Let $V=P(\Bbb R)$ and $1 ≤ i$ be the vector space of the polynomials with real coefficients, on the field of real numbers $\Bbb R$. Let $T_i(f)=f^{(i)}$ the $i$th derivate of $f$.

a) I have to show that for any $n \in \Bbb N$, $\{T_1, T_2,..., T_n\}$ is a linearly independent subset of $L(V)$ (the set of oprators of $V$.)

b) If $f\in V$ of degree $n$ then for every $g \in P_n(\Bbb R)$ there exist scalars $c_0,c_1,..,c_n$ such that $g = c_0f + c_1f'+ ... + c_nf^{(n)}$.

For a) I tried to consider a specific polynomials, $x^n$ for various $n$. Thenthat $x' = 1$ and $x'' = 0$, so

$$(c_1 T_1 + ... + c_n T_n) x = c_1 \cdot 1 + 0 + ... + 0 = 0$$

implies that $c_1 = 0$.


$$(c_1 T_1 + c_2 T_2 + ... + c_n T_n) x^2 = 0 + c_2 (x^2)'' + c_3 (x^2)''' + ... + c_n (x^2)^{(n)} = 2c_2 = 0$$

then $c_2 = 0$.but i can't prove it general o don't know how to prove this more directly.

Can you help me to solve this both parts, i really will appreciate it, please.


For a), assume that $\sum_{j=1}^nc_jT_j=0$. Using this with the polynomial $x^n$ and noticing that $T_i(X^n)=n!/(n-i)!\cdot X^{n-i}$, we obtain $$\sum_{i=1}^n\frac 1{(n-i)!}c_iX^{n-i}=0$$ hence $c_i=0$ for each $i\in \{1,\dots,n\}$.

For b), we can prove the result by induction on $n$. The statement $\mathcal P(n)$ is: "if $F$ is a polynomial of degree $n$, then $P(n)$ is generated by $F^{(i)}$, $⁰\leqslant i\leqslant n$."

When $n=1$ it is clear. If it's true for $n$, assume that $f$ is of degree $n+1$ and let $g$ be a polynomial of degree $n+1$. If $a_{n+1}$ is the leading coefficient of $f$ and $b_{n+1}$ that of $g$, then $G:=g-\frac{b_{n+1}}{a_{n+1}}f$ is a polynomial of degree at most $n$. Since $F:=f'$ is of degree $n$, using $\mathcal P(n)$ we may express $G$ as a linear combination of $f^{(i)}$, $1\leqslant i\leqslant n+1$.

  • $\begingroup$ For a) is it correct this? : Suppose there exist $c_1,...,c_n \in \Bbb R$ such that $$c_1T_1 + \cdots c_n T_n = 0.$$ Then, for any $f \in P(\Bbb R)$, we have $$(c_1T_1 + \cdots c_n T_n)(f)=c_1 T_1 (f) + \cdots c_n T_n (f) = 0.$$ We must show that $c_1 = c_2 = \cdots = c_n =0.$ Any $f \in P(\Bbb R)$ is of the form $$f=a_m x^m + \cdots + a_1 x + a_0.$$ for some $m \in \Bbb N$. Also, we have that $$f^{(k)}=\sum_{i=k}^m \bigg( \prod_{j=0}^{k-1}(i-j) \bigg)a_i x^{i-k}.$$ $\endgroup$ – Knight Jun 27 '14 at 8:55
  • $\begingroup$ If $m \geq n$, then $T_n(f)=f^{(n)}$ is a linear combination of $1, x, x^2, ..., x^{m-n}$, which is a linearly independent set, and then their coefficients are all zero. Then, we substitute this coefficients in $f^{(n-1)}$, which is a linear combination of $1, x, x^2, ..., x^{m-n}, x^{m-n+1}$ and we obtain that the coefficient of $x^{m-n+1}$ is zero. We continue with this process and we obtain $c_1=c_2=\cdots =c_n = 0$. $\endgroup$ – Knight Jun 27 '14 at 8:55

Hint: continue doing the same. $$T_k x^m=\cdots$$


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