# College Math Teaching

## November 25, 2013

Filed under: differential equations, Laplace transform — Tags: — collegemathteaching @ 10:33 pm

Consider: $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}......$

Now take the Laplace transform of the right hand side: $\frac{1}{s^2} - \frac{3!}{s^4 3!} + \frac{5!}{s^6 5!} .... = \frac{1}{s^2} (1 -\frac{1}{s^2} + \frac{1}{s^4} ...$.

This is equal to: $\frac{1}{s^2} (\frac{1}{1 + \frac{1}{s^2}})$ for $s > 1$ which is, of course, $\frac{1}{1 + s^2}$ which is exactly what you would expect.

This technique works for $e^{x}$ but gives nonsense for $e^{x^2}$.

Update: note that we can get a power series for $e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + ....$ which, on a term by term basis, transforms to $\frac{1}{s} + \frac{2!}{s^3} + \frac{4!}{s^5 2!} + \frac{6!}{s^7 3!} + ... = \frac{1}{s} \sum_{k=0} (\frac{1}{s^2})^k\frac{(2k)!}{k!})$ which only converges at $s = \infty$.

## November 21, 2013

### I am a frigging IDIOT!!!! (aka “Math Professor FAIL”)

Filed under: basic algebra, calculus, series — Tags: , , — collegemathteaching @ 5:17 pm

The topic: power series. I was showing how to manipulate the relation $\frac{1}{1-x} = \sum^{\infty}_{k=0} x^k$ to get power series of various kinds, including series centered at values other than zero. For example, if one wanted the series centered at, say, $x = 2$ we could write $\frac{1}{1-x} = \frac{1}{1-(x-2)-2} = \frac{1}{-1-(x-2)}=\frac{-1}{1-(-1)(x-2)}$ $= -\sum^{\infty}_{k=0} (-1)^k (x-2)^k = \sum^{\infty}_{k=0} (-1)^{k+1} (x-2)^k$.

And so I said: Let’s try to find the series for $\frac{1}{1-x}$ centered at $x = 1$. And. I. Was. Not. Able. To. Do. It.

So I went on to another problem…and it hit me.

O. M. G. (slamming my head on the desk here…dying of embarrassment).

Moral: never get overconfident. 🙂

Yes, I told the class of my blunder as soon as it hit me; they laughed…with me, I think. 😉

## November 19, 2013

### I hear you Secretary Duncan

Filed under: academia, editorial — Tags: — collegemathteaching @ 1:27 pm

I won’t weigh in on the merits of “common core standards” as I have no evidence one way or the other. But I’ll remark on this remark:

U.S. Education Secretary Arne Duncan told a group of state schools superintendents Friday that he found it “fascinating” that some of the opposition to the Common Core State Standards has come from “white suburban moms who — all of a sudden — their child isn’t as brilliant as they thought they were, and their school isn’t quite as good as they thought they were.”
Yes, he really said that. But he has said similar things before. What, exactly, is he talking about?
In his cheerleading for the controversial Common Core State Standards — which were approved by 45 states and the District of Columbia and are now being implemented across the country (though some states are reconsidering) — Duncan has repeatedly noted that the standards and the standardized testing that goes along with them are more difficult than students in most states have confronted.

Emphasis mine.

I don’t know about “white and suburban” but as far as parents having an inflated opinion of their kid’s abilities and the incoming students themselves: oh my goodness yes.

In some ways, university administration doesn’t help. I understand: to GET students to come to your university, you have to massage their egos a bit. No new freshmen means “no jobs for the professors”, at least at the smaller, “undergraduate education oriented” schools.

But…well, many of these students did well at their high schools and yes, they probably think that their high school is “one of the best” and they don’t seem to understand is that most of the incoming class has roughly the same level of talent.

Most have not been around a truly brilliant person and have no conception of what true brilliance is. Hey, if they…gasp…differentiated a polynomial in high school, they “know calculus”, right? 🙂

Rant to follow
It is my opinion (based only on my limited experience, not evidence), that most really don’t want an objective evaluation of their abilities and accomplishments; they want PRAISE, period.

And yes, there have been rare occasions when I’ve thought “so and so won’t make it” when, in fact, they had more talent that I’ve realized. But, surprise to surprise, most students have…uh…average talent. Judging from the boasting that I’ve seen on the social media sites, their respective parents will never believe that in a million years though. 🙂

Note: a reader pointed out that he thinks that the resistance to “common core” comes from the widespread belief that “local control is best”. That may well be true. This post does not mean that I think that Secretary Duncan is correct in the reasons that “common core” is resisted but rather that I agree with his assessment that many overestimate the intelligence of their kids.

## November 18, 2013

### And I get sloppy….divergence of n!x^n

Filed under: calculus, sequences, series — Tags: — collegemathteaching @ 9:29 pm

In class I was demonstrating the various open intervals of absolute convergence and gave the usual $\sum k!x^k$ as an example of a series that converges at $x = 0$ only. I mentioned that “$\sum k!x^k$ doesn’t even pass the divergence test”, which, as it turns out, is true. But why? (yes, it is easier to just use the ratio test and be done with it)

Well, I should have noted: if $x > 0$, then $x > \frac{1}{m}$ for some integer m, then for $k > m$ we have $k!x^k > \frac{1*2*3...*m *(m+1)*(m+2)...*k}{m*m*m...*m*m*m...*m}$ and one can see that this is a finite number times a number which is growing without bound. Hence the sequence of terms of the series grows without bound for any positive value of $x$.

## November 14, 2013

### The Daily Jumble: Math version (Foxtrot)

Filed under: calculus, media — Tags: — collegemathteaching @ 8:28 pm

GROAN.

This isn’t the first for Foxtrot though:

## November 12, 2013

### Why I teach multiple methods for the inverse Laplace Transform.

I’ll demonstrate with a couple of examples:

$y''+4y = sin(2t), y(0) = y'(0) = 0$

If we use the Laplace transform, we obtain: $(s^2+4)Y = \frac{2}{s^2+4}$ which leads to $Y = \frac{2}{(s^2+4)^2}$. Now we’ve covered how to do this without convolutions. But the convolution integral is much easier: write $Y = \frac{2}{(s^2+4)^2} = \frac{1}{2} \frac{2}{s^2+4}\frac{2}{s^2+4}$ which means that $y = \frac{1}{2}(sin(2t)*sin(2t)) = \frac{1}{2}\int^t_0 sin(2u)sin(2t-2u)du = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$.

Note: if the integral went too fast for you and you don’t want to use a calculator, use $sin(2t-2u) = sin(2t)cos(2u) - cos(2t)sin(2u)$ and the integral becomes $\frac{1}{2}\int^t_0 sin(2t)cos(2u)sin(2u) -cos(2t)sin^2(2u)du =$

$\frac{1}{2} (sin(2t))\frac{1}{4}sin^2(2u)|^t_0 - cos(2t)(\frac{1}{4})( t - \frac{1}{4}sin(4u)|^t_0 =$

$\frac{1}{8}sin^3(2t) - \frac{1}{4}tcos(2t) +\frac{1}{16}sin(4t)cos(2t) =$

$\frac{1}{8}(sin^3(2t) +sin(2t)cos^2(2t))-\frac{1}{4}tcos(2t)$

$= \frac{1}{8}sin(2t)(sin^2(2t) + cos^2(2t))-\frac{1}{4}tcos(2t) = -\frac{1}{4}tcos(2t) + \frac{1}{8}sin(2t)$

Now if we had instead: $y''+4y = sin(t), y(0)=0, y'(0) = 0$

The Laplace transform of the equation becomes $(s^2+4)Y = \frac{1}{s^2+1}$ and hence $Y = \frac{1}{(s^2+1)(s^2+4)}$. One could use the convolution method but partial fractions works easily: one can use the calculator (“algebra” plus “expand”) or:

$\frac{A+Bs}{s^2+4} + \frac{C + Ds}{s^2+1} =\frac{1}{(s^2+4)(s^2+1)}$. Get a common denominator and match numerators:

$(A+Bs)(s^2+1) + (C+Ds)(s^2+4) = 1$. One can use several methods to resolve this: here we will use $s = i$ to see $(C + Di)(3) = 1$ which means that $D = 0$ and $C = \frac{1}{3}$. Now use $s = 2i$ so obtain $(A + 2iB)(-3) = 1$ which means that $B = 0, A = -\frac{1}{3}$ so $Y = \frac{1}{3} (\frac{1}{s^2+1} - \frac{1}{s^2+4}$ so $y = \frac{1}{3} (sin(t) - \frac{1}{2} sin(2t)) = \frac{1}{3}sin(t) -\frac{1}{6}sin(2t)$

So, sometimes the convolution leads us to the answer quicker than other techniques and sometimes other techniques are easier.

## November 6, 2013

### Some ethical matters for university professors

Filed under: academia, editorial — Tags: , , , — collegemathteaching @ 11:45 pm

This article entitled “When Students Rate Teachers, Standards Drop (NOTE: if you really want to read the article, just google the title and it should pop up free without having to pay; Defunct Adjunct googled it and it popped up the second time s/he searched for it; or read the summary below in Defunct Adjunct’s comment; it’s nothing new we haven’t heard before, which is why it doesn’t matter too much if people read it; I’m using it more as a segue to my main point). It basically states that when we want to get good evals, we–consciously or unconsciously–let our standards drop. Those of us posting on here complain when we see our colleagues doing this. But is it more complicated than that?

On this blog, we have the motto not to care more than the student does, but is another of our unspoken mottos that we care what the students think of us entirely too much? Or is it that we don’t care, but we know administrators care, and that means we feel we are helpless to do anything about it and then become resentful about it, thereby needing a blog where we can fight back?

This is one issue. Now as far as the Wall Street Journal article: it was opinion; the only data supplied is that the number of self-reported study hours per course is dropping as grades are rising. I think that grade inflation is a fact; I am not sure it is brought on by student evaluations…..or some other factor or number of factors.

But the second major point of the College Misery article is this:

I have been fairly content and proud of the kind of work I’ve been doing in my career. I’ve felt accomplished and have felt as if I have been contributing positively to society. But recently, I feel like I’m working for a company that supports unethical business practices (by admitting students who won’t succeed, and whom we can immediately tell won’t succeed), unethical treatment of its employees (by hiring adjuncts in exchange for shiny pebbles and rotten apples, not to mention overworking everyone!), and unethical promises to society (that we are churning out a workforce ready to take over). This makes me feel like I’m contributing to this culture.

Now I have to modify this: many times, we admit students who have a *low probability* of success in college period, but sometimes we admit students into a major who have an infinitesimally small chance of success (e. g. admitting “math majors” with a math ACT of 17!!!!).

So, when is administration just padding the tuition numbers (“hey, you want a job, don’t you?”) or when is the administration giving someone grossly under qualified a shot? After all, low probability is not zero probability, but then again SOMEONE wins the lottery every time, right? I’m sure that someone, somewhere died from being struck by a falling meteor, right? What probability is too low?

It is hard to tell, isn’t it?

### Inverse Laplace transform example: 1/(s^2 +b^2)^2

Filed under: basic algebra, differential equations, Laplace transform — Tags: — collegemathteaching @ 11:33 pm

I talked about one way to solve $y''+y = sin(t), y(0) =y'(0) = 0$ by using Laplace transforms WITHOUT using convolutions; I happen to think that using convolutions is the easiest way here.

Here is another non-convolution method: Take the Laplace transform of both sides to get $Y(s) = \frac{1}{(s^2+1)^2}$.

Now most tables have $L(tsin(at)) = \frac{2as}{(s^2 + a^2)^2}, L(tcos(at)) = \frac{s^2-a^2}{(s^2+a^2)^2}$

What we have is not in one of these forms. BUT, note the following algebra trick technique:

$\frac{1}{s^2+b^2} = (A)(\frac{s^2-b^2}{(s^2 + b^2)^2} - \frac{s^2+b^2}{(s^2+b^2)^2})$ when $A = -\frac{1}{2b^2}$.

Now $\frac{s^2-b^2}{(s^2 + b^2)^2} = L(tcos(bt))$ and $\frac{s^2+b^2}{(s^2+b^2)^2} = \frac{1}{(s^2+b^2)} = L(\frac{1}{b}sin(bt))$ and one can proceed from there.

### A weird Laplace Transform (a resonance equation)

Filed under: applied mathematics, calculus, differential equations, Laplace transform — collegemathteaching @ 12:01 am

Ok, we have $y" + y = sin(t), y(0) =0, y'(0) = 0$. Now we can solve this by, say, undetermined coefficients and obtain $y = \frac{1}{2}sin(t) -\frac{1}{2}tcos(t)$

But what happens when we try Laplace Transforms? It is easy to see that the Laplace transform of the equation yields $(s^2+1)Y(s)=\frac{1}{s^2+1}$ which yields $Y(s) =\frac{1}{(s^2+1)^2}$

So, how do we take the inverse Laplace transform of $\frac{1}{(s^2+1)^2}$?

Here is one way: we recognize $L(tf(t)) = -1\frac{d}{ds}F(s)$ where $L(f(t)) = F(s)$.

So, we might try integrating: $\int \frac{1}{(s^2+1)^2} ds$.

(no cheating with a calculator! 🙂 )

In calculus II, we do: $s = tan(\theta), ds = sec^2(\theta) d\theta$.

Then $\int \frac{1}{(s^2+1)^2} ds$ is transformed into $\int \frac{sec^2(\theta)}{sec^4 \theta} d\theta = \int cos^2(\theta) d \theta = \int \frac{1}{2} + \frac{1}{2}cos(2 \theta) d \theta = \frac{1}{2} \theta + \frac{1}{4}sin(2 \theta)$ (plus a constant, of course).

We now use $sin(2\theta) = 2sin(\theta)cos(\theta)$ to obtain $\frac{1}{2} \theta + \frac{1}{4}sin(2 \theta) = \frac{1}{2} \theta + \frac{1}{2} sin(\theta)cos(\theta) + C$.

Fair enough. But now we have to convert back to $s$. We use $tan(\theta) = s$ to obtain $cos(\theta) = \frac{1}{\sqrt{s^2+1}}, sin(\theta) = \frac{s}{\sqrt{s^2+1}}$

So $\frac{1}{2} \theta + \frac{1}{2} sin(\theta)cos(\theta)$ converts to $\frac{1}{2}arctan(s) + C +\frac{1}{2}\frac{s}{s^2+1} = \int Y(s) ds$. Now we use the fact that as $s$ goes to infinity, $\int Y(s)$ has to go to zero; this means $C = -\frac{\pi}{2}$.

So what is the inverse Laplace transform of $\int Y(s) ds$?

Clearly, $\frac{1}{2}\frac{s}{s^2+1}$ gets inverse transformed to $\frac{1}{2}cos(t)$, so the inverse transform for this part of $Y(s)$ is $-\frac{t}{2}cos(t)$.

But what about the other part? $\frac{d}{ds} (arctan(s) - \frac{\pi}{2}) = \frac{1}{1+s^2}$ so $\frac{1}{1+s^2} = -L(tf(t))$ which implies that $tf(t) = -sin(t)$ so $-tf(t) = sin(t)$ and so the inverse Laplace transform for this part of $Y(s)$ is $\frac{1}{2} sin(t)$ and the result follows.

Put another way: $L(\frac{sin(t)}{t}) =- arctan(s) + C$ but since we want $0$ when $s = \infty, C = \frac{\pi}{2}$ and so $L(\frac{sin(t)}{t}) = \frac{\pi}{2}- arctan(s) = arctan(\frac{1}{s})$ .