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

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