College Math Teaching

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.

December 16, 2015

And I deducted points for a “Merry Christmas” math joke

Filed under: pedagogy, recreational mathematics — Tags: , — collegemathteaching @ 11:12 pm

From a student’s final exam in my “Life Contingencies” class (and no, I have no actuarial mathematics training…more on that later)

mathpoetry

The student completely ignored the domain considerations for the log function and therefore lost points.

OK, not really. But it makes a better meme to say that.

July 2, 2015

Knowledge that can’t be communicated is worthless

Filed under: editorial, pedagogy — Tags: — collegemathteaching @ 10:31 pm

arf

In the past, I’ve passed out this cartoon to my students. Too many times, I’ve heard “I understand how to do the problem, but I can’t do the problem on the exam.”

Well, I suppose that is a bit like saying:

“I know how to swim, but when I jump in the pool, I drown.”

“I know how to fly the plane, but when I try, I crash.”

May 11, 2015

The hypervolume of the n-ball enclosed by a standard n-1 sphere

I am always looking for interesting calculus problems to demonstrate various concepts and perhaps generate some interest in pure mathematics.
And yes, I like to “blow off some steam” by spending some time having some non-technical mathematical fun with elementary mathematics.

This post uses only:

1. Integration by parts and basic reduction formulas.
2. Trig substitution.
3. Calculation of volumes (and hyper volumes) by the method of cross sections.
4. Induction
5. Elementary arithmetic involving factorials.

The quest: find a formula that finds the (hyper)volume of the region \{(x_1, x_2, x_3,....x_k) | \sum_{i=1}^k x_i^2 \leq R^2 \} \subset R^k

We will assume that the usual tools of calculus work as advertised.

Start. If we done the (hyper)volume of the k-ball by V_k  we will start with the assumption that V_1 = 2R ; that is, the distance between the endpoints of [-R,R] is 2R.

Step 1: we show, via induction, that V_k =c_kR^k where c_k is a constant and R is the radius.

Our proof will be inefficient for instructional purposes.

We know that V_1 =2R hence the induction hypothesis holds for the first case and c_1 = 2 . We now go to show the second case because, for the beginner, the technique will be easier to follow further along if we do the k = 2 case.

Yes, I know that you know that V_2 = \pi R^2 and you’ve seen many demonstrations of this fact. Here is another: let’s calculate this using the method of “area by cross sections”. Here is x^2 + y^2 = R^2 with some y = c cross sections drawn in.

crosssections

Now do the calculation by integrals: we will use symmetry and only do the upper half and multiply our result by 2. At each y = y_c level, call the radius from the center line to the circle R(y) so the total length of the “y is constant” level is 2R(y) and we “multiply by thickness “dy” to obtain V_2 = 4 \int^{y=R}_{y=0} R(y) dy .

But remember that the curve in question is x^2 + y^2 = R^2 and so if we set x = R(y) we have R(y) = \sqrt{R^2 -y^2} and so our integral is 4 \int^{y=R}_{y=0}\sqrt{R^2 -y^2}  dy

Now this integral is no big deal. But HOW we solve it will help us down the road. So here, we use the change of variable (aka “trigonometric substitution”): y = Rsin(t), dy =Rcos(t) to change the integral to:

4 \int^{\frac{\pi}{2}}_0 R^2 cos^2(t) dt = 4R^2 \int^{\frac{\pi}{2}}_0  cos^2(t) dt therefore

V_2 = c_2 R^2 where:

c_2 = 4\int^{\frac{\pi}{2}}_0  cos^2(t)

Yes, I know that this is an easy integral to solve, but I first presented the result this way in order to make a point.

Of course, c_2 = 4\int^{\frac{\pi}{2}}_0  cos^2(t) = 4\int^{\frac{\pi}{2}}_0 \frac{1}{2} + \frac{1}{2}cos(2t) dt = \pi

Therefore, V_2 =\pi R^2 as expected.

Exercise for those seeing this for the first time: compute c_3 and V_3 by using the above methods.

Inductive step: Assume V_k = c_kR^k Now calculate using the method of cross sections above (and here we move away from x-y coordinates to more general labeling):

V_{k+1} = 2\int^R_0 V_k dy = 2 \int^R_0 c_k (R(x_{k+1})^k dx_{k+1} =c_k 2\int^R_0 (R(x_{k+1}))^k dx_{k+1}

Now we do the substitutions: first of all, we note that x_1^2 + x_2^2 + ...x_{k}^2 + x_{k+1}^2 = R^2 and so

x_1^2 + x_2^2 ....+x_k^2 = R^2 - x_{k+1}^2 . Now for the key observation: x_1^2 + x_2^2 ..+x_k^2 =R^2(x_{k+1}) and so R(x_{k+1}) = \sqrt{R^2 - x_{k+1}^2}

Now use the induction hypothesis to note:

V_{k+1} = c_k 2\int^R_0 (R^2 - x_{k+1}^2)^{\frac{k}{2}} dx_{k+1}

Now do the substitution x_{k+1} = Rsin(t), dx_{k+1} = Rcos(t)dt and the integral is now:

V_{k+1} = c_k 2\int^{\frac{\pi}{2}}_0 R^{k+1} cos^{k+1}(t) dt = c_k(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt)R^{k+1} which is what we needed to show.

In fact, we have shown a bit more. We’ve shown that c_1 = 2 =2 \int^{\frac{\pi}{2}}_0(cos(t))dt, c_2 = 2 \cdot 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt = c_1 2\int^{\frac{\pi}{2}}_0 cos^2(t) dt and, in general,

c_{k+1} = c_{k}c_{k-1}c_{k-2} ....c_1(2 \int^{\frac{\pi}{2}}_0 cos^{k+1}(t) dt) = 2^{k+1} \int^{\frac{\pi}{2}}_0(cos^{k+1}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k}(t))dt \int^{\frac{\pi}{2}}_0(cos^{k-1}(t))dt .....\int^{\frac{\pi}{2}}_0(cos(t))dt

Finishing the formula

We now need to calculate these easy calculus integrals: in this case the reduction formula:

\int cos^n(x) dx = \frac{1}{n}cos^{n-1}sin(x) + \frac{n-1}{n} \int cos^{n-2}(x) dx is useful (it is merely integration by parts). Now use the limits and elementary calculation to obtain:

\int^{\frac{\pi}{2}}_0 cos^n(x) dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_0 cos^{n-2}(x)dx to obtain:

\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{3}{4})\frac{\pi}{4} if n is even and:
\int^{\frac{\pi}{2}}_0 cos^n(x) dx = (\frac{n-1}{n})(\frac{n-3}{n-2})......(\frac{4}{5})\frac{2}{3} if n is odd.

Now to come up with something resembling a closed formula let’s experiment and do some calculation:

Note that c_1 = 2, c_2 = \pi, c_3 = \frac{4 \pi}{3}, c_4 = \frac{(\pi)^2}{2}, c_5 = \frac{2^3 (\pi)^2)}{3 \cdot 5} = \frac{8 \pi^2}{15}, c_6 = \frac{\pi^3}{3 \cdot 2} = \frac{\pi^3}{6} .

So we can make the inductive conjecture that c_{2k} = \frac{\pi^k}{k!} and see how it holds up: c_{2k+2} = 2^2 \int^{\frac{\pi}{2}}_0(cos^{2k+2}(t))dt \int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}

= 2^2 ((\frac{2k+1}{2k+2})(\frac{2k-1}{2k})......(\frac{3}{4})\frac{\pi}{4})((\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......\frac{2}{3})\frac{\pi^k}{k!}

Now notice the telescoping effect of the fractions from the c_{2k+1} factor. All factors cancel except for the (2k+2) in the first denominator and the 2 in the first numerator, as well as the \frac{\pi}{4} factor. This leads to:

c_{2k+2} = 2^2(\frac{\pi}{4})\frac{2}{2k+2} \frac{\pi^k}{k!} = \frac{\pi^{k+1}}{(k+1)!} as required.

Now we need to calculate c_{2k+1} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt c_{2k} = 2\int^{\frac{\pi}{2}}_0(cos^{2k+1}(t))dt \frac{\pi^k}{k!}

= 2 (\frac{2k}{2k+1})(\frac{2k-2}{2k-1})......(\frac{4}{5})\frac{2}{3}\frac{\pi^k}{k!} = 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k-1)...(5)(3)} \frac{\pi^k}{k!}

To simplify this further: split up the factors of the k! in the denominator and put one between each denominator factor:

= 2 (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(k)(2k-1)(k-1)...(3)(5)(2)(3)(1)} \pi^k Now multiply the denominator by 2^k and put one factor with each k-m factor in the denominator; also multiply by 2^k in the numerator to obtain:

(2) 2^k (\frac{(2k)(2k-2)(2k-4)..(4)(2)}{(2k+1)(2k)(2k-1)(2k-2)...(6)(5)(4)(3)(2)} \pi^k Now gather each factor of 2 in the numerator product of the 2k, 2k-2…

= (2) 2^k 2^k \pi^k \frac{k!}{(2k+1)!} = 2 \frac{(4 \pi)^k k!}{(2k+1)!} which is the required formula.

So to summarize:

V_{2k} = \frac{\pi^k}{k!} R^{2k}

V_{2k+1}= \frac{2 k! (4 \pi)^k}{(2k+1)!}R^{2k+1}

Note the following: lim_{k \rightarrow \infty} c_{k} = 0 . If this seems strange at first, think of it this way: imagine the n-ball being “inscribed” in an n-cube which has hyper volume (2R)^n . Then consider the ratio \frac{2^n R^n}{c_n R^n} = 2^n \frac{1}{c_n} ; that is, the n-ball holds a smaller and smaller percentage of the hyper volume of the n-cube that it is inscribed in; note the 2^n corresponds to the number of corners in the n-cube. One might see that the rounding gets more severe as the number of dimensions increases.

One also notes that for fixed radius R, lim_{n \rightarrow \infty} V_n = 0 as well.

There are other interesting aspects to this limit: for what dimension n does the maximum hypervolume occur? As you might expect: this depends on the radius involved; a quick glance at the hyper volume formulas will show why. For more on this topic, including an interesting discussion on this limit itself, see Dave Richardson’s blog Division by Zero. Note: his approach to finding the hyper volume formula is also elementary but uses polar coordinate integration as opposed to the method of cross sections.

April 1, 2015

My topology class April Fools me…

Filed under: academia, pedagogy — Tags: — collegemathteaching @ 9:41 pm

My topology class has 6 students and all usually sit in the first or second row of the class room. They are NEVER late.

As I walked to the classroom about 1-2 minutes before the start, the lights were out..and of the desks I could see…empty.

I walked in, looked at the back and said “Happy April 1”.

I LOVED it!

I joked that today, we were having the final exam and they had to prove the Tychonoff Theorem from scratch…full version (infinite product of compact spaces is compact in the product topology).

I gave them all 1 extra point on their homework assignments.

March 13, 2015

Moving from “young Turk” to “old f***”

Filed under: calculus, class room experiment, editorial, pedagogy — Tags: , , — collegemathteaching @ 9:09 pm

Today, one of our hot “young” (meaning: new here) mathematicians came to me and wanted to inquire about a course switch. He noted that his two course load included two different courses (two preparations) and that I was teaching different sections of the same two courses…was I interested in doing a course swap so that he had only one preparation (he is teaching 8 hours) and I’d only have two?

I said: “when I was your age, I minimized the number of preparations. But at my age, teaching two sections of the same low level course makes me want to bash my head against the wall”. That is, by my second lesson of the same course in the same day; I just want to be just about anywhere else on campus; I have no interest, no enthusiasm, etc.

I specifically REQUESTED 3 preparations to keep myself from getting bored; that is what 24 years of teaching this stuff does to you.

COMMENTARY
Every so often, someone has the grand idea to REFORM the teaching of (whatever) and the “reformers” usually get at least a few departments to go along with it.

The common thing said is that it gets professors to reexamine their teaching of (whatever).

But I wonder if many try these things….just out of pure boredom. Seriously, read the buzzwords of the “reform paper” I linked to; there is really nothing new there.

March 11, 2015

Old curmudgeon’s rant: non-seriousness of today’s textbooks

Filed under: academia, pedagogy — Tags: , — collegemathteaching @ 12:38 am

In our department we pass around the “business calculus” course; in a semester we are typically given one “less desirable” course (remedial, business calculus, baby-stats, math for poets or for middle school teachers), one “calculus/differential equations” and one “upper division/linear algebra” type course.

Regardless of level, I’d like to see my students take the course seriously.

So, I look at my “business calculus” text (which I played no role in choosing) and I see stuff like this:

nonsense1

nonsense2

nonsense3

Yes, every few pages, there is a photograph of some young person in business attire (sometimes holding a chart, sometimes not). Yes fans of PC-ness, young people of both sexes and several races are represented; there is even an old, gray person somewhere.

Please tell me what pedagogical purpose this serves.

I swear…our text books are looking more and more like fluff…every year.

Older Posts »

Blog at WordPress.com.