College Math Teaching

December 21, 2018

Over-scheduling of senior faculty and lower division courses: how important is course prep?

It seems as if the time faculty is expected to spend on administrative tasks is growing exponentially. In our case: we’ve had some administrative upheaval with the new people coming in to “clean things up”, thereby launching new task forces, creating more committees, etc. And this is a time suck; often more senior faculty more or less go through the motions when it comes to course preparation for the elementary courses (say: the calculus sequence, or elementary differential equations).

And so:

1. Does this harm the course quality and if so..
2. Is there any effect on the students?

I should first explain why I am thinking about this; I’ll give some specific examples from my department.

1. Some time ago, a faculty member gave a seminar in which he gave an “elementary” proof of why \int e^{x^2} dx is non-elementary. Ok, this proof took 40-50 minutes to get through. But at the end, the professor giving the seminar exclaimed: “isn’t this lovely?” at which, another senior member (one who didn’t have a Ph. D. had had been around since the 1960’s) asked “why are you happy that yet again, we haven’t had success?” The fact that a proof that \int e^{x^2} dx could not be expressed in terms of the usual functions by the standard field operations had been given; the whole point had eluded him. And remember, this person was in our calculus teaching line up.

2. Another time, in a less formal setting, I had mentioned that I had given a brief mention to my class that one could compute and improper integral (over the real line) of an unbounded function that that a function could have a Laplace transform. A junior faculty member who had just taught differential equations tried to inform me that only functions of exponential order could have a Laplace transform; I replied that, while many texts restricted Laplace transforms to such functions, that was not mathematically necessary (though it is a reasonable restriction for an applied first course). (briefly: imagine a function whose graph consisted of a spike of height e^{n^2} at integer points over an interval of width \frac{1}{2^{2n} e^{2n^2}} and was zero elsewhere.

3. In still another case, I was talking about errors in answer keys and how, when I taught courses that I wasn’t qualified to teach (e. g. actuarial science course), it was tough for me to confidently determine when the answer key was wrong. A senior, still active research faculty member said that he found errors in an answer key..that in some cases..the interval of absolute convergence for some power series was given as a closed interval.

I was a bit taken aback; I gently reminded him that \sum \frac{x^k}{k^2} was such a series.

I know what he was confused by; there is a theorem that says that if \sum a_k x^k converges (either conditionally or absolutely) for some x=x_1 then the series converges absolutely for all x_0 where |x_0| < |x_1| The proof isn’t hard; note that convergence of \sum a_k x^k means eventually, |a_k x^k| < M for some positive M then compare the “tail end” of the series: use |\frac{x_0}{x_1}| < r < 1 and then |a_k (x_0)^k| = |a_k x_1^k (\frac{x_0}{x_1})^k| < |r^k|M and compare to a convergent geometric series. Mind you, he was teaching series at the time..and yes, is a senior, research active faculty member with years and years of experience; he mentored me so many years ago.

4. Also…one time, a sharp young faculty member asked around “are there any real functions that are differentiable exactly at one point? (yes: try f(x) = x^2 if x is rational, x^3 if x is irrational.

5. And yes, one time I had forgotten that a function could be differentiable but not be C^1 (try: x^2 sin (\frac{1}{x}) at x = 0

What is the point of all of this? Even smart, active mathematicians forget stuff if they haven’t reviewed it in a while…even elementary stuff. We need time to review our courses! But…does this actually affect the students? I am almost sure that at non-elite universities such as ours, the answer is “probably not in any way that can be measured.”

Think about it. Imagine the following statements in a differential equations course:

1. “Laplace transforms exist only for functions of exponential order (false)”.
2. “We will restrict our study of Laplace transforms to functions of exponential order.”
3. “We will restrict our study of Laplace transforms to functions of exponential order but this is not mathematically necessary.”

Would students really recognize the difference between these three statements?

Yes, making these statements, with confidence, requires quite a bit of difference in preparation time. And our deans and administrators might not see any value to allowing for such preparation time as it doesn’t show up in measures of performance.


October 4, 2018

When is it ok to lie to students? part I

Filed under: calculus, derivatives, pedagogy — collegemathteaching @ 9:32 pm

We’ve arrived at logarithms in our calculus class, and, of course, I explained that ln(ab) = ln(a) + ln(b) only holds for a, b > 0 . That is all well and good.
And yes, I explained that expressions like f(x)^{g(x)} only makes sense when f(x) > 0

But then I went ahead and did a problem of the following type: given f(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1} by using logarithmic differentiation,

f'(x) = \frac{x^3 e^{x^2} cos(x)}{x^4 + 1} (\frac{3}{x} + 2x -tan(x) -\frac{4x^3}{x^4+ 1})

And you KNOW exactly what I did. Right?

Note that f is differentiable for all x and, well, the derivative *should* be continuous for all x it? Well, up to inessential singularities, it is. You see: the second factor is not defined for x = 0, x = \frac{\pi}{2} \pm k \pi , etc.

Well, let’s multiply it out and obtain:
f'(x) = \frac{3x^2 e^{x^2} cos(x)}{x^4 + 1} + \frac{2x^4 e^{x^2} cos(x)}{x^4 + 1} - \frac{x^3 e^{x^2} sin(x)}{x^4 + 1}-\frac{4x^6 e^{x^2} cos(x)}{(x^4 + 1)^2}

So, there is that. We might induce inessential singularities.

And there is the following: in the process of finding the derivative to begin with we did:

ln(\frac{x^3 e^{x^2} cos(x)}{x^4 + 1}) = ln(x^3) + ln(e^{x^2}) + ln(cos(x)) - ln(x^4 + 1) and that expansion is valid only for
x \in (0, \frac{\pi}{2}) \cup (\frac{5\pi}{2}, \frac{7\pi}{2}) \cup .... because we need x^3 > 0 and cos(x) > 0 .

But the derivative formula works anyway. So what is the formula?

It is: if f = \prod_{j=1}^k f_j where f_j is differentiable, then f' = \sum_{i=1}^k f'_i \prod_{j =1, j \neq i}^k f_j and verifying this is an easy exercise in induction.

But the logarithmic differentiation is really just a motivating idea that works for positive functions.

To make this complete: we’ll now tackle y = f(x)^{g(x)} where it is essential that f(x) > 0 .

Rewrite y = e^{ln(f(x)^{g(x)})} = e^{g(x)ln(f(x))}

Then y' = e^{g(x)ln(f(x))} (g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)}) =  f(x)^{g(x)}(g'(x) ln(f(x)) + g(x) \frac{f'(x)}{f(x)})

This formula is a bit of a universal one. Let’s examine two special cases.

Suppose g(x) = k some constant. Then g'(x) =0 and the formula becomes y = f(x)^k(k \frac{f'(x)}{f(x)}) = kf(x)^{k-1}f'(x) which is just the usual constant power rule with the chain rule.

Now suppose f(x) = a for some positive constant. Then f'(x) = 0 and the formula becomes y = a^{g(x)}(ln(a)g'(x)) which is the usual exponential function differentiation formula combined with the chain rule.

September 8, 2018

Proving a differentiation formula for f(x) = x ^(p/q) with algebra

Filed under: calculus, derivatives, elementary mathematics, pedagogy — collegemathteaching @ 1:55 am

Yes, I know that the proper way to do this is to prove the derivative formula for f(x) = x^n and then use, say, the implicit function theorem or perhaps the chain rule.

But an early question asked students to use the difference quotient method to find the derivative function (ok, the “gradient”) for f(x) = x^{\frac{3}{2}} And yes, one way to do this is to simplify the difference quotient \frac{t^{\frac{3}{2}} -x^{\frac{3}{2}} }{t-x} by factoring t^{\frac{1}{2}} -x^{\frac{1}{2}} from both the numerator and the denominator of the difference quotient. But this is rather ad-hoc, I think.

So what would one do with, say, f(x) = x^{\frac{p}{q}} where p, q are positive integers?

One way: look at the difference quotient: \frac{t^{\frac{p}{q}}-x^{\frac{p}{q}}}{t-x} and do the following (before attempting a limit, of course): let u= t^{\frac{1}{q}}, v =x^{\frac{1}{q}} at which our difference quotient becomes: \frac{u^p-v^p}{u^q -v^q}

Now it is clear that u-v is a common factor..but HOW it factors is essential.

So let’s look at a little bit of elementary algebra: one can show:

x^{n+1} - y^{n+1} = (x-y) (x^n + x^{n-1}y + x^{n-2}y^2 + ...+ xy^{n-1} + y^n)

= (x-y)\sum^{n}_{i=0} x^{n-i}y^i (hint: very much like the geometric sum proof).

Using this:

\frac{u^p-v^p}{u^q -v^q} = \frac{(u-v)\sum^{p-1}_{i=0} u^{p-1-i}v^i}{(u-v)\sum^{q-1}_{i=0} u^{q-1-i}v^i}=\frac{\sum^{p-1}_{i=0} u^{p-1-i}v^i}{\sum^{q-1}_{i=0} u^{q-1-i}v^i} Now as

t \rightarrow x we have u \rightarrow v (for the purposes of substitution) so we end up with:

\frac{\sum^{p-1}_{i=0} v^{p-1-i}v^i}{\sum^{q-1}_{i=0} v^{q-1-i}v^i}  = \frac{pv^{p-1}}{qv^{q-1}} = \frac{p}{q}v^{p-q} (the number of terms is easy to count).

Now back substitute to obtain \frac{p}{q} x^{\frac{(p-q)}{q}} = \frac{p}{q} x^{\frac{p}{q}-1} which, of course, is the familiar formula.

Note that this algebraic identity could have been used for the old f(x) = x^n case to begin with.

August 27, 2018

On teaching limits poorly

Filed under: calculus, pedagogy — Tags: — collegemathteaching @ 4:52 pm

I will be talking about teaching limits in a first year calculus class.

The textbook our department is using does the typical:

It APPEARS to be making the claim that the limit of the given function is 4 as x approaches 2 because, well, 4 is between f(2.001) and f(1.999) . But, there are an uncountable number of numbers between those two values; one really needs that the function in question “preserves integers” in order to give a good reason to “guess” that the limit is indeed 4.

I think that the important thing here is that the range is being squeezed as the domain gets squeezed, and, in my honest opinion, THAT is the point of limits: the limit exists when one can tighten the range tolerance by sufficiently tightening the domain tolerance.

But, in general, it is impossible to guess the limit without extra information about the function (e. g. maps integers to integers, etc.)

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.

February 22, 2018

What is going on here: sum of cos(nx)…

Filed under: analysis, derivatives, Fourier Series, pedagogy, sequences of functions, series, uniform convergence — collegemathteaching @ 9:58 pm

This started innocently enough; I was attempting to explain why we have to be so careful when we attempt to differentiate a power series term by term; that when one talks about infinite sums, the “sum of the derivatives” might fail to exist if the sum is infinite.

Anyone who is familiar with Fourier Series and the square wave understands this well:

\frac{4}{\pi} \sum^{\infty}_{k=1} \frac{1}{2k-1}sin((2k-1)x)  = (\frac{4}{\pi})( sin(x) + \frac{1}{3}sin(3x) + \frac{1}{5}sin(5x) +.....) yields the “square wave” function (plus zero at the jump discontinuities)

Here I graphed to 2k-1 = 21

Now the resulting function fails to even be continuous. But the resulting function is differentiable except for the points at the jump discontinuities and the derivative is zero for all but a discrete set of points.

(recall: here we have pointwise convergence; to get a differentiable limit, we need other conditions such as uniform convergence together with uniform convergence of the derivatives).

But, just for the heck of it, let’s differentiate term by term and see what we get:

(\frac{4}{\pi})\sum^{\infty}_{k=1} cos((2k-1)x) = (\frac{4}{\pi})(cos(x) + cos(3x) + cos(5x) + cos(7x) +.....)...

It is easy to see that this result doesn’t even converge to a function of any sort.

Example: let’s see what happens at x = \frac{\pi}{4}: cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) =0

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) + cos(5\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}

cos(\frac{\pi}{4}) + cos(3\frac{\pi}{4}) + cos(5\frac{\pi}{4}) + cos(7\frac{\pi}{4}) = 0

And this repeats over and over again; no limit is possible.

Something similar happens for x = \frac{p}{q}\pi where p, q are relatively prime positive integers.

But something weird is going on with this sum. I plotted the terms with 2k-1 \in \{1, 3, ...35 \}

(and yes, I am using \frac{\pi}{4} csc(x) as a type of “envelope function”)

BUT…if one, say, looks at cos(29x) + cos(31x) + cos(33x) + cos(35x)

we really aren’t getting a convergence (even at irrational multiples of \pi ). But SOMETHING is going on!

I decided to plot to cos(61x)

Something is going on, though it isn’t convergence. Note: by accident, I found that the pattern falls apart when I skipped one of the terms.

This is something to think about.

I wonder: for all x \in (0, \pi), sup_{n \in \{1, 3, 5, 7....\}}|\sum^{n}_{k \in \{1,3,..\}}cos(kx)| \leq |csc(x)| and we can somehow get close to csc(x) for given values of x by allowing enough terms…but the value of x is determined by how many terms we are using (not always the same value of x ).

October 7, 2016

Now what is a linear transformation anyway?

Filed under: linear albegra, pedagogy — Tags: , — collegemathteaching @ 9:43 pm

Yes, I know, a linear transformation L: V \rightarrow W is a function between vector spaces such that L(V \oplus W) = L(V) \oplus L(W) and L(a \odot V) = a \odot L(V) where the vector space operations of vector addition and scalar multiplication occur in their respective spaces.

Previously, I talked about this classical example:

Consider the set R^+ = \{x| x > 0 \} endowed with the “vector addition” x \oplus y = xy where xy represents ordinary real number multiplication and “scalar multiplication r \odot x = x^r where r \in R and x^r is ordinary exponentiation. It is clear that \{R^+, R | \oplus, \odot \} is a vector space with 1 being the vector “additive” identity and 0 playing the role of the scalar zero and 1 playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Then L(x) = ln(x) is a vector space isomophism between R^+ and R (the usual addition and scalar multiplication) and of course, L^{-1}(x) = exp(x) .

Can we expand this concept any further?

Question: (I have no idea if this has been answered or not): given any, say, non-compact, connected subset of R, is it possible to come up with vector space operations (vector addition, scalar multiplication) so as to make a given, say, real valued, continuous one to one function into a linear transformation?

The answer in some cases is “yes.”

Consider L(x): R^+ \rightarrow R^+ by L(x) = x^r , r any real number.

Exercise 1: L is a linear transformation.

Exercise 2: If we have ANY linear transformation L: R^+ \rightarrow R^+ , let L(e) = e^a .
Then L(x) = L(e^{ln(x)}) = L(e)^{ln(x)} = (e^a)^{ln(x)} = x^a .

Exercise 3: we know that all linear transformations L: R \rightarrow R are of the form L(x) = ax . These can be factored through:

x \rightarrow e^x \rightarrow (e^x)^a = e^{ax} \rightarrow ln(e^{ax}) = ax .

So this isn’t exactly anything profound, but it is fun! And perhaps it might be a way to introduce commutative diagrams.

October 4, 2016

Linear Transformation or not? The vector space operations matter.

Filed under: calculus, class room experiment, linear albegra, pedagogy — collegemathteaching @ 3:31 pm

This is nothing new; it is an example for undergraduates.

Consider the set R^+ = \{x| x > 0 \} endowed with the “vector addition” x \oplus y = xy where xy represents ordinary real number multiplication and “scalar multiplication r \odot x = x^r where r \in R and x^r is ordinary exponentiation. It is clear that \{R^+, R | \oplus, \odot \} is a vector space with 1 being the vector “additive” identity and 0 playing the role of the scalar zero and 1 playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Now consider the function L(x) = ln(x) with domain R^+ . (here: ln(x) is the natural logarithm function). Now ln(xy) = ln(x) + ln(y) and ln(x^a) = aln(x) . This shows that L:R^+ \rightarrow R (the range has the usual vector space structure) is a linear transformation.

What is even better: ker(L) =\{x|ln(x) = 0 \} which shows that ker(L) = \{1 \} so L is one to one (of course, we know that from calculus).

And, given z \in R, ln(e^z) = z so L is also onto (we knew that from calculus or precalculus).

So, R^+ = \{x| x > 0 \} is isomorphic to R with the usual vector operations, and of course the inverse linear transformation is L^{-1}(y) = e^y .

Upshot: when one asks “is F a linear transformation or not”, one needs information about not only the domain set but also the vector space operations.

February 10, 2016

Vector subspaces: two examples

Filed under: linear albegra, pedagogy — Tags: — collegemathteaching @ 8:41 pm

I am teaching linear algebra our of the book by Fraleigh and Beauregard. We are on “subspaces” (subsets of R^n for now) and a subspace is defined to be a set of vectors that is closed under both vector addition and scalar multiplication. Here are a couple of examples of non-subspaces:

1. W= \{(x,y)| xy = 0 \} . Now this space IS closed under scalar multiplication, note that this space IS closed under additive inverses. But it is not closed under addition as [x,0] + [0,y]=[x,y] \notin W for x \neq 0, y \neq 0 .

2. (this example is in the book): the vectors \{(n, m) | n, m \in Z \} are closed under vector addition but not under scalar multiplication.

February 9, 2016

An economist talks about graphs

Filed under: academia, economics, editorial, pedagogy, student learning — Tags: , — collegemathteaching @ 7:49 pm

Paul Krugman is a Nobel Laureate caliber economist (he won whatever they call the economics prize).
Here he discusses the utility of using a graph to understand an economic situation:

Brad DeLong asks a question about which of the various funny diagrams economists love should be taught in Econ 101. I say production possibilities yes, Edgeworth box no — which, strange to say, is how we deal with this issue in Krugman/Wells. But students who go on to major in economics should be exposed to the box — and those who go on to grad school really, really need to have seen it, and in general need more simple general-equilibrium analysis than, as far as I can tell, many of them get these days.

There was, clearly, a time when economics had too many pictures. But now, I suspect, it doesn’t have enough.

OK, this is partly a personal bias. My own mathematical intuition, and a lot of my economic intuition in general, is visual: I tend to start with a picture, then work out both the math and the verbal argument to make sense of that picture. (Sometimes I have to learn the math, as I did on target zones; the picture points me to the math I need.) I know that’s not true for everyone, but it’s true for a fair number of students, who should be given the chance to learn things that way.

Beyond that, pictures are often the best way to convey global insights about the economy — global in the sense of thinking about all possibilities as opposed to small changes, not as in theworldisflat. […]

And it probably doesn’t hurt to remind ourselves that our students are, in general, NOT like us. What comes to us naturally probably does not come to them naturally.

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