Question

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

Answer 1

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

Method 1 is clearly superior for this example.