Question

Find the Maclaurin series for the specified function:

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

Short answer

The Maclaurin series is,

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

Step-by-step solution

Step 1. Given Information.

The function is,

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

Step 2. Formulae for the Maclaurin series.

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

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

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

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

localid="1649653268867">f(x)=∑n=0∞fn(0)n!xn.

Step 3. Finding the Maclaurin series.

So, let us first construct the table of the Maclaurin series for the function $f\left(x\right)=\sin \left(x\right)$,

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

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

Or, we can write as,

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