Integral formulas for Fourier coefficients
n cos(nx) = ˇ 2 + X1 k=0 4 ˇ(2k + 1)2 cos((2k + 1)x) = ˇ 2 4 ˇ X1 k=0 cos((2k + 1)x) (2k + 1)2: Remarks: Since k is simply an index of summation, we are free to replace it with n again, yielding ˇ 2 4 ˇ X1 n=0 cos((2n + 1)x) (2n + 1)2: Because f(x) is continuous