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having trouble completing the proof for this question

Let $D:\mathbb{R}[X] \to \mathbb{R}[X]$ be the differentiation operator $D(f(X))=f'(X) .$ Prove that $e^{tD}(f(X)) = f(X+t)$ for $t \in \mathbb{R}$

Im having trouble making sense of the question. At first i tried Taylor's theorem to try and make sense of, and equate the two sides of the equation. This approach hasn't really worked. But could i go a more algebraic route using the fact that there exists a matrix D that represents this operator, and we know that $X^n$ spans D.

Seirios
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user65972
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  • Just to clarify: you don't mean $\mathbb{R}_{\leq n}[x]$, i.e. polynomials with bounded degrees, right? Then I'd suggest adding the [functional-analysis] tag, and mentioning the norm you use on $\mathbb{R}[x]$. If it's complete in this norm, so is the space of bounded linear operators on it, and we can therefore define the exponential of an operator using the Taylor series. – Jonathan Y. Oct 24 '13 at 16:34
  • Why didn't it work? Isn't the equation exactly Taylor's theorem? – user7530 Oct 24 '13 at 16:37
  • My thoughts exactly but my Taylor Series isn't up to scratch – user65972 Oct 24 '13 at 16:43
  • It should work just fine; but without the preliminaries I asked for, how do you know that the operator $e^{tD}$ exists? – Jonathan Y. Oct 24 '13 at 16:44
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    Well, since you are working on $\mathbb{R}[X]$, only a finite number of the $D$'s in $e^{tD}$ are relevant, so there is no need to topologise here. Think of $e^{tD}$ as a notational convenience. – copper.hat Oct 24 '13 at 16:46
  • @copper.hat you've got me, sir! – Jonathan Y. Oct 24 '13 at 16:54

2 Answers2

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You are right in the sense that it is the Taylor Theorem. To be more accurate and to prove it :

For every polynomial $P(X)$ of degree n : $$P(X+t)=\sum_{k=0}^n \frac{P^{(k)}(X)}{k!} t^k=\left(\sum_{k=0}^n \frac{D^k}{k!} t^k\right) P(X)$$

This precisely means that the polyomial in X $P(X+t)$ is equal to the polynomial (in X still !) on the right of the previous expression.

By definition $e^{tD}=\sum_{k=0}^\infty \frac{D^k}{k!} t^k$. Because $D^{n+1}P=0$ for P of degree less than n, the exponential is reduced to the previous expression which shows that for any $P$ of degree n

$$P(X+t)=e^{tD}P(X)$$

(Once again, be careful about the fact that these two quantities should be considered as polynomial in X)

Bertrand R
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Noting (thanks go to @copper.hat) that the Taylor series of $e^{tD}$ is a sum on polynomials (as mentioned in the comments above), for every polynomial $p(X)=\sum_{k=0}^m a_kX^k$ one has $$e^{tD}p(X) = \sum_{n=0}^\infty \frac{1}{n!}t^nD^n\left(\sum_{k=0}^m a_kX^k\right) = \sum_{n=0}^m \frac{1}{n!}t^nD^n\left(\sum_{k=0}^m a_kX^k\right) = \sum_{k=0}^m a_k\left(\sum_{n=0}^m \frac{1}{n!}t^nD^nX^k\right) = \sum_{k=0}^m a_k\left(\sum_{n=0}^k \frac{1}{n!}t^nD^nX^k\right)$$

Can you complete the proof?

Jonathan Y.
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