Given a function f(x), its Maclaurin series will be as follows:
f(x) = f(0) + f'(0)x + f''(0)x^2/(2!) + f'''(0)x^3/(3!) + ...
In order to find the Maclaurin series, we have to calculate the coefficients a_n = f^((n))(0)/(n!) and f(0).
i) Calculating f(0)
f(0) = e^0cos(0) = 1.
ii) a_n coefficients
First derivative
f^((1))(x) = 3e^(3x)cosx - e^(3x)sinx .
f^((1))(x) = e^(3x)(3cosx - sinx).
Then:
a_1 = f^((1))(0)/(1!) = 3/(1!).
We proceed to calculate the next derivatives.
f^((2))(x) = 3e^(3x)(3cosx - sinx) - e^(3x)(3sinx + cosx);
f^((2))(x) = e^(3x)(9cosx - 3sinx - 3sinx - cosx);
f^((2))(x) = e^(3x)(8cosx - 6 sinx);
f^((2))(x) = 2e^(3x)(4cosx - 3sinx).
Then:
a_2 = f^((2))(0)/(2!) = 8/(2!).
Third derivative.
f^((3))(x) = 6e^(3x)(4cosx - 3sinx) - 2e^(3x)(4sinx + 3cosx);
f^((3))(x) = 2e^(3x)(12cosx - 9sinx - 4sinx - 3cosx);
f^((3))(x) = 2e^(3x)(9cosx - 13sinx).
Then:
a_3 = f^((3))(0)/(3!) = 18/(3!).
Fourth derivative.
f^((4))(x) = 6e^(3x)(9cosx - 13sinx) - 2e^(3x)(9sinx + 13cosx);
f^((4))(x) = 2e^(3x)(27cosx - 39sinx - 9sinx - 13cosx);
f^((4))(x) = 2e^(3x)(15cosx - 48sinx);
f^((4))(x) = 6e^(3x)(5cosx - 16sinx).
Then:
a_4 = f^((4))(0)/(4!) = 30/(4!).
Writting our series:
f(x) = 1 + 3/(1!)x + 8/(2!)x^2 + 18/(3!)x^3 + 30/(4!)x^4 + ...
I am sorry, but I was not able to identify a clear pattern to write it as a general sum. But I hope I could help you somehow, anyway!
By the way, you could continue this calculation until the order you want.
