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Recently, I have been looking at operational calculus, integral transforms and so on. If you check my profile, you will see lots of questions related to these topics and no satisfatory answer or good book recommendations that explained the thing to me. So I have browsed some questions on StackExchange and this one caught my attention: Laplace transform for dummies.

Despite I got every single step (algebraically/computationaly speaking), one thing still remains unexplained to me: What is the meaning of $e^{-\tau \frac{d}{dx}}$? More specifically, what should mean $e^{\frac{d}{dx}}$? Is this the so called operator?

Appreciate any book recommendation.

Thanks

Mr. N
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You might know that $$ e^t = \sum_{k=0}^{\infty} \frac{1}{k!} t^k = 1 + t + \frac12 t^2 + \frac16 t^3 + \cdots $$

With $e^{\frac{d}{dx}}$ one means the operator such that $$ (e^{\frac{d}{dx}}f)(a) = \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^k f}{dx^k}(a) = f(a) + f'(a) + \frac12 f''(a) + \frac16 f'''(a) + \cdots $$

md2perpe
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  • However, it does only make sense when you apply it to $f$, doesn't it? How can we convert it into na integral transform, such as Laplace in the example?Also, is there a way of "creating" other operators? – Mr. N Feb 12 '20 at 21:56
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    @Mr.N. Do you also think that a function $f$ only makes sense when it is applied to some argument $a$, or does a function make sense on its own? – md2perpe Feb 13 '20 at 18:57
  • Hmm I see, it makes sense on its own. Thanks for this comparation. – Mr. N Feb 13 '20 at 19:00