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Let $V$ = $P_{n}(\Bbb{R})$ be a vector space of polynomials with real coefficients up to degree $n$.

Let $W = \{ p(x)\in V\mid p(a) = p'(a) = p''(a)=\ldots=p^{(r)}(a) = 0 \}$

What is the dimension of $W$?

I can notice that if $p(x)$ belongs to $W$ then $x-a$ will be a factor of each of $p(x),p'(x),p''(x),\ldots, p^{(r)}(x)$ but still I am unable to explicitly write this polynomial to find the dimension of the subspace.

Please help.

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    $\begingroup$ Indeed, we have the stronger statement that $(x-a)^r$ will be a factor of $p(x)$ ... does that help? $\endgroup$
    – Greg Martin
    Jun 13, 2020 at 20:29
  • $\begingroup$ (n+1)-(r+1)?..isn't it? $\endgroup$
    – Gitika
    Jun 13, 2020 at 20:37

3 Answers 3

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The space $V$ consists of those polynomials $p(x)$ of the form $(x-a)^{r+1}q(x)$, with $\deg q(x)\leqslant n-(r+1)$ (I am assuming that $r<n$). Therefore, $\dim V=n-r$.

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  • $\begingroup$ How do we get that factor? $\endgroup$
    – Gitika
    Jun 13, 2020 at 20:33
  • $\begingroup$ @Bernard Right! I've edited my answer. Thank you. $\endgroup$
    – José Carlos Santos
    Jun 13, 2020 at 20:33
  • $\begingroup$ @Gitika: With Taylor's formula at order $n$, which is an exact formula for polynomials. $\endgroup$
    – Bernard
    Jun 13, 2020 at 20:37
  • $\begingroup$ @Gitika Write your $p(x)$ as a polynomial in $x-a$:$$p(x)=a_0+a_1(x-a)+a_2(x-a)^2+\cdots+a_n(x-a)^n.$$Then $p(a)=0\iff a_0=0$, $p'(a)=0\iff a_1=0$ and so on. So,\begin{align}p(x)&=a_{r+1}(x-a)^{r+1}+\cdots+a_n(x-a)^n\\&=(x-a)^{r+1}(a_{r+1}+\cdots+a_n(x-a)^{n-r}).\end{align} $\endgroup$
    – José Carlos Santos
    Jun 13, 2020 at 20:38
  • $\begingroup$ Why is p(x) of that form? $\endgroup$
    – Gitika
    Jun 13, 2020 at 20:43
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hint

Assuming $r<n$, by Taylor-Lagrange formula,

$$p(x)=\sum_{k=0}^n\frac{(x-a)^k}{k!}p^{(k)}(a)$$

$$=\sum_{k=r+1}^n\frac{(x-a)^k}{k!}p^{(k)}(a)$$

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Here is an ad hoc way:

Let $b_k(x) = (x-a)^k$, $k=r+1,...,n$, note that the $b_k$ are linearly independent.

Note that $b_k \in W $ for $k=r+1,...,n$ hence $\dim W \ge n-r$.

Now suppose $p \in W$. Using synthetic division is it straightforward to show that $p(x) = (x-r)^{r+1} g(x)$. Since $p \in P_n(\mathbb{R})$ we see that $\partial g \le n-r-1$ and we can write $g(x) = g_0+g_1(x-r)+\cdots + g_{n-r-1}(x-a)^{n-r-1}$ and so $p = g_0 b_{r+1}+g_1 b_{r+2} + \cdots +g_{n-r-1}b_n$. Hence $W = \operatorname{sp} \{ b_{r+1}, \cdots , b_n \}$ and so $\dim W = n-r$.

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