# Algebraic transformations to continuously extend functions

Lately I was browsing through my analysis lecture notes (since right know I'm somewhat rusty in analysis) and the proof that $x \mapsto \frac{1}{x}$ is differentiable at every $x'\neq 0$ captured my attention. The easy proof is based on the following algebraic manipulation $$\frac{\frac{1}{x+h}-\frac{1}{x}}{h}=\frac{1}{h}\frac{-h}{x^{2}+xh}=\frac{-1}{x^{2}+xh}.$$ Letting $h$ tend to zero, we get the limit that we sought.

What aroused my interest was the idea that we can (continuously) extend a (continuous) function to a larger domain by simple algebraic manipulations (since in essence this is what we do, when calculating a derivative -- continuously extending a function): The second equality from above is key: By dividing by $h$ the domain changes from $\mathbb{R}\setminus \{0\}$ to $\mathbb{R}$ (if we consider the functions $h\mapsto \frac{1}{h}\frac{-h}{x^{2}+xh}$ and $h\mapsto \frac{-1}{x^{2}+xh}$, for $x\in \mathbb{R}\setminus \{0\}$).

Questions:
1. Is there some general theory of "algebraic transformations" that allow the extension of functions (even if it extends the functions only for one single point, as for in the case of calculating derivatives) ?
(I don't know what algebraic geometry deals with, but to the uneducated ear it sounds like this would be it)
2. What other kinds of purely algebraic tricks, like above, do you know, that allow you (in the context of finding derivatives) to continuously extend functions $h\mapsto \frac{f(x+h)-f(x)}{h}$ ?
[From the examples of my notes (and 2 books I browsed through) the only tricks seems to be:
$\quad$- Manipulating the numerator long enough, until you can factor an $h$ out, so that you can write $\frac{f(x+h)-f(x)}{h}=\frac{h}{h}\cdot s_x(h)$ for some function $s_x$, since the functions $h\mapsto \frac{h}{h}$ is preventing you from continuously extending $h\mapsto \frac{f(x+h)-f(x)}{h}$ from $\mathbb{R}\setminus \{0\}$ to $\mathbb{R}$ (and the function $h\mapsto \frac{h}{h}$ is trival to continuously extend to $\mathbb{R}$).
$\quad$- Writing $f$ as the product of $f=p_1 \cdot p_2$ and then adding and substracting again $p_1(x)\cdot p_2 (x+h)$ as in the proof that $(p_1 \cdot p_2)'=p_1' p_2 + p_1 p_2'$.

(Estimating $\frac{f(x+h)-f(x)}{h}$ etc. - anything that does not involve pure algebraic transformations to extend it - I'm not interested in. Also, very simple algebraic manipulations, like rearranging terms as in the proof that $(\frac{1}{f})'=\frac{-f}{f^2}$, don't count.]

• I don't know what you mean by "purely algebraic". Would you count transforming $1+x+x^2+\dots$, which is only defined for $-1\lt x\lt1$, into $1/(1-x)$, which is defined for all $x\ne1$, as "purely algebraic"? – Gerry Myerson May 12 '13 at 6:49
• @GerryMyerson No, Taylor expansions don't count, since we use analytic tools to transform it into $\frac{1}{1-x}$. A metaphoric explanation of "purely algebraic" would be: These transformation have to have a "finite character"; since $1+x+x^2+\cdots$ already has in "infinity" in it, I don't see how we could manipulate these functions purely algebraically, as long as they are in this form. The purely algebraic treatment, as far as I can see, could only begin with $\frac{1}{1-x}$. – temo May 12 '13 at 7:45
• With "purely algebraic" I also mean, besides manipulating the "internals" of the function, as I did with the $$D(h):=\frac{\frac{1}{x+h}-\frac{1}{x}}{h},$$applying some algebraic transformation to the whole function $D$ (like rotating it, skewing it etc. that somehow continuously extends my functions. (Yes, I'm aware that this is only intuition/hope, but perhaps there is a theory that enables this) – temo May 12 '13 at 7:52
• Come on guys, a week has passed and although I put up a bounty, all I got was a single (!) comment ? There's not much time for me left now to award the bounty... – temo May 19 '13 at 10:26
• I've (temporarily) added the tag "algebraic geometry" in hope that a geometer may stumble upon this question and may tell me, if question 1. makes sense (this shall of course not prevent him from giving his two cents concerning the second question). – temo May 19 '13 at 14:54

In the context of finding derivatives (Question 2), the only problem with $$f(x+h)-f(x)\over h$$ is that at $h=0$ it gives you $0/0$. And the only way to fix that is to find an $h$ in the numerator to cancel that $h$ in the denominator; or, to finesse the problem by doing manipulations to bring it to a form already handled, as in the proof of the product rule.
As an example of what you may have to do to get cancelation of $h$, if you have $$\sqrt{x+h}-\sqrt x\over h$$ you have to multiply top and bottom by $\sqrt{x+h}+\sqrt x$.
• Removable singularity in complex analysis generally refers to something like $(1/z)\sin z$ where you remove the singularity at zero by taking a limit, but I don't think you'd call it an algebraic technique. – Gerry Myerson May 20 '13 at 10:31