# College Math Teaching

## 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.

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)

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

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.”

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