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What are some interesting mathematical things you have learned while grading student work (or marking, if you prefer)?

It is final exams time here, so if anyone can help cast a more positive light on the grading experience, it would be most welcome.

Answers can be things that students wrote, or inspired by something a student wrote, or just something we learned during the grading process in some way. For example, clever proofs that students came up with; nice counterexamples or insights; interesting new questions inspired while grading; even just something you looked up to find out if a student's work was valid. However, for an answer to be interesting, it should be something beyond just a different way to solve a problem.

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I once asked students to find the derivative of $x^x$ (with respect to $x$). One student figured that if the exponent were a constant then the answer would be $xx^{x-1}$ which is to say $x^x$, while if the base were constant the answer would be $x^x\log x$, so she added the two together to get $x^x+x^x\log x$. I was just about to mark the answer as wrong, when I realized that she had arrived at the correct answer – and, later, realized that it wasn't a coincidence, her unorthodox method actually works in a more general setting.

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    $\begingroup$ So basically she is using df(x,x)/dx=partial_1 f+ partial_2 f . nice $\endgroup$
    – lalala
    Commented May 8, 2020 at 16:14
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    $\begingroup$ Right. More generally if $f$ is given by an expression in which $x$ appears several times, and $F(x_1,\dotsc,x_n)$ is defined by replacing each appearance of $x$ in that expression with successive new variables, then $df/dx = \left( \sum_{i=1}^n \partial F/\partial x_i \right)|_{x_1=x_2=\dotsb=x_n=x}$. Once you see it, it pretty much falls out from the "Chain Rule for Paths" in multivariable calculus. But that doesn't take away that it's a really cool idea! $\endgroup$
    – Zach Teitler
    Commented May 8, 2020 at 17:39
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    $\begingroup$ But how did you mark it? Was there any way to tell if she just tried a guess and was lucky, or did she actually have an insight into multivariable calculus? And in the (maybe more likely) first case, did you still give points for the right answer? Full points even? $\endgroup$
    – Torsten Schoeneberg
    Commented May 9, 2020 at 21:07
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    $\begingroup$ “The last act is the greatest treason. To do the right deed for the wrong reason.” T.S. Eliot, Murder in the Cathedral $\endgroup$
    – Nicola Ciccoli
    Commented May 12, 2020 at 15:22
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    $\begingroup$ @Torsten, this happened about 40 years ago, so you'll forgive me if I can't remember how I marked it. $\endgroup$
    – Gerry Myerson
    Commented Jun 19, 2021 at 10:40
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Possibly not what you're looking for, but: the things I've learned while grading are mostly pedagogical, not new mathematical facts (in fact, teaching at a community college as I do, I'm not sure that's ever happened).

One of the main things that sticks with me is this: The rather incredible kaleidoscope of ways that students can misunderstand or be wrong about a thing. Generally the faculty in my department push the thesis that all-multiple-choice testing is fine (even required) for most courses up to calculus, say. The instinct is that it's "obvious" what the common mistakes might be, and these can be covered in a set of 3 or 4 distractor options.

Now, I'm one of the very few instructors (maybe the only one now) who insists on at least a few open-ended questions on any of my tests to see what student work is actually like (and give feedback on it). In doing so, I've discovered a whole lot more "ways to be wrong" then I'd ever imagine, or that could be covered in a multiple-choice test. Looked at from another perspective: any multiple-choice test is an enormous safety net, because it actually rules out the great majority of student responses.

One example: On a college algebra test a few years back, I asked: "Write the equation of the line, with the given properties, in slope-intercept form: Through (-2, 6) and (2, -7)". Out of 40 test submissions, I found there were 26 different unique responses. (!) More specifically: 14 students got the right answer, 2 students duplicated a certain wrong answer, and 24 students each had a unique wrong answer, duplicated by no one else. (Which brings to mind Tolstoy's adage, "All happy families are alike; each unhappy family is unhappy in its own way.")

Second example: For the first time this semester I'm giving programming tests on an actual computer in our lab. (For 20 years I gave programming tests on paper; transitioned to online tests for the COVID pandemic; and found enough advantages that I wanted to keep that as we switched back to in-person teaching.) Coincidentally, the lab has screen-monitoring software that's always on, so without planning it I found myself watching students write code in real-time on a test for the first time ever. I was amazed at how many of my second-semester CS majors couldn't write even basic structures; several were taking many shotgun attempts at simply declaring an array, or couldn't even write a simple for loop, for instance (e.g., mixing up bits of syntax between while, for, and do-while loops, taking as many as 10 minutes of iterations fired at the compiler trying to get it right). One student apparently actually memorized the entire practice test solution, typed that in first (with great difficulty and many compiler errors), and only once that was running tried to modify it to match the actual test question.

Pretty fascinating stuff which I'd have never known if I didn't get to see the students' actual work process.

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    $\begingroup$ The programming example is a horror story. $\endgroup$
    – Steven Gubkin
    Commented May 13, 2022 at 10:37
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    $\begingroup$ @Steven Gubkin: The programming example has me wondering how they would have handled the only programming course I've taken (PL/1), which was in Fall 1977. After working out how you think to do it (an hour for the first couple of weekly assignments, and 10+ hours for the last 5 or 6 of the semester's weekly assignments) and writing the code lines on scrap paper (after the first assignment you accumulate a lot of computer scrap paper), you walk to the computer lab (if you live on campus or within 2 kilometers), bringing your various attempts (continued) $\endgroup$
    – Dave L Renfro
    Commented May 13, 2022 at 14:20
  • $\begingroup$ (because what you think is going to work might not work, but maybe you’ll figure out that something else you tried can be used) and of course bringing your text (this was our text) for all the absolutely essential “grammar rules” that, unlike math formulas, you can’t derive or reason out what a missing detail must be. Then you wait for a keypunch station to become unoccupied so that you can type the punch cards, one card for each line of code. There might be a long outside the keypunch room if you go between about 8 AM and 10 PM on a day (continued) $\endgroup$
    – Dave L Renfro
    Commented May 13, 2022 at 14:20
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    $\begingroup$ @DaveLRenfro I have a theory about how students can succeed in writing programs but have great difficulty writing proofs. If a program is written sloppily, the compiler will say "syntax error" and you can't argue with it. If a proof is written sloppily the professor (or TA) will say something more polite and students can and will argue. $\endgroup$
    – Andreas Blass
    Commented May 13, 2022 at 15:18
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    $\begingroup$ @shoover: Thanks, and I thought I'd double-checked, too. Per my records, looks like there was one more student who answered correctly, and one who didn't get past copying the point-slope formula, so didn't generate a tally for a response. Fixed it now. $\endgroup$
    – Daniel R. Collins
    Commented May 13, 2022 at 16:39
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Reading the answer posted by Daniel R Collins reminded me of something else I learned while marking student work. Not exactly something mathematical, more something about constructing math exams.

I had decided it wasn't fair to have the "distractors" in a multiple choice exam be answers that a student could come up with by making a simple error such as a sign error, the kind of error for which I would give partial credit if it weren't a multiple choice exam. So I wrote a test where all the distractors were crazy things that no student could possibly come up with.

To my horror, the scores on this test were awful.

I figured out why (I think). If a student makes a simple error, and arrives at an answer that is one of the choices, then the student marks that choice, and moves on to the next question. So, the student gets that one question wrong, but spends no more time on it, loses no confidence, and may well get some of the later questions right.

But if a student makes a simple error, and arrives at an answer that is not one of the choices supplied, then the student goes back over the work, perhaps starts the problem from scratch, makes the same error, or maybe a new one, still doesn't arrive at an answer among the choices supplied, maybe panics, but in any event spends a lot of fruitless time on that one problem, has less time to spend on the rest of the test, and less confidence about tackling the rest of the test.

So I learned the reason why distractors should be "plausible" answers. Maybe everyone else already knew that, but no one ever told me, and I had to learn it the hard way.

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    $\begingroup$ Great answer. Could you also include an example? I think if the distractors are completely off (for example not even in the right category), one could go for the principle of exclusion. $\endgroup$
    – Jasper
    Commented May 15, 2022 at 8:22
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    $\begingroup$ @Jasper, that's what I thought, that any student with a bit of insight could see that only one of the choices could possibly be an answer, and then mark that one without even doing any math. But it seems my students didn't approach tests that way. As for specific examples, this was 40 years and four jobs ago, I'm afraid I don't have memories or records of actual examples. $\endgroup$
    – Gerry Myerson
    Commented May 15, 2022 at 10:02
  • $\begingroup$ Was it like asking for the inverse of a matrix with distractors "42", some integral, and a vector? or was it not "that bad"? $\endgroup$
    – Jasper
    Commented May 15, 2022 at 13:57
  • $\begingroup$ @Jasper, maybe not quite that bad, maybe more like a question where a moment's thought would show the answer had to be an integer between one and ten, and the distractors were $2000000$, $-3$, and $\sqrt2$. But, really, I don't have access to any notes from way back then, so I'm just guessing. $\endgroup$
    – Gerry Myerson
    Commented May 16, 2022 at 2:03
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    $\begingroup$ Just go ahead and add this example to the answer, it doesn't matter if it was exactly this one, it is still helpful $\endgroup$
    – Jasper
    Commented May 16, 2022 at 5:23
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I gave an advanced course on Probability that contained some ergodic theory. In exercises, I outlined the usual proof of the equidistribution of $e^{in\theta}$ on the circle, for $\theta/\pi$ irrational. The proof I knew was generalizing equidistribution from indicators of intervals to arbitrary (say, continuous) functions and then using Fourier transform.

Then one of the students pointed out the following elementary solution. Assume that $I,J$ are half-open intervals on the circle, and $I$ is longer than $J$. Then, you can write $I=I_1\sqcup I_2$, where $I_2$ is a translation of $J$ that follows $I_1$ counterclockwise. Let $n_1$ be the first time $\exp(i\theta n)$ belongs to $I_1$, and $n_2$ is the first time after $n_1$ that it belongs to $J$. Then, $\exp(i(n+n_2)\theta)\in J$ implies $\exp(i(n+n_1)\theta)\in I$, which readily implies $$ \frac{1}{N}\#\{n\leq N:\exp(in\theta)\in J\}=\frac{1}{N}\#\{n_2\leq n\leq N:\exp(in\theta)\in J\}+o(1)\leq \frac{1}{N}\#\{n\leq N:\exp(in\theta)\in I\}+o(1). $$ This means that $\liminf$ of the quantity on the right is greater than $\limsup$ of the quantity on the left. From this and additivity of density the result easily follows.

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  • $\begingroup$ You mean for $\theta$ a fixed number irrational w.r.t. $\pi$, yes? (i.e., such that the ratio $\theta/\pi$ be irrational) $\endgroup$
    – Vandermonde
    Commented May 29, 2020 at 18:48
  • $\begingroup$ @Vandermonde, yes, of course. Thanks! $\endgroup$
    – Kostya_I
    Commented May 29, 2020 at 19:05
  • $\begingroup$ I could not get this proof to work. There are examples where $exp(i\theta n_2)$ is very close to the right endpoint of $J$, and $\exp(i\theta n_1)$ is very close to the left endpoint of $I_1$. Choose some $n$ such that $\exp(i\theta(n+n_2)$ is very close to the left endpoint of $J$. Then $\exp(i\theta(n+n_1))$ will be outside of $I$. $\endgroup$
    – Yongyi Chen
    Commented May 18, 2022 at 6:09
  • $\begingroup$ @YongyiChen, I'm no longer sure what I meant two years ago... how about this: let $x=e^{i\theta n_0}$ be the first point hit on $J$. Let $y\in I$ be such that rotating $x$ to $y$ shifts $J$ strictly inside $I$, with an $\epsilon$ of room on both sides. Let $n_1$ be the first time we hit $(y-\epsilon,y+\epsilon)$. Then, $e^{i\theta(n+n_0)}\in J$ implies $e^{i\theta(n+n_1)}\in I$. $\endgroup$
    – Kostya_I
    Commented May 18, 2022 at 21:51
  • $\begingroup$ @Kostya_I That setup makes more sense. $\endgroup$
    – Yongyi Chen
    Commented May 19, 2022 at 16:08
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An answer I saw a few times while marking a particular question was $$\ln(x+1)=\ln(x)+\ln.$$ I think this explains the 'everything is linear' phenomenon: everything is linear because everything is multiplication.

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  • $\begingroup$ I'm surprised it didn't continue $= \ln x + \ln 1 = \ln x$, though to be sure I don't know what they were trying for. $\endgroup$
    – Vandermonde
    Commented Jun 2, 2020 at 0:53
  • $\begingroup$ @Vandermonde Those students didn't do that. A couple of others did, and more put $\ln(x)+\ln(1)$ without then cancelling. $\endgroup$
    – Jessica B
    Commented Jun 3, 2020 at 6:24
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    $\begingroup$ There's even a thread here on this phenomenon. $\endgroup$
    – Ruslan
    Commented Jun 23, 2020 at 19:30
  • $\begingroup$ Noooooooooooooo. At least write ln(1)! $\endgroup$
    – DKNguyen
    Commented Aug 1, 2022 at 13:37
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    $\begingroup$ @DKNguyen That's the point. They haven't identified it as a function (which suggests they may have almost no concept of functions at all). $\endgroup$
    – Jessica B
    Commented Oct 23, 2022 at 8:14
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Question was:

Evaluate $\int{(\cos^3{x}-\cos^5{x})}\text{d}x$.

Student (desperately weak at math) simply wrote: $\dfrac{1}{3}\sin^3{x}-\dfrac{1}{5}\sin^5{x}+c$.

I thought, "She's using a wrong method, badly"... then I realized that her answer was correct.

It turns out that if you replace the 3 and the 5 (in both the question and answer) with any other numbers, it doesn't work.

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  • $\begingroup$ So they'd used Wolfram/somesuch ? $\endgroup$
    – ryang
    Commented Jul 25, 2022 at 16:37
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    $\begingroup$ @ryang It was an exam question, under normal exam conditions, so no internet. She didn't show any working. I think she just mangled up the power rule together with the integrals of cos and sin, and her answer was correct by coincidence. Either that or it was a stroke of genius. $\endgroup$
    – Dan
    Commented Jul 25, 2022 at 16:44
  • $\begingroup$ This could be reformulated as a seemingly intimidating homework problem. Very cool. $\endgroup$
    – James S. Cook
    Commented Jul 25, 2022 at 21:39
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This, from an exam I proctored for an absent teacher.

$y=32.5+27.5\cos\frac{\pi}{10}\left(x-8\right)$

Students were given a real life word problem, the movement of a rider on a ferris wheel.

A quick side note, one student came up to me to ask what a ferris wheel was, and I was reminded we often make assumptions about basic knowledge that might not be true. One can get to age 16 and never seen a ferris wheel.

The equation above was the correct equation for the rider height at time x. The next question was to give the rider's height at time x=2. The correct answer was 24.002 which rounds nicely to 24. Multiple students answered -124.42. Because they entered the numbers into their calculator without proper parenthesis.

For me, there are two issues. Students are using calculators from grade school, but not being taught proper use. We can use a bit of time each year to walk kids through the required keystrokes to get good results.

Second, and most concerning. The students graphed the equation. The graphs were beautiful, minimum 5, maximum 60. I struggle to understand how they can so easily get such a result (a negative height!) and not return to the equation to track down their error.

TL:DR

In general, we need to address calculator skills with students. We also need to mindfully teach the skill of checking one's answer.

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    $\begingroup$ how did they get the negative number exactly? $\endgroup$
    – Jasper
    Commented May 16, 2022 at 15:30
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    $\begingroup$ +1. That said: I can teach estimating and double-checking basic equations every day to my remedial students and it just comes off as useless blathering to them. (Literally: On the last day I've had students say, "I never got this estimation thing".) Sometimes I think students need to work on a physical construction and have it really break before they might care about this. $\endgroup$
    – Daniel R. Collins
    Commented May 16, 2022 at 16:07
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    $\begingroup$ I think a lot of students regard "math" as separate from "real world". After all, it is only in "math" that people will buy 300 watermelons and so on. So you get students thinking that if a sweater on 20% sale cost \$80, then its original price could perfectly well have been \$60, after all 80-20=60, why not; and if they wonder why the original price is higher than the sale price, they probably think, well, it's math, it's not supposed to make sense... $\endgroup$
    – Zach Teitler
    Commented May 16, 2022 at 19:55
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    $\begingroup$ In regards to the negative height not triggering a sense of "something's wrong", I blame poorly-designed "real life" problems. I recently raised an issue regarding an example exam problem about the behavior of a plucked string. The math all worked out, but the equations described a one-meter string with the elasticity of Silly Putty, "plucked" by pulling the center almost four kilometers to one side. If you want students to sanity-check their answers, you need to consistently provide questions with sane answers. (Negative height just means the wheel is built in a trench.) $\endgroup$
    – Mark
    Commented May 16, 2022 at 21:56
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    $\begingroup$ Mark - to be clear, multiple students (3) had this incorrect answer. All three had beautiful graphs that had proper min/max. I’m less startled by a subterranean depth than a result at odds with the graph they produced. $\endgroup$
    – JTP - Apologise to Monica
    Commented May 16, 2022 at 22:42
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I once met a student who computed $\sqrt{29}$ to $4$ digits precisions manually. At first I suspected cheating with computer since it's an at home exam. But the student showed me they can use bisection method to compute this so I let it go.

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    $\begingroup$ Computing square roots by hand used to be taught in secondary school 100+ years ago. The method I learned -- from an 1899 math text -- resembles long division and is not much more complex than that. The method generalizes to cube, fourth, etc. roots. $\endgroup$
    – shoover
    Commented May 8, 2022 at 5:23
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    $\begingroup$ 1897, not 1899. Here's the 1904 edition: archive.org/details/arithmeticmensu00schogoog/page/n104/mode/… $\endgroup$
    – shoover
    Commented May 13, 2022 at 15:11
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    $\begingroup$ @shoover I was taught the method when I was 11 in middle school, around some 40-odd years ago... No need to go back one century! And I actually used that or series expansion whenever I forgot the calculator at a university exam (something that happened frequently). $\endgroup$
    – Massimo Ortolano
    Commented May 14, 2022 at 11:30
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    $\begingroup$ I also was taught manual square root finding a little less than 40 years ago. More information about it here. $\endgroup$
    – JRN
    Commented May 15, 2022 at 5:16
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    $\begingroup$ @MassimoOrtolano That looks like the same method I learned, only yours is illustrated better. $\endgroup$
    – shoover
    Commented Jul 25, 2022 at 23:04
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An interesting one I saw on a student exam many years ago.

The sequence diverges because the Cauchy criterion is dissatisfied.

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Question was:

"Find the Maclaurin series of $\ln{(2+2x)}$ up to the term in $x^3$.

Student A: $\ln{(2+2x)}=\ln{(2(1+x))}=\ln2+x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$

Student B: $\ln{(2+2x)}=\ln{(1+(1+2x))}=(1+2x)-\dfrac{(1+2x)^2}{2}+\dfrac{(1+2x)^3}{3}+...$

Of course, Student A is correct. (Student B has found a Taylor series but not a Maclaurin series.)

But it is interesting to equate the coefficients in the two answers (temporarily ignoring the fact that the two series have different intervals of convergence). We get the following results:

Equate coefficients of $x^1$: $2-2+2-2+2-...=1$, that is, $1-1+1-1+1-...=\dfrac12$

Equate coefficients of $x^2$: $-2+4-6+8-10+...=-\dfrac{1}{2}$, that is, $1-2+3-4+5-...=\dfrac14$

Equate coefficients of $x^3$: $\dfrac43(2-6+12-20+30)$, that is, $1-3+6-10+15-...=\dfrac18$

And so on.

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  • $\begingroup$ I once designed a (U.S. based) calculus 2 homework problem based on this idea, where to get things to agree for a particular Maclaurin series term (probably the linear or quadratic term, but I don't remember now) you have to correctly account for all the subsequent contributing coefficients in the non-Maclaurin expansion, and in doing this you'd get an infinite series whose explicit sum can be determined (with hints I provided). $\endgroup$
    – Dave L Renfro
    Commented Jul 25, 2022 at 13:44

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