The Maclaurin series is a special case of the Taylor series, which is used to expand functions around the point x=0. To find the Maclaurin series expansion of the function f(x) = cos(3x), we will use the formula for the Taylor series:
f(x) = f(0) + f'(0)x + rac{f”(0)}{2!}x^2 + rac{f”'(0)}{3!}x^3 + … + rac{f^{(n)}(0)}{n!}x^n
First, we need to find the derivatives of f(x) = cos(3x) and evaluate them at x = 0:
- f(0) = cos(0) = 1
- f'(x) = -3sin(3x)
ightarrow f'(0) = -3sin(0) = 0 - f”(x) = -9cos(3x)
ightarrow f”(0) = -9cos(0) = -9 - f”'(x) = 27sin(3x)
ightarrow f”'(0) = 27sin(0) = 0 - f””(x) = 81cos(3x)
ightarrow f””(0) = 81cos(0) = 81
Now we can construct the Maclaurin series using the derivatives:
The series will include only the non-zero derivatives:
f(x) = 1 + 0 imes x + rac{-9}{2!}x^2 + 0 imes x^3 + rac{81}{4!}x^4 + …
This simplifies to:
f(x) = 1 – rac{9}{2}x^2 + rac{81}{24}x^4 + …
The general form for the Maclaurin series of cos(kx) is:
cos(kx) = 1 – rac{(kx)^2}{2!} + rac{(kx)^4}{4!} – rac{(kx)^6}{6!} + …
So, substituting k = 3, we find:
f(x) = 1 – rac{(3x)^2}{2!} + rac{(3x)^4}{4!} – rac{(3x)^6}{6!} + … = 1 – rac{9x^2}{2} + rac{81x^4}{24} – rac{729x^6}{720} + …
This series converges for all x and is the Maclaurin series expansion of f(x) = cos(3x).