0
$\begingroup$

There is linear transformation $S,T : \mathbb{R}_n[X] \to\mathbb{R}_n[X]$ . $T: T( p(x))=p(2x)$. $S:S(p(x)) = p'(x)$?

Where $\mathbb{R}_n[X]$ denotes polynomials with power up to $n$.

prove that $ST = 2TS$.

how can one prove this equality and does it exist ? tried using standard polynomial equation of $1 +x+x^2$ did not work. any help would be appreciated.

$\endgroup$
  • 2
    $\begingroup$ It would be a good idea to have your question be self-contained in the body of your question and your title be descriptive. Placing half the question (only) in the title is poor practice. $\endgroup$ – Eric Towers Mar 27 '16 at 0:35
  • $\begingroup$ Does your "$R$" mean "$\Bbb{R}$" ($\Bbb{R}$)? What is "$R[x]n$"? (Maybe you mean $(\Bbb{R}[x])^n$, the set of $n$-element vectors with elements from the polynomials on $x$ with real coefficients? If so, do you really mean to take component-wise vector derivatives?) $\endgroup$ – Eric Towers Mar 27 '16 at 0:40
  • $\begingroup$ sorry if it wasnt clear, but by R[x]n i meant polynomials with power up to n. $\endgroup$ – Ancient Dragon Mar 27 '16 at 0:43
1
$\begingroup$

You have, for $p(x)=\sum_{k=0}^na_kx^k$, $$ STp(x)=S\left(\sum_{k=0}^na_k(2x)^k\right)=S\left(\sum_{k=0}^n2^ka_kx^k\right) =\sum_{k=1}^nk\,2^ka_kx^{k-1}. $$ And $$ TSp(x)=T\left(\sum_{k=1}ka_kx^{k-1}\right)=\sum_{k=1}^nka_k(2x)^{k-1}=\sum_{k=1}^nk\,2^{k-1}a_kx^{k-1}. $$ The two expressions differ by a factor of $2$, and $p$ was arbitrary. So $$ ST=2TS. $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This is more targetted at OP: In what way is this $p(x)$ an element of "$R[x]n$"? $\endgroup$ – Eric Towers Mar 27 '16 at 0:42
  • $\begingroup$ @EricTowers Target missed. :-( $\endgroup$ – Did May 15 '16 at 11:57
1
$\begingroup$

This is a really cute problem! When you think about it some more it becomes clear that the seemingly curious fact $ST=2TS$ is more or less obvious, and it should be possible to provide an abstract nonsense proof. Here it is:

Write $p=\bigl(x\mapsto p(x)\bigr)$. Then $Tp=\bigl(x\mapsto p(2x)\bigr)$, and the chain rule then gives $$STp=\bigl(x\mapsto p(2x)\bigr)'=\bigl(x\mapsto 2p'(2x)\bigr)\ .\tag{1}$$ On the other hand, $Sp=\bigl(x\mapsto p'(x)\bigr)$, so that $$TSp=\bigl(x\mapsto Sp(2x)\bigr)=\bigl(x\mapsto p'(2x)\bigr)\ .\tag{2}$$ Since $(1)$ and $(2)$ hold for all differentiable $p$ ($p$ doesn't have to be a polynomial) it follows that $ST=2TS$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.