1ok....... got............
Show dat there is no real constant c>0 such dat cos√(x+c)=cos√x for real no.s x ≥ 0.
-
UP 0 DOWN 0 0 3
3 Answers
62cos√(x+c)=cos√x
so
√(x+c)=√x + 2nÎ
so
√(x+c)-√x = 2nÎ
for all x!
LHS is a continuous function (which is not constant! unless c=0) ... RHS is a discrete function! they cant be equal!
62If the previous proof is not convincing enuf..
√(x+c)-√x = 2nΠfor all x
hence √(x+c)-√x = 2nΠfor 0 and a fixed n=k
hence
√(0+c)-√0 = 2kÎ
√c = 2kÎ
c=(2kÎ )2
so we have limited the value of c to this value..
But this does not hold for x=c bcos
at x=c
√(x+c)-√x = √2c-√c = √c(√2 - 1) = 2kΠ(√2 - 1)
but this has to also be equal to 2nπ for some other n .. let for n=t
so
2kÎ (√2 - 1) = 2tπ where t is integer..
so k(√2 - 1) = t
√2 = t/k+1
Lhs is irrational.. RHS rational.. contradiction...
also,