# College Math Teaching

## August 20, 2018

### Algebra for Calculus I: equations and inequalities

Filed under: basic algebra, calculus, pedagogy — collegemathteaching @ 9:24 pm

It seems simple enough: solve $3x+ 4 = 7$ or $\frac{2}{x-5} \leq 3$.

So what do we tell our students to do? We might say things like “with an equation we must do the same thing to both sides of the equation (other than multiply both sides by zero)” and with an inequality, “we have to remember to reverse the inequality if we, say, multiply both sides by a negative number or if we take the reciprocal”.

And, of course, we need to check afterwards to see if we haven’t improperly expanded the solution set.

But what is really going on? A moment’s thought will reveal that what we are doing is applying the appropriate function to both sides of the equation/inequality.

And, depending on what we are doing, we want to ensure that the function that we are applying is one-to-one and taking note if the function is increasing or decreasing in the event we are solving an inequality.

Example: $x + \sqrt{x+2} = 4$ Now the standard way is to subtract $x$ from both sides (which is a one to one function..subtract constant number) which yields $\sqrt{x+2} = 4-x$. Now we might say “square both sides” to obtain $x+2 = 16-8x+x^2 \rightarrow x^2-9x+ 14 = 0 \rightarrow (x-7)(x-2) = 0$ but only $x = 2$ works. But the function that does that, the “squaring function”, is NOT one to one. Think of it this way: if we have $x = y$ and we then square both sides we now have $x^2 = y^2$ which has the original solution $x = y$ and $x = -y$. So in our example, the extraneous solution occurs because $(\sqrt{7+2})^2 = (4-7)^2$ but $\sqrt{7+2} \neq -3$.

If you want to have more fun, try a function that isn’t even close to being one to one; e. g. solve $x + \frac{1}{4} =\frac{1}{2}$ by taking the sine of both sides. 🙂

(yes, I know, NO ONE would want to do that).

As far as inequalities: the idea is to remember that if one applies a one-to-one function on both sides, one should note if the function is increasing or decreasing.

Example: $2 \geq e^{-x} \rightarrow ln(2) \geq -x \rightarrow ln(\frac{1}{2}) \leq x$. We did the switch when the function that we applied ($f(x) = -x$ was decreasing.)

Example: solving $|x+9| \geq 8$ requires that we use the conditional definition for absolute value and reconcile our two answers: $x+ 9 \geq 8$ and $-x-9 \geq 8$ which leads to the union of $x \geq -1$ or $x \leq -17$

The fun starts when the function that we apply is neither decreasing nor increasing. Example: $sin(x) \geq \frac{1}{2}$ Needless to say, the $arcsin(x)$ function, by itself, is inadequate without adjusting for periodicity.

## October 23, 2014

### Being part of the problem…

Filed under: academia, basic algebra, editorial — Tags: , — collegemathteaching @ 10:55 am

An eight-month probe has estimated that the “shadow curriculum” that existed at the University of North Carolina from 1993 to 2011 offered a grade-point boost from phony coursework to more than 3,100 students, including a disproportionately high percentage of student-athletes.[…]

A number of the students’ papers included brief introductions and closings with only “fluff” in between, according to the report.

“Hundreds” of the courses, the report reads, were independent-study. Yet when the university tightened standards on the amount of independent study a student could undertake, Crowder altered her program, creating courses she identified as lecture courses, but which mirrored independent study in that lectures never happened. The Wainstein report found 188 such courses between 1999 and 2011, in which 47.4 percent of the enrollments in these “paper classes” were student-athletes, who generally comprise 4 percent of the student population. Once Crowder retired in 2009, Nyang’oro sustained the practices for two more years until his retirement in 2011, albeit less voluminously.

In all the “paper classes,” the report found an average issued grade of 3.62, set against 3.28 for the regular classes in the department. Among student-athletes, though, the report found an average issued grade of 3.55, further above the 2.84 among student-athletes in the regular department classes.

The grade was based on one paper, which wasn’t merely scanned. But note the following: the average issued grade IN THE DEPARTMENT was 3.28???

Now this is far from the first public scandal involving university athletes, but university athletes aren’t the focus of this post.

I am thinking about what *I* have been complicit in over the years.

No, I’ve never given credit for incorrect work. And no, I’ve never falsified a grade.

But….well, let me start with a story from my youth.

When I was young, my dad wanted me to learn how to hit a baseball. So, he’d toss a ball at me when I was holding a bat. BUT, he’d often..deliberately hit my bat with the ball, so I got the impression that I was hitting the ball.

And, at times, it appears that I do that with my exams; I hit their bats.

If I gave what I considered *honest* exams in calculus and assigned partial credit honestly, I’d say that 70 percent of my “C” students would get D’s or F’s. About 50 percent of my B students would make C’s and perhaps 50 percent of my A students could get B’s. And I’d get called on the carpet…get complaints and accused of “not being student centered”…of “lacking compassion.”

What is going on?

Well, though we have a smaller percentage of students going to college now than in, say, 2009, check out the data:

When I was an undergraduate, about 50 percent of high school graduates went to college. When I started teaching college, it was about 60 percent…it rose to roughly 70 percent and is now about 65 percent. I think that we are seeing at least a little bit of “regression to the mean” effects.

Perhaps as a result, we are seeing things like this:

Algebra is an onerous stumbling block for all kinds of students: disadvantaged and affluent, black and white. In New Mexico, 43 percent of white students fell below “proficient,” along with 39 percent in Tennessee. Even well-endowed schools have otherwise talented students who are impeded by algebra, to say nothing of calculus and trigonometry.

California’s two university systems, for instance, consider applications only from students who have taken three years of mathematics and in that way exclude many applicants who might excel in fields like art or history. Community college students face an equally prohibitive mathematics wall. A study of two-year schools found that fewer than a quarter of their entrants passed the algebra classes they were required to take.

“There are students taking these courses three, four, five times,” says Barbara Bonham of Appalachian State University. While some ultimately pass, she adds, “many drop out.”

ALGEBRA? (and no, I am not talking about group theory or ring theory!)

Yes, that algebra.

So there are times when I wonder if I am participating in a fraud of a type.

## September 19, 2014

### Freshman calculus: they don’t always know the basics….

Filed under: basic algebra, calculus — Tags: , — collegemathteaching @ 5:43 pm

Example one: many students don’t know that $\frac{\frac{a}{b}}{c} \ne \frac{a}{\frac{b}{c}}$ (of course, assume that $a, b, c, \ne 0$. ) Where this came up: when we computed $lim_{h \rightarrow 0} \frac{\frac{1}{1+h} - 1}{h}$ we obtained $lim_{h \rightarrow 0} \frac{\frac{-h}{1+h}}{h}$ and a student didn’t understand why this was equal to $lim_{h \rightarrow 0} \frac{-1}{1+h}$

I ended up asking the student to simplify $\frac{\frac{2}{3}}{2} =$ and ….asking: ok, “if I have 2/3’rds of a pie and I give two people an equal piece of that, how much pie does each person get?

Example two: I gave “find the domain of $\frac{1}{\sqrt{x^2 - 9}}$ and two students didn’t understand why an answer of $-3 > x > 3$ was logically impossible. One of them told me: “my calculus teacher in high school told me to do it this way”: I am 99.99 percent that this isn’t true, but, well, I stayed with it until the student understood why such a statement was logically impossible. Oh yes, this same “I had calculus in high school” student was sure that I was wrong when I told him that “the derivative of a constant function is zero”; he was SURE that it is “1”.

## August 27, 2014

### Nice collection of Math GIFs

Filed under: basic algebra, calculus, elementary mathematics, pedagogy — Tags: — collegemathteaching @ 12:25 am

No, these GIF won’t explain better than an instructor, but I found many of them to be fun.

Via: IFLS

## August 21, 2014

### Calculation of the Fourier Transform of a tent map, with a calculus tip….

I’ve been following these excellent lectures by Professor Brad Osgood of Stanford. As an aside: yes, he is dynamite in the classroom, but there is probably a reason that Stanford is featuring him. 🙂

And yes, his style is good for obtaining a feeling of comradery that is absent in my classroom; at least in the lower division “service” classes.

This lecture takes us from Fourier Series to Fourier Transforms. Of course, he admits that the transition here is really a heuristic trick with symbolism; it isn’t a bad way to initiate an intuitive feel for the subject though.

However, the point of this post is to offer a “algebra of calculus trick” for dealing with the sort of calculations that one might encounter.

By the way, if you say “hey, just use a calculator” you will be BANNED from this blog!!!! (just kidding…sort of. 🙂 )

So here is the deal: let $f(x)$ represent the tent map: the support of $f$ is $[-1,1]$ and it has the following graph:

The formula is: $f(x)=\left\{\begin{array}{c} x+1,x \in [-1,0) \\ 1-x ,x\in [0,1] \\ 0 \text{ elsewhere} \end{array}\right.$

So, the Fourier Transform is $F(f) = \int^{\infty}_{-\infty} e^{-2 \pi i st}f(t)dt = \int^0_{-1} e^{-2 \pi i st}(1+t)dt + \int^1_0e^{-2 \pi i st}(1-t)dt$

Now, this is an easy integral to do, conceptually, but there is the issue of carrying constants around and being tempted to make “on the fly” simplifications along the way, thereby leading to irritating algebraic errors.

So my tip: just let $a = -2 \pi i s$ and do the integrals:

$\int^0_{-1} e^{at}(1+t)dt + \int^1_0e^{at}(1-t)dt$ and substitute and simplify later:

Now the integrals become: $\int^{1}_{-1} e^{at}dt + \int^0_{-1}te^{at}dt - \int^1_0 te^{at} dt.$
These are easy to do; the first is merely $\frac{1}{a}(e^a - e^{-a})$ and the next two have the same anti-derivative which can be obtained by a “integration by parts” calculation: $\frac{t}{a}e^{at} -\frac{1}{a^2}e^{at}$; evaluating the limits yields:

$-\frac{1}{a^2}-(\frac{-1}{a}e^{-a} -\frac{1}{a^2}e^{-a}) - (\frac{1}{a}e^{a} -\frac{1}{a^2}e^a)+ (-\frac{1}{a^2})$

Add the first integral and simplify and we get: $-\frac{1}{a^2}(2 - (e^{-a} -e^{a})$. NOW use $a = -2\pi i s$ and we have the integral is $\frac{1}{4 \pi^2 s^2}(2 -(e^{2 \pi i s} -e^{-2 \pi i s}) = \frac{1}{4 \pi^2 s^2}(2 - cos(2 \pi s))$ by Euler’s formula.

Now we need some trig to get this into a form that is “engineering/scientist” friendly; here we turn to the formula: $sin^2(x) = \frac{1}{2}(1-cos(2x))$ so $2 - cos(2 \pi s) = 4sin^2(\pi s)$ so our answer is $\frac{sin^2( \pi s)}{(\pi s)^2} = (\frac{sin(\pi s)}{\pi s})^2$ which is often denoted as $(sinc(s))^2$ as the “normalized” $sinc(x)$ function is given by $\frac{sinc(\pi x)}{\pi x}$ (as we want the function to have zeros at integers and to “equal” one at $x = 0$ (remember that famous limit!)

So, the point is that using $a$ made the algebra a whole lot easier.

Now, if you are shaking your head and muttering about how this calculation was crude that that one usually uses “convolution” instead: this post is probably too elementary for you. 🙂

## August 7, 2014

### Letting complex algebra make our calculus lives easier

Filed under: basic algebra, calculus, complex variables — Tags: , — collegemathteaching @ 1:37 am

If one wants to use complex arithmetic in elementary calculus, one should, of course, verify a few things first. One might talk about elementary complex arithmetic and about complex valued functions of a real variable at an elementary level; e. g. $f(x) + ig(x)$. Then one might discuss Euler’s formula: $e^{ix} = cos(x) + isin(x)$ and show that the usual laws of differentiation hold; i. e. show that $\frac{d}{dx} e^{ix} = ie^{ix}$ and one might show that $(e^{ix})^k = e^{ikx}$ for $k$ an integer. The latter involves some dreary trigonometry but, by doing this ONCE at the outset, one is spared of having to repeat it later.

This is what I mean: suppose we encounter $cos^n(x)$ where $n$ is an even integer. I use an even integer power because $\int cos^n(x) dx$ is more challenging to evaluate when $n$ is even.

Coming up with the general formula can be left as an exercise in using the binomial theorem. But I’ll demonstrate what is going on when, say, $n = 8$.

$cos^8(x) = (\frac{e^{ix} + e^{-ix}}{2})^8 =$

$\frac{1}{2^8} (e^{i8x} + 8 e^{i7x}e^{-ix} + 28 e^{i6x}e^{-i2x} + 56 e^{i5x}e^{-i3x} + 70e^{i4x}e^{-i4x} + 56 e^{i3x}e^{-i5x} + 28e^{i2x}e^{-i6x} + 8 e^{ix}e^{-i7x} + e^{-i8x})$

$= \frac{1}{2^8}((e^{i8x}+e^{-i8x}) + 8(e^{i6x}+e^{-i6x}) + 28(e^{i4x}+e^{-i4x})+ 56(e^{i2x}+e^{-i2x})+ 70) =$

$\frac{70}{2^8} + \frac{1}{2^7}(cos(8x) + 8cos(6x) + 28cos(4x) +56cos(2x))$

So it follows reasonably easily that, for $n$ even,

$cos^n(x) = \frac{1}{2^{n-1}}\Sigma^{\frac{n}{2}-1}_{k=0} (\binom{n}{k}cos((n-2k)x)+\frac{\binom{n}{\frac{n}{2}}}{2^n}$

So integration should be a breeze. Lets see about things like, say,

$cos(kx)sin(nx) = \frac{1}{(2)(2i)} (e^{ikx}+e^{-ikx})(e^{inx}-ie^{-inx}) =$

$\frac{1}{4i}((e^{i(k+n)x} - e^{-i(k+n)x}) + (e^{i(n-k)x}-e^{-i(n-k)x}) = \frac{1}{2}(sin((k+n)x) + sin((n-k)x)$

Of course these are known formulas, but their derivation is relatively simple when one uses complex expressions.

## July 17, 2014

### Parabolas have interesting properties

And this is one of them:

## July 15, 2014

### A quickie from Mathematics Magazine

Filed under: basic algebra, editorial, elementary mathematics, recreational mathematics — Tags: — collegemathteaching @ 9:35 pm

It has been a long time since I’ve posted; I’ve spent time doing various things, including revising a paper that a journal editor wanted revised.

I’ll speak more about that later.

I got the latest Mathematics Magazine in the mail (volume 87, No. 3, June 2014), and the article “Surprises” by Felix Lazebnik is chock full of delightful tidbits, many of which I didn’t know.

Here is a fun one that you can share with your non-mathematically inclined friends (other tidbits there require some mathematics background).

(Surprise 7): A watermelon is 99 percent water (by weight). One ton of watermelons was shipped and during shipment some water evaporated. The watermelons that arrived were made up of 98 percent water (by weight). What was the weight of the shipment when it arrived?

The answer is in the article, but I suggest you give it a go. It will take, at most, a minute or two (if that). This shows the power of basic algebra to discipline our thinking and how our intuition can deceive us.

## 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 6, 2013

### 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.

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