- Consider the function $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ given by \begin{equation*} f(x,y) := e^{x^2+y^2}-1. \end{equation*}
Find the directional derivative of $f$ in the direction of $\mathbf{u} = (u_1,u_2)$ at the point $(0,0)$.
To get the directional derivative of $f$ in the direction of $\mathbf{u} = (u_1,u_2)$ we need to normalise $\mathbf{u}$ and take the dot product with $\nabla{f}$. So $||\mathbf{u}|| = \sqrt{u_1^2+u_2^2}$, so $\hat{\mathbf{u}} = \frac{1}{\sqrt{u_1^2+u_2^2}}(u_1,u_2) = \left(\frac{u_1}{\sqrt{u_1^2+u_2^2}},\frac{u_2}{\sqrt{u_1^2+u_2^2}}\right)$. Hence, the directional derivative of $f$ at the point $(0,0)$ in the direction of $\mathbf{u} = (u_1,u_2)$ is \begin{equation*} \hat{\mathbf{u}}\cdot \nabla{f} = \begin{bmatrix} \frac{u_1}{\sqrt{u_1^2+u_2^2}} & \frac{u_2}{\sqrt{u_1^2+u_2^2}} \end{bmatrix}\cdot \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0. \end{equation*}
- The error function $\text{erf} : \mathbb{R}\mapsto \mathbb{R}$ is defined by \begin{equation*} \text{erf}(x) := \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} \; dt. \end{equation*}
Explain why erf is infinitely differentiable and find the quadratic Taylor polynomial of erf about $a = 0$.
Is question 1 good and for question 2 do we show there is an nth derivative for erf(x) and not too sure about the Taylor polynomial.
