Let $a_1, a_2, ... a_n$ and $b_1, b_2, ... b_n$ be given numbers. If $x_1, x_2, ... x_n$ are distinct numbers, prove that there is a polynomial function $f$ of degree $2n - 1$, such that $f(x_j) = f'(x_j) = 0$ for $i \ne j$ and $f(x_i) = a_i$ and $f'(x_i) = b_i$ Hint: Use the fact that if $a$ is a double root of $f(x)$ then, $f(a) = f'(a) = 0$.

This is a very tough question.

If the conditions are sufficed then:

$$f(x) = (x - x_j)^2 g(x)$$ where $g(x)$ is some other function.

As well as:

$$f(x_i) = a_i, f'(x_i) = b_i$$

But I do not know how to continue

  • $\begingroup$ You wrote "$f(x_j) = f'(x_j) = 0$ for $i \neq j$". What does this mean? There is no $i$ specified here. $\endgroup$
    – Alex Zorn
    Commented Feb 26, 2015 at 16:30

1 Answer 1


Continuing your line of thinking, since $f$ must have $(x-x_j)^2$ as a factor for each $j\neq i$, the unique factorization of polynomials implies that $f$ has the form $$f(x)=\left(\prod_{j\neq i} (x-x_j)^2\right)g(x)$$ Then consider what degree $g$ must have.

  • $\begingroup$ why the product? $\endgroup$
    – Amad27
    Commented Feb 28, 2015 at 4:36
  • $\begingroup$ We want $f$ to be divisible by $(x-x_j)^2$ for each $j$; hence it must be divisible by their least common multiple, which in this case is their product, since they have no common factors. $\endgroup$ Commented Feb 28, 2015 at 4:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .