Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A capital delta ($\Delta$) is commonly used to indicate a difference (especially an incremental difference). For example, $\Delta x = x_1 - x_0$

My question is: is there an analogue of this notation for ratios?

In other words, what's the best symbol to use for $[?]$ in $[?]x = \dfrac{x_1}{x_0}$?

share|cite|improve this question
What you have there is $1+\dfrac{\Delta x}{x_0}$, i.e. one more than the proportionate change, which tends to $1$ as the change gets smaller. I suspect that in most cases you may actually want the proportionate change itself. – Henry Jun 11 '11 at 13:58
@Henry if you'd like some context, I'm using this to express changes in a product in terms of changes of its individual terms. The reason I'm using ratios is that the product of ratios is the ratio of the products. – trutheality Jun 11 '11 at 14:08
up vote 7 down vote accepted

Not entirely standard, but in Peter Henrici's discussion of the (justly famous) quotient-difference (QD) algorithm in the books Elements of Numerical Analysis (see p. 163) and Essentials of Numerical Analysis (see p. 155), he defines the quotient operator as


in complete analogy with the (forward) difference operator $\Delta$.

Henrici's a pretty sharp mathematician, so I wouldn't mind borrowing notation from him if I were in your shoes...

Here's a screenshot of the relevant page of the first book (sorry, I don't have a digital copy of the other book):


share|cite|improve this answer

The best symbol to use is $\exp\Delta\log$: $$\exp\Delta\log x = \exp(\log x_1-\log x_0) = \frac{x_1}{x_0}.$$ The point is that this operation isn't "qualitatively different" from $\Delta$, so it may be reduced to $\Delta$. So far, I haven't used any new symbols but if you want some multiplicative new creative symbols, see e-percentages and units of evidence:

There is no compact symbol for $\exp\Delta\log$. If you want an ally who would endorse the idea to introduce such a symbol, you may count on me. What about $\Delta^\times$?

share|cite|improve this answer
What if $x_1$ and $x_0$ are negative? Also, $\Delta^{\times}$ doesn't seem particularly good, as it mixes the D in difference with multiplication. – t.b. Jun 11 '11 at 15:06
Dear Theo, $\log(-1)=\pi i$ and $\exp(\pi i)=-1$: is there any problem with that? Something's changing multiplicatively but changing sign would be a discontinuous process in the real numbers, anyway, so it only makes good sense in the complex realm. Concerning the notation, not sure whether I understand which $D$ you mean. – Luboš Motl Jun 11 '11 at 18:01
Dear Luboš, 1) no, no problem with that, but it might be worth pointing out. The process is not really discontinuous, but simply undefined at $0$, as it should be. 2) The $\Delta$ is a Greek D(elta) for the D in difference. – t.b. Jun 11 '11 at 18:08
Dear Theo, what I mean by "discontinuous" is that there is no continuous $f(x)$ such that $f(0)$ is positive and $f(1)$ is negative so that $f(k)/f(0)$ which is $\Delta^\times x$ at some moment would be well-defined for all $k$ between $0$ and $1$. That's a warning sign - if one uses $\Delta^\times$ for things that change sign, it could be an unnatural thing that can go awry at moment... I know that $\Delta$ is the Greek counterpart of $D$ but I think it's a good idea to distinguish them. Did your $D$ mean $\Delta$? – Luboš Motl Jun 12 '11 at 4:02
Thank you for the input, I was hoping there was something more compact than $\exp \Delta \log$ and while $\Delta^{\times}$ has some appeal it may confuse some readers because of the association of $\Delta$ with differences. I think I might just end up switching all the expressions to log-space or coming up with some other symbol (perhaps an unambiguous placement of $\div$?) – trutheality Jun 12 '11 at 23:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.