24i drew the graph too..but i ma not getting it...can u post ur graph ?
the statemnet is
sin4π{x}=sin(4πx) for all x εR
its given to be true...
but i dont think it to be....
what do u say ?
-
UP 0 DOWN 0 0 6
6 Answers
24
1it will be true
see sinx lies between -1 to 1
as sinx=sin(pie-x) therefore when the value of sinx is from -1 to 0
the graph of both will come out to be same
and for 0 to 1
{x} will be equal to x so both graphs will be same
and for x=1
{x}=0
so sin4Ï€{x}=0=sin(4Ï€x)=sin4Ï€=0
as the function is periodic and as the graph is coming same in(-Ï€,Ï€)
therfore it will be same for all x εR
24but area enclosed isnt coming same...i used a software on my friends PC ...it agave diif results for both
1i think it is true...coz sin ( 4Î {x}) can be written as sin ( 4Î (x-[x]) )
=> sin (4Î x -4Î [x] ) .........since [x] is an integer due to greatest integer function...then its equivalent to
=> sin (4Î x)....
can ne1 say whether i am right or wrong??
3let a function be f(x)=sin4Ï€x-sin(4Ï€{x})=2sin4Ï€[x-{x}]cos4Ï€[x+{x}]
and sin (npi)=0 so f(x)=0