IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sandeepramesh

okay i accept that i had some other sum n some other long proof in mind for this :)

elastiboysai

neways it all boils down t ow at me n shrids said rt?

hash_include

@ ananth:
in case you didnt know, 2/

2 =

2
:)

sandeepramesh

never mind :) try the algebra sums i posted

PS: i posted this sum here by mistake n then decided not to change it as it had a good proof from the basics :)

anandghegde

$\cos \sqrt{2x} = \cos \sqrt{2(x+T)}$

is not possible...but how to prove this??? maybe

$\sqrt{2x} \pm \sqrt{2(x+T)} =2\pi k$ is not possible right??

sandeepramesh

u are arriving along the sidelines of my proof mate

anandghegde

oh sorry i noted down the question wrongly....

elastiboysai

$\cos \sqrt{2x} = \cos \sqrt{2(x+T)}$
now rt the genl xprn
square it n u'll find that T is dependent on n or sth else.
that is jus the outline.

sandeepramesh

lol youve also noted the question wrongly rotfl

elastiboysai

samthin appl here also
jus assume its periodic T per
write cosx+cos[2x=cos(x+t)+cos[2(x+t) [- root
do sum jugglery,
go on to solv for a period (it ll be depdt on n or sth lik dat)

sandeepramesh

no actually ull arrive at a nice contradiction :)

pardesi

f(0)=2,,,,say the function has a period T then
f(T)=2
so at that poit T both Cos T=1 and Cos (sqrt{2}T)=1
the first gives T=2mpi for some integre m
and the second gives sqrt{2}T=2npi for soe integer n
which is a contradiction unless both m=n=0 ...so the function can't be periodic

sandeepramesh

nice :) there you will otw arrive at a contradiction that sqrt(2) has to be rational :)

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