# Show that there is a unique polynomial of degree at most $2n+1$ such that $q^{[k]}(x_1)=a_k,$ $q^{[k]}(x_2)=b_k$ for $k=0, \dots, n$.

Let $x_1, x_2 \in \mathbb R$ and let $(a_0, a_1, \dots, a_n), (b_0, b_1, \dots, b_n)$ be $(n+1)$-tuples of real numbers. Show that there is a unique polynomial of degree at most $2n+1$ such that $$q^{[k]}(x_1)=a_k,$$$$q^{[k]}(x_2)=b_k$$ for $k=0, \dots, n$.

Any hints on how to get started with this exercise?

• Does $[k]$ mean the $k$th derivative? – астон вілла олоф мэллбэрг Jan 19 '18 at 23:24
• @астонвіллаолофмэллбэрг I believe so – Al Jebr Jan 20 '18 at 0:07
• Suppose you want to perform induction. Can you do the base case $k=1$? – астон вілла олоф мэллбэрг Jan 20 '18 at 0:11
• @астонвіллаолофмэллбэрг Base case is $k=0.$ – saulspatz Jan 20 '18 at 0:39
• @saulspatz You are right, but it is too easy, I thought, so it is better to assess the complexity of the question by trying a harder base case. – астон вілла олоф мэллбэрг Jan 20 '18 at 0:42

EDIT There was an essential step left out in the proof. Thanks to Al Jebr for pointing it out.

Since I've muddled the matter with my comments, I'll post an answer, although you only asked for a hint. I'll try to phrase it as a sequence of hints, so you can stop reading when you want.

First, this seems to have something to do with Taylor polynomials. After all, they allow us to find a polynomial that satisfies the criteria for one of these tuples.

In particular, there is a polynomial $$p_1(x)= \sum_{k=0}^n{\frac{a_k}{k!}(x-x_1)^k}$$ that satisfies the criteria for the first tuple, and furthermore every polynomial $q$ that satisfies these criteria is of the form $q(x)=p_1(x)+(x-x_1)^{n+1}p_2(x)$ for some polynomial $p_2$.

Now the $a_k$ don't tell us any more. Any polynomial will work for $p_2,$ so we ask how the second tuple can help us find the coefficients of $p_2.$

By now, it's easy to guess that Taylor's theorem should help here, too. For convenience in what follows, I'll write $F(x) = (x-x_1)^{n+1}.$ so that our formula becomes $q(x)=p_1(x)+F(x)p_2(x).$

Since all we know about is what happens at $x_2,$ it's plain that we must evaluate $q$ at $x_2$. Notice that $F^{(k)}(x_2) \ne 0$ for $0\le k \le n,$ since we are given $x_1 \ne x_2.$

EDIT This is where I left out a step. I said that we have to use Taylor's theorem a second time, but then I didn't do it!

Now we look at the Taylor polynomial of $p_2$ at $x_2.$
$$p_2(x)= \sum_{k=0}^n{\frac{c_k}{k!}(x-x_2)^k,}$$ for some constants $c_k$. By Leibniz's formula for the derivative of a product,$$q^{(i)}(x_2) = p_1^{(i)}(x_2) + \sum_{j=0}^i{\binom{i}{j}F^{(i-j)}(x_2)}p_2^{(j)}(x_2)\text{ for } i=0,...,n.$$

Since none of the derivatives of $F$ vanish, and all the terms involving $c_k$ vanish, except the term involving $c_i,$ we can compute the coeffcients of $p_2$ one by one. Again, by Taylor's theorem, $p_2$ is unique up to term of degree higher than $n.$

• What is $p_2$? Is it the same as $p_1$ but with coefficient equal to $b_k$ instead of $a_k$? – Al Jebr Jan 21 '18 at 16:42
• Also, I am seeing that $q(x_2) \ne b_0$. – Al Jebr Jan 21 '18 at 16:51
• @AlJebr $p2$ is defined in the answer, $q(x)=p1(x)+(x−x1)n+1p2(x)q(x)=p1(x)+(x−x1)n+1p2(x)$ for some polynomial $p2.$ Where are you seeing $q(x_2) \ne b_0?$ By assumption, $q$ satisfies the conditions. – saulspatz Jan 21 '18 at 16:59
• $p_1(x_2) \ne 0$ so how can $q(x_2) = b_0$? – Al Jebr Jan 21 '18 at 17:11
• @AlJebr I'll try to edit my answer to make it plainer to you. Just give me a few minutes. – saulspatz Jan 21 '18 at 17:13

To start write out the general $2n+1$- degree polynomial..$q(x)=c_{2n+1}x^{2n+1}+c_{2n}x^{2n}+\dots +c_1x+c_0$ . We have $q^0(x_1)=q(x_1)=c_{2n+1}x_1^{2n+1}+\dots +c_0=a_0$, $q^1(x_1)=(2n+1)c_{2n+1}{x_1}^{2n}+{2n}{x_1}^{2n-1}+\dots +c_1=a_1$ etc...

Hint: It is easy to take the derivatives of the polynomial...

You get a system of $2n+2$ equations in the $2n+2$ coefficients of the polynomial...

Proceed to (try to) solve for those coefficients ...

Recall that by the fundamental theorem of algebra, a polynomial is identically zero only if the polynomial is the zero polynomial (that is, the coefficients are all zero)...

As a hint, I would use the following fact:

Let $V$ be a finite-dimensional vector space, and $\phi_1, \ldots, \phi_n \in V^*$ a collection of functionals on $V$. Then

(a) $\{ \phi_1, \ldots, \phi_n \}$ spans $V^*$ if and only if for each $x \in V$, $\phi_1(x) = \cdots = \phi_n(x) = 0$ implies $x = 0$.

(b) $\{ \phi_1, \ldots, \phi_n \}$ is linearly independent if and only if for each $a_1, \ldots, a_n \in \mathbb{R}$, there exists $x \in V$ such that $\phi_1(x) = a_1, \ldots, \phi_n(x) = a_n$.

(c) $\{ \phi_1, \ldots, \phi_n \}$ is a basis for $V^*$ if and only if for each $a_1, \ldots, a_n \in \mathbb{R}$, there exists a unique $x \in V$ such that $\phi_1(x) = a_1, \ldots, \phi_n(x) = a_n$.

In your case, the functionals $\cdot^{}(x_1), \ldots, \cdot^{[n]}(x_1), \cdot^{}(x_2), \ldots, \cdot^{[n]}(x_2)$ on $P_{2n+1}(\mathbb{R})$ have count exactly equal to the dimension of the dual space. Therefore, they form a basis of the dual space (equivalent to the desired conclusion) if and only if they span the dual space.

This reduces the original problem to proving that if $p^{}(x_1) = \cdots = p^{[n]}(x_1) = p^{}(x_2) = \cdots = p^{[n]}(x_2) = 0$ for $p \in P_{2n+1}(\mathbb{R})$, then $p=0$.

• Thanks for the answer, but it's a bit too advanced for me. – Al Jebr Jan 20 '18 at 1:36