## A mathematical puzzle (of sorts)

June 28, 2012 by lukebarnes

I happened across the following mathematical titbit, which I will set as a question to the reader. Let,

(1)

You are required to prove that

.

Striking, no? Obviously, sticking it into a calculator doesn’t count. I know a rather indirect way, which unfortunately involves the phrase “by inspection”. I’ll share it with you below the fold. There must be a nicer way!

The **short story** is that equation (1) is a solution of the cubic equation,

which has only one real solution, which *by inspection* is . But seriously, there has to be a deeper reason than that, doesn’t there?

My **long story** starts with the probability of a photon being Rayleigh-scattered through an angle cosine ,

If we want to draw a random deviate from this distribution (using the transformation method), then we solve:

which leads to the following cubic equation,

where . I then waged war with the cubic solution, and emerged both victorious and surprised with this:

(2)

where . I then tried to check the solution by making sure that when , which leads to equation (1). I was hoping that there would be a way to simplify (2) – it’s bogging down my Monte Carlo code.

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on June 29, 2012 at 10:03 pm |Stephen SerjeantOne useful trick is that (sqrt(5)+2)*(sqrt(5)-2) = 5 – 4 = 1. If you cube both sides of equation (1) and use that trick it quickly reduces to mu^3 = 4-3mu or 4=mu*(3+mu^2) and by inspection mu=1 is the only real solution.

on July 1, 2012 at 11:03 pm |AlexI have had the same problem with solving cubics in the past. I never could figure out what to do about it

on July 3, 2012 at 9:39 pm |gThe thing to notice is that 2+sqrt(5) = [(1+sqrt(5))/2]^3 — which you can find by writing (a+b.sqrt(5))^3 = 2+sqrt(5), hoping that a,b come out rational, and solving for the rational and rational.sqrt(5) components; or by remembering some stuff related to Fibonacci numbers.

Now, of course, the same thing holds with “-” in place of “+” on both sides, and your expression for mu is just (1+sqrt5)/2 + (1-sqrt5)/2, which really *is* obviously 1 :-).

on July 4, 2012 at 7:47 am |lukebarnesHurray! I think you’ve nailed it. Now to attack equation (2) with the same idea…

on July 8, 2012 at 8:33 am |gI don’t think there’s much to do with your expression for mu in general, beyond the observation that (s+c)(s-c) = ss-cc = (cc+1)-cc = 1, so that the two terms in the expression are reciprocals. So if you’re currently taking two cube roots, you can get by with just one. (And indeed you should — the one with the positive sign — to avoid significance loss. Though it won’t make much difference when |c| never goes above 2.)

One other note: mu in (1) and in (2) differ by a change of sign. I prefer the choice in (1) and think it’s better to write it as (c+s)^1/3 + (c-s)^1/3.