I wonder if there is a way to calculate the
Sn=sinx+sin2x+…+sinnx
but using only derivatives ?
Answer
Using telescopic sums:
sin(mx)sin(x/2)=12(cos((m−1/2)x)−cos((m+1/2)x))
Hence: Snsinx2=12(cosx2−cos((n+12)x))=sinnx2⋅sin(n+1)x2.
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
No comments:
Post a Comment