Question

How to calculate cos x from its nth term in Maclaurin series??

Answer 1

The Maclaurin series of $f(x)=\cos x$ is

 

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

 

Let us look at some details.

The Maclaurin series for f(x) in general can be found by

$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n}$

Let us find the Maclaurin series for $f(x)=\cos x.$

By taking the derivatives,

 $\begin{aligned} &f(x)=\cos x \Rightarrow f(0)=\cos (0)=1 \\ &f^{\prime}(x)=-\sin x \Rightarrow f^{\prime}(0)=-\sin (0)=0 \\ &f^{\prime \prime}(x)=-\cos x \Rightarrow f^{\prime \prime}(0)=-\cos (0)=-1 \\ &f^{\prime \prime \prime}(x)=\sin x \Rightarrow f^{\prime \prime \prime}(0)=\sin (0)=0 \\ &f^{(4)}(x)=\cos x \Rightarrow f^{(4)}(0)=\cos (0)=1 \end{aligned}$

Since $f(x)=f^{(4)}(x),$ the cycle of $\{1,0,-1,0\}$ repeats itself.

So, we have the series

$f(x)=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\cdots=\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !}$