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

1 Answer
Sep 18, 2015

#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.