# Non-trivial values of error function $\operatorname{erf}(x)$?

The so called error function $\operatorname{erf}(x)$ is defined as $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt,$$ and it is well known that $\operatorname{erf}(\infty)=1$.

Are there any other known closed-form values of $\operatorname{erf}(x)$, except for $\operatorname{erf}(0)$ and $\operatorname{erf}(\pm\infty)$?

• What do you mean by "known"? – Qiaochu Yuan Apr 8 '13 at 22:19
• Wolfram only knows additional "values" at erf($\pm i \infty$) (which are $\pm i\infty$), see functions.wolfram.com/GammaBetaErf/Erf/03. It's not a proof of nonexistence but it's rather discouraging – Cocopuffs Apr 8 '13 at 22:25
• What do you mean by "actual example"? In what sense is, say, $\pi$ a known number but $\int_0^1 e^{-t^2} \, dt$ an unknown number? After all, $\pi$ is just $4 \int_0^1 \sqrt{1 - t^2} \, dt$. – Qiaochu Yuan Apr 8 '13 at 22:35
• @QiaochuYuan: I believe I mean something like this: Does there exist any $x$ such that both $x$ and $\operatorname{erf}(x)$ can be expressed as values of some elementary function? – Mårten W Apr 8 '13 at 22:44
• @Mårten: this is not what you want. Every real number is the value of some elementary function (namely the constant elementary function with value that real number). But see www-math.mit.edu/~tchow/closedform.pdf for another suggestion (with the definition in that paper I would be extremely surprised if the answer was yes, but I would also be extremely surprised if you could prove it). – Qiaochu Yuan Apr 9 '13 at 0:16

If the values aren't listed on the Wolfram function page, I would be surprised if you found them anywhere else. The only listed closed form values are for $$0$$, $$\pm\infty$$, and $$\pm i\infty$$. However, you can find various equivalent formulations, continued fractions, and the like on that page. A good reference, generally, for most well-known functions.

Note: all summations in this answer are implied to range from n=0 to n=∞, with Δn set to equal 1.

The error function is an integral of the bell curve between 0 and x, as we all know. The bell curve is equal to e^(-x^2)*2/sqrt(π).

1. e^x= Σ (x^n)/(n!)

2. e^(-x^2)= Σ ((-x^2)^n)/(n!)

3. e^(-x^2)2/sqrt(π)=2/sqrt(π) Σ ((-x^2)^n)/(n!)

4. To integrate this function, we simply multiply each term in the polynomial this summation represents by x/(exponent in term+1). The exponent in each term is 2*n, so we multiply each term by x/(n+1). Since x is consistent throughout the summation, though, we can just multiply the entire summation by x and divide each term by (2n+1).

erf(x)= 2x/sqrt(π)* Σ ((-x^2)^n)/(n!*(2n+1))

And there you have it, an exact representation of the error function. If you don't believe me, test it out on a graphing calculator. You'll notice for each term you add (n=0, n=1, n=2), your graph will get closer and closer to the error function.

• Welcome to the site. You can refer to this post for tips on how to format your formula. However, I believe the Original Poster (OP) is not asking for an exact series representation for the error function which carries a truncation error for any given number of terms $n$, but asking for an elementary closed form number. In a certain sense, it's like finding easily memorable values to $\sin(x)$ such as $\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$. – Frenzy Li Aug 23 '16 at 17:05
• I know about the series representations, but those are indeed not what I am looking for. – Mårten W Aug 23 '16 at 18:43
• I hate to break it to you, but according to Wikipedia, the error function cannot be evaluated in closed form. As well, all nontrancendental x values (except of course for 0 and, depending on definition, infinity) have a transcendental y value, even when you divide said y value by the square root of pi. sorry I kinda misunderstood the question the first time. – Math Machine Aug 25 '16 at 18:53

There are other exact values of erf(x) besides 0 and plus or minus infinity this graph there are horizontal asymptotes at y=1 and y=-1 because

The Z-score is a probability that certain data fall within a certain interval. For example a Z-score of 1 is represented by erf(1) = 0.682 which means that plus or minus 1 standard deviation from the mean will encompass approximately 68.2% of the population based upon a sample population. A Z-score of 2 erf(2) = 0.955, erf(3)= 0.9995. In order to be sure that you have covered the entire population or be 100% certain about your data you have to cover an infinite number of standard deviations, which is why it would be impossible to do such a thing.

Example mean IQ is 100 and SD is 15, since erf(3) = 0.9995 then 99.95% of the population will have an IQ within 3 standard deviations from the mean.

STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score. https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable

• This doesn't answer the question at all. OP is asking for which values do we have an explicit expression for the erf function in terms of known constants, not a table of approximate values. – Alex R. Apr 10 '15 at 3:18
• The first paragraph is also confused. The limit values $\operatorname{erf}(\pm \infty) := \lim_{x \to \pm \infty} = \pm 1$ are precisely two of the three mentioned in the question, and so are not "other exact values". You seem to have thought that $\operatorname{erf}$ achieves the values $\pm \infty$ somewhere, but that does not happen on either the extended real line or the extended complex plane (aka the Riemann sphere). – epimorphic Jun 14 '15 at 22:08
• @Aethelred Please delete this answer. It does not answer the question. – Franklin Pezzuti Dyer Sep 29 '18 at 19:53
• Frpzzd: The answer to the question is no. There are no non-trivial solutions to the error function, except where x=0 or x=+-∞, or where x is trivial itsself. If you’re looking for someone to tell you there are non trivial solutions without lying, you’re not going to get it. – Math Machine Oct 2 '18 at 22:49
• And PS, you can downvote people’s answers on alt accounts all you want, it won’t change the truth. This is why I switched to reddit, and I recommend everyone else here does too. Their community guidelines actually make sense. – Math Machine Oct 2 '18 at 22:50