Consider the inner product space C[0,1] with the inner product . Let f(x)=1, g(x)=x2. Find cos Solution f(x)=1 g(x)=x^2 so im using [ ] for inner product [f,g]= limits in each integration are from 0 to 1 integrate x^2dx=1/3 ||f||=sqrt([f,f])=sqrt(1)=1 ||g||=sqrt[g,g]=sqrt0.20 now cos()=[f,g]/||f||||g||=1/3sqrt0.20 now ||f||cos()=1/3sqrt0.20 unit vector along x^2=g/||g||=x^2/sqrt0.20 so p=||f||cos()x^2/sqrt0.20=x^2/0.60 now[f-p,p] = integrate (1x2/0.60)(x2/0.60)dx=0 hence f-p is orthogonal to p ||1-p||=4/9 ||1||=1 ||p||=5/9 clearly ||1-p||+||p||=||1|| hence pythagoran theorem is verified.