20
$\begingroup$

Following some thoughts in real analysis and in analogy with the classical definition of the derivative $$ f^\prime(x) = \lim_{h\rightarrow 0}{f(x+h)-f(x)\over h} $$ I considered the following hyper-derivative definition: $$ f^\nabla(x) = \lim_{h\rightarrow 1}\log_h{f(x\cdot h)\over f(x)} $$ It is easily proved then that $$ (f\cdot g)^\nabla = f^\nabla + g^\nabla $$

I would like please if someone can give some expert opinion in the following:

  • Is the hyper derivative definition well defined? or somedody can see a potential problem?
  • Can someone provide a geometric interpretation for the hyper derivative definition?
$\endgroup$
11
  • $\begingroup$ Can you show the easy proof? $\endgroup$ Commented Nov 7, 2016 at 15:12
  • 3
    $\begingroup$ @gaffney It's just the facr that $$\log_h\frac{f(xh)g(xh)}{f(x)g(x)}=\log_h\frac{\lvert f(xh)\rvert }{\lvert f(x)\rvert}+\log_h\frac{\lvert g(xh)\rvert }{\lvert g(x)\rvert}$$ though I must say that this definition needs $f(x)$ not to change sign in a neighbourhood of $x$. $\endgroup$
    – user228113
    Commented Nov 7, 2016 at 15:15
  • 2
    $\begingroup$ Did the word hyper come from somewhere? $\endgroup$
    – snulty
    Commented Nov 7, 2016 at 16:09
  • 7
    $\begingroup$ This operation was studied around 1901-1902 by Robert Edouard Moritz (1868-1940), who called it quotientiation, and was rediscovered by Michael Grossman and Robert Katz in the early 1970s, who called it Non-Newtonian calculus. Also, both Moritz and Grossman/Katz considered versions for other operations as well. See the references I posted in this 24 December 2005 sci.math post. $\endgroup$ Commented Nov 7, 2016 at 19:47
  • 1
    $\begingroup$ The link to the 24 December 2005 sci.math post I gave in a previous comment (Nov 7, 2016) no longer works. However, that post is still online at the google sci.math archive. $\endgroup$ Commented Nov 13, 2022 at 7:27

3 Answers 3

31
$\begingroup$

What is $\boldsymbol{\,f^\nabla(x)}$? $$ \begin{align} f^\nabla(x) &=\lim_{h\to1}\log_h\left(\frac{f(xh)}{f(x)}\right)\\ &=\lim_{h\to1}\frac{\log(f(xh))-\log(f(x))}{\log(h)}\\ &=\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\log(1+\delta)}\\ &=\lim_{\delta\to0}\frac{\delta\,x}{\log(1+\delta)}\,\lim_{\delta\to0}\frac{\log(f(x+\delta\,x))-\log(f(x))}{\delta \,x}\\ &=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\ &=x\,\frac{f'(x)}{f(x)} \end{align} $$ which is well-defined as long as $\,f'(x)$ exists and $f(x)\ne0$.

Furthermore, if $f'(0)$ exists and $f(0)=0$, then $f^\nabla(0)=1$.


Geometric Interpretation

Extending the result above, $$ \begin{align} f^\nabla(x) &=x\,\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))\\ &=\frac{\mathrm{d}\log(f(x))}{\mathrm{d}\log(x)} \end{align} $$ which would be the slope of the graph of $f(x)$ in a log-log plot.

$\endgroup$
16
$\begingroup$

Assuming that $f$ is differentiable in the usual sense in $x$, $$ f(xh)=f(x)+(h-1)xf'(x)+o(h-1). $$ If $f(x)\neq 0$, then $ f(xh)/f(x)=1+(h-1)xf'(x)/f(x)+o(h-1)$. Moreover, $\ln h=h-1+o(h-1) $, hence $$ \log_h\frac{f(xh)}{f(x)}=\ln\frac{f(xh)}{f(x)}\cdot (\ln\,h)^{-1}=\frac{(h-1)xf'(x)/f(x)+o(h-1)}{h-1+o(h-1) }=\\ =x\frac{f'(x)}{f(x)}+o(h-1). $$ Therefore $$ f^\nabla(x)=x\cdot (\ln f)'(x). $$ This yields easily $(f\cdot g)^\nabla = f^\nabla + g^\nabla$ and provides the geometrical meaning in terms of the one of the derivative.

$\endgroup$
4
$\begingroup$

Let's say $f$ is continuous, $x>0$ and $f(x)>0$. Then, you are basically calculating $$\lim_{h\to 1} \frac{\ln f(xh)-\ln f(x)}{\ln h}=\lim_{u:=\ln h\to 0}\frac{\ln f(x\cdot e^u)-\ln f(x)}{u}=\lim_{u\to 0}\frac{\ln f(e^{u+\ln x})-\ln f(e^{\ln x})}{u}=\frac{d(\ln\circ f\circ\exp)}{dy}(\ln x)$$

So, for $x>0$ and $f(x)\ne 0$, you have that $f^\nabla(x)=g'(\ln x)$, where $g(y)=\ln \lvert f(e^y)\rvert$

Similarly, for $x<0$ and $f(x)\ne0$, then $f^\nabla (x)=g'(\ln x)$, where $g(y)=\ln\lvert f(-e^y)\rvert$.

If $f(x)=0$, then the $f^\nabla$ is not defined. And, clearly, if $f(0)\ne 0$, then $f^\nabla(0)=0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .