# How to get the boolean value of an inequality involving an InterpolatingFunction?

Here's the code:

yan = FunctionInterpolation[x^2, {x, -1, 1}];
FullSimplify[yan[x] > -1, -1 < x < 1]


Needless to say, what I expect to see in the output is "True", but FullSimplify doesn't seem to work. What function should I turn to?

@J.M. @belisarius @acl

yan = FunctionInterpolation[x^2, {x, -1, 1}];
MinValue[{yan[x], -1 < x < 1},x]>-1

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Huh? "seems not to work" is correct? I just delete it because of my language sense…OK, let me add it back. –  xzczd Aug 6 '12 at 5:55
"doesn't seem to work" as you added is more standard, but in my opinion "seems not to work" is also acceptable and understandable. –  Mr.Wizard Aug 6 '12 at 6:10
In fact I've become confused after I searched the Internet, so I turned to the standard form to be on the safe side 囧. –  xzczd Aug 6 '12 at 6:47

The following is basically the same @acl did, but using the package InterpolatingFunctionAnatomy which (in principle) will behave better than peeking at the internal structures when Mma version changes.

Needs["DifferentialEquationsInterpolatingFunctionAnatomy"];
yan = FunctionInterpolation[x^2, {x, -1, 1}];

yin = InterpolatingPolynomial[Transpose[Flatten /@
{InterpolatingFunctionCoordinates@yan,
InterpolatingFunctionValuesOnGrid@yan}], x];
FullSimplify[yin > -1, -1 < x < 1]
(*
True
*)

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very nice. how come I have less votes than you though, despite being first and explaining details?! Not fair! :) –  acl Aug 5 '12 at 19:38
@acl That was because I forgot to upvote your answer. Easy to correct! :D –  belisarius Aug 5 '12 at 19:40
really, I was joking! –  acl Aug 5 '12 at 19:48
In fact, you don't need to load the package if you remember the actual syntax being used internally; in this case, it's yan["Coordinates"] and yan["ValuesOnGrid"] that can be used directly. Still, this works only because the original function was well approximated by a polynomial. In general, a polynomial interpolant can be more oscillatory than the piecewise polynomial interpolant used by InterpolatingFunction[]; be careful! –  Ｊ. Ｍ. Aug 5 '12 at 23:48
@J.M. yes, it was intended just for this case –  belisarius Aug 5 '12 at 23:52

You can do this by explicitly constructing the InterpolatingPolynomial corresponding to yan, and then using FullSimplify:

yin = InterpolatingPolynomial[Transpose[Flatten /@ {yan[[3]],yan[[4]]}],x];
FullSimplify[yin > -1, -1 < x < 1]
(*True*)


Why does this work? Because yan actually has a list of points:

FullForm[yan]


so I can extract them with Transpose[Flatten /@ {yan[[3]],yan[[4]]}] and use them to construct a polynomial, which does the same thing as the interpolation function but which FullSimplify can now handle.

Maybe there's a better way to construct the InterpolatingPolynomial but this works.

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