f(x) = sin(x) + cos(2x)

f'(x) = cos(x) - 2 sin(2x)

f''(x) = -sin(x) -4 cos(2x)

f'''(x) = -cos(x) +8 sin(2x)

f''''(x) = sin(x) + 16 cos(2x)

so if we are going to expand as a Taylor series about 0, we need:

f(0) = 1,

f'(0) = 1

f''(0) = -4

f'''(0) = -1

f''''(0) = 16

and so our series is:

f(x) = f(0) + f'(0) x + f''(0) x^2/2! + f'''(0) x^3/3! + f''''(0) x^4/4! + ..

.....= 1 + x - 2x^2 - x^3/6 + (2/3) x^4 + ...

(note this has gone too far as this takes us up to the 5th non-zero term)

RonL