Welcome to Math 257

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Week of...

Notes and Links

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; MondayTuesday 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 57PM; 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 1028

The Final Exam on Thursday April 20 (and some office hours sessions before).

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Riddle Repository

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$\int _{M}d\omega =\int _{\partial M}\omega$



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)