Data Table 7
New member
A Taylor series for f(x) = cos x, expanded about x = pi/2, will have the form
sum [g(x)] (x - pi/2)^(2k+1). Given this, g(k) =
My answer: (-1)^(k+1) / (2k + 1)!
sum [g(x)] (x - pi/2)^(2k+1). Given this, g(k) =
My answer: (-1)^(k+1) / (2k + 1)!