A scalar field on R^{n} is a function f : R^{n} → R.

Visualized by drawing graph in R^{n} × R or by drawing level sets in domain R^{n}.

A vector field on R^{n} is a function F : R^{n} → R^{n}.

Visualized by drawing vectors in domain R^{n}. Vector Fields can represent many things Example 1: multiple paths. Vectors represent velocity (tangent) vectors of paths. Defn: A flow line of a vector field F : R^{n} → R^{n} is a differentiable path x : I ⊂ R → R^{n} such that

x′(t) = F (x(t))

That is that tangent vector to x at time t is x′(t) = F (x(t)).

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ex 8 (p. 212): F : R^{2} → R^{2}, F (x, y) = (−y, x) To find flow line(s) need x : R → R^{n} such that x′(t) = F (x(t))

I.e, need (x′(t), y′(t)) = F (x(t), y(t))

I.e, need (x′(t), y′(t)) = (−y(t), x(t))

That is need to solve x′(t) = −y(t) and y′(t) = x(t). Solution: x(t) = (acost − bsint, asint + bcost) Check: x′(t) = (−asint − bcost, acost − bsint) F (x(t)) = F (acost − bsint, asint + bcost) = (−asint − bcost, acost − bsint)

Example 2: A vector field can represent a gradient field. Vectors = ∇f, for some scalar field f. Defn: A gradient field on R^{n} is a vector field F : R^{n} → R^{n} such that F (x, y) = ∇f, for some scalar field f : R^{n} → R f is called the potential function for F.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Ex: f : R^{2} → R, f(x, y) = x^{2} + 3y^{2}, ∇f = (2x, 6y). F : R^{2} → R^{2}, F (x, y) = (2x, 6y) is a gradient field with potential function f. Equipotential set of F = level set of f. Thus the vector F (x, y) is perpendicular to an equipotential set of F = level set of f since F (x, y) = ∇f points in the direction of steepest ascent for the terrain described by the graph of f in R^{2} × R. Magnitude of the vector F (x, y) indicate the steepness of the slope Calc 1 Review: Taylor’s Thm for f : R → R. Suppose f ∈ C^{k}, Let p_{k}(x) = f(a) + f′(a)(x − a) + ... + \frac{f^{(k)}(a)}{k!}(x − a)^{k} Then f(x) = p_{k}(x) + R_{k}(x, a) where lim_{x→a} \frac{R_{k}(x,a)}{(x−a)^{k}} = 0. Prop 1.2: If f^{(k+1)} exists, then there exists c between a and x such that R_{k}(x, a) = \frac{f^{(k+1)}(c)}{(k+1)!}(x − a)^{k+1} Estimate ln(2) using degree 3 Taylor polynomial for f(x) = ln(x) about a = 1 f(x) = ln(x) f′(x) = x^{−1} f′′(x) = −x^{−2} f′′′(x) = 2x^{−3} p_{3}(x) =

Thus ln(2) ∼ p_{3}(2) =

ln(x) = p_{3}(x) + R_{3}(x − 1) f^{(4)}(x) = −6x^{−4} = −\frac{6}{x^{4}} R_{3}(x − 1) = \frac{f^{(4)}(c)}{(4)!}(x − 1)^{4} =

where c is btwn 1 and 2.

ln(2) = p_{3}(2) + R_{3}(2, 1) = where c ∈ (1, 2)

Multivariable version: Taylor’s Thm for f : R → R. Suppose f ∈ C^{k}, Let p_{k}(x) = f(a) + Σ_{i=1}^{n}f_{x_{i}}(a)(x_{i} − a_{i}) + ...

+\frac{1}{k!}Σ_{i_{1},...,i_{k}=1}^{n}f_{i_{1}...i_{k}}(a)(x_{i_{1}} − a_{i_{1}})...(x_{i_{k}} − a_{i_{k}})

Then f(x) = p_{k}(x) + R_{k}(x, a) where lim_{x→a} \frac{R_{k}(x,a)}{||x−a||^{k}} = 0. If f : R^{n} → R ∈ C^{2}, then R_{1}(x, a) = \frac{1}{2}Σ_{i,j=1}^{n}f_{x_{i}x_{j}}(c)(x_{i} − a_{i})(x_{j} − a_{j}) for some c on the line segment joining a and x. If f : R^{n} → R ∈ C^{k+1}, then (x_{i_{k+1}} − a_{i_{k+1}}) R_{k}(x, a) = \frac{1}{(k+1)!}Σ_{i_{1},...,i_{k+1}=1}^{n}f_{x_{i_{1}} ...x_{i_{k+1}}} (c)(x_{i_{1}}−a_{i_{1}})...(x_{i_{k+1}}−a_{i_{k+1}}) for some c on the line segment joining a and x.