This is where I'll be putting class notes, homework, other announcements.
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Wednesday, January 21, 2009
Friday, June 13, 2008
Final exam with answers
Here.
Here's how grades work: add up your homework percentage, midterms, twice the final, and divide by 5.
Then find the range:
-----A
70
-----B
55
-----C
40
-----D
25
-----F
with + and - in those ranges too.
Here's how grades work: add up your homework percentage, midterms, twice the final, and divide by 5.
Then find the range:
-----A
70
-----B
55
-----C
40
-----D
25
-----F
with + and - in those ranges too.
Thursday, June 12, 2008
Monday, June 09, 2008
Neeraj's office hours
"I plan on having OH on Wed 2:30-3:30 so that they can come by to pick up exams and such. (2:30 is when one finals time-slot ends, so even if students have exams that day, they can drop by between 2:30-3.)"
Office hours
I'll be in my office (7450 APM) Tuesday 11-4 and Wednesday 11-2. I'll be giving preference to my Math 109 students on Tuesday and my Ma 20b students on Wednesday, but anyone will be welcome any time.
If you want to call ahead to check "Are you already tied up with the other class?", feel free; my office number is 858-534-6450.
(Don't get 7450 and 6450 confused!)
If you want to call ahead to check "Are you already tied up with the other class?", feel free; my office number is 858-534-6450.
(Don't get 7450 and 6450 confused!)
Some practice questions (now with answers)
Of course you should look again at the practice questions for the midterms. Here are some new ones. "^" means superscript, so x^2 means x squared, etc.
1. Write out the first ten terms (up to x^9) in the Taylor series of
a) f(x) = 1 / (1-x^2)
A. We did this for 1/(1-x) and got 1 + x + x^2 + x^3 + ...
So plug in x^2, and you get
1/(1-x^2) = 1 + x^2 + x^4 + x^6 + x^8 + ... omitting terms higher than x^9.
b) (1-x^4) f(x).
A. You could take the answer above and multiply by 1-x^4. There'd be a lot of cancelation.
Or you could notice that 1 - x^4 = (1+x^2)(1-x^2), so the answer to (b) is 1 + x^2 on the nose.
2. For which values of b does
\integral_{x=0}^{infinity} (e^{bx} + e^{-bx}) dx
converge?
A. None.
If b>0, then the first term goes to infinity, and the second to zero, so the integrand doesn't go to zero (it goes to infinity).
If b<0, then the second term goes to infinity, and the first to zero, so the integrand doesn't go to zero (it goes to infinity).
If b=0, then both terms are 1, so the integrand doesn't go to zero (it's always 2).
3. You've invented the Next Big Thing. It will sell increasingly well,
until the Next Next Big Thing comes out, at which point sales will collapse.
But that won't happen until N months from now.
A representative of MicrohooTM offers you $1000 up front, plus
a pay rate of $100/month at time t. Alternately, she can offer
you a pay rate of $t/month (paid continuously), with nothing up front.
Which deal should you take? You answer should depend on N.
A. This was a lousy question (which is why I kicked it out of the final, where no question is as badly worded as this one).
What it was supposed to get across was the idea that if you take deal #1, then by time S (S < N) you've accumulated 1000 + 100*S dollars.
Whereas if you take deal #2, by time S (S < N) you've accumulated integral_{t=0}^S t dt dollars.
For S small, 1000 + 100*S > S^2/2. Whereas for S large, deal #2 is better. You can figure out when deal #2 starts to beat deal #1 by solving this quadratic equation for S. If N is less than that number, take deal #1; if N is more, take deal #2.
1. Write out the first ten terms (up to x^9) in the Taylor series of
a) f(x) = 1 / (1-x^2)
A. We did this for 1/(1-x) and got 1 + x + x^2 + x^3 + ...
So plug in x^2, and you get
1/(1-x^2) = 1 + x^2 + x^4 + x^6 + x^8 + ... omitting terms higher than x^9.
b) (1-x^4) f(x).
A. You could take the answer above and multiply by 1-x^4. There'd be a lot of cancelation.
Or you could notice that 1 - x^4 = (1+x^2)(1-x^2), so the answer to (b) is 1 + x^2 on the nose.
2. For which values of b does
\integral_{x=0}^{infinity} (e^{bx} + e^{-bx}) dx
converge?
A. None.
If b>0, then the first term goes to infinity, and the second to zero, so the integrand doesn't go to zero (it goes to infinity).
If b<0, then the second term goes to infinity, and the first to zero, so the integrand doesn't go to zero (it goes to infinity).
If b=0, then both terms are 1, so the integrand doesn't go to zero (it's always 2).
3. You've invented the Next Big Thing. It will sell increasingly well,
until the Next Next Big Thing comes out, at which point sales will collapse.
But that won't happen until N months from now.
A representative of MicrohooTM offers you $1000 up front, plus
a pay rate of $100/month at time t. Alternately, she can offer
you a pay rate of $t/month (paid continuously), with nothing up front.
Which deal should you take? You answer should depend on N.
A. This was a lousy question (which is why I kicked it out of the final, where no question is as badly worded as this one).
What it was supposed to get across was the idea that if you take deal #1, then by time S (S < N) you've accumulated 1000 + 100*S dollars.
Whereas if you take deal #2, by time S (S < N) you've accumulated integral_{t=0}^S t dt dollars.
For S small, 1000 + 100*S > S^2/2. Whereas for S large, deal #2 is better. You can figure out when deal #2 starts to beat deal #1 by solving this quadratic equation for S. If N is less than that number, take deal #1; if N is more, take deal #2.
Saturday, June 07, 2008
Sections to study for the final
5.2,3,4,5,6
6.1,2,3
7.2,6,7
9.1,2,3,5
10.1,2,3,5,7
The final will cover all of those (or, as much as can practicably be tested in 3 hours), pretty evenly. It will NOT be just stuff since the second midterm!
Same basic rules for the final -- closed book, bring one double-sided sheet of notes, no calculator, no blue book necessary -- with the extra proviso that you bring your UCSD ID. Do remember to turn off your phone... or else.
6.1,2,3
7.2,6,7
9.1,2,3,5
10.1,2,3,5,7
The final will cover all of those (or, as much as can practicably be tested in 3 hours), pretty evenly. It will NOT be just stuff since the second midterm!
Same basic rules for the final -- closed book, bring one double-sided sheet of notes, no calculator, no blue book necessary -- with the extra proviso that you bring your UCSD ID. Do remember to turn off your phone... or else.
Monday, June 02, 2008
Thursday, May 29, 2008
Last HW, due Friday June 6
10.3 #5,7,10,14 but you can use whatever test you like
#41,47
10.5 #1,5,6,24
10.7 #5,13,21,79
#41,47
10.5 #1,5,6,24
10.7 #5,13,21,79
Wednesday, May 28, 2008
Wednesday May 28
Some criteria for convergence of infinite series:
If it "absolutely converges", it converges.
Comparison theorem; if a series with positive terms is bounded above by a convergent series, then it too converges.
Integral test; if our series comes from a decreasing positive function, then the series converges if and only if the improper integral converges. We actually proved this!
If it "absolutely converges", it converges.
Comparison theorem; if a series with positive terms is bounded above by a convergent series, then it too converges.
Integral test; if our series comes from a decreasing positive function, then the series converges if and only if the improper integral converges. We actually proved this!
Monday, May 19, 2008
Practice problems for midterm #2 (this Friday)
7.2 # 13,23,29,37,49
7.6 # 17,22,24
7.7 # 15,28,34
9.1 # 9,20,33,43
9.2 # 6
9.5 # 9,13
7.6 # 17,22,24
7.7 # 15,28,34
9.1 # 9,20,33,43
9.2 # 6
9.5 # 9,13
Thursday, May 15, 2008
Monday and Wednesday May 12 & 14
5/12
9.2 Exponential growth and decay. Newton's law of cooling.
Nibbling on basil example of exponential growth (possibly negative).
9.4 What the logistic equation means, above carrying capacity.
9.5 First-order linear homogeneous equations.
A derivation of the integrating factor, based on the idea that one solves the homogeneous first and uses it as a stepping-stone.
5/14
Rederivation of the general solution.
Applied to a nonmotivated example from the book.
Then we thought about filling up a bathtub, initially half-full of cold water, with warm water, while it's draining (more slowly than it fills). During this process the water draining becomes warmer and warmer. What's the temperature at the time the bath is full?
This turned out to be a linear inhomogeneous first-order DE.
10.1 Infinite sequences. The definition of "this sequence converges to x", with epsilons and large Ns.
9.2 Exponential growth and decay. Newton's law of cooling.
Nibbling on basil example of exponential growth (possibly negative).
9.4 What the logistic equation means, above carrying capacity.
9.5 First-order linear homogeneous equations.
A derivation of the integrating factor, based on the idea that one solves the homogeneous first and uses it as a stepping-stone.
5/14
Rederivation of the general solution.
Applied to a nonmotivated example from the book.
Then we thought about filling up a bathtub, initially half-full of cold water, with warm water, while it's draining (more slowly than it fills). During this process the water draining becomes warmer and warmer. What's the temperature at the time the bath is full?
This turned out to be a linear inhomogeneous first-order DE.
10.1 Infinite sequences. The definition of "this sequence converges to x", with epsilons and large Ns.
Wednesday, May 14, 2008
Friday, May 09, 2008
Wednesday & Friday May 7 & 9
9.1 Differential equations.
Separability.
Straight lines through the origin.
Circles around the origin.
Initial value problems.
Growth of a cell.
The logistic equation.
Linear homogeneous & inhomogeneous.
The spring equation (2nd order homogeneous).
The spring equation in presence of gravity (inhomogeneous).
Separability.
Straight lines through the origin.
Circles around the origin.
Initial value problems.
Growth of a cell.
The logistic equation.
Linear homogeneous & inhomogeneous.
The spring equation (2nd order homogeneous).
The spring equation in presence of gravity (inhomogeneous).
Tuesday, May 06, 2008
Monday May 5
Comparison theorem for improper integrals.
With proper integrals, we don't have to worry about whether the answer is actually finite; it is. With improper ones sometimes we can't calculate them exactly, but we still want to know the answer is finite. The comparison theorem lets us do that; it's much more important than in the proper-integral situation.
9.1 Diff-eqs.
We just started this, and solved y' = y/x (the solutions are straight lines through the origin, which makes sense).
With proper integrals, we don't have to worry about whether the answer is actually finite; it is. With improper ones sometimes we can't calculate them exactly, but we still want to know the answer is finite. The comparison theorem lets us do that; it's much more important than in the proper-integral situation.
9.1 Diff-eqs.
We just started this, and solved y' = y/x (the solutions are straight lines through the origin, which makes sense).
Friday, May 02, 2008
Friday May 2
I pointed out that step 1 of the partial fractions expansion algorithm -- long division of the numerator by the denominator, leaving a remainder term -- can be thought of as "peeling off the terms that blow up at x = infinity". Which can be done almost the same way as we do the rest of the algorithm; divide by the highest power of x, then look at the limit as x->infinity.
Not that there's much reason to do it that way; I was just sayin'.
Then we did improper integrals. I spent a long time on one example: integral0picot(x) dx. This is improper at both ends, and there's no best way to evaluate it, in that different approaches give different answers. So one should declare this to have no answer.
Not that there's much reason to do it that way; I was just sayin'.
Then we did improper integrals. I spent a long time on one example: integral0picot(x) dx. This is improper at both ends, and there's no best way to evaluate it, in that different approaches give different answers. So one should declare this to have no answer.
Wednesday April 2
We saw the general rule for doing partial fraction expansion, i.e. what to do if the polynomial in the denominator has repeated roots.
Then we touched upon improper integrals.
Then we touched upon improper integrals.
Thursday, May 01, 2008
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