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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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    $\begingroup$ Are you going to write an article or a blog post based on the answers? $\endgroup$ – Rusty Core May 8 at 23:15
  • $\begingroup$ No, I’m not planning to write an article or post based on this. Why? Do you think I should think about it? I could. It hadn’t occurred to me till you asked. I’ll think about it. But for now, no, no plans to use answers for my own writings. $\endgroup$ – Zach Teitler May 9 at 5:06
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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 May 8 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 May 8 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 May 9 at 21:07
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    $\begingroup$ @TorstenSchoeneberg I think all but the harshest graders would give full points to that. $\endgroup$ – MathematicsStudent1122 May 11 at 7:17
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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 May 12 at 15:22
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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 May 29 at 18:48
  • $\begingroup$ @Vandermonde, yes, of course. Thanks! $\endgroup$ – Kostya_I May 29 at 19:05
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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 Jun 2 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 Jun 3 at 6:24
  • $\begingroup$ There's even a thread here on this phenomenon. $\endgroup$ – Ruslan Jun 23 at 19:30
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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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