Question

How do you find the taylor series for `e^(x^2)`?

Answer 1

`1+x^2+(x^4)/(2!)+(x^6)/(3!)+(x^8)/(4!)+cdots`. This converges to `e^(x^2)` for all values of `x`.

Explanation:

This can be obtained most simply by taking the well-known Taylor series for `e^(x)` centered at 0, which is `1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+cdots`, and replacing `x` with `x^2`.

You can also try using the formula `f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+cdots`, but that is a much less pleasant approach.