WIERD
Hey folks,
Figure out the problem in this sequence of reasoning.
e^{2π.i }= 1
then for any x
e^{ix} = e^{ix}.e^{2π.i}= e^{i(x+2π)}
=> (e^{ix})^{i} = {e^{i(x + 2π)}}^{i }
^{=> e-x = e-x . e-2π=> e-2π = 1=> -2π = 0 Swetabh}^{}
Hey folks,
Figure out the problem in this sequence of reasoning.
e^{2π.i }= 1
then for any x
e^{ix} = e^{ix}.e^{2π.i}= e^{i(x+2π)}
=> (e^{ix})^{i} = {e^{i(x + 2π)}}^{i }
^{=> e-x = e-x . e-2π=> e-2π = 1=> -2π = 0 Swetabh}^{}
Author : bravura at 5/30/2006 02:51:00 PM
9 comments:
Hii pal
The problem is when to power the sides to 'i' ...
Its not against logic, but it adds certain conditions. Ill write more about it later, g2g rite now.
~ Twish ~
hi swetabh,
what my FUSED mind thinks is that u are trying to equate two undefined terms.if u see it in this way,
e^(ix)^i=cos(ix)+i sin(ix).
ix is a complex term. so it does not come under the domain of cos or sine functions, which become undefined there. so the two terms RHS & LHS in ur equation, both become undefined. therefore they can't be equated..
bhagyashree.
hi swetabh,
i wanted 2 ask something else 2. what's the great logic behind ur name:bravura?
bhagyashree
Hey Bhagyashree,
Don't u think the name sounds really wierd and ferocious. I think it does but it also means something. I kinda just wanted my name to be different.
Swetabh
You've to ask yourself the question... wat happens when we raise something to a complex power ? Its wrong to say that e^(-x) is the only result of e6^(ix)^i.
How ? take m^n.Here while n is a natural nember, we have one simple solution. But as soon as we allow n to be fractions, things get messed up we get many but a finite solution. Similarly talkin of complex powers , when you raise to a complex power , you don't get 1 or 2 or a finite number of answers but an infinite set ! For examlpe, over here... e^[(ix)*i] is equal to e ^[2nπ - x] and similarly e^[(x+2π)*i*i] is equal to e^[ 2mπ - x]. Now here you've taken the principal values of the arguements & hence arises the falacy. Equating the above two and putting suitable values for n & m the fallacy vanishes! As for using principal arguements, it doesn't work out for its like saying that the cuberoots of 1 are equal just because 1^(1/3) = w & w^2 & 1 ....
If you want I can still post further explaining man.
~ Twish ~
I reckon twish is rite... u messed up with the principal and general arguments - hope u got your answer!
Hey every1,
the problem with this bit of reasoning is that here i have raised something to the power i.
this statement in itself is undefined. it is like saying that 4=2^2 therefore 4^i =(2^2)^i. This statement is not valid.
Swetabh
Hey SWETABH !!!
YOU'RE WRONG HERE PAL ! THE STATEMENT IS VERY MUCH DEFINED AND COMLPLEX POWERS ARE COMMON IN HIGHER MATHS. HOWEVER, ALONG WITH THAT THERE ARE INFINITE SOLUTIONS WHEN RAISED TO COMPLEX POWERS. GO READ SOME TEXT ON THIS TOPIC FROM SOMEWHERE YAAR.
~ TWISH ~
Hey Twish,
I certainly didn't know about that.
I'll study a little and then answer back. thanks for the info.
Swetabh
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