1oh ok then...................
Pun in title intended :D
This is easy, but good for starters:
If f(x+a) = \frac{1+f(x)}{1-f(x)}, prove that f is periodic.
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19 Answers
1[100][101][102]....................[98]........[99].........[109]...........
9put f(x) = tan(g(x))
tan(g(x+a)) = tan( Î /4 + g(x))
tan(g(x+4a)) =tan(Î + g(x))
ie f(x+4a) = f(x)
learn this approach it makes a mockery of the long related substitutions u need to otherwise make
1bhaiyya.............is ther any mistake???? y did u depink us..............????
plzz tell naa wat is d mistake........
1akand....ur great yaar..........and uur not a starter naa......[3][3][3].......
1wel........
we got f(x)=-1/f(x+2a)
by doing sum more things we also get
f(x)=-1/f(x-2a)
so....
f(x-2a)=f(x+2a)
put x as x+2a
so
f(x)=f(x+4a)
so now its periodic with period 4a..
1NOPE................wat he proved is not periodic......ther is a lot more to it.............
1after doing sum stuff..............
we also get
f(x+a)=-1/f(x-a).......................better for starters hehe
1dont dude..........its already pinked..............and this thread is closed...........hehe
1dude amit.............u r not a starter i guess hehehe.........dont solve da....they wer meant for 11thies...............let them also think hehe [3][3][3][3]
1Put x=(x+a)
then f(x+2a)= 1+f(x+a)/1-f(x+a)
1+f(x+a)=2
1-f(x+a)=-2f(x)
therefore f(x+2a)=2/-2f(x)
1/-f(x)=f(x+2a)
hecne f is a priodic function
1ok d period is 4a...................atleast sum1 prove or ill post my solution.