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
Thursday, May 29, 2008
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
Tuesday, April 29, 2008
Monday, April 28, 2008
Monday April 28
More partial fractions.
Long division of polynomials (i.e. the usual long division algorithm, but with polynomials).
Factoring the denominator using complex roots.
We haven't yet addressed the possibility that the denominator has repeated factors; that will be next time.
Long division of polynomials (i.e. the usual long division algorithm, but with polynomials).
Factoring the denominator using complex roots.
We haven't yet addressed the possibility that the denominator has repeated factors; that will be next time.
Friday, April 25, 2008
Friday April 25
Section 7.5. Hyperbolic trig functions.
We talked a bit about why the word "hyperbolic" is there. Both circles and hyperbolae are conic sections.
Section 7.6. Partial fractions.
So far we did a couple of examples. The basic trick, to figure out the coefficient on some term, is to multiply through by that denominator THEN set the once-denominator equal to 0. All other terms die, exposing the one you're trying to compute.
We talked a bit about why the word "hyperbolic" is there. Both circles and hyperbolae are conic sections.
Section 7.6. Partial fractions.
So far we did a couple of examples. The basic trick, to figure out the coefficient on some term, is to multiply through by that denominator THEN set the once-denominator equal to 0. All other terms die, exposing the one you're trying to compute.
Midterm #1 breakdown
In the extremely unlikely event that your final letter grade were to be determined from only the midterm #1 grade, it would be as follows:
80-100 A
60-80 B
45-60 C
30-45 D
<30 F
We will not be figuring out your final grade using this letter, but rather the raw number, so don't worry if e.g. you got 79 rather than 80.
80-100 A
60-80 B
45-60 C
30-45 D
<30 F
We will not be figuring out your final grade using this letter, but rather the raw number, so don't worry if e.g. you got 79 rather than 80.
Wednesday, April 23, 2008
HW #3 due Wednesday April 30
6.3 #9,15,23,25
7.2 #2,3,7,20,36,45,66
Read 7.3 and looking heavenward, reflect internally upon the value of complex exponentials.
Integrate exp(2x) sin(x).
7.2 #2,3,7,20,36,45,66
Read 7.3 and looking heavenward, reflect internally upon the value of complex exponentials.
Integrate exp(2x) sin(x).
Wednesday April 23
Last time I introduced complex numbers, and how to think about multiplying them: they scale and rotate.
This time we studied pure rotations (i.e. scaling by 1), and showed that if there's any justice, the complex number z such that multiplying by z implements rotation by theta, should be z = exp(i theta).
This gave us formulae for sin and cos in terms of complex exponentials, which is good because the latter are much easier to work with.
This time we studied pure rotations (i.e. scaling by 1), and showed that if there's any justice, the complex number z such that multiplying by z implements rotation by theta, should be z = exp(i theta).
This gave us formulae for sin and cos in terms of complex exponentials, which is good because the latter are much easier to work with.
Tuesday, April 22, 2008
Office hours Thursday
I am traveling and will be back tomorrow; the Tuesday office hour is moved to Thursday.
Answers to the first midterm will be up soon.
Answers to the first midterm will be up soon.
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