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Offline omorr  
#1 Posted : 27 December 2010 22:25:48(UTC)
omorr


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Hello,

See the picture, please:

As you could see, all the three available interpolating functions
(linterp, cinterp, ainterp as lines) are passing through the data points (symbols-"x" ). If you could not see them at once, just use CTRL, SHIFT + scrol wheel to change the axis scales and change them in accordance to the data ranges.

I hope this helped.

Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
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Offline omorr  
#2 Posted : 28 December 2010 14:13:59(UTC)
omorr


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Hello,

1/T was regarded as matrix inverse, therefore T mast be a square matrix (it is a vector in your case). See the example please.
M←mat(3,4,1,2,3,1,1,1,-8,3,3)1/M=mat(3.12,-4.12,-0.12,-2.12,3.12,0.12,0.12,-0.12,-0.12,3,3)M^{-1}=mat(3.12,-4.12,-0.12,-2.12,3.12,0.12,0.12,-0.12,-0.12,3,3)
If you want a new vector with the reciprocal values of its elements - you have to use loops. See the example please:
T:mat(298,500,1000,1500,2000,2500,3000,7,1)
for(i,range(1,length(T)),el(Trec,i)←1/el(T,i))
Trec—mat(1/298,1/500,1/1000,1/1500,1/2000,1/2500,1/3000,7,1)Trec=mat(3.36*10^{-3},2*10^{-3},1*10^{-3},6.67*10^{-4},5*10^{-4},4*10^{-4},3.33*10^{-4},7,1)

Regards,
Radovan
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