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View Full Version : Great lucky guess by a student

Chronos
02-02-2010, 09:32 PM
The other day, I was grading quantum homeworks, and for one of the problems, the students needed the value of the infinite sum 1/1 + 1/9 + 1/25 + 1/49 + ... (the sum over odd n of 1/n2). Well, this is a nontrivial sum to work out from scratch, and it's a physics class, not a math class, so the expectation was that the students would just look it up, or plug it into Mathematica, or the like. One student, though, apparently couldn't find it that way, so he guessed a value for it.

He guessed pi2/8 .

Because after all, that's the first thing you'd guess, right?

hajario
02-02-2010, 09:34 PM
How do you know it was a guess?

Inner Stickler
02-02-2010, 09:34 PM
Are you sure he's not having a go at you?

nivlac
02-02-2010, 09:49 PM
I really doubt it was a guess.

Chronos
02-02-2010, 10:30 PM
I know it was a guess because he said so on his paper. To be fair, it wasn't a completely blind guess: He did record his thought process. He knew that a a few other similar sums come out to pi squared over something, so he added up the first few terms and picked one that looked close.

Saganist
02-02-2010, 10:40 PM
So... it wasn't really a lucky guess at all. Also, I thought that sum was pi^2 / 6?

e: Never mind, that's for all n, not just odd.

billfish678
02-02-2010, 10:54 PM
In my high school calc class I'd do shit like that. The teacher was pretty bad and the book wasnt much better, so I really didnt grok what calc really was besides some set of rules.

And I am horrible at details, which doesnt really help when you are doing math by the rules without really understanding what its all about. But often times I kinda had a clue what the answer should be and would do some back of the envelope calc to decide if perhaps it was say 2e or pi/4 or whatever else I suspected it to be. And if I didnt use it to guess an answer, I'd use it to help confirm an answer I'd worked out.

Which reminds me of a physics test problem once. The problem involved finding where the electric field of several charges was equal. Figuring it out was an algebraic mess. Most folks, including me, missed that problem. Not because we didnt know what we were doing. Just because the math was messy and time was short. One smart ass said they were all equal at infinity without doing any math. That wasnt what the prof had in mind but he gave em credit anyway.

02-03-2010, 04:47 AM
I remember loving the quote the last time this came up:

Integrating Sqrt(tan). This one's easy. Use the method of guess-and-check. Guess a function, differentiate it, and see if you get the original function back. If you do, you're done. If not, guess again. The key to the method of guess and check is to make good guesses. Here, I would recommend guessing 1/2*tan(x)^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x))/(cos(x)*sin(x))^(1/2)-1/2*2^(1/2)*ln(cos(x)+2^(1/2)*tan(x)^(1/2)*cos(x)+sin(x)).

Sure enough, when we check d/dx [1/2*tan(x)^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x))/(cos(x)*sin(x))^(1/2)-1/2*2^(1/2)*ln(cos(x)+2^(1/2)*tan(x)^(1/2)*cos(x)+sin(x))] , we get sqrt(tan(x)). It's simple!

Harmonious Discord
02-03-2010, 08:01 AM
An intuitive guess. Is the student by any chance a visual learner (http://en.wikipedia.org/wiki/Visual_learning). I could see them visualizing the progression and giving an answer they can't explain to you.

Giles
02-03-2010, 08:27 AM
He guessed pi2/8 .

So... it wasn't really a lucky guess at all. Also, I thought that sum was pi^2 / 6?

e: Never mind, that's for all n, not just odd.
If you know the sum for all n, then it's easy to calculate for odd n.

If the sum for all n is pi2/6, then the sum for even n is pi2/24 (each term is 1/4 of the corresponding term for all n), and the difference between 1/6 and 1/24 is 1/8.

So, could he have known the sum for all n?

Chronos
02-04-2010, 12:45 PM
If you know the sum for all n, then it's easy to calculate for odd n.

If the sum for all n is pi2/6, then the sum for even n is pi2/24 (each term is 1/4 of the corresponding term for all n), and the difference between 1/6 and 1/24 is 1/8.

So, could he have known the sum for all n? That's an elegant derivation, but I don't think it's the one he used. The sums he was using as his starting point were still over odd n, but had different exponents in the denominator (1/n^4 and 1/n^6, IIRC). And no, I don't have any idea how he knew those.

Hello Again
02-04-2010, 12:52 PM
It's not a guess its more of a Fermi Solution (http://iws.ccccd.edu/mbrooks/demos/fermi_questions.htm). He took certain information he did know and applied it to a problem that seemed to big too answer directly.

I'm also an inductive reasoner and sometimes just "see" the answer (not often enough! but it happens). When dealing with math I sometimes can't explain how I know the answer. It just see the whole thing at once, not as a series of steps. It can be incredibly frustrating -- basically there are two kind of math problems: "easy" and "impossible" - "easy" if I see the complete answer at once and "impossible" if not.

Ludovic
02-04-2010, 01:28 PM
One time when we were practising for our AP Lit AP exam we had to read a poem from our book and take an AP-test style multiple choice quiz on it of 5 questions. I didn't bring my book that day so I just guessed based on context, and got 4 out of 5 right. The class average was 2 or 3 out of 5 and they had their books.

Ludovic
02-04-2010, 01:31 PM
Oh, and this guess is more lucky, in college I was taking World History and was taking a quiz on Indian civilization. I had no clue for one question and just wrote in a plausible-sounding Indian word: Dharma, which happened to be right. I had no idea what dharma was beyond probably being a real Indian word.

lissener
02-04-2010, 02:01 PM
Yeah I'm going with "not a guess." A guess would be "4." Or "giraffe."