Answer to Differential Calculus problem. Here is a photo

Anne Louise Cabungcal

Answered question

2022-09-07

Eliza Beth13

Skilled2023-05-31Added 130 answers

To find the Maclaurin series for the function

f (x) = \cos (2 x)

, we can use the known Maclaurin series expansion for the cosine function.
The Maclaurin series expansion for

\cos (x)

is given by:

\cos (x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - \dots

To find the Maclaurin series for

f (x) = \cos (2 x)

, we substitute

2 x

for

x

in the above series:

\cos (2 x) = 1 - \frac{(2 x)^{2}}{2!} + \frac{(2 x)^{4}}{4!} - \frac{(2 x)^{6}}{6!} + \frac{(2 x)^{8}}{8!} - \dots

Simplifying further:

\cos (2 x) = 1 - \frac{4 x^{2}}{2!} + \frac{16 x^{4}}{4!} - \frac{64 x^{6}}{6!} + \frac{256 x^{8}}{8!} - \dots

Therefore, the Maclaurin series for

f (x) = \cos (2 x)

is:

f (x) = 1 - \frac{4 x^{2}}{2!} + \frac{16 x^{4}}{4!} - \frac{64 x^{6}}{6!} + \frac{256 x^{8}}{8!} - \dots

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