I was playing around with SymPy, a symbolic math package for Python, and ran across `nsimplify`

. It takes a floating point number and tries to simplify it: as a fraction with a small denominator, square root of a small integer, an expression involving famous constants, etc.

For example, suppose some calculation returned 4.242640687119286 and you suspect there’s something special about that number. Here’s how you might test where it came from.

>>> from sympy import * >>> nsimplify(4.242640687119286) 3*sqrt(2)

Maybe you do a calculation numerically, find a simple expression for the result, and that suggests an analytical solution.

I think a more common application of `nsimplify`

might be to help you remember half-forgotten formulas. For example, maybe you’re rusty on your trig identities, but you remember that cos(π/6) is something special.

>>> nsimplify(cos(pi/6)) sqrt(3)/2

Or to take a more advanced example, suppose that you vaguely remember that the gamma function takes on recognizable values at half integer values, but you don’t quite remember how. Maybe something involving π or *e*. You can suggest that `nsimplify`

include expressions with π and *e* in its search.

>>> nsimplify(gamma(3.5), constants=[pi, E]) 15*sqrt(pi)/8

You can also give `nsimplify`

a tolerance, asking it to find a simple representation within a neighborhood of the number. For example, here’s a way to find approximations to π.

>>> nsimplify(pi, tolerance=1e-5) 355/113

With a wider tolerance, it will return a simpler approximation.

>>> nsimplify(pi, tolerance=1e-2) 22/7

Finally, here’s higher precision approximation to π that isn’t exactly simple:

>>> nsimplify(pi, tolerance=1e-7) exp(141/895 + sqrt(780631)/895)

* * *

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This is just quite unreasonably cool!

Obligatory XKCD reference: https://xkcd.com/1047/

Several of those are marks on my slipstick. Can’t get much simpler than that.

Maple has a similar function called

`identify`

.Also interesting are Plouffe’s inverter and integer relation algorithms.

This is also really really cool! :) –JamesMills / prologic

This reminded me of the Inverse Symbolic Calculator: http://isc.carma.newcastle.edu.au/.

Ain’t sin(pi/6) == 1/2 ?

You probably meant sin(pi/3)

Odd. On my calculator (and matching my trig knowledge) sin(pi/6) = 0.5, not sqrt(3)/2.

I have absolutely no use for such a thing.

However, I’ve bookmarked it anyway.

Quite interesting.

Doesn’t anybody find this slightly dangerous? I can imagine people finding a result numerically, and giving an analytical form just to make it look like they found it analytically. – Yes, there’s people who would do that.

Frank and Joe: Thanks. I meant to say cosine. I’ve corrected the post.

So cool! Thanks for the info!

wonderful. Never imagined that one day software will do this!! How fast are such calculations?

Alok: The calculations are quick. Most of the time you get a rational approximation, and algorithms for approximation with the smallest possible denominators are well known and fast. I don’t know how extensive it is in considering square roots etc.

Great post and great technique! Recently, I saw two clever uses of it (up to the CAS utilized) that is worth sharing:

[1] http://blog.wolfram.com/2013/05/01/after-100-years-ramanujan-gap-filled/

[2] http://blog.wolfram.com/2013/03/28/from-close-to-perfect-a-triangle-problem/

Nice find, John!

@mrpaladin

@Alex, I was thinking that myself. For some reason, I am reminded of the fine structure constant.

Such methods played a part in Euler’s solution of the famous Basel Problem: find the exact value of 1 + 1/2^2 + 1/3^2 + 1/4^2 + …

He computed the value of the series to six decimal places, then somehow recognized that it seemed to be equal to pi^2 / 6. This was a clue which later led him to a proof that this was indeed the correct value of the series.

So cool! Thanks for sharing.

PS, I find 0.7182818284590453(e-2) which is not well simplified(143656365691809/200000000000000),:)

nineright: If you suggest e is a constant to consider, it’ll work.

`nsimplify(0.7182818284590453, constants=[E])`

returns e-2.exp((141 + sqrt(780631))/895) is not so bad for an automatic approximation to π. Still, 12 digits and two functions yield only 11 correct digits, which is not quite a good deal.

The following gives 10 correct digits at the cost of 5 digits and one transcendental function reused twice plus two calls of the same function. Not a good deal either, but at least it is more beautiful. Also, this one was found by hand:

log(16*log(878)/log(16*log(878)))