|| Welcome to the Blog managed by the KVPY 2005 Batch || Twish asks members to comment on the blog MaKeOvEr!! || RG says: Looks like a famine situation here || Blog glows in bright shades || KVamPys tame their own minds... with new mysterious posts on TP ?! || What is TP after all ? || KVamPYs start thinking about their Summer Projects as the Entrances are about to end.. || IIT? IISc? IISER? KVamPYs wonder where to enjoy this summer.. || Obiwan and Sunita in ISSER || Arun awaiting replies to his letter. || Swetabh ( Bhakt ) and Abhilash trying for IIT Kanpur ( Along with Twish ) || What about the next year ? Apply again for KVPY ? || Bulbs light up as blog fills up with posts. || Latest News brought to u by Twish (Twishmay) and Gr81 (RaSh) (EMAIL US NEWS) || EvErY bOdY KnOwS..... KVamPYs RoCk!! ||

Tuesday, May 30, 2006

WIERD

Hey folks,

Figure out the problem in this sequence of reasoning.


e2π.i = 1
then for any x
eix = eix.e2π.i= ei(x+2π)

=> (eix)i = {ei(x + 2π)}i


=> e-x = e-x . e-2π
=> e-2π = 1
=> -2π = 0


Swetabh

9 comments:

TwIsTeR said...

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 ~

Anonymous said...

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.

bhagyashree said...

hi swetabh,
i wanted 2 ask something else 2. what's the great logic behind ur name:bravura?
bhagyashree

bravura said...

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

TwIsTeR said...

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 ~

Gr81 said...

I reckon twish is rite... u messed up with the principal and general arguments - hope u got your answer!

bravura said...

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

TwIsTeR said...

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 ~

bravura said...

Hey Twish,
I certainly didn't know about that.
I'll study a little and then answer back. thanks for the info.
Swetabh