Calculating directional derivatives and Taylor series

multivariable-calculus

  1. 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*}

  1. 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.

Best Answer

  1. Your answer to 1 looks good to me.
  2. Yes, to explain why a function $\ f\ $ is infinitely differentiable, you need to explain why it has an $\ n^\mathrm{th}\ $ derivative for every positive integer $\ n\ $.
  3. Presumably the "quadratic Taylor polynomial" of a function $\ f\ $ means the quadratic polynomial formed by the first three terms of its Taylor series: $\ f\left(0\right) + f'\hskip{-0.2em}\left(0\right)x +f''\hskip{-0.2em}\left(0\right)x^2\ $, so you'd have to determine what that is for $\ f=\mathrm{erf}\ $.
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