Symbol for "probably equal to" (barring pathology)?

Question

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 $$ f(x) \texttt{ ¯\_(ツ)_/¯ } \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a)$$

Answer 1

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.

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I would avoid:

-Unclear notation or phrases.
-Unclear, loaded, or advanced terminology.

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...")

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

Here's the symbol I use.

≈ you can type it using alt+2248 in MS Windows.

Answer 7

What about:

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

Answer 8

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: $$x\,\overset{?}{{=}}\,y+z.$$

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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 $$f(x)\overset{*}=\sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a).$$ $\scriptsize{^{*}\text{Terms and conditions apply.}}$

Answer 9

What about:

$$ \mathrm{P} \left( f(x) \approx \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a) \right) \approx 1$$

Answer 10

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