My students probably struggled with the last few topics covered in the semester as my coverage on those topics was not as thorough as my coverage on the rest of the topics covered in the semester.
On a related note, I occasionally ponder the possibility of covering less, but in greater detail.
On a different note, I think it's strange how many students ask for review sessions, but when I bring up the idea of extending class time or a separate hour (temporarily ignoring the issue of increasing credit) to my peers, there is the idea that students wouldn't agree to it. The problem with office hours is that students either have other classes schedule during that time or are afraid to come. Office hours by appointment is fine, as long as it isn't abused.
Maybe there should be a game. Played by a professor. Every time the professor says something that might be on the exam, students should say some key phrase like "That's magic!" Perhaps every once in a while, and especially the first week, this should be aided with a pause or a hand signal or a sign (much like an "Applause" sign).
Wednesday, May 16, 2012
Next TA Meeting
This general idea of removing bias crossed my mind. How should we, if at all, deal with section to section discrepancies? I like the idea of mixing up the exams and having the person's name on the very back page of the last problem. That way, you can keep the scoring page on the front, and when you hand back the exams, you just turn the pile over and they can look for their names without seeing each other's scores. It also reduces section bias. But undoubtedly there is differences for the homework portion and emphasis of material from section to section. Maybe one of the TA emphasizes a certain type of problem and grades with respect to that emphasis. So even unbiased grading on midterm and final exams will be lopsided. I definitely would either scale each section so that all sections are comparable, or assign grades based on individual sections. TAs might expect a certain student to do well or do poorly.
See also next post.
See also next post.
Other Ideas to Focus On
$\lim a_n=0$ does not imply convergence.
Students use the ratio/root test on the endpoints of a power series. It's hard for them to grasp that those are exactly the points where the ratio/root test gives 1, and hence inconclusive. Apply the ratio/root test is a waste of time. Just as bad is when they apply the ratio/root test to the endpoints and get a value other than 1.
$\lim\left(1+\frac{x}{n}\right)^n=e^x$
Students use the ratio/root test on the endpoints of a power series. It's hard for them to grasp that those are exactly the points where the ratio/root test gives 1, and hence inconclusive. Apply the ratio/root test is a waste of time. Just as bad is when they apply the ratio/root test to the endpoints and get a value other than 1.
$\lim\left(1+\frac{x}{n}\right)^n=e^x$
Saturday, May 12, 2012
Mistakes
THE FOLLOWING IS FALSE:
$$\frac{c}{a+b}=\frac{c}{a}+\frac{c}{b}$$
FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\frac{x^2}{1+x^2}=\frac{x^2}{1\vphantom{x^2}}+\frac{x^2}{x^2}$$
TRY IT WITH NUMBERS:
$$\frac{2}{1+1}=1$$
$$\frac{2}{1}+\frac{2}{1}=4$$
$$\frac{c}{a+b}=\frac{c}{a}+\frac{c}{b}$$
FAKE EXAMPLE OF THIS ERROR BEING MADE:
Student writes:
$$\frac{x^2}{1+x^2}=\frac{x^2}{1\vphantom{x^2}}+\frac{x^2}{x^2}$$
TRY IT WITH NUMBERS:
$$\frac{2}{1+1}=1$$
$$\frac{2}{1}+\frac{2}{1}=4$$
Tuesday, May 8, 2012
Practice Exam 5
I don't feel it is necessary for me to make another practice test. At this point you should have realized they're just homework problems in disguise. Now it's time for you to make your own practice test. Below is a rough guideline.
Instructions: Do the following. Exceptions to the rule are anything mentioned by Dr. Brown to focus on or not focus on. Please show your work and explain whenever possible. It's better to say what's on your mind (provided it's correct), because the graders are not mind readers. Please simplify answers when reasonable, such as $\sin{2\pi/3}$ or $\log(1)$.
1) Midterm 1
2) Midterm 2
3) One problem similar to every different type of "easy" homework problem. By "easy" homework problem, I mean the less conceptual and abstract ones. Do odd-numbered problem so you can check your solution with the back of the book. Alternate about every twenty minutes between easier and harder sections (different for every student).
4) My four practice tests
5) Learn to find exams in other locations. Skip problems when they don't apply. Let me get you started: http://www.math.jhu.edu/~wsw/F11/.
Bonus (for students looking for the extra challenge): One problem similar to every different type of "hard" homework problem.
Instructions: Do the following. Exceptions to the rule are anything mentioned by Dr. Brown to focus on or not focus on. Please show your work and explain whenever possible. It's better to say what's on your mind (provided it's correct), because the graders are not mind readers. Please simplify answers when reasonable, such as $\sin{2\pi/3}$ or $\log(1)$.
1) Midterm 1
2) Midterm 2
3) One problem similar to every different type of "easy" homework problem. By "easy" homework problem, I mean the less conceptual and abstract ones. Do odd-numbered problem so you can check your solution with the back of the book. Alternate about every twenty minutes between easier and harder sections (different for every student).
4) My four practice tests
5) Learn to find exams in other locations. Skip problems when they don't apply. Let me get you started: http://www.math.jhu.edu/~wsw/F11/.
Bonus (for students looking for the extra challenge): One problem similar to every different type of "hard" homework problem.
Monday, May 7, 2012
Compute the Taylor series for x^4-3x^2+1 around a=1
Compute the Taylor series for $f(x) = x^4-3x^2+1$ around $a=1$.
Method 1 Take derivatives.
Step 1: Determine the series coefficients.
Step 1a: Evaluate the function at 1.
$f(1) = -1$
Step 1b: Take the first derivative and evaluate it at 1.
$f^{\prime}(x) = 4x^3-6x$
$f^{\prime}(1) = -2
Step 1c: Take the second derivative and evaluate it at 1.
$f^{\prime\prime}(x) = 12x^2-6$
$f^{\prime\prime}(1) = 6
Step 1d: And so on...
$f^{\prime\prime\prime}(x) = 24x$
$f^{\prime\prime\prime}(1) = 24$
$f^{\prime\prime\prime\prime}(x) = 24$
$f^{\prime\prime\prime\prime}(1) = 24$
Step 2: Lay out the coefficients in front of the appropriate power of $(x-a)$ and divide by the corresponding factorial.
For example, the coefficient of $(x-1)^2$ is the evaluation of the second derivative (2) divided by the factorial of two.
In tabular form we have the following:
At the end of the day we have $$-1-2(x-1)+3(x-1)^2+4(x-1)^3+(x-1)^4$$
[20120507]
Method 2 Potentially the easiest method provided it's allowed by the instructor.
First we recognize that $f$ can be written as $((x-1)+1)^4-3((x-1)+1)^2+1$.
Then we expand the terms to obtain $((x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$ from the first term, $-3((x-1)^2+2(x-1)+1)$ from the second term, and $1$.
Simplifying we get $(x-4)^4+4(x-1)^3+3(x-1)^2-2(x-1)-1$.
[20191223]
Method 3 Theoretical method; arguably harder.
We know the coefficient of $(x-1)^4$ must be 1, because it is the only term that contributes to the highest degree term of $f$.
We expand $(x-1)^4$ to get x^4-4x^3+6x^2-4x+1 (I computed the expansion with the help of Pascal's triangle).
There are no powers of $x^3$ in $f$ so we need to offset it with $4(x-1)^3$. The expansion here is $4(x^3-3x^2+3x-1)$ or $4x^3-12x^2+12x-4$.
Keeping track we have $-6x^2+8x-3$ to worry about. So we add $3(x-1)^2$. The expansion here is $3(x^2-2x+1)$ or $3x^2-6x+3$.
The remaining amount to worry about is $2x$. So we add $-2(x-1)$ or $-2x+2$.
The remaining about to worry about is $2$. So we add $-1$.
Remark: Throughout this method, I use the term "add." It's useful to think in terms of adding negative "x" instead of subtracting "x." My high school teacher taught my fellow students and me that subtraction (Satan) and division (Devil) are evil and so we should instead "add a negative" and "multiply by the inverse," respectively.
[20191223]
Method 1 Take derivatives.
Step 1: Determine the series coefficients.
Step 1a: Evaluate the function at 1.
$f(1) = -1$
Step 1b: Take the first derivative and evaluate it at 1.
$f^{\prime}(x) = 4x^3-6x$
$f^{\prime}(1) = -2
Step 1c: Take the second derivative and evaluate it at 1.
$f^{\prime\prime}(x) = 12x^2-6$
$f^{\prime\prime}(1) = 6
Step 1d: And so on...
$f^{\prime\prime\prime}(x) = 24x$
$f^{\prime\prime\prime}(1) = 24$
$f^{\prime\prime\prime\prime}(x) = 24$
$f^{\prime\prime\prime\prime}(1) = 24$
Step 2: Lay out the coefficients in front of the appropriate power of $(x-a)$ and divide by the corresponding factorial.
For example, the coefficient of $(x-1)^2$ is the evaluation of the second derivative (2) divided by the factorial of two.
In tabular form we have the following:
| Coefficient | Factorial | x term | All Together |
|---|---|---|---|
| $f^{(0)}(1)=f(1)$ | 0! | $(x-1)^0$ | $\frac{-1}{0!}$ |
| $f^{(1)}(1)=f^{\prime}(1)$ | 1! | $(x-1)^1$ | $\frac{-2}{1!}(x-1)$ |
| $f^{(2)}(1)=f^{\prime\prime}(1)$ | 2! | $(x-1)^2$ | $\frac{6}{2!}(x-1)^2$ |
| $f^{(3)}(1)=f^{\prime\prime\prime}(1)$ | 3! | $(x-1)^3$ | $\frac{24}{3!}(x-1)^3$ |
| $f^{(4)}(1)=f^{\prime\prime\prime\prime}(1)$ | 4! | $(x-1)^4$ | $\frac{24}{4!}(x-1)^4$ |
At the end of the day we have $$-1-2(x-1)+3(x-1)^2+4(x-1)^3+(x-1)^4$$
[20120507]
Method 2 Potentially the easiest method provided it's allowed by the instructor.
First we recognize that $f$ can be written as $((x-1)+1)^4-3((x-1)+1)^2+1$.
Then we expand the terms to obtain $((x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$ from the first term, $-3((x-1)^2+2(x-1)+1)$ from the second term, and $1$.
Simplifying we get $(x-4)^4+4(x-1)^3+3(x-1)^2-2(x-1)-1$.
[20191223]
Method 3 Theoretical method; arguably harder.
We know the coefficient of $(x-1)^4$ must be 1, because it is the only term that contributes to the highest degree term of $f$.
We expand $(x-1)^4$ to get x^4-4x^3+6x^2-4x+1 (I computed the expansion with the help of Pascal's triangle).
There are no powers of $x^3$ in $f$ so we need to offset it with $4(x-1)^3$. The expansion here is $4(x^3-3x^2+3x-1)$ or $4x^3-12x^2+12x-4$.
Keeping track we have $-6x^2+8x-3$ to worry about. So we add $3(x-1)^2$. The expansion here is $3(x^2-2x+1)$ or $3x^2-6x+3$.
The remaining amount to worry about is $2x$. So we add $-2(x-1)$ or $-2x+2$.
The remaining about to worry about is $2$. So we add $-1$.
Remark: Throughout this method, I use the term "add." It's useful to think in terms of adding negative "x" instead of subtracting "x." My high school teacher taught my fellow students and me that subtraction (Satan) and division (Devil) are evil and so we should instead "add a negative" and "multiply by the inverse," respectively.
[20191223]
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