Arctan(f(x)) is almost the same as Erf(f(x)) for many f(x). Is the just coincidence or is there a reason?

Question

For example:

Arctan(x) is almost Erf(x) (subtle differences in absolute value and curve)

Arctan(x^50) is almost Erf(x^50) (difference in absolute value)

and many others, so we can conclude:

Arctan(f(x)) ~ Erf(f(x)) (~ meaning Is a poor approximation of)

Is there a reason for this or is this just a strange coincidence for the 10 equasion I tested (I listed only a few)?

Answer 1

I do not think this is an accurate observation. Note that $$\text{erf}(x) = \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{n!(2n+1)}$$ and $$\arctan(x) = \sum_{n=1}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$$ While there are some similarities between the Taylor Series of the two functions in question, I do not think it is enough to claim they are almost the same.

Answer 2

It is for sure that one needs to scale $\arctan(x)$ for comparison. It is very likely to exist intrinsic relationships we have not found enough. But here seems one of them $$\arctan(x)=\sqrt{\pi}\int_0^\infty e^{-t^2}\text{erf}(xt)\;dt$$ References can be found here https://arxiv.org/pdf/1603.03310.pdf

Answer 3

There is no reason to consider the argument $f(x)$ rather than $x$.

What you call "almost the same" is in reality "they are both S-shaped curves with horizontal asymptotes" and the resemblance stops here. In practice, they are very different, in particular in the way they rejoin the asymptotes. There are million S-shaped curves with horizontal asymptotes.

https://www.wolframalpha.com/input/?i=erf%28x%29%3Barctan%28x%29

Sorry to say, but your intuition is pretty naive and would be like claiming that $\pi e\approx\sqrt{73}$ is an interesting relation.

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Regarding the case $x^{50}$, I guess that you just looked at small values of $x$ ($|x|<1$), so that $x^{50}$ is tiny and observed that any smooth function with $f(0)=0$ and $f'(0)=1$ is well approximated by $f(x)\approx x$. Take $x=2$, and the dream vanishes.