# How do you find the taylor series of sin (x^2)?

Jun 11, 2015

The answer (about $x = 0$) is x^2-x^6/(3!)+x^10/(5!)-x^14/(7!)+\cdots

#### Explanation:

You can get this answer in two ways:

1) Replace $x$ with ${x}^{2}$ in the well-known Taylor series for sin(x)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+\cdots (about $x = 0$).

2) Let $f \left(x\right) = \sin \left({x}^{2}\right)$ and use the formula f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots. Note that $f ' \left(x\right) = 2 x \cos \left({x}^{2}\right)$, $f ' ' \left(x\right) = 2 \cos \left({x}^{2}\right) - 4 {x}^{2} \sin \left({x}^{2}\right)$, etc... $f \left(0\right) = 0$, $f ' \left(0\right) = 0$, $f ' ' \left(0\right) = 2$, etc...

Method 1 is clearly superior for this example.