Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have following points (-|b-a|,a), (0,b), (|c-b|,c) with a, b and c as two-dimensional vectors. These should be interpolated component-by-component with a second-degree polynomial p.

My problem now is, that I don't know how the polynomial should look like: $p(-|b-a|)= \alpha(-|b-a|)^2 + \beta(-|b-a|)+\gamma = a$

$p(-|b_x - a_x|) = \alpha(-|b_x-a_x|)^2 + \beta(-|b_x-a_x|)+\gamma = a_x$ $p(-|b_y - a_y|) = \alpha(-|b_y-a_y|)^2 + \beta(-|b_y-a_y|)+\gamma = a_y$

$\cdots$

Is my conversion correct?

share|improve this question
    
20 percent accept rate? Could you do something about that? –  Gerry Myerson Jun 8 '12 at 6:54
add comment

1 Answer 1

No. The polynomial you need is:

$$p(t) = \frac{t(t-|c-b|)}{|b-a|(|b-a|+|c-b|)}a + \frac{(t+|b-a|)(t-|c-b|)}{-|b-a||c-b|} b + \frac{t(t+|b-a|)}{|c-b|(|c-b|+|b-a|)} c.$$

share|improve this answer
    
Thank you. Could you please explain how do you got this polynomial? –  monoid Jun 8 '12 at 7:41
    
It is called the Lagrange polynomial. See en.wikipedia.org/wiki/Lagrange_polynomial. In fact, it is two scalar polynomials, one for the x-component and one for the y-component. I just stacked them together for compactness. –  copper.hat Jun 8 '12 at 7:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.