1617-257/Classnotes for Friday October 14: Difference between revisions

Latest revision as of 21:50, 19 October 2016

DBN: Classes: 2016-17: 1617-257 / Navigation

#	Week of...	Notes and Links
Welcome to Math 257
Edits to the Math 257 web pages no longer count for the purpose of good deed points.
1	Sep 12	About This Class; Day 1 Handout (pdf, html); Monday, Wednesday, Friday, Day 2 Handout (pdf, html); First Week Notes.
2	Sep 19	Monday, Wednesday, Tutorial 2 Handout, Friday, Second week notes, HW1, HW1 Solutions.
3	Sep 26	Monday, Wednesday, Tutorial 3 Handout, Friday, Third week notes, Class Photo, HW2, HW2 Solutions.
4	Oct 3	Monday, Wednesday, Friday, 4th week notes, HW3, HW3 Solutions.
5	Oct 10	Monday is Thanksgiving, no class; Wednesday, Friday, 5th week notes, HW4, HW4 Solutions.
6	Oct 17	Monday, Wednesday, Friday, 6th week notes, HW5, HW5 Solutions.
7	Oct 24	Monday, Wednesday, Friday, 7th week notes
8	Oct 31	Monday; Term test 1; Wednesday, HW6, HW6 pdf, HW6 Solutions, Friday, 8th week notes.
9	Nov 7	Monday is last day to switch to MAT 237; Monday-Tuesday is UofT Fall Break; Wednesday, HW7, HW7 Solutions, Friday, 9th week notes.
10	Nov 14	Monday, Wednesday, HW8, HW8 pdf, HW8 Solutions, Friday, Lecture recordings, 10th week notes.
11	Nov 21	Monday, Wednesday, HW9, HW9 pdf, HW9 Solutions, Friday, Lecture recordings, 11th week notes.
12	Nov 28	Monday, Wednesday, HW10, HW10 Solutions, makeup class on Thursday at GB 120 at 5PM, no class and no DBN office hours Friday! 12th week notes.
13	Dec 5	Monday, Wednesday, 13th week notes Semester ends on Wednesday - no class Friday.
B	Dec 12,19,26	No classes: other classes' finals, winter break.
14	Jan 2	Class resumes Friday at RS211, no tutorials or office hours this week, Friday, Friday notes.
15	Jan 9	Monday, Wednesday, Friday, Weekly notes
16	Jan 16	Monday, Term test 2; Wednesday, HW11, HW11 inline pdf,HW11 solutions, Friday, Weekly notes
17	Jan 23	Monday; Hour 44 Handout (pdf, html); Wednesday, HW12, HW12 inline pdf, HW12 Solutions, Friday, Weekly notes
18	Jan 30	Monday, Wednesday, HW13, HW13 inline pdf, HW13 Solutions,Friday, Weekly Notes
19	Feb 6	Monday, Wednesday, HW14, HW14 inline pdf, Friday, Weekly Notes
20	Feb 13	Monday, Wednesday, HW15, HW15 inline pdf, HW15 solutions, Friday,Weekly Notes, UofT examination table posted on Friday.
R	Feb 20	Reading week - no classes; Tuesday is the last day to drop this class.
21	Feb 27	Monday, Wednesday, HW16, HW16 inline pdf, Friday, Weekly Notes
22	Mar 6	Monday, Wednesday, Friday, Weekly Notes
23	Mar 13	Monday, Term test 3 on Tuesday at 5-7PM; Wednesday, HW17, HW17 inline pdf, HW17 Solutions, Friday, Weekly Notes,Orientation_notes
24	Mar 20	Monday, Wednesday, HW18, HW18 inline pdf, Friday
25	Mar 27	No Monday class, Wednesday, Thursday (makeup for Monday), HW19, HW19 inline pdf, Friday, Weekly Notes
26	Apr 3	Monday, Wednesday; Semester ends on Wednesday - no tutorials Wednesday and Thursday, no class Friday.
F	Apr 10-28	The Final Exam on Thursday April 20 (and some office hours sessions before).
Register of Good Deeds
Riddle Repository
Add your name / see who's in!
$\int _{M}d\omega =\int _{\partial M}\omega$

Dror's notes above / Students' notes below

Symmetry of Second Partial Derivatives

(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)

Theorem

Let $A$ be an open subset of $\mathbb {R} ^{2}$ , and let $f:A\rightarrow \mathbb {R}$ be of class $C^{2}$ . Then, for all $c\in A$ , we have that $D_{1}D_{2}f(c)=D_{2}D_{1}f(c)$ .

Proof

Step 1

We begin by defining a function that will help us in our proof. Let $(a,b)\in A\subseteq \mathbb {R} ^{2}$ be an arbitrary point. We then define the function $\lambda$ as follows:

$\lambda (h,k)=f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)$ .

Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of $\lambda$ :

${\begin{aligned}D_{1}D_{2}f(a,b)&={\frac {\partial }{\partial x}}{\frac {\partial f}{\partial y}}(a,b)\\&=\lim _{h\to 0}({\frac {1}{h}})[{\frac {\partial f}{\partial y}}(a+h,b)]-[{\frac {\partial f}{\partial y}}(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)]-[\lim _{k\to 0}({\frac {1}{k}})f(a,b+k)-f(a,b)]\\&=\lim _{h\to 0}({\frac {1}{h}})[\lim _{k\to 0}({\frac {1}{k}})f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})[f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b)]\\&=\lim _{h\to 0}\lim _{k\to 0}({\frac {1}{hk}})\lambda (h,k)\end{aligned}}$

And similarly for $D_{2}D_{1}f(a,b)$ , but with the two limits taken in the opposite order.

Step 2

Now, let's make use of this function. Let us consider the square $Q=[a,a+h]\times [b,b+k]$ , where $h$ and $k$ are so small that $Q\subseteq A$ . We will show that there exist points $p,q\in Q$ such that:

${\begin{aligned}&\lambda (h,k)=D_{2}D_{1}f(p)\cdot hk\\&\lambda (h,k)=D_{1}D_{2}f(q)\cdot hk\end{aligned}}$

The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.

Consider the function $\phi$ , defined such that $\phi (s)=f(s,b+k)-f(s,b)$ . Then $\phi (a+h)-\phi (a)=\lambda (h,k)$ . By hypothesis, $D_{1}f$ exists at all points of $A$ , so we can differentiate $\phi$ with respect to the first variable of $f$ on the interval $[a,a+h]$ . By the mean value theorem, this means that there exists a point $s_{0}\in [a,a+h]$ such that $\lambda (h,k)=\phi (a+h)-\phi (a)=\phi ^{\prime }(s_{0})\cdot h=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h$ .

Now we want to apply the mean value theorem one more time. Consider the function $\gamma$ , defined such that $\gamma (t)=D_{1}f(s_{0},t)$ . By hypothesis, $D_{2}D_{1}f$ exists at all points of $A$ , so we can differentiate $\gamma$ with respect to the second variable of $f$ on the interval $[b,b+k]$ . So, by the mean value theorem, there exists a point $t_{0}\in [b,b+k]$ such that $\gamma (b+k)-\gamma (b)=\gamma ^{\prime }(t_{0})\cdot k$ . Thus:

${\begin{aligned}\lambda (h,k)&=[D_{1}f(s_{0},b+k)-D_{1}f(s_{0},b)]\cdot h\\&=[\gamma (b+k)-\gamma (b)]\cdot h\\&=\gamma ^{\prime }(t_{0})\cdot hk\\&=D_{2}D_{1}f(s_{0},t_{0})\cdot hk\end{aligned}}$

Step 3

Now we can prove the theorem. Let $c=(a,b)\in A$ , and let $t>0$ be so small that $Q_{t}=[a,a+t]\times [b,b+t]\subseteq A$ . By what we have just shown, $\lambda (t,t)=D_{2}D_{1}f(p_{t})\cdot t^{2}$ , for some $p_{t}\in Q_{t}$ .

$t$ is the length of the sides of the rectangle $Q_{t}$ , so as $t\rightarrow 0$ , $p_{t}\rightarrow c$ . By hypothesis, $D_{2}D_{1}f$ is continuous, and so as $p_{t}\rightarrow c$ , $D_{2}D_{1}f(p_{t})\rightarrow D_{2}D_{1}f(c)$ . Thus:

$\lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{2}D_{1}f(c)$ .

We can use the same argument, and the second equality from step 2, to show that:

$\lim _{t\to 0}{\frac {\lambda (t,t)}{t^{2}}}=D_{1}D_{2}f(c)$ .

Therefore, by the uniqueness of limits, the two quantities must be equal. $\square$

Handwritten Lecture Notes in PDF

MAT257 - Lecture14 (Oct 14)

@@ Line 2: / Line 2: @@
 {{1617-257:Dror/Students Divider}}
+== Symmetry of Second Partial Derivatives ==
+'''(Note: This is based off of the proof in the textbook, and may be slightly different from how it was presented in lecture.)'''
+=== Theorem ===
+Let <math>A</math> be an open subset of <math>\mathbb{R}^2</math>, and let <math>f: A \rightarrow \mathbb{R}</math> be of class <math>C^2</math>. Then, for all <math>c \in A</math>, we have that <math>D_1 D_2 f(c) = D_2 D_1 f(c)</math>.
+=== Proof ===
+==== Step 1 ====
+We begin by defining a function that will help us in our proof. Let <math>(a, b) \in A \subseteq \mathbb{R}^2</math> be an arbitrary point. We then define the function <math>\lambda</math> as follows:
+<math>\lambda(h, k) = f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)</math>.
+Why do we define such a function? In fact, we can define both of the partial derivatives in question in terms of <math>\lambda</math>:
+<math>
+\begin{align}
+D_1 D_2 f(a, b) & = \frac{\partial}{\partial x} \frac{\partial f}{\partial y} (a, b) \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\frac{\partial f}{\partial y} (a + h, b)] - [\frac{\partial f}{\partial y} (a, b)] \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b)] - [\lim_{k \to 0} (\frac{1}{k}) f(a, b + k) - f(a, b)] \\
+& = \lim_{h \to 0} (\frac{1}{h}) [\lim_{k \to 0} (\frac{1}{k}) f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
+& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) [f(a + h, b + k) - f(a + h, b) - f(a, b + k) + f(a, b)] \\
+& = \lim_{h \to 0} \lim_{k \to 0} (\frac{1}{hk}) \lambda(h, k)
+\end{align}
+</math>
+And similarly for <math>D_2 D_1 f(a, b)</math>, but with the two limits taken in the opposite order.
+==== Step 2 ====
+Now, let's make use of this function. Let us consider the square <math>Q = [a, a + h] \times [b, b + k]</math>, where <math>h</math> and <math>k</math> are so small that <math>Q \subseteq A</math>. We will show that there exist points <math>p, q \in Q</math> such that:
+<math>
+\begin{align}
+& \lambda(h, k) = D_2 D_1 f(p) \cdot hk \\
+& \lambda(h, k) = D_1 D_2 f(q) \cdot hk
+\end{align}
+</math>
+The proofs of the two equalities are symmetric, and thus we only explicitly prove the first one. We will prove this through a double application of the mean value theorem.
+Consider the function <math>\phi</math>, defined such that <math>\phi(s) = f(s, b + k) - f(s, b)</math>. Then <math>\phi(a + h) - \phi(a) = \lambda(h, k)</math>. By hypothesis, <math>D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\phi</math> with respect to the first variable of <math>f</math> on the interval <math>[a, a + h]</math>. By the mean value theorem, this means that there exists a point <math>s_0 \in [a, a + h]</math> such that <math>\lambda(h, k) = \phi(a + h) - \phi(a) = \phi^\prime(s_0) \cdot h = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h</math>.
+Now we want to apply the mean value theorem one more time. Consider the function <math>\gamma</math>, defined such that <math>\gamma(t) = D_1 f(s_0, t)</math>. By hypothesis, <math>D_2 D_1 f</math> exists at all points of <math>A</math>, so we can differentiate <math>\gamma</math> with respect to the second variable of <math>f</math> on the interval <math>[b, b + k]</math>. So, by the mean value theorem, there exists a point <math>t_0 \in [b, b + k]</math> such that <math>\gamma(b + k) - \gamma(b) = \gamma^\prime(t_0) \cdot k</math>. Thus:
+<math>
+\begin{align}
+\lambda(h, k) & = [D_1 f(s_0, b + k) - D_1 f(s_0, b)] \cdot h \\
+& = [\gamma(b + k) - \gamma(b)] \cdot h \\
+& = \gamma^\prime(t_0) \cdot hk \\
+& = D_2 D_1 f(s_0, t_0) \cdot hk
+\end{align}
+</math>
+==== Step 3 ====
+Now we can prove the theorem. Let <math>c = (a, b) \in A</math>, and let <math>t > 0</math> be so small that <math>Q_t = [a, a + t] \times [b, b + t] \subseteq A</math>. By what we have just shown, <math>\lambda(t, t) = D_2 D_1 f(p_t) \cdot t^2</math>, for some <math>p_t \in Q_t</math>.
+<math>t</math> is the length of the sides of the rectangle <math>Q_t</math>, so as <math>t \rightarrow 0</math>, <math>p_t \rightarrow c</math>. By hypothesis, <math>D_2 D_1 f</math> is continuous, and so as <math>p_t \rightarrow c</math>, <math>D_2 D_1 f(p_t) \rightarrow D_2 D_1 f(c)</math>. Thus:
+<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_2 D_1 f(c)</math>.
+We can use the same argument, and the second equality from step 2, to show that:
+<math>\lim_{t \to 0} \frac{\lambda(t, t)}{t^2} = D_1 D_2 f(c)</math>.
+Therefore, by the uniqueness of limits, the two quantities must be equal. <math>\square</math>

1617-257/Classnotes for Friday October 14: Difference between revisions

Latest revision as of 21:50, 19 October 2016

Contents

Symmetry of Second Partial Derivatives

Theorem

Proof

Step 1

Step 2

Step 3

Handwritten Lecture Notes in PDF

Navigation menu

Page actions

Page actions

Personal tools

Navigation

Search

Tools