Description
Given four points and suppose it's necessary to fit a cubic through them.
Let $y(x)=ax^3+bx^2+cx+d$ in which
- $y(-3)=\alpha$
- $y(-1)=\beta$
- $y(1)=\gamma$
- $y(3)=\delta$
Then:
- $a=\frac{1}{48}((\delta-\alpha)-3(\gamma-\beta))$
- $b=\frac{1}{16}((\delta+\alpha)-(\gamma+\beta))$
- $c=\frac{1}{48}(27(\gamma-\beta)-(\delta-\alpha))$
- $d=\frac{1}{16}(9(\gamma+\beta)-(\delta+\alpha))$
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