Question

Obtain the Taylor series expansion of cos²z about z = 0.

Answer 1

Answer on Question #84499 – Math – Complex Analysis

Question

Obtain the Taylor series expansion of $\cos^2 z$ about $z = 0$.

Solution

Apply trigonometric identity

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

Then we use the Taylor series expansion of $\cos z$

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

Substitution $2z$ instead of $z$ gives

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

Substitute (2) into (1), we get

$$\begin{aligned} \cos^2 z &= \frac{1}{2} \left(1 + \sum_{n=0}^{\infty} \frac{(-1)^n (2z)^{2n}}{(2n)!}\right) = \frac{1}{2} \left(1 + 1 + \sum_{n=1}^{\infty} \frac{(-1)^n (2z)^{2n}}{(2n)!}\right) = \\ &= 1 + \frac{1}{2} \sum_{n=1}^{\infty} (-1)^n 2^{2n} \frac{z^{2n}}{(2n)!} = 1 + \sum_{n=1}^{\infty} (-1)^n 2^{2n-1} \frac{z^{2n}}{(2n)!} \end{aligned}$$

Answer: $\cos^2 z = 1 + \sum_{n=1}^{\infty} (-1)^n 2^{2n-1} \frac{z^{2n}}{(2n)!}$.

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