A) If $$x_1, ..., x_n$$are distinct numbers, find a polynomial function $$f_i$$of degree $$n-1$$ which is 1 at $$x_i$$ and 0 at $$x_j$$ for $$j \neq i$$
B) Now find a polynomial function $$f$$ of degree $$n-1$$ such that $$f(x_i) = a_i$$where $$a_1, ..., a_n$$are given numbers. You should use the functions $$f_i$$ from part a

Question

asnaseer · Answer

I think solution to A is:$$f_i=(x-x_1)(x-x_2)...(x-x_i+1)(x-x_{i+1})...(x-x_n)$$

asnaseer · Answer

sorry:$$f_i=\frac{(x-x_1)(x-x_2)...(x-x_i+1)(x-x_{i+1})...(x-x_n)}{(x_i-x_1)(x_i-x_2)...(x_i-x_j)...(x_i-x_n)}$$where no x_j term in the denominator is equal to x_i

asnaseer · Answer

then B would just be:$$f(x_i)=\frac{(x-a_1)(x-a_2)...(x)(x-a_{i+1})...(x-a_n)}{(x_i-a_1)(x_i-a_2)...(a_i-a_j)...(a_i-x_n)}$$

zarkon · Answer

wouldn't (B) be 

$$f(x)=\sum_{i=1}^{n}a_{i}f_{i}(x)$$

asnaseer · Answer

of course - now why didn't I spot that |-)