 5, 2009 Okay the answer is as follows: 2 pi needs to be expressed as 0 to create an equivalent expression due to the fact that you are working in polar coordinates with imaginary numbers. You would end up with 0*i = 0 which is true. Quote Link to comment Share on other sites More sharing options...
0 unreality Posted January 5, 2009 Author Report Share Posted January 5, 2009 actually - as I understand it ( ) - nobody has gotten it yet ;D Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 5, 2009 Report Share Posted January 5, 2009 (edited) actually - as I understand it ( ) - nobody has gotten it yet ;D I'm pretty sure Scrappy is right, though his answer is a bit hard to follow. In polar coordinates, an angle of 2*pi radians is the same as 0 radians, and the 2*pi must be reduced to 0 to simplify the expression, just as 4*pi and 6*pi, and any 2*n*pi, where n is an integer...no matter how many times you go around the circle, you still end up at 0. When you enter ln(e^(2*n*pi*i)) for any n into a calculator, it gives you an answer of 0 because it automatically reduces the expression. edit: btw, welcome to the Den, Scrappy! Edited January 5, 2009 by lazboy Quote Link to comment Share on other sites More sharing options...
0 unreality Posted January 6, 2009 Author Report Share Posted January 6, 2009 (edited) yes, so the trig functions reduce it on their own - my calculator doesn't do what you're saying So there is at least two fallacies then the problem occurs when you square it. When you square something, the information is lost. Ie: 5^2 = 25 = -5^2 for that reason, when you take the square root of 25, you're not sure whether it's positive or negative 5, so you write ±5 well in this case, when you square -1, if you leave it as (-1)2 instead of +1, then it works: eiπ = -1 square e2iπ = (-1)2 take ln of both sides 2iπ = ln((-1)2) = 2*ln(-1) 2iπ = 2*ln(-1) drop the 2s iπ = ln(-1) Which is a true statement of course edit-typo Edited January 6, 2009 by unreality Quote Link to comment Share on other sites More sharing options...
0 unreality Posted January 6, 2009 Author Report Share Posted January 6, 2009 I know a couple more good paradoxes, but I'm waiting to see if someone else posts one first Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 6, 2009 Report Share Posted January 6, 2009 You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents. Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can Quote Link to comment Share on other sites More sharing options...
0 unreality Posted January 6, 2009 Author Report Share Posted January 6, 2009 you're mistaking it, I think. the π stands for pi, 3.14159... It looks kind of like an 'n' letter but I was trying to use the pi symbol Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 7, 2009 Report Share Posted January 7, 2009 (edited) You actually have a fallacy in the very first statement. I don't know how to do the funky formatting though, so I'm going to use ^ to represent exponents. Instead of e^(in) = -1, you should be using e^i(2n+1). After all, if n is an integer, then n can be any integer, including the even ones. The way to represent odd integers isn't n, it's 2n+1. This would modify the proof from there on out, but I can't do the formatting, and don't want to do the math, so someone else can As unreality said, the n isn't an n its a π (or pi). If you change the font, then its a bit more legible: π π π π e^(iπ) = -1 (or e^(iπ) + 1 = 0) is a famous equation*, and, for me at least, is a pretty surprising one linking so many fundamental numbers. I first read the equation as e ^ ln (or natural log) and thought unreality had made a different blunder, meaning to put ln(e) = 1. *eulers equation, i think Edited January 7, 2009 by armcie Quote Link to comment Share on other sites More sharing options...
0 unreality Posted January 7, 2009 Author Report Share Posted January 7, 2009 (edited) it's called Euler's Identity, and is a special case of Euler's Equation (I guess it's actually called Euler's Formula not Euler's Equation, whatever ) To me, it's just as surprising as a formula for i^x I discovered by accident a few months ago: i^x = cos(x * pi/2) + i*sin(x * pi/2) that's off the top of my head btw, but I'm pretty sure it's correct. Of course, the special cases give i^2 = -1, i^3 = -i, i^4 = 1, etc, likewise i^(1/2), the square root of i, gives sqrt(2)/2 + i*sqrt(2)/2 edit: more surprising is that i^i is a real number lol Edited January 7, 2009 by unreality Quote Link to comment Share on other sites More sharing options...
Question
unreality
This is a good one
eiπ = -1
[square both sides]
e2iπ = 1
[take the natural logarithm of both sides]
2iπ = ln(1) = 0
But something is wrong! 2*pi*i does not equal 0, it equals 2pi*i, ie, ~6.28i. What went wrong?
Post answers or your own mathematical fallacies
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