If f(x)= e^x^2 show that f(2n)(0)= (2n)!/ n! . (f(2n)(x) is Differentiation of f(x)--2nth).
Thanks.
The power series for $\displaystyle e^{x^2}$ is $\displaystyle \sum_{k=0}^\infty\frac{x^{2k}}{k!}$ (just substitute $\displaystyle x^2$ for x in the usual power series for $\displaystyle e^x$). Compare that with the Taylor series formula $\displaystyle f(x) = \sum_{r=0}^\infty\frac{f^{(r)}(0)}{r!}x^r$, and you see that, for the function $\displaystyle f(x) = e^{x^2}$, $\displaystyle f^{(r)}(0) = 0$ if r is odd. If $\displaystyle r=2k$ is even, then compare the coefficients of $\displaystyle x^{2k}$ in the two series to see that $\displaystyle \frac1{k!} = \frac{f^{(2k)}(0)}{(2k)!}$.