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My consideration might be total nonsense (as a high school student, I lack the mathematical knowledge to really check my idea), but I was just wondering whether one could find a continuous generalization of the Taylor series comparable to the continuous Fourier transform.

Just like the Fourier transform decomposes a function $f(t)$ into a sum of trigonometric functions with a result $\hat{f}(\omega)=\mathcal{F}(f)(\omega)$ with the domain of frequencies, could one define a Taylor transform $\mathcal{T}(f)$ that would operate on $n$th derivatives?

Instead of the trigonometric functions in Fourier transform, $f$ would be decomposed into a sum of polynomials of form $\frac{1}{n!}x^n$.

So is it a valid operation to generalize the Taylor series for continuous values of $n$ (i.e. using integrals instead of sums)? My naive approach would be

$$f(t) = \intop_0^{\infty} \mathcal{T}(f)(n)\cdot \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$

So in fact, the transform would define (or at least require) non-integer derivatives of a function. Do these exist (like a non-integer iterates of functions can be defined too)?

Example: Assuming that $\frac{\mathrm{d}^ne^x}{\mathrm{d}x^n}(0) = 1$ for any $x\in \mathbb{R}$, I'd come up with

$$e^t = \intop_0^{\infty} \frac{t^n}{\Gamma(n+1)} \mathrm{d}n$$ as a continuous version of

$$e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!}$$

Thus $$\mathcal{T}(e^t)(n) = 1,\,n\in\mathbb{R}$$

Is a transform like I explained possible or - in case it is - even somehow useful?

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    You might want to look into using different notation for that last infinite integral, because that's definitely not the exponential function (it's real only for nonnegative $t$).2010-11-21
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    http://en.wikipedia.org/wiki/Fractional_calculus2010-11-21
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    Ah, thanks. Fractional calculus seems to answer most of the question. Yet, would a transform like the one I asked about be useful in fractional calculus?2010-11-21
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    I like this question because it covers some of the same ground as my previous question. That having been said, I don't think the machinery of fractional derivatives is necessarily required here, because frequently one can come up with continuous generalizations of the Taylor (or MacLaurin) series coefficients -- I gave the example of Catalan numbers in my question -- the Fibonacci and Lucas numbers too have closed form expressions.2010-11-21
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    @deoxygerbe: Ah interesting, the questions fit together really well. But - in the context of my question - couldn't your generalized continuous coefficients be interpreted as fractional derivatives? You'd just have gone the other way round.2010-11-21
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    I've tried poking around on wikipedia's transforms category and I can't find anything appropriate. I think that in the absence of someone chirping up and saying "yes, I know what that is", the best strategy would be to start computing this transform for various functions.2010-11-22

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