pattern matching - Can I get Mathematica to recognize common series expansions of trigonometric functions?
I have some complicated functions from numerical procedures that go to simple trig functions in the limit of small time-steps. For example, I can calculate that one of these functions goes to:
x - 2x^3 / 3 + 2x^5 / 15
This is $\sin{x} \cos{x}$ minus terms of $O(x^7)$ and higher. Is there any way for Mathematica to recognize such a pattern automatically? I have other similar series expansions and I don't want to have to find all of them manually.
---Edit---
Thank you Szabolcs and corey979! You asked for more examples/background so here it is. I wanted to convert a 2x2 matrix of polynomials to trigonomic functions. I can obtain this matrix to arbitrary order in some small parameter h. To first order that looks like:
{{1 + h (t - (2 t^3)/3 + (2 t^5)/15), h (t^2 - t^4/3 + (2 t^6)/45)}, {h (t^2 - t^4/3 + (2 t^6)/45), 1 + h (-t + (2 t^3)/3 - (2 t^5)/15)}}
To second order that looks like:
{{1 + h^2 (t^2/2 - t^4/6) + h (t - (2 t^3)/3 + (2 t^5)/15), h (t^2 - t^4/3 + (2 t^6)/45)}, {h (t^2 - t^4/3 + (2 t^6)/45), 1 + h^2 (t^2/2 - t^4/6) + h (-t + (2 t^3)/3 - (2 t^5)/15)}}
etc
For a single matrix element, I was able to collect the coefficients of various orders of h and use With the data and FindFunction approach to turn those coefficents into trigonomic functions. It seems to work okay if I have the coefficients have enough terms, but I can get pretty wacky numbers if it doesn't have enough terms. I use this code:
findtrigfunc[inseries_] := Module[{outseries=inseries},
outseries = outseries //CoefficientList[#,h]&;
For[i=1, i <= Length[outseries],i++,
{data = Table[Evaluate@{x, outseries[[i]] //. t->x}, {x, 0, 1, 0.01}];
outseries[[i]] = FindFormula[data, x, TargetFunctions -> {Times,Sin, Cos}]}];
Sum[outseries[[i]]*h^(i- 1) , {i,1,Length[outseries]}]//.x->t]
Here's what that produces for the first matrix element up to second order in h and order 7 in t:
1. + 1.0043 h Cos[t] Sin[t] + h^2 (-0.00618535 + 0.596221 Sin[Sin[t]]^2)
When I increase the order in t from 7 to 11, I get extra terms that get me a more reasonable answer:
1. + h^2 (0.49996 - 0.499954 Cos[t]^2) + 1.00001 h Cos[t] Sin[t]
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