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.
  1. For a function f and distinct points $\alpha$, $\beta$, $\gamma$; what is meant by $f[\alpha,\beta,\gamma]$?
  2. Find the Lagrange form for the polynomial $P(x)$ that interpolates $f(x) = \frac{4x}{x+1}$ at $0$, $1$ and $3$.

For (1), I can say that we have a difference divided:$$ \frac{f[\gamma]-f[\beta]}{\gamma - \beta}$$ but a little lost on handling for three.

For (2): $$(x-0) \times f(1) \times f(3) + (x-1) \times f(0) \times f(3) + (x-3) \times f(0) \times f(1)$$ Will this work?

share|improve this question
    
What textbook are you using? Has your book already discussed divided differences? There should be a mention of the recursive formula for generating them. –  J. M. May 3 '12 at 8:08
    
burden and faires, but I'm here to ask question –  mary May 3 '12 at 8:09
    
@mary: As for your first question, it's called "Newton's interpolating polynomial". –  Gigili May 3 '12 at 9:40
add comment

1 Answer 1

up vote 2 down vote accepted

It is:

$$f[\alpha,\beta,\gamma]=\frac{f[\alpha,\beta]-f[\beta,\gamma]}{\alpha-\gamma}$$

Indeed, it works like that for $n$ points as well.

As for the number $2$,

$$P(x)=f(0) \times L_0(x) + f(1) \times L_1(x)+ f(3) \times L_2(x)=f(1) \times L_1(x)+ f(3) \times L_2(x) $$

Where:

$$L_j(x)=\frac{(x-x_0)(x-x_1) \dots (x-x_{j-1})(x-x_{j+1}) \dots (x-x_n)}{(x_j-x_0)(x_j-x_1) \dots (x_j-x_{j-1})(x_j-x_{j+1}) \dots (x_j-x_n)}$$

And $n=2$ (the number of points). Therefore:

$$L_0(x)=\frac{(x-x_1)(x-x_2)}{(x_j-x_1)(x_j-x_2)}=\frac{(x-1)(x-3)}{(0-1)(0-3)}=\frac{(x-1)(x-3)}{3}$$

(Actually, you don't need to calculate $L_0$)

$$L_1(x)=\frac{(x-x_0)(x-x_2)}{(x_j-x_0)(x_j-x_2)}=\frac{(x-0)(x-3)}{(1-0)(1-3)}=$$

$$L_2(x)=\frac{(x-x_0)(x-x_1)}{(x_j-x_0)(x_j-x_1)}=\frac{(x-0)(x-1)}{(3-0)(3-1)}=\frac{x(x-1)}{6}$$

Finally, we have:

$$P(x)=f(1) \times L_1(x)+ f(3) \times L_2(x)= 2 \times \frac{x(x-3)}{-2} + 3 \times \frac{x(x-1)}{6}=\frac{x(x-1)}{2} -x(x-3)$$

$$P(x)=\frac{-x^2+5x}{2}$$

share|improve this answer
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.