College Math Teaching

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 $x$ but..is 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.

March 12, 2018

And I embarrass myself….integrate right over a couple of poles…

Filed under: advanced mathematics, analysis, calculus, complex variables, integrals — Tags: — collegemathteaching @ 9:43 pm

I didn’t have the best day Thursday; I was very sick (felt as if I had been in a boxing match..chills, aches, etc.) but was good to go on Friday (no cough, etc.)

So I walk into my complex variables class seriously under prepared for the lesson but decide to tackle the integral

$\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt$

Of course, you know the easy way to do this, right?

$\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt =\frac{1}{2} \int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt$ and evaluate the latter integral as follows:

$sin(t) = \frac{1}{2i}(z-\frac{1}{z}), dt = \frac{dz}{iz}$ (this follows from restricting $z$ to the unit circle $|z| =1$ and setting $z = e^{it} \rightarrow dz = ie^{it}dt$ and then obtaining a rational function of $z$ which has isolated poles inside (and off of) the unit circle and then using the residue theorem to evaluate.

So $1+sin^2(t) \rightarrow 1+\frac{-1}{4}(z^2 -2 + \frac{1}{z^2}) = \frac{1}{4}(-z^2 + 6 -\frac{1}{z^2})$ And then the integral is transformed to:

$\frac{1}{2}\frac{1}{i}(-4)\int_{|z|=1}\frac{dz}{z^3 -6z +\frac{1}{z}} =2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1}$

Now the denominator factors: $(z^2 -3)^2 -8$ which means $z^2 = 3 - \sqrt{8}, z^2 = 3+ \sqrt{8}$ but only the roots $z = \pm \sqrt{3 - \sqrt{8}}$ lie inside the unit circle.
Let $w = \sqrt{3 - \sqrt{8}}$

Write: $\frac{z}{z^4 -6z^2 +1} = \frac{\frac{z}{((z^2 -(3 + \sqrt{8})}}{(z-w)(z+w)}$

Now calculate: $\frac{\frac{w}{((w^2 -(3 + \sqrt{8})}}{(2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}}$ and $\frac{\frac{-w}{((w^2 -(3 + \sqrt{8})}}{(-2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}}$

Adding we get $\frac{-1}{2 \sqrt{8}}$ so by Cauchy’s theorem $2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1} = 2i 2 \pi i \frac{-1}{2 \sqrt{8}} = \frac{2 \pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}$

Ok…that is fine as far as it goes and correct. But what stumped me: suppose I did not evaluate $\int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt$ and divide by two but instead just went with:

\$latex $\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt \rightarrow i \int_{\gamma}\frac{zdz}{z^4 -6z^2 +1}$ where $\gamma$ is the upper half of $|z| = 1$? Well, $\frac{z}{z^4 -6z^2 +1}$ has a primitive away from those poles so isn’t this just $i \int^{-1}_{1}\frac{zdz}{z^4 -6z^2 +1}$, right?

So why not just integrate along the x-axis to obtain $i \int^{-1}_{1}\frac{xdx}{x^4 -6x^2 +1} = 0$ because the integrand is an odd function?

This drove me crazy. Until I realized…the poles….were…on…the…real…axis. ….my goodness, how stupid could I possibly be???

To the student who might not have followed my point: let $\gamma$ be the upper half of the circle $|z|=1$ taken in the standard direction and $\int_{\gamma} \frac{1}{z} dz = i \pi$ if you do this property (hint: set $z(t) = e^{it}, dz = ie^{it}, t \in [0, \pi]$. Now attempt to integrate from 1 to -1 along the real axis. What goes wrong? What goes wrong is exactly what I missed in the above example.

August 28, 2017

Integration by parts: why the choice of “v” from “dv” might matter…

We all know the integration by parts formula: $\int u dv = uv - \int v du$ though, of course, there is some choice in what $v$ is; any anti-derivative will do. Well, sort of.

I thought about this as I’ve been roped into teaching an actuarial mathematics class (and no, I have zero training in this area…grrr…)

So here is the set up: let $F_x(t) = P(0 \leq T_x \leq t)$ where $T_x$ is the random variable that denotes the number of years longer a person aged $x$ will live. Of course, $F_x$ is a probability distribution function with density function $f$ and if we assume that $F$ is smooth and $T_x$ has a finite expected value we can do the following: $E(T_x) = \int^{\infty}_0 t f_x(t) dt$ and, in principle this integral can be done by parts….but…if we use $u = t, dv = f_x(t), du = dt, v = F_x$ we have:

\

$t(F_x(t))|^{\infty}_0 -\int^{\infty}_0 F_x(t) dt$ which is a big problem on many levels. For one, $lim_{t \rightarrow \infty}F_x(t) = 1$ and so the new integral does not converge..and the first term doesn’t either.

But if, for $v = -(1-F_x(t))$ we note that $(1-F_x(t)) = S_x(t)$ is the survival function whose limit does go to zero, and there is usually the assumption that $tS_x(t) \rightarrow 0$ as $t \rightarrow \infty$

So we now have: $-(S_x(t) t)|^{\infty}_0 + \int^{\infty}_0 S_x(t) dt = \int^{\infty}_0 S_x(t) dt = E(T_x)$ which is one of the more important formulas.

December 28, 2016

Commentary: our changing landscape and challenges

Filed under: calculus, editorial — collegemathteaching @ 10:34 pm

Yes, I haven’t written anything of substance in a while; I hope to remedy that in upcoming weeks. I am teaching differential equations this next semester and that is usually good for a multitude of examples.

Our university is undergoing changes; this includes admitting students who are nominally STEM majors but who are not ready for even college algebra.

Our provost wants us to reduce college algebra class sizes…even though we are down faculty lines and we cannot find enough bodies to cover courses. Our wonderful administrators didn’t believe us when we explained that it is difficult to find “masters and above” part time faculty for mathematics courses.

And so: with the same size freshmen class, we have a wider variation of student abilities: those who are ready for calculus III, and those who cannot even add simple fractions (yes, one of these was admitted as a computer science major!). Upshot: we need more people to teach freshmen courses, and we are down faculty lines!

Then there is the pressure from the bean-counters in our business office. They note that many students are avoiding our calculus courses and taking them at community colleges. So, obviously, we are horrible teachers!

Here is what the administrators will NOT face up to: students frequently say that passing those courses at a junior college is much easier; they don’t have to study nearly as much. Yes, engineering tells us that students with JC calculus don’t do any worse than those who take it from the mathematics department.

What I think is going on: at universities like ours (I am NOT talking about MIT or Stanford!), the mathematics required in undergraduate engineering courses has gone down; we are teaching more mathematics “than is necessary” for the engineering curriculum, at least the one here.

So some students (not all) see the extra studying required to learn “more than they need” as wasted effort and they resent it.

The way we get these students back: lower the mathematical demands in our calculus courses, or at least lower the demands on studying the more abstract stuff (“abstract”, by calculus standards).

Anyhow, that is where we are. We don’t have the resources to offer both a “mathematical calculus” course and one that teaches “just what you need to know”.

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.

June 15, 2016

Elementary Math in the news: elections

Filed under: calculus, elementary mathematics, news — Tags: — collegemathteaching @ 9:11 pm

Ok, mostly I am trying to avoid writing up the painful details of a proposed mathematics paper.
But I do follow elections relatively closely. In the California Democratic primary, CNN called the election for Hillary Clinton late on June 7; at the time she lead Bernie Sanders 1,940,588-1,502,043, which is a margin of 438,537 votes. Percentage wise, the lead was 55.8-43.2, or 12.6 percentage points.

But due to mail in balloting and provisional ballot counting, there were still many votes to count. As of this morning, the totals were:

2,360,266-1,887,178 for a numerical lead of 473,088 votes. Percentage wise, the lead was 55.1-44.0, or 11.1 percentage points.

So, the lead grew numerically, but shrunk percentage wise.

“Big deal”, you say? Well, from reading social media, it is not obvious (to some) how a lead can grow numerically but shrink as a percentage.

Conceptually, it is pretty easy to explain: suppose one has an election involving 1100 voters who MUST choose between candidates. Say the first 100 votes that are counted happened to come from a strongly pro-Hillary group, and the tally after 100 was 90 Hillary, 10 Bernie. Then suppose the next 1000 was closer, say 550 for Hillary and 450 for Bernie. Then the lead grew by 100 votes (80 to 180) but the percentage lead shrunk from 80 percentage points to a 16.36 percentage point lead (58.18 to 41.82 percent). And it is easy to see that if the rest of the vote was really 55 percent Hillary, her percent of the vote would asymptotically shrink to close to 55 percent as the number of votes counted went up.

So, how might one have students model it? Let $H(t), B(t)$ be increasing functions of $t$ which represent the number of votes for Hillary and Bernie as a function of time. Assume no mistakes, hence $H(t), B(t)$ can be assumed to be increasing functions. So we want a case there $D(t) = H(t)-B(t)$ is an increasing function but $P(t) = \frac{H(t)}{H(t)+ B(t)}$ decreases with time.

Without calculus: rewrite $P(t) = \frac{1}{1+\frac{B(t)}{H(t)}}$ and note that $P(t)$ decreases as $\frac{B(t)}{H(t)}$ increases; that is, as $B(t)$ outgrows $H(t)$. But $H(t)$ must continue to outgrow $B(t)$. That is, the new ballots must still include more Hillary Bernie ballots, but the ratio of Bernie ballots to Hillary ballots must be going down.

If we use some calculus, we see that $H'(t)$ must exceed $B'(t)$ but to make $P(t)$ decrease, use the quotient rule plus a tiny bit of algebra to conclude that $H'(t)B(t)-B'(t)H(t)$ must be negative, or that $\frac{B'(t)}{B(t)} > \frac{H'(t)}{H(t)}$. That is, the Bernie ballots must be growing at a higher percentage rate than the Hillary ballots are.

None of this is surprising, but it might let the students get a feel of what derivatives are and what proportional change means.

June 7, 2016

Pop-math: getting it wrong but being close enough to give the public a feel for it

Space filling curves: for now, we’ll just work on continuous functions $f: [0,1] \rightarrow [0,1] \times [0,1] \subset R^2$.

A curve is typically defined as a continuous function $f: [0,1] \rightarrow M$ where $M$ is, say, a manifold (a 2’nd countable metric space which has neighborhoods either locally homeomorphic to $R^k$ or $R^{k-1})$. Note: though we often think of smooth or piecewise linear curves, we don’t have to do so. Also, we can allow for self-intersections.

However, if we don’t put restrictions such as these, weird things can happen. It can be shown (and the video suggests a construction, which is correct) that there exists a continuous, ONTO function $f: [0,1] \rightarrow [0,1] \times [0,1]$; such a gadget is called a space filling curve.

It follows from elementary topology that such an $f$ cannot be one to one, because if it were, because the domain is compact, $f$ would have to be a homeomorphism. But the respective spaces are not homeomorphic. For example: the closed interval is disconnected by the removal of any non-end point, whereas the closed square has no such separating point.

Therefore, if $f$ is a space filling curve, the inverse image of a points is actually an infinite number of points; the inverse (as a function) cannot be defined.

And THAT is where this article and video goes off of the rails, though, practically speaking, one can approximate the space filling curve as close as one pleases by an embedded curve (one that IS one to one) and therefore snake the curve through any desired number of points (pixels?).

So, enjoy the video which I got from here (and yes, the text of this post has the aforementioned error)

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