So I’ve been working on one of my papers where we need to compute some numbers. They end up being determined by a cubic equation.
However, one often finds surprising identities when Mathematica spits out a bunch of numbers expressed in algebraic form. Here is one of them:
![\frac{1}{42} \sqrt[3]{\frac{1}{2} \left(90720 \sqrt{7446}-2859138\right)}-\frac{7893}{7\ 2^{2/3} \sqrt[3]{2859138+90720 \sqrt{7446}}}=0](http://s0.wp.com/latex.php?latex=%5Cfrac%7B1%7D%7B42%7D+%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B2%7D+%5Cleft%2890720+%5Csqrt%7B7446%7D-2859138%5Cright%29%7D-%5Cfrac%7B7893%7D%7B7%5C+2%5E%7B2%2F3%7D+%5Csqrt%5B3%5D%7B2859138%2B90720+%5Csqrt%7B7446%7D%7D%7D%3D0&bg=ffffff&fg=333333&s=0 "\frac{1}{42} \sqrt[3]{\frac{1}{2} \left(90720 \sqrt{7446}-2859138\right)}-\frac{7893}{7\ 2^{2/3} \sqrt[3]{2859138+90720 \sqrt{7446}}}=0")
PS: I use italics in the word surprising above only because if one does not know the origin of these identities they might seem surprising.
So take pity on an old man and tell us from whence this came. Please.
Hi Tom:
It results from looking for the (unique) real roots of two related cubic equations with integer coefficients. The two real roots are such that their sum must be an integer. Thus the non rational part has to cancel between them in a nice way.
There is an analytic formula for these roots, so when I put them in my computer using Mathematica (without simplifying), it spits them out in the form above, because the coefficients of the two polynomials are different. The computer dos not automatically understand that they are the same, but if tested numerically one gets that their difference is as small as one wants it to be.
Dear David, I am not sure whether you know the command “Simplify”. When I write down your formula and make Simplify[...], it returns a clear zero in my Mathematica 7.0.1. Here is a screenshot:
https://picasaweb.google.com/lubos.motl/TheReferenceFrame4#5634297936280697410
If you need it, I will show you a human-readable proof that it’s exactly zero that doesn’t require the undoubtedly fancy context in which that identity has emerged.
I noticed you probably know “simplifying”, so let me make this point. It is very sensible that Mathematica doesn’t try to simplify at absolutely every step because the proof that things like this vanish and simplify are pretty complicated and the checks whether the simplification would help would only be helpful in a very small percentage of expressions.
HI Lubos:
It didn’t seem to work on Mathematica 8, which is what I was using. I wouldn’t be surprised if they tweaked the algorithm.
That’s bizarre. Could you please retype the command exactly as in the screenshot – especially without decimal points that make the calculation numerical – and confirm that it won’t derive that it’s zero?
I would urge one of the contacts at Wolfram because this is a bad regression if it is real.
Mathematica 8.0.1.0 returns zero, too.
I tried again and this time it worked. It was not working last time. I’ve always find that the way the program acts depends on the history of how things are done.
By the way, I was trying to derive what you’re working on, by guessing the cubic equation. The depressed cubic was nice and simple, with some rational coefficients, having powers of 14 in the denominators, but I suppose that you have an even simpler cubic equation which is not depressed – which has a nonzero coefficient of the quadratic term – and I wasn’t able to get a nice form for that. Maybe some numerical mistakes…