math 3

QL

It is easy to see that f(-1)=f(1) = 0   Make the substitution: x = cos(y), then the equation becomes: f(cos(2y)) = 2 cos (y) f( cos(y)), y  in [0 pi]   Let g(y) = f(cosy), so we have g(2y) = 2 cos(y) g(y)  -- call this equation 1www.ddhw.com hence g(2(pi-y)) = 2 cos(pi-y) g(pi-y) = -2 cos(y) g(pi-y) Since g(2y) = g(2(pi-y)), we have so  2cos(y)g(y) = -2 cos(y) g(pi-y), hence g(pi-y) = - g(y)    for any y != pi/2. By continuity, g(pi/2) = - g(pi/2), hence g(pi/2) = 0, hence for any y in [0 pi], we have g(pi-y) = -g(y) -- call this equation 2    Now, assume the conclusion does not hold, then the set www.ddhw.com A = { y|g(y)!=0 } must be non-empy and open, so it can be represented as union of non overlapping intervals. We can pick one of intervals with the largest length.  Since g(pi/2) = 0, this interval is either a subset of [0 pi/2] or [pi/2 pi]. By equation 1, it can not be a subset of [0 pi/2] By equation 1 and 2, it can not be a subset of [pi/2 pi] Contradiction.         www.ddhw.com