Resolved: "What is the taylor series of cos2(x)?"

Whothoromapyfb

Answered question

2023-01-07

What is the taylor series of

\cos^{2} (x)

tjaran12f

Beginner2023-01-08Added 13 answers

I'll assume you're referring to the Maclaurin series, also known as the Taylor series about 0.
The Taylor series for

\cos x

about 0 is:

\sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{x^{2 n}}{(2 n)!} = \frac{1}{0!} - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots

Rather than trying to multiply this series by itself to get the series for

\cos^{2} x

note that:

\cos 2 x = 2 \cos^{2} x - 1

and hence:

\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)

\cos^{2} x = \frac{1}{2} (1 + \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{{(2 x)}^{2 n}}{(2 n)!})

\cos^{2} x = \frac{1}{2} (2 + \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{{(2 x)}^{2 n}}{(2 n)!})

\cos^{2} x = 1 + \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{{(2 x)}^{2 n}}{2 (2 n)!}

\cos^{2} x = 1 + \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{2^{2 n - 1} x^{2 n}}{(2 n)!}

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