College Math Teaching

October 7, 2021

A “weird” implicit graph

Filed under: calculus, implicit differentiation, pedagogy — oldgote @ 12:46 am

I was preparing some implicit differentiation exercises and decided to give this one:

If sin^2(y) + cos^2(x) =1 find {dy \over dx} That is fairly straight forward, no? But there is a bit more here than meets the eye, as I quickly found out. I graphed this on Desmos and:

What in the world? Then I pondered for a minute or two and then it hit me:

sin^2(y) = 1-cos^2(x) \rightarrow sin^2(y) = sin^2(x) \rightarrow \pm(y \pm 2k \pi ) = \pm (x +2 k \pi) which leads to families of lines with either slope 1 or slope negative 1 and y intercepts multiples of \pi

Now, just blindly doing the problem we get 2sin(x)cos(x) = 2 {dy \over dx} cos(y)sin(y) which leads to: {sin(x)cos(x) \over sin(y)cos(y)} = {dy \over dx} = \pm {\sqrt{1-cos^2(y)} \sqrt{1-sin^2(x)} \over \sqrt{1-cos^2(y)} \sqrt{1-sin^2(x)}}  = \pm 1 by both the original equation and the circle identity.

September 13, 2021

Integrals of functions with nice inverses

This idea started as a bit of a joke:

Of course, for readers of this blog: easy-peasy. u =\sqrt{tan(x)} \rightarrow u^2 =tan(x) \rightarrow x = arctan(u^2), dx = {2udu \over 1+u^4} so the integral is transformed into \int {2u^2 \over 1+u^4} du and so we’ve entered the realm of rational functions. Ok, ok, there is some work to do.

But for now, notice what is really doing on: we have a function under the radical that has an inverse function (IF we are careful about domains) and said inverse function has a derivative which is a rational function

More shortly: let f(x) be such that {d \over dx} f^{-1}(x) = q(x) then:

\int (f(x))^{1 \over n} dx gets transformed: u^n = f(x) \rightarrow x =f^{-1}(u^n) and then dx = nu^{n-1}q(u^n) and the integral becomes \int n u^n q(u^n) du which is a rational function integral.

Yes, yes, we need to mind domains.

August 23, 2021

Vaccine efficacy wrt Hospitalization

I made a short video; no, I did NOT have “risk factor”/”age group” breakdown, but the overall point is that vaccines, while outstanding, are NOT a suit of perfect armor.

Upshot: I used this local data:

The vaccination rate of the area is slightly under 50 percent; about 80 percent for the 65 and up group. But this data doesn’t break it down among age groups so..again, this is “back of the envelope”:

{100-23 \over 100} = .77 or about 77 percent efficacy with respect to hospitalization, and {32-2 \over 30} =.9375 or 93.75 percent with respect to ending up in the ICU.

Again, the efficacy is probably better than that because of the lack of risk factor correction.

Note: the p-value for the statistical test of H_0 vaccines have no effect on hospitalization” vs. “effect” is 6 \times 10^{-13}

The video:

June 26, 2021

So, you want our tenure-track academic math job…

Filed under: academia, editorial, mathematician, mathematics education — Tags: , — collegemathteaching @ 8:39 pm

Background: we are a “primarily undergraduate” non-R1 institution. We do not offer math master’s degrees but the engineering college does.

Me: old full professor who has either served on or chaired several search committees.

I’ll break this post down into the two types of jobs we are likely to offer:
Tenure Track lecturer

Tenure Track Assistant Professor.

Lecturer

No research requirement; this job consists of teaching 12 hour loads of lower division mathematics classes, mostly “business calculus and below”; college algebra and precalculus will be your staples. There will be some service work too.

What we are looking for:

Evidence that you have taught lower division courses (college algebra, precalculus, maybe “baby stats”) successfully. Yes, it is great that you were the only postdoc asked to teach a course on differentiable manifolds or commutative ring theory but that is not relevant to this job.

So hopefully you have had taught these courses in the past (several times) and your teaching references talk about how well you did in said courses; e. g. students did well in said courses, went on to the next course prepared, course was as well received as such a course can be, etc. If you won a teaching award of some kind (or nominated for one), that is good to note. And, in this day and age..how did the online stuff go?

Teaching statement: ok, I am speaking for myself, but what I look for is: did you evaluate your own teaching? What did you try? What problems did you notice? Where could you have done better, or what could you try next time? Did you discuss your teaching with someone else? All of those things stand out to me. And yes, that means recognizing that what you tried didn’t work this time…and that you have a plan to revise it..or DID revise it. This applies to the online stuff too.

Diversity Statement Yes, that is a relatively new requirement for us. What I look for: how do you adjust to having some cultural variation in your classroom? Here are examples of what I am talking about:

We usually get students from the suburbs who are used to a “car culture.” So, I often use the car speedometer as something that gives you the derivative of the car’s position. But I ended up with a student from an urban culture and she explained to me that she and her friends took public transportation everywhere…I had to explain what a speedometer was. It was NOT walking around knowledge.

Or: there was a time when I uploaded *.doc files to our learning management system. It turns out that not all students have Microsoft word; taking a few seconds to make them *.pdf files made it a LOT easier for them.

Other things: not everyone gets every sports analogy, gambling analogy (cards, dice, etc.) so be patient when explaining the background for such examples.

Also: a discussion on how one adjusts for the “gaps” in preparation that students have is a plus; a student can place into a course but have missing topics here and there. And the rigor of the high school courses may well vary from student to student; some might expect to be given a “make up” exam if they do poorly on an exam; another might have been used to be given credit for totally incorrect work (I’ve seen both).

Also: if you’ve tutored or volunteered to help a diverse group of students, be sure to mention that (e. g. maternity homes, sports teams, urban league, or just the tutoring center, etc.)

Transcript: yes, we require it, but what we are looking for is breath for the lecturer’s job: the typical is to have three of the following covered: “algebra, analysis, topology, probability, statistics, applied math”

Cover letter: Something that shows that you know the type of job we are offering is very helpful; if you state that you “want to direct undergraduate research”, well, our lecturer job will be a huge letdown.

Assistant Professor

This job will involve 9-12 hours teaching; 10-11 is typical and we do have a modest research requirement. 2-3 papers in solid journals will be sufficient for tenure; you might not want to have your heart set on an Annals of Math publication. If you do get one, you won’t be with us for long anyway. There is also advising and service work.

What we are looking for: teaching: we want some evidence that you can teach the courses typically taught by our department. This means some experience in calculus/business calculus for our math track, and statistics for our statistics track. For this job, some evidence for upper division is a plus, but not required nor even expected; is is an extra “nice to have.”

But it is all but essential that your teaching references talks about your performance in teaching lower division classes (calculus or below); if all you have is “the functional analysis students loved him/her”, that is not helpful. Being observed while teaching a lower division course is all but essential.

Teaching and Diversity statement : same as for the lecturer job. An extra: did you have any involvement with the math club?

Research: the thing we are looking for is: will you “die on the vine” or not? Having a plan: “I intend to move from my dissertation in this direction” is a plus, as is having others to collaborate with (though collaboration isn’t necessary). Also, a statement from your advisor that you can work INDEPENDENTLY ..that is, you can find realistic problems to work on and do NOT need hand holding, is a major plus. You are likely to be somewhat isolated here. And of course, loving mathematics is essential with us. Not all candidates do..if you see your dissertation as a task you had to do to get the credential then our job isn’t for you.

Another plus: having side projects that an undergraduate can work on is a plus. We do have some undergraduate research but that won’t be the bulk of the job.

Transcript: same as the lecturer job.

May 25, 2021

Power series: review for inexperienced calculus teachers

Filed under: infinite series, Power Series, series, Taylor Series — oldgote @ 5:59 pm

This is “part 2” of a previous post about series. Once again, this is not designed for students seeing the material for the first time.

I’ll deal with general power series first, then later discuss Taylor Series (one method of obtaining a power series).

General Power Series Unless otherwise stated, we will be dealing with series such as \sum a_n x^n though, of course, a series can expanded about any point in the real line (e. g. \sum a_n (x-c)^n ) Note: unless it is needed, I’ll be suppressing the index in the summation sign.

So, IF we have a function f(x) = \sum a_n x^n = a_0 + a_1 x + a_2 x^2 + ....a_k x^k + a_{k+1} x^{k+1} ..

what do we mean by this and what can we say about it?

The first thing to consider, of course, is convergence. One main fact is that the open interval of convergence of a power series centered at 0 is either:
1. Non existent …series converges at x = 0 only (e. g. \sum_{k=0} (k!)x^k ; try the ratio test) or

2. Converges absolutely on (-r, r) for some r >0 and diverges for all |x| > r (like a geometric series) (anything possible for |x| = r ) or

3. Converges absolutely on the whole real line.

Many texts state this result but do not prove it. I can understand as this result is really an artifact of calculus of a complex variable. But it won’t hurt to sketch out a proof at least for the “converge” case for (-r, r) so I’ll do that.

So, let’s assume that \sum a_n c^n converges (either absolute OR conditional ) So, by the divergence test, we know that the sequence of terms a_n c^n \rightarrow 0 . So we can find some index M such that n > M \rightarrow |a_n c^n| < 1

So write: |a_n x^n| = |a_n||c^n||| {x^n \over c^n} | < | {x^n \over c^n} | and \sum  | {x^n \over c^n} |  converges absolutely for |x| < |c| . The divergence follows from a simple “reversal of the inequalities (that is, if the series diverged at x = c ).

Though the relation between real and complex variables might not be apparent, it CAN be useful. Here is an example: suppose one wants to find the open interval of convergence of, say, the series representation for {1 \over 3+2x^2 } ? Of course, the function itself is continuous on the whole real line. But to find the open interval of convergence, look for the complex root of 3+2 x^2 that is closest to zero. That would be x = \sqrt{{3 \over 2}} i so the interval is |x| <  \sqrt{{3 \over 2}}

So, what about a function defined as a power series?

For one, a power series expansion of a function at a specific point (say, x =0 ) is unique:

Say a_0 + a_1 x + a_2 x^2 + a_3 x^3.....  = b_0 + b_1 x + b_2 x^2 +  b_3 x^3 .... on the interval of convergence, then substituting x=0 yields a_0 = b_0 . So subtract from both sides and get: a_1 x + a_2 x^2 + a_3 x^3..  =  b_1 x + b_2 x^2 +  b_3 x^3 .... Now assuming one can “factor out an x” from both sides (and you can) we get a_1 = b_1 , etc.

Yes, this should have property been phrased as a limit and I’ve yet to show that f(x) = \sum a_n x^n is continuous on its open interval of absolute convergence.

Now if you say “doesn’t that follow from uniform convergence which follows from the Weierstrass M test” you’d be correct…and you are probably wasting your time reading this.

Now if you remember hearing those words but could use a refresher, here goes:

We want |f(x) -f(t)| < \epsilon for x, t sufficiently close AND both within the open interval of absolute convergence, say (-r, r) Choose s < r where |s| > max(|x|, |t|) and M such that {\epsilon \over 4} > \sum_{k=M} |a_k s^k| and \delta >0 where |x-t| < \delta \rightarrow {\epsilon \over 2 }  >| \sum_{k=0} ^{M-1} a_k x^k  -  \sum_{k=0} ^{M-1} a_k t^k | (these are just polynomials).

The rest follows: |f(x) -f(t)| = | \sum_{k=0} ^{M-1} a_k x^k  -  \sum_{k=0} ^{M-1} a_k t^k  + \sum_{k=M} |a_k x^k| - \sum_{k=M} |a_k t^k| | \leq   | \sum_{k=0} ^{M-1} a_k x^k  -  \sum_{k=0} ^{M-1} a_k t^k | + 2 |  \sum_{k=M} |a_k s^k| | \leq \epsilon

Calculus On the open interval of absolute convergence, a power series can be integrated term by term and differentiated term by term. Neither result, IMHO, is “obvious.” In fact, sans extra criteria, for series of functions, in general, it is false. Quickly: if you remember Fourier Series, think about the Fourier series for a rectangular pulse wave; note that the derivative is zero everywhere except for the jump points (where it is undefined), then differentiate the constituent functions and get a colossal mess.

Differentiation: most texts avoid the direct proof (with good reason; it is a mess) but it can follow from analysis results IF one first shows that the absolute (and uniform) convergence of \sum a_n x^n implies the absolute (and uniform) convergence of \sum n a_n x^{n-1}

So, let’s start here: if \sum a_n x^n is absolutely convergent on (-r, r) then so is \sum a_n nx^{n-1} .

Here is why: because we are on the open interval of absolute convergence, WLOG, assume x > 0 and find s > 0 where \sum a_n (x+s)^n is also absolutely convergent. Now, note that (x+s)^n = x^n + s nx^{n-1}  + ...s^n > snx^{n-1} so the series s \sum a_n nx^{n-1} converges by direct comparison on (-r, r) which establishes what we wanted to show.

Of course, this doesn’t prove that the expected series IS the derivative; we have a bit more work to do.

Again, working on the open interval of absolute convergence, let’s look at:

lim_{x \rightarrow t} {f(x) -f(t) \over x-t } =  lim_{x \rightarrow t}  {1 \over x-t} (\sum a_n x^n -\sum a_n t^n)

Now, we use the fact that \sum a_n n x^{n-1} is absolutely convergent on (-r, r) and given any \epsilon > 0 we can find M > 0 so that n > m \rightarrow \sum_{n} a_k k s^{k-1} for s between x and t

So let’s do that: pick \epsilon >0 and then note:

lim_{x \rightarrow t}  {1 \over x-t} (\sum a_n x^n -\sum a_n t^n)  =

lim_{x \rightarrow t}  {1 \over x-t} (\sum_{k=0}^{n-1}a_k( x^k-t^k) ) +\sum_{k=n} a_k (x^k-t^n) ) =

lim_{x \rightarrow t}  \sum_{k=0}^{n-1} a_k {x^k-t^k \over x-t} + \sum_{k=n} a_k {x^k-t^n \over x-t}

=  \sum_{k=1}^{n-1} a_k k x^{k-1}  + lim_{x \rightarrow t}    \sum_{k=n} a_k {x^k-t^n \over x-t}

Now apply the Mean Value Theorem to each term in the second term:

=  \sum_{k=1}^{n-1} a_k k x^{k-1}  +  lim_{x \rightarrow t}  \sum_{k=n} a_k k(s_k)^{k-1} where each s_k is between x and t . By the choice of n the second term is less than \epsilon , which is arbitrary, and the first term is the first n terms of the “expected” derivative series.

So, THAT is “term by term” differentiation…and notice that we’ve used our hypothesis …almost all of it.

Term by term integration

Theoretically, integration combines easier with infinite summation than differentiation does. But given we’ve done differentiation, we can then do anti-differentiation by showing that \sum a_k {x^{k+1} \over k+1} converges and then differentiating.

But let’s do this independently; it is good for us. And we’ll focus on the definite integral.

\int^b_a \sum a_k x^k dx = (of course, [a,b] \subset (-r, r) )

Once again, choose n so that \sum_{k=n} a_k x^k < \epsilon

Then \int^b_a  \sum a_k x^k dx  =   \int^b_a \sum_{k=0}^{n-1} a_kx^k dx + \int^b_a \sum_{k=n} a_k x^k dx and this is less than (or equal to)

\sum_{k=0}^{n-1} \int^b_a a_kx^k dx + \epsilon (b-a)

But \epsilon is arbitrary and so the result follows, for definite integrals. It is an easy exercise in the Fundamental Theorem of Calculus to extract term by term anti-differentiation.

Taylor Series

Ok, now that we can say stuff about a function presented as a power series, what about finding a power series representation for a function, PROVIDED there is one? Note: we’ll need the function to have an infinite number of derivatives on an open interval about the point of expansion. We’ll also need another condition, which we will explain as we go along.

We will work with expanding about x = 0 . Let f(x) be the function of interest and assume all of the relevant derivatives exist.

Start with \int^x_0 f'(t) dt  = f(x) -f(0) \rightarrow f(x) =f(0) + \int ^x_0 f'(t) dt which can be thought of as our “degree 0” expansion plus remainder term.

But now, let’s use integration by parts on the integral with u = f'(t), dv = dt, du = f''(t) and v = (t-x)  (it is a clever choice for v ; it just has to work.

So now we have: f(x) = f(0) + |^x_0 f'(t)(t-x) -  \int^x_0f''(t) (t-x) dt = f(0) + xf'(0) -    \int^x_0f''(t) (t-x) dt

It looks like we might run into sign trouble on the next iteration, but we won’t as we will see: do integration by parts again:

u =f''(t), dv = (t-x), du = f'''(t), v = {1 \over 2} (t-x)^2 and so we have:

f(x) =f(0) + xf'(0) -( |^x_0 f''(t) {1 \over 2}(t-x)^2 -{1 \over 2} \int^x_0 f'''(t)(t-x)^2 dt

This turns out to be f(0) +xf'(0) +{1 \over 2} f''(0)x^2 + \int^x_0 f'''(t) (t-x)^2 dt .

An induction argument yields

f(x) = \sum^n_{k=0}  f^{(k)}{1 \over k!}x^k +{1 \over n} \int^x_0 f^{(n+1)}(t) (t-x)^n dt

For the series to exist (and be valid over an open interval) all of the derivatives have to exist and;

lim_{n \rightarrow \infty}  {1 \over n!} \int^x_0 f^{(n+1)}(t) (t-x)^n dt  = 0 .

Note: to get the Lagrange remainder formula that you see in some texts, let M = max \{|f^{(n+1)} (t) \} for t \in [0,x] and then |  {1 \over n!} \int^x_0 f^{(n+1)}(t) (t-x)^n dt | \leq {1 \over n!} M \int^x_0 (t-x)^n dt| = {M \over (n+1)!}| |^x_0(t-x)^{(n+1)} | =M{x^{(n+1)} \over (n+1)!}

It is a bit trickier to get the equality f^{(n+1)}(\eta) {x^{n+1} \over (n+1)!} as the error formula; it is a Mean Value Theorem for integrals calculation.

About the remainder term going to zero: this is necessary. Consider the classic counterexample:

f(x) = \begin{cases} e^{-1 \over x^2} & \text{ for} x \neq 0  \\ 0 & \text{ otherwise} \end{cases}

It is an exercise in the limit definition and L’Hopital’s Rule to show that f^{(k)} (0) = 0 for all k \in \{0, 1, 2, ... \} and so a Taylor expansion at zero is just the zero function, and this is valid at 0 only.

As you can see, the function appears to flatten out near x =0 $ but it really is NOT constant.

Note: of course, using the Taylor method isn’t always the best way. For example, if we were to try to get the Taylor expansion of {1 \over 1+x^2} at x = 0 it is easier to use the geometric series for {1 \over 1-u} and substitute u= -x^2; the uniqueness of the power series expansion allows for that.

May 21, 2021

Introduction to infinite series for inexperienced calculus teachers

Filed under: calculus, mathematics education, pedagogy, Power Series, sequences, series — oldgote @ 1:26 pm

Let me start by saying that this is NOT: this is not an introduction for calculus students (too steep) nor is this intended for experienced calculus teachers. Nor is this a “you should teach it THIS way” or “introduce the concepts in THIS order or emphasize THESE topics”; that is for the individual teacher to decide.

Rather, this is a quick overview to help the new teacher (or for the teacher who has not taught it in a long time) decide for themselves how to go about it.

And yes, I’ll be giving a lot of opinions; disagree if you like.

What series will be used for.

Of course, infinite series have applications in probability theory (discrete density functions, expectation and higher moment values of discrete random variables), financial mathematics (perpetuities), etc. and these are great reasons to learn about them. But in calculus, these tend to be background material for power series.

Power series: \sum^{\infty}_{k=0} a_k (x-c)^k , the most important thing is to determine the open interval of absolute convergence; that is, the intervals on which \sum^{\infty}_{k=0} |a_k (x-c)^k | converges.

We teach that these intervals are *always* symmetric about x = c (that is, at x = c only, on some open interval (c-\delta, c+ \delta) or the whole real line. Side note: this is an interesting place to point out the influence that the calculus of complex variables has on real variable calculus! These open intervals are the most important aspect as one can prove that one can differentiate and integrate said series “term by term” on the open interval of absolute convergence; sometimes one can extend the results to the boundary of the interval.

Therefore, if time is limited, I tend to focus on material more relevant for series that are absolutely convergent though there are some interesting (and fun) things one can do for a series which is conditionally convergent (convergent, but not absolutely convergent; e. g. \sum^{\infty}_{k=1} (-1)^{k+1} {1 \over k} .

Important principles: I think it is a good idea to first deal with geometric series and then series with positive terms…make that “non-negative” terms.

Geometric series: \sum ^{\infty}_{k =0} x^k ; here we see that for x \neq 1 , \sum ^{n}_{k =0} x^k= {1-x^{n+1} \over 1-x } and is equal to n+1 for n = 1 ; to show this do the old “shifted sum” addition: S = 1 + x + x^2 + ...x^n , xS = x+x^2 + ...+x^{n+1} then subtract: S-xS = (1-x)S = 1-x^{n+1} as most of the terms cancel with the subtraction.

Now to show the geometric series converges, (convergence being the standard kind: \sum^n_{k = 0} c_k = S_n the “n’th partial sum, then the series \sum^{\infty}_{k = 0} c_k  converges if an only if the sequence of partial sums S_n converges; yes there are other types of convergence)

Now that we’ve established that for the geometric series, S_n =  {1-x^{n+1} \over 1-x }  and we get convergence if |x^{n+1}| goes to zero, which happens only if |x| < 1 .

Why geometric series: two of the most common series tests (root and ratio tests) involve a comparison to a geometric series. Also, the geometric series concept is used both in the theory of improper integrals and in measure theory (e. g., showing that the rational numbers have measure zero).

Series of non-negative terms. For now, we’ll assume that \sum a_k has all a_k \geq 0 (suppressing the indices).

Main principle: though most texts talk about the various tests, I believe that most of the tests involved really involve three key principles, two of which the geometric series and the following result from sequences of positive numbers:

Key sequence result: every monotone bounded sequence of positive numbers converges to its least upper bound.

True: many calculus texts don’t do that much with the least upper bound concept but I feel it is intuitive enough to at least mention. If the least upper bound is, say, b , then if a_n is the sequence in question, there has to be some N  > 0 such that a_n > b-\delta for any small, positive \delta . Then because a_n is monotone, b> a_{m} > b-\delta for all m > n

The third key principle is “common sense” : if \sum c_k converges (standard convergence) then c_k \rightarrow 0 as a sequence. This is pretty clear if the c_k are non-negative; the idea is that the sequence of partial sums S_n cannot converge to a limit unless |S_n -S_{n+1}| becomes arbitrarily small. Of course, this is true even if the terms are not all positive.

Secondary results I think that the next results are “second order” results: the main results depend on these, and these depend on the key 3 that we just discussed.

The first of these secondary results is the direct comparison test for series of non-negative terms:

Direct comparison test

If 0< c_n \leq b_n  and \sum b_n converges, then so does \sum c_n . If \sum c_n diverges, then so does \sum b_n .

The proof is basically the “bounded monotone sequence” principle applied to the partial sums. I like to call it “if you are taller than an NBA center then you are tall” principle.

Evidently, some see this result as a “just get to something else” result, but it is extremely useful; one can apply this to show that the exponential of a square matrix is defined; it is the principle behind the Weierstrass M-test, etc. Do not underestimate this test!

Absolute convergence: this is the most important kind of convergence for power series as this is the type of convergence we will have on an open interval. A series is absolutely convergent if \sum |c_k| converges. Now, of course, absolute convergence implies convergence:

Note 0 < |c_k| -c_k \leq 2|c_k| and if \sum |c_k| converges, then \sum |c_k|-c_k converges by direct comparison. Now note c_k = |c_k|-(|c_k| -c_k) \rightarrow \sum c_k is the difference of two convergent series: \sum |c_k| -\sum (|c_k|-c_k ) and therefore converges.

Integral test This is an important test for convergence at a point. This test assumes that f is a non-negative, non-decreasing function on some [1, \infty) (that is, a >b \rightarrow f(a) \geq f(b) ) Then \sum f(n) converges if and only if \int_1^{\infty} f(x)dx converges as an improper integral.

Proof: \sum_{n=2} f(n) is just a right endpoint Riemann sum for \int_1^{\infty} f(x)dx and therefore the sequence of partial sums is an increasing, bounded sequence. Now if the sum converges, note that \sum_{n=1} f(n) is the right endpoint estimate for \int_1^{\infty} f(x)dx so the integral can be defined as a limit of a bounded, increasing sequence so the integral converges.

Yes, these are crude whiteboards but they get the job done.

Note: we need the hypothesis that f is decreasing (or non-decreasing). Example: the function f(x) = \begin{cases}  x , & \text{ if } x \notin \{1, 2, 3,...\} \\ 0, & \text{ otherwise} \end{cases} certainly has \sum f(n) converging but \int^{\infty}_{1} f(x) dx diverging.

Going the other way, defining f(x) = \begin{cases}  2^n , & \text{ if }  x \in [n, n+2^{-2n}] \\0, & \text{ otherwise} \end{cases} gives an unbounded function with unbounded sum \sum_{n=1} 2^n but the integral converges to the sum \sum_{n=1} 2^{-n} =1 . The “boxes” get taller and skinnier.

Note: the above shows the integral and sum starting at 0; same principle though.

Now wait a minute: we haven’t really gone over how students will do most of their homework and exam problems. We’ve covered none of these: p-test, limit comparison test, ratio test, root test. Ok, logically, we have but not practically.

Let’s remedy that. First, start with the “point convergence” tests.

p-test. This says that \sum {1 \over k^p} converges if p> 1 and diverges otherwise. Proof: Integral test.

Limit comparison test Given two series of positive terms: \sum b_k and \sum c_k

Suppose lim_{k \rightarrow \infty} {b_k \over c_k} = L

If \sum c_k converges and 0 \leq L < \infty then so does \sum b_k .

If \sum c_k diverges and 0 < L \leq \infty then so does \sum b_k

I’ll show the “converge” part of the proof: choose \epsilon = L then N such that n > N \rightarrow  {b_n \over c_n } < 2L This means \sum_{k=n} b_k \leq \sum_{k=n} c_k and we get convergence by direct comparison. See how useful that test is?

But note what is going on: it really isn’t necessary for lim_{k \rightarrow \infty} {b_k \over c_k}  to exist; for the convergence case it is only necessary that there be some M for which M >  {b_k \over c_k}  ; if one is familiar with the limit superior (“limsup”) that is enough to make the test work.

We will see this again.

Why limit comparison is used: Something like \sum {1 \over 4k^5-2k^2-14} clearly converges, but nailing down the proof with direct comparison can be hard. But a limit comparison with \sum {1 \over k^5} is pretty easy.

Ratio test this test is most commonly used when the series has powers and/or factorials in it. Basically: given \sum c_n consider lim_{k \rightarrow \infty} {c_{k+1} \over c_{k}} = L (if the limit exists..if it doesn’t..stay tuned).

If L < 1 the series converges. If L > 1 the series diverges. If L = 1 the test is inconclusive.

Note: if it turns out that there is exists some N >0 such that for all n > N we have {c_{n+1} \over c_n } < \gamma < 1 then the series converges (we can use the limsup concept here as well)

Why this works: suppose there exists some N >0 such that for all n > N we have {c_{n+1} \over c_n } < \gamma < 1 Then write \sum_{k=n} c_k = c_n + c_{n+1} + c_{n+2} + ....

now factor out a c_n to obtain c_n (1 + {c_{n+1} \over c_n} + {c_{n+2} \over c_n} + {c_{n+3} \over c_{n}} +....)

Now multiply the terms by 1 in a clever way:

c_n (1 + {c_{n+1} \over c_n} + {c_{n+2} \over c_{n+1}}{c_{n+1} \over c_n} + {c_{n+3} \over c_{n+2}}  {c_{n+2} \over c_{n+1}}  {c_{n+1} \over c_{n}}   +....) See where this is going: each ratio is less than \gamma so we have:

\sum_{k=n} c_k \leq c_n \sum_{j=0} (\gamma)^j which is a convergent geometric series.

See: there is geometric series and the direct comparison test, again.

Root Test No, this is NOT the same as the ratio test. In fact, it is a bit “stronger” than the ratio test in that the root test will work for anything the ratio test works for, but there are some series that the root test works for that the ratio test comes up empty.

I’ll state the “lim sup” version of the ratio test: if there exists some N such that, for all n>N we have (c_n)^{1 \over n} < \gamma < 1 then the series converges (exercise: find the “divergence version”).

As before: if the condition is met, \sum_{k=n} c_n \leq \sum_{k=n} \gamma^k so the original series converges by direction comparison.

Now as far as my previous remark about the ratio test: Consider the series:

1 + ({1 \over 3}) + ({2 \over 3})^2 + ({1 \over 3})^3 + ({2 \over 3})^4 +...({1 \over 3})^{2k-1} +({2 \over 3})^{2k} ...

Yes, this series is bounded by the convergent geometric series with r = {2 \over 3} and therefore converges by direct comparison. And the limsup version of the root test works as well.

But the ratio test is a disaster as {({2 \over 3})^{2k}  \over  ({1 \over 3})^{2k-1} } ={2^{2k} \over 3 } which is unbounded..but {({1 \over 3})^{2k+1}  \over  ({2 \over 3})^{2k} }  ={1 \over (2^{2k} 3) } .

What about non-absolute convergence (aka “conditional convergence”)

Series like \sum_{k=1} (-1)^{k+1} {1 \over k} converges but does NOT converge absolutely (p-test). On one hand, such series are a LOT of fun..but the convergence is very slow and unstable and so might say that these series are not as important as the series that converges absolutely. But there is a lot of interesting mathematics to be had here.

So, let’s chat about these a bit.

We say \sum c_k is conditionally convergent if the series converges but \sum |c_k| diverges.

One elementary tool for dealing with these is the alternating series test:

for this, let c_k >0 and for all k, c_{k+1} < c_k .

Then \sum_{k=1} (-1)^{k+1} c_k converges if and only if c_k \rightarrow 0 as a sequence.

That the sequence of terms goes to zero is necessary. That it is sufficient in this alternating case: first note that the terms of the sequence of partial sums are bounded above by c_1 (as the magnitudes get steadily smaller) and below by c_1 - c_2 (same reason. Note also that S_{2k+2} = S_{2k} -c_{2k+1} + c_{2k+2} < S_{2k} so the sequence of partial sums of even index are an increasing bounded sequence and therefore converges to some limit, say, L . But S_{2k+1} = S_{2k} + c_{2k+1} and so by a routine “epsilon-N” argument the odd partial sums converge to L as well.

Of course, there are conditionally convergent series that are NOT alternating. And conditionally convergent series have some interesting properties.

One of the most interesting properties is that such series can be “rearranged” (“derangment” in Knopp’s book) to either converge to any number of choice or to diverge to infinity or to have no limit at all.

Here is an outline of the arguments:

To rearrange a series to converge to L , start with the positive terms (which must diverge as the series is conditionally convergent) and add them up to exceed L ; stop just after L is exceeded. Call that partial sum u_1. Note: this could be 0 terms. Now use the negative terms to go of the left of L and stop the first one past. Call that l_1 Then move to the right, past L again with the positive terms..note that the overshoot is smaller as the terms are smaller. This is u_2 . Then go back again to get l_2 to the left of L . Repeat.

Note that at every stage, every partial sum after the first one past L is between some u_i, l_i and the u_i, l_i bracket L and the distance is shrinking to become arbitrarily small.

To rearrange a series to diverge to infinity: Add the positive terms to exceed 1. Add a negative term. Then add the terms to exceed 2. Add a negative term. Repeat this for each positive integer n .

Have fun with this; you can have the partial sums end up all over the place.

That’s it for now; I might do power series later.

May 10, 2021

Series convergence tests: the “harder to use in calculus 1” tests may well be the most useful.

I talked about the root and ratio test here and how the root test is the stronger of the two tests. What I should point out that the proof of the root test depends on the basic comparison test.

And so..a professor on Twitter asked:

Of course, one proves the limit comparison test by the direct comparison test. But in a calculus course, the limit comparison test might appear to be more readily useful..example:

Show \sum {1 \over k^2-1} converges.

So..what about the direct comparison test?

As someone pointed out: the direct comparison can work very well when you don’t know much about the matrix.

One example can be found when one shows that the matrix exponential e^A where A is a n \times n matrix.

For those unfamiliar: e^A = \sum^{\infty}_{k=0} {A^k \over k!} where the powers make sense as A is square and we merely add the corresponding matrix entries.

What enables convergence is the factorial in the denominators of the individual terms; the i-j’th element of each A^k can get only so large.

But how does one prove convergence?

The usual way is to dive into matrix norms; one that works well is |A| = \sum_{(i,j)} |a_{i,j}| (just sum up the absolute value of the elements (the Taxi cab norm or l_1 norm )

Then one can show |AB| \leq |A||B| and |a_{i,j}| \leq |A| and together this implies the following:

For any index k where a^k_{i,j} is the i-j’th element of A^k we have:

| a^k_{i,j}  | \leq |A^k| \leq |A|^k

It then follows that | [ e^A ]_{i,j} | \leq \sum^{\infty}_{k=0} {|A^k |\over k!} \leq  \sum^{\infty}_{k=0} {|A|^k \over k!} =e^{|A|} . Therefore every series that determines an entry of the matrix e^A is an absolutely convergent series by direct comparison. and is therefore a convergent series.

August 29, 2020

Commentary: life with Webassign

Filed under: editorial, pedagogy — oldgote @ 5:45 pm

Since I am online, I decided to use Webassign for homework.
Well, of course, some students have had trouble with it..in particular the graphing application.
I am not saying that their graphing tool is hard to learn; in fact I might play with it myself. BUT, in this particular class, I want my students to focus on learning the material and NOT have to spend hours learning to get good with their graphing tools. For graphing: Desmos is outstanding.

Normally, I am not sympathetic to student complaints or frustrations, but here, I can see it.

July 14, 2020

An alternative to trig substitution, sort of..

Ok, just for fun: \int \sqrt{1+x^2} dx =

The usual is to use x =tan(t), dx =sec^2(t) dt which transforms this to the dreaded \int sec^3(t) dt integral, which is a double integration by parts.
Is there a way out? I think so, though the price one pays is a trickier conversion back to x.

Let’s try x =sinh(t) \rightarrow dx = cosh(t) dt so upon substituting we obtain \int |cosh(t)|cosh(t) dt and noting that cosh(t) > 0 alaways:

\int cosh^2(t)dt Now this can be integrated by parts: let u=cosh(t) dv = cosh(t) dt \rightarrow du =sinh(t), v = sinh(t)

So \int cosh^2(t)dt = cosh(t)sinh(t) -\int sinh^2(t)dt but this easily reduces to:

\int cosh^2(t)dt = cosh(t)sinh(t) -\int cosh^2(t)-1 dt \rightarrow 2\int cosh^2(t)dt  = cosh(t)sinh(t) -t + C

Division by 2: \int cosh^2(t)dt = \frac{1}{2}(cosh(t)sinh(t)-t)+C

That was easy enough.

But we now have the conversion to x: \frac{1}{2}(cosh(t)sinh(t) \rightarrow \frac{1}{2}x \sqrt{1+x^2}

So far, so good. But what about t \rightarrow   arcsinh(x) ?

Write: sinh(t) = \frac{e^{t}-e^{-t}}{2} =  x \rightarrow e^{t}-e^{-t} =2x \rightarrow e^{t}-2x -e^{-t} =0

Now multiply both sides by e^{t} to get e^{2t}-2xe^t -1 =0 and use the quadratic formula to get e^t = \frac{1}{2}(2x\pm \sqrt{4x^2+4} \rightarrow e^t = x \pm \sqrt{x^2+1}

We need e^t > 0 so e^t = x + \sqrt{x^2+1} \rightarrow t = ln|x + \sqrt{x^2+1}| and that is our integral:

\int \sqrt{1+x^2} dx = \frac{1}{2}x \sqrt{1+x^2} + \frac{1}{2} ln|x + \sqrt{x^2+1}| + C

I guess that this isn’t that much easier after all.

July 12, 2020

Logarithmic differentiation: do we not care about domains anymore?

Filed under: calculus, derivatives, elementary mathematics, pedagogy — collegemathteaching @ 11:29 pm

The introduction is for a student who might not have seen logarithmic differentiation before: (and yes, this technique is extensively used..for example it is used in the “maximum likelihood function” calculation frequently encountered in statistics)

Suppose you are given, say, f(x) =sin(x)e^x(x-2)^3(x+1) and you are told to calculate the derivative?

Calculus texts often offer the technique of logarithmic differentiation: write ln(f(x)) = ln(sin(x)e^x(x-2)^3(x+1)) = ln(sin(x)) + x + 3ln(x-2) + ln(x+1)
Now differentiate both sides: ln((f(x))' = \frac{f'(x)}{f(x)}  = \frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1}

Now multiply both sides by f(x) to obtain

f'(x) = f(x)(\frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1}) =

\

(sin(x)e^x(x-2)^3(x+1)(\frac{cos(x)}{sin(x)} + 1 + \frac{3}{x-2} + {1 \over x+1})

And this is correct…sort of. Why I say sort of: what happens, at say, x = 0 ? The derivative certainly exists there but what about that second factor? Yes, the sin(x) gets cancelled out by the first factor, but AS WRITTEN, there is an oh-so-subtle problem with domains.

You can only substitute x \in \{ 0, \pm k \pi \} only after simplifying ..which one might see as a limit process.

But let’s stop and take a closer look at the whole process: we started with f(x) = g_1(x) g_2(x) ...g_n(x) and then took the log of both sides. Where is the log defined? And when does ln(ab) = ln(a) + ln(b) ? You got it: this only works when a > 0, b > 0 .

So, on the face of it, ln(g_1 (x) g_2(x) ...g_n(x)) = ln(g_1(x) ) + ln(g_2(x) ) + ...ln(g_n(x)) is justified only when each g_i(x) > 0 .

Why can we get away with ignoring all of this, at least in this case?

Well, here is why:

1. If f(x) \neq 0 is a differentiable function then \frac{d}{dx} ln(|f(x)|) = \frac{f'(x)}{f(x)}
Yes, this is covered in the derivation of \int {dx \over x} material but here goes: write

|f(x)| =   \begin{cases}      f(x) ,& \text{if } f(x) > 0 \\      -f(x),              & \text{otherwise}  \end{cases}

Now if f(x) > 0 we get { d \over dx} ln(f(x)) = {f'(x) \over f(x) } as usual. If f(x) < 0 then |f(x)| = =f(x), |f(x)|' = (-f(x))' = -f'(x) and so in either case:

\frac{d}{dx} ln(|f(x)|) = \frac{f'(x)}{f(x)} as required.

THAT is the workaround for calculating {d \over dx } ln(g_1(x)g_2(x)..g_n(x)) where g_1(x)g_2(x)..g_n(x) \neq 0 : just calculate {d \over dx } ln(|g_1(x)g_2(x)..g_n(x)|) . noting that |g_1(x)g_2(x)..g_n(x)| = |g_1(x)| |g_2(x)|...|g_n(x)|

Yay! We are almost done! But, what about the cases where at least some of the factors are zero at, say x= x_0 ?

Here, we have to bite the bullet and admit that we cannot take the log of the product where any of the factors have a zero, at that point. But this is what we can prove:

Given g_1(x) g_2(x)...g_n(x) is a product of differentiable functions and g_1(a) g_2(a)...g_k(a) = 0 k \leq n then
(g_1(a)g_2(a)...g_n(a))' = lim_{x \rightarrow a}  g_1(x)g_2(x)..g_n(x) ({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...{g_n'(x) \over g_n(x})

This works out to what we want by cancellation of factors.

Here is one way to proceed with the proof:

1. Suppose f, g are differentiable and f(a) = g(a) = 0 . Then (fg)'(a) = f'(a)g(a) + f(a)g'(a) = 0 and lim_{x \rightarrow a} f(x)g(x)({f'(x) \over f(x)} + {g'(x) \over g(x)}) = 0
2. Now suppose f, g are differentiable and f(a) =0 ,  g(a) \neq 0 . Then (fg)'(a) = f'(a)g(a) + f(a)g'(a) = f'(a)g(a) and lim_{x \rightarrow a} f(x)g(x)({f'(x) \over f(x)} + {g'(x) \over g(x)}) = f'(a)g(a)
3.Now apply the above to g_1(x) g_2(x)...g_n(x) is a product of differentiable functions and g_1(a) g_2(a)...g_k(a) = 0 k \leq n
If k = n then (g_1(a)g_2(a)...g_n(a))' = lim_{x \rightarrow a}  g_1(x)g_2(x)..g_n(x) ({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...{g_n'(x) \over g_n(x}) =0 by inductive application of 1.

If k < n then let g_1...g_k = f, g_{k+1} ..g_n  =g as in 2. Then by 2, we have (fg)' =  f'(a)g(a) Now this quantity is zero unless k = 1 and f'(a) neq 0 . But in this case note that lim_{x \rightarrow a} g_1(x)g_2(x)...g_n(x)({g_1'(x) \over g_1(x)} + {g_2'(x) \over g_2(x)} + ...+ {g_n'(x) \over g_n(x)})  = lim_{x \rightarrow a} g_2(x)...g_n(x)(g_1'(x)) =g(a)g_1(a)

So there it is. Yes, it works ..with appropriate precautions.

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