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In my symmetries of classical mechanics course we have looked at taylor expansions. Our notes claim that; $$ f(x + a) = \sum_{n=0}^\infty \frac{1}{n!} f^{(n)}(x)a^n ≡ \exp{\left( a \frac{d}{dx}\right)} f(x) $$ I am happy with the first equality, its explained quite nicely in this question. The second equality is puzzling though. I dont even know what to make of the derivative inside the exponent.

How is the second equality derived?

Community
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Jekowl
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1 Answers1

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It applies the exponential function $\exp(y) = \sum \frac{1}{n!}y^n$ to the differential operator $a\frac{d}{dx}$ to give $$\exp\left(a\frac{d}{dx}\right) = \sum \frac{1}{n!}a^n \frac{d^n}{dx^n}.$$ This, when applied to $f(x)$, gives your middle expression.

Henry
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    My question might seem stupid, but what is the meaning of $\exp(D)$ ? It is natural to say that : $\displaystyle \exp(D) ; \mathop{=} \limits^{\mathrm{def}} ; \sum_{n=0}^{+\infty} \frac{D^{n}}{n!}$ but I'm wondering in which space this series is converging, for which topology. – pitchounet Apr 22 '15 at 12:30
  • I second jibounet's question! – GPerez Apr 22 '15 at 12:37