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When differentiating we usually take a limit and drop the infinitesimal terms.

But what if not to drop anything?

First, we extend the real numbers with an infinitesimal element $\varepsilon$ which has its own inverse $1/\varepsilon=\omega$.

And define the full derivative of a function formally as follows:

$$D_{full}[f(x)]=\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$

Now we can compute full derivatives of polynomials in closed form:

$$D_{full}[a]=0$$ $$D_{full}[ax]=a$$ $$D_{full}[x^2]=2x+\varepsilon$$ $$D_{full}[x^3]=3 x^2+3 \varepsilon x+\varepsilon ^2$$

etc.

We also can find a function that remains invariant against full differentiation. It is not exponent with base $e$ though. To find it we solve the equation:

$$\frac{f(x+\varepsilon)-f(x)}{\varepsilon}=f(x)$$

The solution is a set of functions

$$C (\varepsilon +1)^{\frac{x}{\varepsilon }}$$

of which the most simple is

$$(\varepsilon +1)^{\frac{x}{\varepsilon }}$$

We can call it "full exponent" and re-define trigonometric and inverse trigonometric functions accordingly. For instance, full logarithm, sine and cosine become

$$\operatorname{flog}\,\,x=\frac{\varepsilon \ln(x)}{\ln(\varepsilon + 1)}$$

$$\operatorname{fsin}\,\,x=\frac{ (1+i\varepsilon)^{x/\varepsilon }-(1-i\varepsilon )^{x/\varepsilon }}{2i}$$

$$\operatorname{fcos}\,\,x=\frac{ (1+i\varepsilon)^{x/\varepsilon }+(1-i\varepsilon )^{x/\varepsilon }}{2}$$

etc (these full sine and full cosine satisfy the equation $f''=-f$ with full derivative).

The same expressions for differentiation occurs in time scale calculus with a scale parameter. I wonder whether anybody ever considered such operation of "full differentiation" either in the framework of non-standard analysis or time scales or otherwise and whether it has any established name?

Note that we can also in a similar way define its inverse operator, "full integral" that would be

$$\int_{full} f(x)dx=\varepsilon \lim_{t\to x/\varepsilon} \sum_t f(\varepsilon t)$$

where $\sum_t$ is indefinite sum.

Thus we get

$$\int_{full} a \,dx=ax$$

$$\int_{full} x \,dx=\frac{x^2}{2}-\frac{\varepsilon x}{2}$$

$$\int_{full} x^2 \,dx=\frac{x^3}{3}-\frac{\varepsilon x^2}{2}+\frac{\varepsilon ^2 x}{6}$$

$$\int_{full} a^x \,dx=\frac{\varepsilon a^x}{a^{\varepsilon }-1}$$

$$\int_{full} \sin x \,dx=-\frac{1}{2} \varepsilon \sin (x)-\frac{1}{2} \varepsilon \cot \left(\frac{\varepsilon }{2}\right) \cos (x)$$

etc.

Note also that we can define full derivative in a more symmetric way:

$$D_{sym}[f(x)]=\frac{f(x+\varepsilon)-f(x-\varepsilon)}{2\varepsilon}$$

With this definition some formulas become simplier:

$$D_{sym}[e^x]=\frac{e^x \sinh (\varepsilon )}{\varepsilon }$$

$$D_{sym}[\sin x]=\frac{\sin (\varepsilon ) \cos (x)}{\varepsilon }$$

$$D_{sym}[1/x]=\frac{1}{\varepsilon ^2-x^2}$$

The invariant function for this operation, playing the role of exponent will be

$$f(x)=\left(\sqrt{\varepsilon ^2+1}+\varepsilon \right)^{x/\varepsilon }$$

Anixx
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    See https://en.wikipedia.org/wiki/Q-derivative – kjetil b halvorsen Mar 25 '15 at 10:20
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    In non-standard (hyperreal) analysis, the derivative is defined as the standard part of your "full derivative". So in a sense, yes, it has been considered, but more interest seems to be on the "real part". It doesn't have a name, so far as I'm aware (judging from Keisler's book). – Hayden Mar 25 '15 at 10:23
  • @kjetil b halvorsen this is not q-derivative, totally different thing. – Anixx Mar 25 '15 at 10:27
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    It works perfectly well if you don't drop the infinitesimals - but you just end up studying a curve approximated by a series of connected straight lines. It's easiest to do this with smooth infinitesimal analysis. –  Mar 25 '15 at 10:35
  • @Hayden in usual non-standard analysis they usually do not introduce distinguished elements $\varepsilon$ and $\omega$, thus they cannot uniquely define full derivative. – Anixx Mar 25 '15 at 11:01
  • @Anixx This is true; through taking the standard part one removes any dependencies on the choice of $\epsilon$. You could think of your process as giving a family of functions, indexed by the set of infinitesimals. Then the ambiguity no longer appears (of course, I'm not sure if anyone actually does this in practice). – Hayden Mar 25 '15 at 12:58
  • @Hayden I am about introducing one element, $\varepsilon$, similarly to how complex $i$ introduced. One can argue such system would be undefinable, but then complex numbers are also undefinable because $i$ is indistinguishable from $-i$. By the way, in the context of surreals, hyperreal numbers can be considered a subfield $No(\omega)$. Then this $\omega$ has definite meaning: it is considered equal to the first infinite ordinal. – Anixx Mar 25 '15 at 13:02
  • @Anixx I understand, but I just mean it is natural to extend your definition to look at the set of all such "full derivatives". And yes, sadly I don't think one could expect an isomorphism interchanging any two infinitesimals. – Hayden Mar 25 '15 at 13:15
  • http://math.stackexchange.com/questions/822664/could-we-assign-a-numerical-value-to-an-infinitesimal –  Mar 25 '15 at 13:49
  • A side comment that has no bearing on this question: @Anixx, you can take $No(\omega)$, but as I said in answer to http://math.stackexchange.com/questions/1193422/why-hyperreal-numbers-are-built-so-complicatedly/ that won't give you the hyperreals. – Mark S. Mar 25 '15 at 18:38
  • @Mark S. $No(\omega)$ is a hyperreal system, subfield of surreals http://ohio.edu/people/ehrlich/Unification.pdf – Anixx Mar 25 '15 at 18:41
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    I think you misunderstand the notation in that PDF. $No(\omega)$ is actually the dyadic rationals thanks to the tree rank (see Theorem 15). $ No(\omega_1) $ is a hyperreal system assuming CH, but that has a lot of different flavors of infinitesimals, not just what you can get with reals and $\omega $. – Mark S. Mar 25 '15 at 19:07
  • @Mark S. so basically if to add to rationals $\varepsilon$ as in this post we get Levi-Civita field? – Anixx Mar 25 '15 at 19:15
  • @Anixx you're working with the minimal hyperreal field $\mathbb{R}(\epsilon)$, in which you can indeed uniquely define this "full derivative." – Kevin Carlson Mar 25 '15 at 20:57
  • @Kevin Carlson is this field truly hyperreal? – Anixx Mar 25 '15 at 20:58
  • @Anixx You're right, I shouldn't call it hyperreal, which means your approach is also not identical to that of nonstandard analysis. – Kevin Carlson Mar 25 '15 at 23:02
  • @Anixx If you add an infinitesimal to rationals, you get formal Laurent series with rational coefficients. The Levi-Civita field is different in that it allows real coefficients and rational powers of $\epsilon$. But to define your full derivative in cases that aren't polynomials, you need a method for defining things like $sin(\epsilon)$. Hyperreal fields make this work perfectly, but http://www.physics.umanitoba.ca/~khodr/Publications/RS-Overview-offprints.pdf suggests that the Levi-Civita field is probably good enough, at least for analytic functions (I haven't thought about it much). – Mark S. Mar 26 '15 at 01:07
  • @Mark S. what a problem in defining $\sin \varepsilon$? It just can be represented as a series or in closed form... Where the problem is? – Anixx Mar 26 '15 at 09:16
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    Interesting how, with the symmetric variant, your version of $e$ is $(\sqrt{\varepsilon^2+1}+\varepsilon)^{1/\varepsilon}=e-\dfrac{e\varepsilon^2}6 +\dfrac{4e\varepsilon^4}{45}-\dotsb$ – Akiva Weinberger Mar 26 '15 at 20:04
  • @columbus8myhw I think e can be expressed in closed form by modifying the formula. – Anixx Apr 01 '15 at 13:18
  • @Anixx Is that not "closed form"? $(\sqrt{\varepsilon^2+1}+\varepsilon)^{1/\varepsilon}$? (Also, weirdly, that's an even function of $\varepsilon$. I didn't expect that.) – Akiva Weinberger Apr 01 '15 at 13:19
  • @columbus8myhw it is not equal to e. – Anixx Apr 01 '15 at 13:41
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    @Anixx $\displaystyle\lim_{\varepsilon\to0} (\sqrt{\varepsilon^2+1} +\varepsilon)^{1/\varepsilon}=e$ – Akiva Weinberger Apr 01 '15 at 13:42
  • @columbus8myhw this is not closed form... – Anixx Apr 01 '15 at 14:05

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As I understand it, this is just the same as h-calculus. The h-derivative is defined as,

$$ D_{h} = \dfrac{f(x+h) - f(x)}{h} $$

, where $h\ne 0$. [1] has a small chapter on it.

[1] Kac, V., & Cheung, P. (2002). Quantum calculus. Springer Science & Business Media.

Eliad
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  • Does it mean time scales? – Anixx Mar 26 '15 at 19:42
  • h-Calculus is a "subset" of Quantum calculus, which is in turn an example of time-scale calculus. This might also be helpful: ČERMÁK, J., & NECHVÁTAL, L. (2010). On (q, h)-analogue of fractional calculus. Journal of Nonlinear Mathematical Physics, 17(1), 51-68. – Eliad Mar 27 '15 at 09:38