Symbol for “probably equal to” (barring pathology)?

I am writing lecture notes for an applied statistical mechanics course and often need to express the notion that something is very probably true for functional forms found in the wild, without launching into a full digression for pathological exceptions.

For example, I would like to remind students that a function's Taylor series sometimes has the useful property of converging to the function's value.

Is there a mathematical symbol for this type of "equality" that is more dignified than my current options: $$f(x) \stackrel{\textrm{(good odds)}}{=} \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a)$$ or $$f(x) \texttt{ ¯\_(ツ)_/¯ } \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a)$$

• Don't know a symbol myself, but +1 for the smiley face version! – Paul Sinclair Sep 22 '15 at 4:15
• Perhaps an equal in quotes: "="? – Teepeemm Sep 22 '15 at 13:12
• Given how easy it is to make mistakes with fuzzy notions of equality, I propose the notation $a=b+O(sh*t)$. – guest Sep 22 '15 at 21:33
• How about the symbol "$\text{is probably equal to}$"? – Najib Idrissi Sep 23 '15 at 9:38
• What about the = but with ~s ? That means approx right? – Alec Teal Sep 23 '15 at 12:13

I recommend handling this issue simply, such as:

1) "Under mild conditions we get from Fourier theory:"

$$f(x) = \sum_{k=0}^{\infty} a_k \cos(2\pi k x)$$

2) "Under mild conditions we get from Taylor series theory:"

$$f(x) = \sum_{k=0}^{\infty} a_k (x-a)^k$$

It is understood that the precise conditions under which equality holds can be obtained by looking at the details of that theory.

I would avoid:

-Unclear notation or phrases.

For example: It sounds like you are talking about a class of problems for which no probability model is defined (or relevant). Thus, unclear phrases like "strong odds" will confuse. Funky equality signs will evoke laughter, but it will be nervous laughter since nobody will know what you are talking about. Advanced measure theory concepts such as "almost surely" will drive your teaching ratings down ("The professor expects us to know advanced probability which is not a prerequisite for this course...")

• I agree, no need to be inventing notation here. All you're trying to say is that the equation is true for a certain class of functions $f$. So just define that class (even informally, like just saying "for most everyday functions $f$") and move on. – Jack M Sep 22 '15 at 13:19
• I generally agree, too. But there is one advantage to using a special symbol for the equality, particularly for Taylor series: students tend to see the "=" and forget everything you told them about the requirements for it to hold. If introduced with the same care as described, but also with a special equals, it might remind them when looking at the formula later that equality is not automatically guaranteed. – Paul Sinclair Sep 22 '15 at 16:17
• Whenever I see "mild conditions" or similar, I want to punch the writer for not spelling out what they are. Maybe not in too simple cases, but whenever you enter a field with a multitude of approaches it grind my gears. – Forgottenscience Sep 23 '15 at 14:49
• A further thought (years later, looking back): In statistics and probability, students often write crazy equations like: $$X = f_X(x)dx$$ where $X$ is a random variable, $f_X(x)$ a PDF, and $dx$ a differential-without-an-integral. Thus, students often have only a hazy understanding of the material and are struggling just to write equations that make sense. If strange equal signs are introduced, it can make that struggle harder. It may even encourage the idea that “we never really know if an equation holds so it is fine if I write $X=f_X(x)dx$ as long as I use a special equal sign.” – Michael Jun 1 '17 at 16:17

When I’m writing up tentative results that remain yet to prove (perhaps subject to some additional regularity conditions or assumptions), I usually do this: $$x\,\overset{?}{{=}}\,y+z.$$

On a less serious note:

Let $f:\mathbb R\to\mathbb R$ be a function of class $\mathcal C^{\infty}$. Then, one has, for any $x\in\mathbb R$ and $a\in\mathbb R$, that $$f(x)\overset{*}=\sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a).$$ $\scriptsize{^{*}\text{Terms and conditions apply.}}$

• I really really like this solution. – Anubian Noob Sep 23 '15 at 4:46
• What about $$f(x) = \left\{\begin{array}{cc} e^{-1/x^2} & \mbox{if x \neq 0} \\ 0 & \mbox{if x=0}\end{array}\right.$$ This fits your conditions but the Taylor expansion does not work at $x=0$. See macalester.edu/aratra/chapt2/chapt2_6c.html – Michael Sep 23 '15 at 20:21
• @Michael Sir, I apologize for the inconvenience. It seems to be the case that our terms do not apply to $x=0$ for this particular function. Once again, I’m terribly sorry. As a token of our appreciation of your business and in order to compensate you for this inconvenience, I wonder whether you would be willing to accept the opportunity to try our formula at any other point $x\in\mathbb R\setminus\{0\}$. – triple_sec Sep 23 '15 at 22:06
• @triple_sec : Sorry, I did not get the original footnote joke. Your statement in tan seemed written as a theorem, I thought your point was to be fully precise from the start, and I thought the footnote referred back to the "terms and conditions" given in the theorem. I gather from your response that the joke was to speak in ambiguous legaleze/businesseze. I am slow. Anyway I want my money back. – Michael Sep 24 '15 at 4:10
• @Michael We regret that our service has not met your satisfaction. You can expect a full refund within $\lim_{N\to\infty}\sum_{n=1}^Nn^{-1}$ days. In the meanwhile, we will be working very hard to retain your business. Here at Taylor & Maclaurin’s, our ultimate goal is your uniform convergence on unbounded radii. – triple_sec Sep 24 '15 at 4:23

How about $f(x)\quad\overline{--}\quad etc.$ , indicating that there is a hole in the floor that you have to occasionally be careful not the step through?

Edit: better interpretation: call it "leaky equals", because occasionally the equality leaks out of it.

• It has to be "Leaquals". – Thomas Cleberg Sep 22 '15 at 16:09
• The upper line also gets leaky when selected. – jingyu9575 Sep 23 '15 at 9:17
• I feel like that would be too easy to overlook. – Teepeemm Oct 2 '15 at 18:51

I have often seen the symbol $"="$ (including the quotation marks, as in $a "="b$ for instance) to stand for 'is sort of equal to, but not quite'. The problem could be that we're missing an error term, or some hypothesis, or the equality is only on a large set but not everywhere, etc. Hence, your equation would come out as $$f(x) =” \sum_{k=0}^\infty \frac{(x-a)^k}{k!}f^{(k)}(a).$$

Personally, I find it quite good, because it both keeps the standard equality symbol $=$ which is familiar to everybody, and signifies clearly that something could go wrong.

The downside is that the symbol is rather vague. It may not be obvious if $$f(x) =” \sum_{k} a_k x^k$$ means that there is an error term missing, that the equality does not hold everywhere because the rhs diverges, or if indeed the function $f$ needs to be assumed to be analytic. But I suppose that if you're in a context where such confusion is possible, you should spell things out in more detail, e.g. as suggested in Michael's answer.

• as a side note, I'd be careful with using quotes, as they are AFAIR commonly used for real-life language statements used in logical/mathematical contexts, $p="z\ is\ provable"$ and for linguistic parts in formal language analysis, e.g. $q=\{"I","ate","my","cat"\}$, as seen in e.g. en.wikipedia.org/wiki/Formal_language#Examples or en.wikipedia.org/wiki/Quasi-quotation – vaxquis Sep 23 '15 at 16:49
• @vaxquis: You did what now to your cat? You monster!! – Vandermonde Sep 23 '15 at 18:59

Non-seriously: $$f(x) = \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a) \;\;\approx\;\; \text{true}$$

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Tom-Tom Sep 23 '15 at 22:05
• @Tom-Tom, yes it does. You need to look past the title question and read the actual body. – goblin Sep 23 '15 at 22:07
• @Tom-Tom I kind of disagree. While not a real (serious) answer, it is an answer along the same lines as the "* Terms and conditions apply" and "$\;\mathrm{P}(\ldots) \approx 1\;$" answers. While not a direct answer (by providing a special symbol like $\;\overset{*}=\;$) it provides a more general indicator of probably-truthness, which the OP was looking for. – Marnix Klooster Sep 23 '15 at 22:10
• @MarnixKlooster. You put a disclaimer "non-seriously" and you repeat that it is not serious in your comment. I still believe, this contribution would better fitted as comment rather than an answer, regardless of its content, just because of this. – Tom-Tom Sep 24 '15 at 7:17

I have seen the $\sim$ symbol used for this purpose; e.g., writing $$f(x) \sim \sum_{n=-\infty}^\infty a_n e^{inx/2\pi L}$$ to signify that the $a_n$ are the Fourier coefficients of $f$ without making any definite convergence claims.

• I frequently see this symbol used to mean "is asymptotic to", which is not what OP intends. – Austin Mohr Sep 23 '15 at 3:05
• $\sim$ is one of the most overloaded binary relations in the pack. I think the last thing it needs is more meanings! – Chappers Sep 23 '15 at 22:36

$$\overset{p}{=}$$

• If you like to live dangerously and don’t mind upsetting your probability-theorist colleagues, you could also try $$\overset{\text{a.s.}}=$$ – triple_sec Sep 23 '15 at 3:16

You could write $$f(x)=\sum_{k=0}^{\infty}\frac{(x-a)^k}{k!}f^{(k)}(a)\quad \text{a.s.},$$ which stands for "almost surely". This is used by mathematical analysists to mean that the statement holds for a set of objects with full measure. You could even define a measure "for fun" which makes this statement literally true.

• This is too mistakable for a genuine mathematical statement - in your example, it could easily be read as saying the equality is true for a set of $x$'s of full measure. For the purposes of this situation, more whimsical notations or phrases seem preferable. – Zev Chonoles Sep 22 '15 at 5:03
• Ironically, under any of the notions of random (periodic) function I'm thinking of, a Fourier or Taylor series would almost surely agree with a given function nowhere. (Admittedly, they often work for compositions of closed-form expressions and the ones of practical interest, but that's selection bias in action.) – Vandermonde Sep 23 '15 at 19:07
• I should be a little more explicit with what I meant by "a measure that makes this statement literally true". Put a measure on the space of all functions $f\colon \mathbb R\to\mathbb R$ which is supported on the analytic functions. – pre-kidney Sep 25 '15 at 2:30
• Ehh... To do so is to say 'functions are analytic, except when they aren't'. – Vandermonde Sep 25 '15 at 20:45

$$\mathrm{P} \left( f(x) \approx \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a) \right) \approx 1$$
• I didn't downvote, but if you make a statement about the probability of something happening, then you are assuming some sort of probability space. Defining that space would be far more work than you save by introducing the notation. Besides, $\approx$ takes care of "something strange about this equality" all by itself. – Teepeemm Oct 2 '15 at 18:49