Gradient

$\pi _{1}(S^{1})=?\,$

This article/section deals with mathematical concepts appropriate for a student in late university or graduate level.

In mathematics, the gradient is a vector associated to a point $p$ of a differentiable function $f(x_{1},...,x_{n})$ which takes real values. Specifically, the gradient at $p$ is a vector in $R^{n}$ which points in the direction in which $f$ increases most rapidly at $p$. The magnitude of the gradient at $p$ is equal to the maximum directional derivative of $f$ at $p$. The gradient is an extension of the idea of derivative to functions with more than one variable.

More precisely, we define the gradient, $\nabla f$ of $f$ to be the vector field:

$\nabla f=({\frac {\partial f}{\partial x_{1}}},...,{\frac {\partial f}{\partial x_{n}}})$

consisting of the various partial derivatives of $f$. If $u$ is a unit vector in $R^{n}$, then, by the chain rule, the directional derivative of $f$ in the direction of $u$ is simply the dot product:

$\nabla f\cdot u$

Evidently by the Cauchy-Schwartz inequality, the directional derivative in the direction $u$ is maximal in the direction of the gradient, and equal to $||\nabla f||$ for $u$ a unit vector in the direction of the gradient.

Properties of the Gradient

If $f$ is a differentiable function with smooth level sets $f^{-1}(c)$, then the gradient vector field $\nabla f$ is perpendicular to the level sets of $f$. For fix a level set $S=f^{-1}(c)$, and let $v$ be a vector tangent to $S$ at $p$. Then we can find a curve $\gamma (t)$ on $S$ with $\gamma '(0)=v$. Now

$f\circ \gamma (t)=c$

since $S$ is a level set. Taking derivatives of both sides and applying the chain rule, we get that

$\nabla f\cdot \gamma '(0)=\nabla f\cdot v=0$

Thus, $\nabla f$ is perpendicular to $v$ at $p$, i.e., the gradient of $f$ is perpendicular to the level sets of $f$.