College Math Teaching

August 29, 2020

Commentary: life with Webassign

Filed under: editorial, pedagogy — oldgote @ 5:45 pm

Since I am online, I decided to use Webassign for homework.
Well, of course, some students have had trouble with particular the graphing application.
I am not saying that their graphing tool is hard to learn; in fact I might play with it myself. BUT, in this particular class, I want my students to focus on learning the material and NOT have to spend hours learning to get good with their graphing tools. For graphing: Desmos is outstanding.

Normally, I am not sympathetic to student complaints or frustrations, but here, I can see it.

July 14, 2020

An alternative to trig substitution, sort of..

Ok, just for fun: \int \sqrt{1+x^2} dx =

The usual is to use x =tan(t), dx =sec^2(t) dt which transforms this to the dreaded \int sec^3(t) dt integral, which is a double integration by parts.
Is there a way out? I think so, though the price one pays is a trickier conversion back to x.

Let’s try x =sinh(t) \rightarrow dx = cosh(t) dt so upon substituting we obtain \int |cosh(t)|cosh(t) dt and noting that cosh(t) > 0 alaways:

\int cosh^2(t)dt Now this can be integrated by parts: let u=cosh(t) dv = cosh(t) dt \rightarrow du =sinh(t), v = sinh(t)

So \int cosh^2(t)dt = cosh(t)sinh(t) -\int sinh^2(t)dt but this easily reduces to:

\int cosh^2(t)dt = cosh(t)sinh(t) -\int cosh^2(t)-1 dt \rightarrow 2\int cosh^2(t)dt  = cosh(t)sinh(t) -t + C

Division by 2: \int cosh^2(t)dt = \frac{1}{2}(cosh(t)sinh(t)-t)+C

That was easy enough.

But we now have the conversion to x: \frac{1}{2}(cosh(t)sinh(t) \rightarrow \frac{1}{2}x \sqrt{1+x^2}

So far, so good. But what about t \rightarrow   arcsinh(x) ?

Write: sinh(t) = \frac{e^{t}-e^{-t}}{2} =  x \rightarrow e^{t}-e^{-t} =2x \rightarrow e^{t}-2x -e^{-t} =0

Now multiply both sides by e^{t} to get e^{2t}-2xe^t -1 =0 and use the quadratic formula to get e^t = \frac{1}{2}(2x\pm \sqrt{4x^2+4} \rightarrow e^t = x \pm \sqrt{x^2+1}

We need e^t > 0 so e^t = x + \sqrt{x^2+1} \rightarrow t = ln|x + \sqrt{x^2+1}| and that is our integral:

\int \sqrt{1+x^2} dx = \frac{1}{2}x \sqrt{1+x^2} + \frac{1}{2} ln|x + \sqrt{x^2+1}| + C

I guess that this isn’t that much easier after all.

July 12, 2020

Logarithmic differentiation: do we not care about domains anymore?

Filed under: calculus, derivatives, elementary mathematics, pedagogy — collegemathteaching @ 11:29 pm

The introduction is for a student who might not have seen logarithmic differentiation before: (and yes, this technique is extensively used..for example it is used in the “maximum likelihood function” calculation frequently encountered in statistics)

Suppose you are given, say, f(x) =sin(x)e^x(x-2)^3(x+1) and you are told to calculate the derivative?

Calculus texts often offer the technique of logarithmic differentiation: write ln(f(x)) = ln(sin(x)e^x(x-2)^3(x+1)) = ln(sin(x)) + x + 3ln(x-2) + ln(x+1)
Now differentiate both sides: ln((f(x))' = \frac{f'(x)}{f(x)}  = \frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1}

Now multiply both sides by f(x) to obtain

f'(x) = f(x)(\frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1}) =


(sin(x)e^x(x-2)^3(x+1)(\frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1})

And this is correct…sort of. Why I say sort of: what happens, at say, x = 0 ? The derivative certainly exists there but what about that second factor? Yes, the sin(x) gets cancelled out by the first factor, but AS WRITTEN, there is an oh-so-subtle problem with domains.

You can only substitute x \in \{ 0, \pm k \pi \} only after simplifying ..which one might see as a limit process.

But let’s stop and take a closer look at the whole process: we started with f(x) = g_1(x) g_2(x) ...g_n(x) and then took the log of both sides. Where is the log defined? And when does ln(ab) = ln(a) + ln(b) ? You got it: this only works when a > 0, b > 0 .

So, on the face of it, ln(g_1 (x) g_2(x) ...g_n(x)) = ln(g_1(x) ) + ln(g_2(x) ) + ...ln(g_n(x)) is justified only when each g_i(x) > 0 .

Why can we get away with ignoring all of this, at least in this case?

Well, here is why:

1. If f(x) \neq 0 is a differentiable function then \frac{d}{dx} ln(|f(x)|) = \frac{f'(x)}{f(x)}
Yes, this is covered in the derivation of \int {dx \over x} material but here goes: write

|f(x)| =   \begin{cases}      f(x) ,& \text{if } f(x) > 0 \\      -f(x),              & \text{otherwise}  \end{cases}

Now if f(x) > 0 we get { d \over dx} ln(f(x)) = {f'(x) \over f(x) } as usual. If f(x) < 0 then |f(x)| = =f(x), |f(x)|' = (-f(x))' = -f'(x) and so in either case:

\frac{d}{dx} ln(|f(x)|) = \frac{f'(x)}{f(x)} as required.

THAT is the workaround for calculating {d \over dx } ln(g_1(x)g_2(x)..g_n(x)) where g_1(x)g_2(x)..g_n(x) \neq 0 : just calculate {d \over dx } ln(|g_1(x)g_2(x)..g_n(x)|) . noting that |g_1(x)g_2(x)..g_n(x)| = |g_1(x)| |g_2(x)|...|g_n(x)|

Yay! We are almost done! But, what about the cases where at least some of the factors are zero at, say x= x_0 ?

Here, we have to bite the bullet and admit that we cannot take the log of the product where any of the factors have a zero, at that point. But this is what we can prove:

Given g_1(x) g_2(x)...g_n(x) is a product of differentiable functions and g_1(a) g_2(a)...g_k(a) = 0 k \leq n then
(g_1(a)g_2(a)...g_n(a))' = lim_{x \rightarrow a}  g_1(x)g_2(x)..g_n(x) ({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...{g_n'(x) \over g_n(x})

This works out to what we want by cancellation of factors.

Here is one way to proceed with the proof:

1. Suppose f, g are differentiable and f(a) = g(a) = 0 . Then (fg)'(a) = f'(a)g(a) + f(a)g'(a) = 0 and lim_{x \rightarrow a} f(x)g(x)({f'(x) \over f(x)} + {g'(x) \over g(x)}) = 0
2. Now suppose f, g are differentiable and f(a) =0 ,  g(a) \neq 0 . Then (fg)'(a) = f'(a)g(a) + f(a)g'(a) = f'(a)g(a) and lim_{x \rightarrow a} f(x)g(x)({f'(x) \over f(x)} + {g'(x) \over g(x)}) = f'(a)g(a)
3.Now apply the above to g_1(x) g_2(x)...g_n(x) is a product of differentiable functions and g_1(a) g_2(a)...g_k(a) = 0 k \leq n
If k = n then (g_1(a)g_2(a)...g_n(a))' = lim_{x \rightarrow a}  g_1(x)g_2(x)..g_n(x) ({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...{g_n'(x) \over g_n(x}) =0 by inductive application of 1.

If k < n then let g_1...g_k = f, g_{k+1} ..g_n  =g as in 2. Then by 2, we have (fg)' =  f'(a)g(a) Now this quantity is zero unless k = 1 and f'(a) neq 0 . But in this case note that lim_{x \rightarrow a} g_1(x)g_2(x)...g_n(x)({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...+ {g_n'(x) \over g_n(x)})  = lim_{x \rightarrow a} g_2(x)...g_n(x)(g_1'(x)) =g(a)g_1(a)

So there it is. Yes, it works ..with appropriate precautions.

July 10, 2020

This always bothered me about partial fractions…

Filed under: algebra, calculus, complex variables, elementary mathematics, integration by substitution — Tags: — collegemathteaching @ 12:03 am

Let’s look at an “easy” starting example: write \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}
We know how that goes: multiply both sides by (x-1)(x+1) to get 1 = A(x+1) + B(x-1) and then since this must be true for ALL x , substitute x=-1 to get B = -{1 \over 2} and then substitute x = 1 to get A = {1 \over 2} . Easy-peasy.

BUT…why CAN you do such a substitution since the original domain excludes x =1, x = -1 ?? (and no, I don’t want to hear about residues and “poles of order 1”; this is calculus 2. )

Lets start with \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} with the restricted domain, say x \neq 1
Now multiply both sides by x-1 and note that, with the restricted domain x \neq 1 we have:

\frac{1}{x+1}  = A + \frac{B(x-1)}{x+1} But both sides are equal on the domain (-1, 1) \cup (1, \infty) and the limit on the left hand side is lim_{x \rightarrow 1} {1 \over x+1 } = {1 \over 2} So the right hand side has a limit which exists and is equal to A . So the result follows..and this works for the calculation for B as well.

Yes, no engineer will care about this. But THIS is the reason we can substitute the non-domain points.

As an aside: if you are trying to solve something like {x^2 + 3x + 2 \over (x^2+1)(x-3) } = {Ax + B \over x^2+1 } + {C \over x-3 } one can do the denominator clearing, and, as appropriate substitute x = i and compare real and imaginary parts ..and yes, now you can use poles and residues.

July 3, 2020

What will happen this fall?

Filed under: COVID19, pedagogy — Tags: — collegemathteaching @ 12:22 am

Yes, I should be doing math but, well, I can tell you what I’ve done since online and many of my summer duties have ended:

1. I’ve purchased some “stuff for hybrid learning” equipment, to wit: drawing board and a document camera.
2. I also have a gallon of hand sanitizer, face shield and 100 masks to hand out to students who “forget” to wear one to class.

Yes, I know, my university announced that we will start in person, go to Thanksgiving and then finish remotely. And exactly how far we get remains to be seen; I know that USC just announced that they are going online from the start.

That might get the dominoes falling.

But here are my plans for my two “hybrid” classes (4 meetings a week, but due to social distance limits, half the class will come M, Th, half on W, F)
a. Stuff in the lessons is mandatory; students are responsible for notes, quizzes, assignments
b. But, I will NOT require in person attendance, ever. All notes will be posted on line (I am working on them right now), all class room sessions will be put on video (maybe even live streamed) and
c. All testing, quizzes, etc. will be online. Yes, they will be open book; that is really the only way to be fair. Hence I’ll have to be creative with my exams.

As far as my actuarial science class: similar, though this class should be able to meet social distancing requirements. My not requiring in person attendance is for the students (e. g. what if they are worried, have some sort of medical condition, etc.)

I did attend a two week session on online learning and have some ideas on how to upload videos and the like. I’ll do some experimenting beforehand.

And yes, I have two papers to finish; I hope that this note gets me inspired to get back to it.

April 12, 2020

A tidbit with respect to Laplace transforms and sin(x)/x

Filed under: complex variables, integrals, Laplace transform, media — collegemathteaching @ 9:01 pm

I’ve discovered the channel “blackpenredpen” and it is delightful.
It is a nice escape into mathematics that, while far from research level, is “fun” and beyond mere fluff.

And that got me to thinking about \int^{\infty}_0 \frac{sin(x)}{x} dx . Yes, this can be done by residues

But I’ll look at this with Laplace Transforms.

We know that \mathcal{L}(sin(x)) = \int^{\infty}_0 e^{-st}sin(t)dt = \frac{1}{s^2+1}
But note that the antiderivative of e^{-st} with respect to s is -\frac{1}{t}e^{-st} That might not seem like much help, but then notice \int^{\infty}_0 e^{-st} ds = \frac{-1}{t}e^{-st}|^{\infty}_0 = \frac{1}{t} (assuming s > 0

So why not: \int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)dt ds = \int^{\infty}_0 \frac{1}{s^2+1} ds =arctan(s)|^{\infty}_0 = \frac{\pi}{2}
Now since the left hand side is just a double integral over the first quadrant (an infinite rectangle) the order of integration can be interchanged:
\int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)dt ds = \int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)ds dt  = \int^{\infty}_0 sin(t) \int^{\infty}_0 e^{-st}ds dt = \int^{\infty}_0 sin(t)\frac{1}{t} dt

and that is equal to \frac{\pi}{2} .

Note: \int_0^x\frac{sin(t)}{t} dt is sometimes called the Si(x) function


April 6, 2020

How I am cutting corners in class

Filed under: complex variables, differential equations, Laplace transform — collegemathteaching @ 12:06 am

Ok, which is more difficult?

1. Solve x'' + 6x' + 13x = sin(t), x(0) = x'(0) = 0 using Laplace transforms or:

2. Given Y = \frac{1}{s^4 + 6s^3 +10s^2 + 6s + 9} find the inverse Laplace transform.

Clearly, 2 is harder and in texts I’ve used, we had to do those prior to doing 1. But, in a way, you have to do 2 in order to do 1:

(s^2 + 6s + 13)X(s) = \frac{1}{s^2+1}  \rightarrow X(s) = \frac{1}{(s^2 + 6s +13)(s^2 + 1)}

But this is already factored and students can be taught to “attempt to factor and if you can’t, complete the square” and this leads immediately to:

X(s) = \frac{1}{(s^2 + 6s + 9 +4)(s^2+1)} = \frac{As + B}{(s+3)^2+4} + \frac{Cs + D}{s^2 +1} which can be resolved by partial fractions.

In our “one less week plus online” I will do much more of 1 than 2.

Of course, there is still some work to do; we still have to solve (As+B)(s^2+1) +(Cs+D)((s+3)^2 +4) =1

I will teach the “eliminate the term method by using complex numbers:

Let s = i to get

(Ci+D)(12+6i) = 12D-6C +(12C + 6D)i = 1 \rightarrow D = -2C \rightarrow -30C = 1 \rightarrow C = -\frac{1}{30}, D = \frac{1}{15}
Let s = -3+2i

\rightarrow s^2+1 = 6-12i \rightarrow (-3A+B +2iA)(6-12i)

= -18A+6B +24A +36iA+12Ai-12Bi = -6A+6B +(48A-12B)i = 1

\rightarrow B=4A, 6A+6B= 1\rightarrow 30A=1

So we have A = \frac{1}{30}, B = \frac{2}{15}, C = -\frac{1}{30}, D = \frac{1}{15}

The particular solution part pulls back to -\frac{1}{30} cos(t) + \frac{1}{15} sin(t)

There is a work to do for the other part:

To get the s+3 shift we have to add and subtract 3; this leads to:

\frac{A(s+3) + B-3A}{(s+3)^2+4} =\frac{A(s+3) }{(s+3)^2+4} + \frac{ B-3A}{(s+3)^2+4}

\frac{1}{30}\frac{(s+3) }{(s+3)^2+4} + \frac{1}{30}\frac{1}{2}\frac{2}{(s+3)^2+4} (adjusting the second term for the 4 = 2^2
And this part pulls back to \frac{1}{30}e^{-3t}cos(2t) +\frac{1}{60} e^{-3t}sin(2t)

Yeah, I know; if you are reading this, you already know this stuff, but I think using i helps speed things up a bit.

And yes, you could have just used the convolution integral and have been done with it, though one would have had to have used
\frac{1}{2}e^{-3t}sin(2t) * sin(t) =\int^t_0\frac{1}{2}e^{-3u}sin(2u)sin(2t-2u)du and been done with it. (you remembered the 1/2, didn’t you? )

March 29, 2020

A change of variable to determine if growth is still exponential

This video is pretty good, and I thought that I’d add some equations to the explanation:

So, in terms of the mathematics, what is going on?

The graph they came up with is “new confirmed cases” on the y-axis (log scale) and total number of cases on the x-axis. Let’s see what this looks like for exponential growth.

Here, letting the total number of cases at time t be denoted by P(t) , the number of new cases is P'(t) , the first derivative.

In the case of exponential growth, P(t) = Ae^{kt} where k is positive.

P'(t) = Ake^{kt} which is what is being plotted on the y-axis. So with the change of variable we are letting u = Ae^{kt} and our new function is F(u) = ku , which, of course, is a straight line through the origin. That is, of course, IF the growth is exponential.

To get a feel for what this looks like, suppose we had polynomial growth; say P(t) = At^k . Then P'(t) =Akt^{k-1} = ak\frac{t^{k}}{t} =ak\frac{u}{u^{\frac{1}{k}}} =aku^{\frac{k-1}{k}} In the case of linear growth we’d have F(u) =ak (constant) and for, say, k = 3, F(u) =3au^{\frac{2}{3}} or a “concave down” function.

Now for the logistic situation in which the number of cases grows exponentially at first and then starts to level out to some steady state value, call it L, the relationship between the number of cases and the new number of cases looks like P'(t) = akP(L-P)) so our F(u) =aku(L-u) which is a quadratic which opens down.

Yes, this gets studied in differential equations class when we study autonomous differential equations.

Now for some graphs:

This is exponential growth vs. logistic growth; we get something similar to the latter when cases start to peak.

Here, I tweaked the logistic model to have the same derivative as the exponential model near t = 0 .

Here: we have linear growth P(t) = 5t vs the F(u) = 5

Here: cubic growth P(t) = 5t^3 vs. F(u) = 5u^{\frac{2}{3}}

March 26, 2020

My review lessons online

Filed under: applications of calculus, COVID19, differential equations, linear albegra — collegemathteaching @ 11:04 am

We had an extra week to prepare to teach online, so I put notes from the previous few weeks up in blog form:

That was quite a bit of work, but I did find some cool videos out there and embedded them in my lessons.

March 24, 2020

My teaching during the COVID-19 Pandemic

My university has moved to “online only” for the rest of the semester. I realize that most of us are in the same boat.
Fortunately, for now, I’ve got some academic freedom to make changes and I am taking a different approach than some.

Some appear to be wanting to keep things “as normal as possible.”

For me: the pandemic changes everything.

Yes, there are those on the beach in Florida. That isn’t most of my students; it could be some of them.

So, here is what will be different for me:
1) I am making exams “open book, open note” and take home: they get it and are given several days to do it and turn it back in, like a project.
Why? Fluid situations, living with a family, etc. might make it difficult to say “you HAVE to take it now…during period X.” This is NOT an online class that they signed up for.
Yes, it is possible that some cheat; that can’t be helped.

Also, studying will be difficult to do. So, getting a relatively long “designed as a programmed text” is, well, getting them to study WHILE DOING THE EXAM. No, it is not the same as “study to put it in your brain and then show you know it” at exam time. But I feel that this gets them to learn while under this stressful situation; they take time aside to look up and think about the material. The exam, in a way, is going through a test bank.

2) Previously, I thought of testing as serving two purposes: a) it encourages students to review and learn and b) distinguishing those with more knowledge from those with lesser knowledge. Now: tests are to get the students to learn..of course diligence will be rewarded. But who does well and who does not..those groups might change a little.

3) Quiz credit: I was able to sign up for webassign, and their quizzes will be “extra credit” to build on their existing grade. This is a “carrot only” approach.

4) Most of the lesson delivery will be a polished set of typeset notes with videos. My classes will be a combination of “live chat” with video where I will discuss said notes and give tips on how to do problems. I’ll have office hours ..some combination of zoom meetings which people can join and I’ll use e-mail to set up “off hours” meetings, either via chat or zoom, or even an exchange of e-mails.

We shall see how it works; I have a plan and think I can execute it, but I make no guarantee of the results.
Yes, there are polished online classes, but those are designed to be done deliberately. What we have here is something made up at the last minute for students who did NOT sign up for it and are living in an emergency situation.

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