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Science journals of interest: Scientific American - Nature - New Scientist - Science AAAS - Science Daily

LAST EDITED: March 17, 2015

Here are some interesting links I had stickied:
Yeano's thread: Proving Something is Independent of our Axioms.
Yeano's thread: Algebraic Topology and Model Theory

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Couple of Putnam Exam Q's that Richard Feynman took in 1939
Posted: Posted May 22nd, 2017
Edited May 22nd, 2017 by The Fly

The celebrated Nobel Laureate physicist Richard Feynman took the 1939 Putnam exam competition (when he was 20 or 21 at MIT) and scored among the top 5 or so students I think. Here are a couple of problems from them that sort of appealed to me.

1. Suppose \(\alpha, \beta, \gamma\) are roots of the cubic equation \(x^3 + ax^2 + bx + c = 0\). Find the cubic equation that has roots \(\alpha^3, \beta^3, \gamma^3\).

2. Solve the integral \(\Large{\int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}}}\).

I haven't solved these, so don't have a solution (at least not as I write this post). There are other interesting problems on that exam too but I thought to pick these two which jumped at me.

My apologies for being away for quite a long time due to my workload having become quite severe in the last few months -- but I want to thank behind the scenes moderators for taking care of my forum here.

All the best,
The Fly


There are 7 Replies

Caution: in the first problem there's the Greek letter alpha \(\alpha\) and the usual letter "a". (In some fonts they could look alike, but they are different here.) In the cubic you have a b and c, but the roots are alpha beta and gamma.

Posted May 22nd, 2017 by The Fly
The Fly

Should the answer to 1 be in terms of a, b, and c?

Posted June 2nd, 2017 by EN

For the second, here's what I got:

$$\Large{\int_1^3 \frac{dx}{\sqrt{(x-1)(3-x)}}}$$

$$\Large{\int_1^3 \frac{dx}{\sqrt{-x^2+4x-3}}}$$

$$\Large{\int_1^3 \frac{dx}{\sqrt{1 - (x-2)^2}}}$$

With \(u\) substitution \(u = x-2\), we get \(du = dx\), giving us

$$\Large{\int_1^3 \frac{du}{\sqrt{1 - u^2}}}$$

Which has a standard antiderivitive of \(\arcsin(u)\), giving us \(\arcsin(x-2)\) from 1 to 3. This gives \(\arcsin(1) - \arcsin(-1) = \pi/2 - (-\pi/2) = \pi\).

Pretty neat!

Edited June 2nd, 2017 by EN

Well done EN, neat and clean integral indeed!

About your first question, the problem didn't say that the requested cubic polynomial has to be in terms of a b and c. So the problem is a lot easier than it appears! One could write the requested cubic in terms of a b and c by writing them in terms of alpha beta and gamma (which isn't hard to do, being linear equations). But this seems too messy!

Posted June 3rd, 2017 by The Fly
The Fly

Yeah, I was just thinking of


Edited June 4th, 2017 by EN

The constant term is easy, at least. The constant term of the original equation is


whereas the constant term of the cubed roots polynomial is


Edited June 4th, 2017 by EN

Yes, that's right.

Posted June 21st, 2017 by The Fly
The Fly
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