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How to verify when $n=1$ and $a=x_0, b=x_1$, then Hermite cubics provide a Hermite interpolation of [,].

I have derived the following $4$ polynomials form a basis for the degree $3$ polynomials on $[\alpha, \beta]$ $$H_0=3(\frac{\beta-x}{\beta-\alpha})^2-2(\frac{\beta-x}{\beta-\alpha})^3,$$ $$H_1=3(\frac{x-\alpha}{\beta-\alpha})^2-2(\frac{x-\alpha}{\beta-\alpha})^3,$$ $$S_0=(\frac{(\beta-x)^2}{\beta-\alpha})-(\frac{(\beta-x)^3}{(\beta-\alpha)^2}),$$ $$S_1=(-\frac{(x-\alpha)^2}{\beta-\alpha})+(\frac{(x-\alpha)^3}{(\beta-\alpha)^2}),$$

But, I am unsure how to show that the two points interpolate given the conditions above.

Kenta S
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Cherry
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1 Answers1

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(Note $a = \alpha = x_0, b = \beta = x_1$.) Check whether the functions you list have the properties that you want, namely: $$ H_0(\alpha) = 1, \quad H_1(\alpha) = 0, \quad S_0(\alpha) = 0,\quad S_1(\alpha) = 0 \\ H_0(\beta) = 0, \quad H_1(\beta) = 1, \quad S_0(\beta) = 0,\quad S_1(\beta) = 0 \\ H_0'(\alpha) = 0, \quad H_1'(\alpha) = 0, \quad S_0'(\alpha) = 1,\quad S_1'(\alpha) = 0 \\ H_0'(\beta) = 0, \quad H_1'(\beta) = 0, \quad S_0'(\beta) = 0,\quad S_1'(\beta) = 1 $$

GEdgar
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  • Hmm, so does that mean, all I have to do is for example, to find that $$_0()=1$$ I take the equation $$_0=3((−)/(−))^2−2((−)/(−))^3$$ and plug in for x. After canceling out and everything, that would give me 1? Then I repeat the process for all of the other points? @GEdgar – Cherry Dec 10 '19 at 03:41
  • Also, my follow up to that would be, why do those numbers you provided provide a Hermite interpolation of an arbitrary interval [a,b]. Does this mean for any given interval, those would be my set properties when n=1? I guess my issue is understanding why n=1 is so special. @GEdgar – Cherry Dec 10 '19 at 03:54