HW 2.1: 2, 8 (due Friday, next week) f : R^{n} → R is differentiable at a if

lim_{x→a} \frac{f(x) − f(a) − T (x − a)}{||x − a||} = 0

lim_{x→a} \frac{f(x) − f(a) − Σb_{i}(x_{i} − a_{i})}{||x − a||} = 0

y = f(a) + Σb_{i}(x_{i} − a_{i}) approximates y = f(x) f(x) = f(a) + T (x − a) + ||x − a||r(x, a)

where lim_{x→a}r(x, a) = 0

Thm 1.1: If f is differentiable at a, then 1.) f is continuous at a. 2.) All partial derivatives exist at a. 3.) b_{i} = (\frac{∂f}{∂x_{i}})_{a} Proof: 1.) lim_{x→a}f(x) = lim_{x→a}f(a)+T (x−a)+||x−a||r(x, a) = 2,3.) \frac{∂f}{∂x_{j}} (a) = lim_{h→0} \frac{f(a+he_{j})−f(a)}{h} = lim_{h→0} \frac{f(a)+T (a+he_{j}−a)+||a+he_{j}−a||r(a+he_{j},a)−f(a)}{h} = lim_{h→0} \frac{T (he_{j})+|h|r(a+he_{j},a)}{h} = lim_{h→0} \frac{hT (e_{j})+|h|r(a+he_{j},a)}{h} Thm 1.3: If \frac{∂f}{∂x_{j}} exist for all j in a nbhd of a and if they are continuous at a, then f is differentiable at a. Defn: Let V be a nonempty open subset of R^{n}, f : V → R^{m}, p ∈ N. i.) f is C^{p} on V is each partial derivative of order k ≤ p exists and is continuous on V. ii.) f is C^{∞} on V if f is C^{p} on V for all p ∈ N (f is smooth). Chain rule 1: Suppose f : (a, b) → R^{n}, g : R^{n} → R, then

\frac{d}{dt}(g ∘ f)_{t_{0}} = D(G)_{f(t_{0})}D(f)_{t_{0}} = (b_{1}, ..., b_{n}) ( \begin{matrix}f′_{1}(t_{0}) \\ f′_{2}(t_{0}) \\ ... \\ f′_{n}(t_{0}) \end{matrix} )

= Σ_{i=1}^{n}(\frac{∂g}{∂x_{i}})_{f(t_{0})}(\frac{df_{i}}{dt})_{t_{0}}

Ex: f(t) = (t^{2}, sin(t)), D(f) = ( \begin{matrix}2t \\ cos(t) \end{matrix} )

g(x, y) = x + y^{3}, D(g) = (1, 3y^{2}) (g ∘ f)(t) = g(t^{2}, sin(t)) = t^{2} + sin^{3}(t) (g ∘ f)′(t) = 2t_{0} + 3sin^{2}(t_{0})cos(t_{0})

D(g)_{f(t_{0})}D(f)_{t_{0}} = (1, 3sin^{2}(t_{0})) ( \begin{matrix}2t_{0} \\ cos(t_{0}) \end{matrix} ),

Defn: U is starlike with respect to a if x ∈ U implies \overline{ax} ⊂ U Thm 1.5 (Mean Value Theorem) Let g by a differentiable function on an open set U ⊂ R^{n}. Let a ∈ U and suppose U is starlike with respect to a. Then given x ∈ U, there exists c ∈ R, 0 < t_{0} < 1 such that g(x) − g(a) = Σ_{i=1}^{n}(\frac{∂g}{∂x_{i}})_{p}(x_{i} − a_{i}) where p = a + t_{0}(x − a)

Cor 1.6: If |\frac{∂g}{∂x_{i}} | < K on U for all i, then for all x ∈ U, |g(x) − g(a)| < K\sqrt{n}||x − a||

Cor 1.7 If f ∈ C^{r} on U, then \frac{∂^{k}g}{∂x_{i_{1}} ∂x_{i_{2}} ...∂x_{i_{k}}} = \frac{∂^{k}g}{∂x_{j_{1}} ∂x_{j_{2}} ...∂x_{j_{k}}} where (j_{1}, j_{2}, ..., j_{k}) is a permutation of (i_{1}, i_{2}, ..., i_{k})

2.2: f : R^{n} → R^{m} Let π_{i} : R^{m} → R, π_{i}(x) = x_{i} f = (f_{1}, ..., f_{m}) where f_{i} = π_{i} ∘ f f continuous iff f_{i} continuous for all i f ∈ C^{r} iff f_{i} ∈ C^{r} for all i f ∈ C^{∞} iff f_{i} ∈ C^{∞} for all i Defn: The Jacobian matrix of f at a is

[ \frac{∂f_{i}}{∂x_{j}} (a) ]_{m×n} = [ \begin{matrix}\frac{∂f_{1}}{∂x_{1}} (a) & & ... & & \frac{∂f_{1}}{∂x_{n}} (a) \\ . & . & & & . \\ . & & . & & . \\ . & & & . & . \\ \frac{∂f_{m}}{∂x_{1}} (a) & & ... & & \frac{∂f_{m}}{∂x_{n}} (a) \end{matrix} ]

2.1 Let V be an open subset of R^{n}, a ∈ V, f : V → R^{m}. Then f is differentiable at a if and only if there is a matrix T and a function ϵ : R^{n} → R^{m} such that lim_{h→0}ϵ(h) = 0 and

f(a + h) − f(a) = T (h) + ||h||ϵ(h)

Or equivalently, there exists an m-tuple, R(x, a) = (r_{1}(x, a), r_{2}(x, a), ..., r_{m}(x, a) such that lim_{x→a}||R(x, a)|| = 0 and

f(x) = f(a) + T (x − a) + ||x − a||R(x, a)

Thm 2.2: Let f by a differentiable function on an open set U ⊂ R^{n}. Let a ∈ U and suppose U is starlike with respect to a. If |\frac{∂f_{i}}{∂x_{i}} | < K on U for all i, j, then for all x ∈ U,

||f(x) − f(a)|| < K\sqrt{nm}||x − a||

Proof: ||f(x) − f(a)|| = \sqrt{Σ_{i=1}^{m}(f_{i}(x) − f_{i}(a))^{2}}

< \sqrt{Σ_{i=1}^{m}(K\sqrt{n}||x − a||)^{2}} = \sqrt{m(K\sqrt{n}||x − a||)^{2}} = K\sqrt{nm}||x − a||

Thm 2.3 (Chain rule): Suppose U ⊂ R^{m} is open and f : U → V ⊂ R^{m}, g : V → R^{p}. Let h = g ∘ f. Suppose f is differentiable at a ∈ U and g is differentiable at f(a) ∈ V. Then h is differentiable at a ∈ U and D(h)_{a} = D(G)_{f(a)}D(f)_{a}.

Cor 2.4: If f, g ∈ C^{r} on U, V respectively, then h = g ∘ f ∈ C^{r}.