# interp1 fails to interpolate correctly

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Nate Ballintyn on 17 Nov 2019
Commented: Vladimir Sovkov on 19 Nov 2019
I have a set of input data which the function interp1 is incorrectly interpolating.
For a statement Vq = interp1(X,V,Xq), I have
X =
-0.0124393747904463
0.0134396235257952
0.0344412390769423
0.0850313641677704
0.0850004675375172
0.126329531171773
V =
-4
-2
0
2
4
6
Clq =
0.0772456107479591
The result interp1 gives me is Vq = 3.3864734865877. This is clearly wrong, but may be related to the face that between V = 4 and V =6, X is not monotonically increasing/decreasing. Can someone help me understand why this is failing?

Matt J on 17 Nov 2019
Edited: Matt J on 17 Nov 2019
X needs to be montonically increasing (or interp1 will pre-sort the data so that it is).

Vladimir Sovkov on 18 Nov 2019
Why do you consider it is wrong? In the plot (see the attachment) this result looks quite reasonable.

Nate Ballintyn on 19 Nov 2019
Your point is taken, and technically there is no generic way to interpolate a double valued function. However, I am nethertheless tasked with doing precisely that. I'm trying to find the angle of attack of an aircraft that, for some inexplicable reason, is generating less lift when the nose is raised form 2 to 4 degrees of pitch in this case. Maybe I'll make a polynomial fit and root solve for the angle....
Walter Roberson on 19 Nov 2019
polynomial fits cannot solve double-valued functions.
Sometimes you should use a function of two variables instead.
Sometimes you can rewrite as a parameteric function.
Sometimes you just need to segment to decide which function is in control. For example hysterisis has to be handled as a pair of functions, not by trying to fit one single function.
Vladimir Sovkov on 19 Nov 2019
Maybe, switching roles of X and V can help, e.g.:
VV=fzero(@(z)(interp1(V,X,z)-Clq),1)
VV=1.69220264208349