# Specify the values of $p$ and $p'$ for a polynomial

Problem 10-26 from Spivak's Calculus, 4th edition:

Let $a_1, \dotsc, a_n$ and $b_1, \dotsc, b_n$ be given numbers. If $x_1, \dotsc, x_n$ 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 $j\neq i$, and $f(x_i)=a_i$ and $f'(x_i)=b_i$.

The way I know (and the way listed in the answer book) is to set $f(x) = \left( \prod_{j \neq i}(x-x_j)^2\right)(ax+b)=g(x)(ax+b)$ and solve for $a$ and $b$ in terms of $g'(x_i)$ and $g(x_i)$.

The structure of the problem led me to wonder if there is an elegant expression of $f$ in terms of integrals, perhaps something like $$f(x)=p(x) \int^x q(\xi)d\xi.$$

Does anyone have any thoughts on this?

• I think you mean $f(x_j)=f'(x_j)=0$, not $+$.
– user147263
Commented Aug 10, 2014 at 4:53

No, I would not expect to find $f$ (or a factor of it) as an antiderivative of something. Antiderivative would help if we only had the conditions $f'(x_j)=\dots$; then of course we could interpolate to find $f'$, and integrate that. But here both $f$ and $f'$ are prescribed.
Let's try anyway and see how far it goes. Part of the task it to have $f(x_j)=0$ for $j\ne i$, and the most natural (and probably the only practical) way to do this is to introduce the factor of $\prod_{j\ne i} (x-x_j)$. So, let's say $$f(x) = \prod_{j\ne i} (x-x_j)\int^x q(\xi)\,d\xi$$ where $q$ is to be found. Now $$f'(x_j) = \prod_{k\ne i,j} (x_j-x_k)\int^{x_j} q(\xi)\,d\xi$$ and we are in trouble: the supposed advantage of looking for $f$ in integral form (integral canceling derivative) does not materialize. Equating $f(x_j)$ to zero imposes an integral condition on $q$ that is not transparent at all.
What is transparent is that $\int^{x} q(\xi)\,d\xi$ must be divisible by $\prod_{j\ne i}(x-x_j)$. So we put that factor in as well, and now it's back to same solution you have.