Question

Find the Maclaurin series for the specified function:

$\cos \left(x\right)$.

Short answer

The Maclaurin series is,

$f\left(x\right)=\sum _{k=0}^{\infty }{(-1)}^k\frac{1}{\left(2k\right)!}{x}^{2k}$.

Step-by-step solution

Step 1. Given Information.

The function is,

$\cos \left(x\right)$.

Step 2. Formulae the Maclaurin series.

Let $f\left(x\right)=\cos \left(x\right)$.

Since for any function $f$with derivatives of all orders at the point $x_0=0$, then the Maclaurin series is,

localid="1649791589053" $f\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4+.....$

Or we can write the general form Maclaurin series of the function is,

$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n$

Step 3. Finding the Maclaurin series.

The table of the Maclaurin series for the function is,

The Maclaurin series for the function$f\left(x\right)=\cos \left(x\right)$is,

$f\left(x\right)=1+0x+\frac{(-1)}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{1}{4!}{x}^4+\frac{0}{5!}{x}^5+\frac{\left(-1\right)}{6!}{x}^6+.....$

Or, we can write it as,

$f\left(x\right)=\sum _{k=0}^n{(-1)}^k\frac{1}{\left(2k\right)!}{x}^{2k}$.