Gradient

Theorem: Gradient

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\mathbf{a}\in \mathcal{D}$.

If $f$‘s directional derivative ${\partial }_{\hat{\mathbf{r}}}f(\mathbf{a})$ along every direction $\hat{\mathbf{r}}$ exists at $\mathbf{a}$, then there exists a unique vector $\nabla f(\mathbf{a})$, which depends on $\mathbf{a}$ but not on $\hat{\mathbf{r}}$, such that ${\partial }_{\hat{\mathbf{r}}}f(\mathbf{a})$ is the dot product of $\nabla f(\mathbf{a})$ and $\hat{\mathbf{r}}$:

$$(\nabla f(\mathbf{a}))\cdot \hat{\mathbf{r}}={\partial }_{\hat{\mathbf{r}}}f(\mathbf{a})$$

PROOF

TODO

Definition: Gradient

The vector $\nabla f(\mathbf{a})$ is known as the gradient of $f$ at $\mathbf{a}\in \mathcal{D}$.

INTUITION

The gradient at $\mathbf{a}$ shows the directions in which small deviations from $\mathbf{a}$ result in the largest increase and the largest decrease in the value of $f$:

• An infinitesimally small deviation from $\mathbf{a}$ in the direction of $\nabla f(\mathbf{a})$ will result in the greatest possible increase in the value of $f$. If the deviation from $\mathbf{a}$ is in any other direction, then the increase in the value of $f$ will necessarily be smaller (closer to 0).
• An infinitesimally small deviation from $\mathbf{a}$ in the direction of $-\nabla f(\mathbf{a})$ will result in the greatest possible decrease in the value of $f$. If the deviation from $\mathbf{a}$ is in any other direction, then the decrease in the value of $f$ will necessarily be smaller (closer to 0).

NOTATION

$$\nabla f(\mathbf{a})\mathrm{grad}f(\mathbf{a})$$

Note: Gradient as a Function

If there is no specific $\mathbf{a}$ mentioned, then the term “gradient” usually refers to the entire vector field $\nabla f:\mathcal{D}\to {\mathbb{R}}^n$ which to each $\mathbf{a}$ assigns $\nabla f(\mathbf{a})$.

Theorem: Gradient in Cartesian Coordinates

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field.

If $f$ is partially differentiable with respect to Cartesian coordinates at $\mathbf{a}$, then the components of its gradient there are precisely its partial derivatives with respect to Cartesian coordinates at $\mathbf{a}$:

$$\nabla f(\mathbf{a})=\left[\begin{array}{c}\frac{\partial f}{\partial x^1}(\mathbf{a})\\ ⋮\\ \frac{\partial f}{\partial x^n}(\mathbf{a})\end{array}\right]$$

PROOF

TODO