Find the Cubic MacLurin expansion of e^{x^2}.
First, I tried the sub $t=x^2$ and used the regular expansion for $e^t$. But that was wrong. Can I not do non-linear substituions?
My calculations:
$ e^t = 1 + t + \dfrac{t^2}{2} + \dfrac{t^3}{3!} + \ldots\\ t = x^2\\ e^{x^2} = 1 + x^2 + \dfrac{x^4}{4} + \dfrac{x^6}{6} + \ldots\\ $
Direct method:
$ \begin{align} &f = e^{x^2} & f(0) = 1\\ &f' = 2xe^{x^2} & f'(0) = 0\\ &f'' = 2(f'x + f) & f''(0) = 2\\ &f''' = 2(f''x + f' + f') & f'''(0) = 0 \end{align}\\ f(x) = e^{x^2} = 1 + 0 + \dfrac{\not 2}{\not 2!}x^2 \ \leftarrow \text{Correct}\\ f(x) = f(0) + f'(0) + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 $
As you are aware already, the Maclaurin expansion of $e^t$ is$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots$$ Replacing $t$ with $x^2$ gives \begin{align}e^{x^2}&=1+(x^2)+\frac{(x^2)^2}{2!}+\frac{(x^2)^3}{3!}+\cdots \\ &= 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\cdots\end{align}