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So, while trying to find something else, it looks like I've found, for many $f(x)$:

$$f(x) + f'(x) + f''(x) + f^{(3)}(x) + \dots + f^{(n)}(x)$$

Assuming that there is an easy way to find this sum above, is there any use for it? I will elaborate a bit. I mean that I believe I have found a method that finds the sum of all of the derivatives above, and is much faster than calculating each derivative. In fact, it seems that calculating the sum above for most functions isn't much harder than calculating $f^{(n)}(x)$, and it also should give a "closed form" of elementary expressions for most $f(x)$.

I have one example that comes to mind: a "closed form" for a partial sum of $e^x$, as in this question. If my ideas work, we would have the closed form that this question asks for.

So I'm wondering, is there anything else that this method is useful for?

IMPORTANT NOTE

I'm assuming that we have use of the "fractional calculus", which gives us the ability to calculate $f^{(n)}$ reasonably well and efficiently, using "differintegrals". This may make the sum above fairly trivial. I'm sorry if I misled anyone.

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Matt Groff
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    Probably useful: Let $g(x)=f(x)+\dotsb+f^{(n)}$. Then $g'-g=f^{(n+1)}-f$. – Akiva Weinberger Jan 05 '16 at 16:52
  • Sounds interesting. Curious about when you say your method "isn't much harder than calculating $f^{(n)}(x)$". I'd calculate that by calculating $f'(x), f''(x), f'''(x)$, etc. which is what you're doing, so presumably your method is even faster than calculating $f^{(n)}(x)$, unless I've misunderstood. – Eli Rose Jan 05 '16 at 17:10
  • @EliRose: I apologize: I'm actually assuming that the method has $f^{(n)}$ already calculated, and I'll modify the question accordingly. I was assuming that I can use the "fractional calculus" and differintegrals to find $f^{(n)}$ reasonably well, but that of course isn't always the case. So assuming that, my method is probably trivial. :-( – Matt Groff Jan 05 '16 at 17:13
  • @AkivaWeinberger: Again, I'm assuming that we already have access to $f^{(n)}$, which could make the method trivial. I modified the question accordingly. – Matt Groff Jan 05 '16 at 17:18
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    This has some interesting interpretations via linear algebra: if you think of D as representing the operation of differentiation, you're looking at a geometric series where the ratio is "differentiation". If you take $n\to \infty$, you are computing $\frac{1}{1-D}f$, and it's worthwhile looking into what this represents. There's also some interesting variations on this - if you define $\Delta f(x):=f(x+1)-f(x)$, then in some formal sense $e^D-1=\Delta$, where $\Delta$ represents that operator. – πr8 Jan 05 '16 at 17:20
  • @πr8: I noticed that too, and I realized that I don't quite have that - I can really only get $\frac{1-D^n}{1-D}f$, but there does seem to be some leeway in the method to easily add additional operators. In other words, It seems to be easy to get: $\frac{1-(x^2\cdot D)^n}{1-(x^2\cdot D)}f$, or something else like: $\frac{1-(x^2\cdot D)^n}{1-D}f$. I'm wondering if we can get additional use out of this. Additionally, I note that we also have a similar operation for integration, for which I use $I$: $\frac{1-I^n}{1-I}f$ – Matt Groff Jan 05 '16 at 17:31
  • Yep - and $DI$ should just be the identity. $\frac{1}{1-D}$ ought to be an inverse of sorts to $1-D$ in the same way that $I$ is a sort of inverse to $D$. For a certain class of functions (the ones for which the sum converges, I think), $id+D+D^2+...$ is equivalent to a suitable integral. I suppose where you go from here depends on if you want to find a concise expression for what the series does, or use the series as an alternative to another task - perhaps carrying out integrations. – πr8 Jan 05 '16 at 17:39
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    Continuing my last comment: Multiply by $e^x$, we get $(ge^x)'=f^{(n+1)}e^x-fe^x$, so:$$g(x)=e^{-x}\int\left(f^{(n+1)}(x);e^x-f(x)e^x\right)dx$$ – Akiva Weinberger Jan 05 '16 at 17:39
  • One clear use of this type of thinking can be seen by looking at the Taylor series for $f(x+h)$, which (for good reason), look a lot like $e^{hD}f(x)$ – πr8 Jan 05 '16 at 17:41
  • @AvikaWeinberger: Thank you for your thoughts! I'm going to be playing around with all of these ideas, to see where I can take this. It looks like the method will allow for any function, which could be important. Amazing! – Matt Groff Jan 05 '16 at 17:54

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Something similar is used in Ivan Niven's proof of the irrationality of $\pi$.

Michael Hardy
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