Gradient vector: Difference between revisions

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Unfortunately, the above notation does not make direct sense because it is not permissible to divide a scalar by a vector. To rectify this, we revisit what the <math>\epsilon-\delta</math> definition of the derivative says. It turns out that that <math>\epsilon-\delta</matH> definition can more readily be generalized to functions of vector variables. The key insight is to use the [[dot product of vectors]].
Unfortunately, the above notation does not make direct sense because it is not permissible to divide a scalar by a vector. To rectify this, we revisit what the <math>\epsilon-\delta</math> definition of the derivative says. It turns out that that <math>\epsilon-\delta</matH> definition can more readily be generalized to functions of vector variables. The key insight is to use the [[dot product of vectors]].
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==Definition as a function==
==Definition as a function==
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If the gradient vector of <math>f</math> exists at all points of the domain of <math>f</math>, we say that <math>f</math> is differentiable everywhere on its domain.
If the gradient vector of <math>f</math> exists at all points of the domain of <math>f</math>, we say that <math>f</math> is differentiable everywhere on its domain.
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==Relation with directional derivatives==
==Relation with directional derivatives==
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| generic point, point-free notation (multiple variables) || Suppose <math>f</math> is a function of a vector variable <math>\overline{x} = \langle x_1,x_2,\dots,x_n \rangle</math>. Suppose <math>\overline{u}</math> is a unit vector. We then have:<br><math>D_{\overline{u}}(f) = \overline{u} \cdot (\nabla f)</math><br>The right side here is a dot product of vector-valued functions (the constant function <math>\overline{u}</math> and the gradient vector of <math>f</math>). The equality holds [[concept of equality conditional to existence of one side|whenever the right side makes sense]].
| generic point, point-free notation (multiple variables) || Suppose <math>f</math> is a function of a vector variable <math>\overline{x} = \langle x_1,x_2,\dots,x_n \rangle</math>. Suppose <math>\overline{u}</math> is a unit vector. We then have:<br><math>D_{\overline{u}}(f) = \overline{u} \cdot (\nabla f)</math><br>The right side here is a dot product of vector-valued functions (the constant function <math>\overline{u}</math> and the gradient vector of <math>f</math>). The equality holds [[concept of equality conditional to existence of one side|whenever the right side makes sense]].
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===Relation with directional derivatives and partial derivatives===
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===Proof of relation===
===Proof of relation===


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Revision as of 04:16, 22 April 2012

Definition at a point

Generic definition

Suppose f is a function of many variables. We can view f as a function of a vector variable. The gradient vector at a particular point in the domain is a vector whose direction captures the direction (in the domain) along which changes to f are concentrated, and whose magnitude is the directional derivative in that direction.

If the gradient vector of f exists at a point, then we say that f is differentiable at that point.

Formal epsilon-delta definition

Suppose f is a function of a vector variable x¯. Suppose c¯ is a point in the interior of the domain of f, i.e., f is defined in an open ball centered at c¯. The gradient vector of f at c¯, denoted (∇f)(c¯), is a vector v¯ satisfying the following:

  • For every ϵ>0
  • there exists δ>0 such that
  • for every x¯ satisfying 0<|x¯−c¯|<δ (in other words, x¯ is in an open ball of radius δ centered at c¯, but not qual to c¯)
  • we have |f(x¯)−f(c¯)−v¯⋅(x¯−c¯)|<ϵ|x¯−c¯|

Note on why the epsilon-delta definition is necessary

Intuitively, we want to define the gradient vector analogously to the derivative of a function of one variable, i.e., as the limit of the difference quotient:

limx¯→c¯f(x¯)−f(c¯)x¯−c¯

Unfortunately, the above notation does not make direct sense because it is not permissible to divide a scalar by a vector. To rectify this, we revisit what the ϵ−δ definition of the derivative says. It turns out that that ϵ−δ definition can more readily be generalized to functions of vector variables. The key insight is to use the dot product of vectors.

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Definition as a function

Generic definition

Suppose f is a function of many variables. We can view f as a function of a vector variable. The gradient vector of f is a vector-valued function (with vector outputs in the same dimension as vector inputs) defined as follows: it sends every point to the gradient vector of the function at the point. Note that the domain of the function is precisely the subset of the domain of f where the gradient vector is defined.

If the gradient vector of f exists at all points of the domain of f, we say that f is differentiable everywhere on its domain.

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Relation with directional derivatives

Statement of relation

Version type Statement
at a point, in vector notation (multiple variables) Suppose f is a function of a vector variable x¯=⟨x1,x2,…,xn⟩. Suppose u¯ is a unit vector and a¯ is a point in the domain of f. Suppose that the gradient vector of f at a¯ exists. We denote this gradient vector by ∇f(a¯). Then, we have the following relationship:
Du¯(f)(a¯)=u¯⋅(∇f(a¯))
The right side here is the dot product of vectors.
generic point, in vector notation (multiple variables) Suppose f is a function of a vector variable x¯=⟨x1,x2,…,xn⟩. Suppose u¯ is a unit vector. We then have:
Du¯(f)(x¯)=u¯⋅(∇f(x¯))
The right side here is a dot product of vectors. The equality holds whenever the right side makes sense.
generic point, point-free notation (multiple variables) Suppose f is a function of a vector variable x¯=⟨x1,x2,…,xn⟩. Suppose u¯ is a unit vector. We then have:
Du¯(f)=u¯⋅(∇f)
The right side here is a dot product of vector-valued functions (the constant function u¯ and the gradient vector of f). The equality holds whenever the right side makes sense.
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{{#widget:YouTube|id=RvponqyVtFU}}

Relation with directional derivatives and partial derivatives

Fill this in later

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Proof of relation

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