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I was going through BERT paper which uses GELU (Gaussian Error Linear Unit) which states equation as $$ GELU(x) = xP(X ≤ x) = xΦ(x).$$ which in turn is approximated to $$0.5x(1 + tanh[\sqrt{ 2/π}(x + 0.044715x^3)])$$

Could you simplify the equation and explain how it has been approximated.

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2 Answers 2

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GELU function

We can expand the cumulative distribution of $\mathcal{N}(0, 1)$, i.e. $\Phi(x)$, as follows: $$\text{GELU}(x):=x{\Bbb P}(X \le x)=x\Phi(x)=0.5x\left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$

Note that this is a definition, not an equation (or a relation). Authors have provided some justifications for this proposal, e.g. a stochastic analogy, however mathematically, this is just a definition.

Here is the plot of GELU:

Tanh approximation

For these type of numerical approximations, the key idea is to find a similar function (primarily based on experience), parameterize it, and then fit it to a set of points from the original function.

Knowing that $\text{erf}(x)$ is very close to $\text{tanh}(x)$

and first derivative of $\text{erf}(\frac{x}{\sqrt{2}})$ coincides with that of $\text{tanh}(\sqrt{\frac{2}{\pi}}x)$ at $x=0$, which is $\sqrt{\frac{2}{\pi}}$, we proceed to fit $$\text{tanh}\left(\sqrt{\frac{2}{\pi}}(x+ax^2+bx^3+cx^4+dx^5)\right)$$ (or with more terms) to a set of points $\left(x_i, \text{erf}\left(\frac{x_i}{\sqrt{2}}\right)\right)$.

I have fitted this function to 20 samples between $(-1.5, 1.5)$ (using this site), and here are the coefficients:

By setting $a=c=d=0$, $b$ was estimated to be $0.04495641$. With more samples from a wider range (that site only allowed 20), coefficient $b$ will be closer to paper's $0.044715$. Finally we get

$\text{GELU}(x)=x\Phi(x)=0.5x\left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)\simeq 0.5x\left(1+\text{tanh}\left(\sqrt{\frac{2}{\pi}}(x+0.044715x^3)\right)\right)$

with mean squared error $\sim 10^{-8}$ for $x \in [-10, 10]$.

Note that if we did not utilize the relationship between the first derivatives, term $\sqrt{\frac{2}{\pi}}$ would have been included in the parameters as follows $$0.5x\left(1+\text{tanh}\left(0.797885x+0.035677x^3\right)\right)$$ which is less beautiful (less analytical, more numerical)!

Utilizing the parity

As suggested by @BookYourLuck, we can utilize the parity of functions to restrict the space of polynomials in which we search. That is, since $\text{erf}$ is an odd function, i.e. $f(-x)=-f(x)$, and $\text{tanh}$ is also an odd function, polynomial function $\text{pol}(x)$ inside $\text{tanh}$ should also be odd (should only have odd powers of $x$) to have $$\text{erf}(-x)\simeq\text{tanh}(\text{pol}(-x))=\text{tanh}(-\text{pol}(x))=-\text{tanh}(\text{pol}(x))\simeq-\text{erf}(x)$$

Previously, we were fortunate to end up with (almost) zero coefficients for even powers $x^2$ and $x^4$, however in general, this might lead to low quality approximations that, for example, have a term like $0.23x^2$ that is being cancelled out by extra terms (even or odd) instead of simply opting for $0x^2$.

Sigmoid approximation

A similar relationship holds between $\text{erf}(x)$ and $2\left(\sigma(x)-\frac{1}{2}\right)$ (sigmoid), which is proposed in the paper as another approximation, with mean squared error $\sim 10^{-4}$ for $x \in [-10, 10]$.

Here is a Python code for generating data points, fitting the functions, and calculating the mean squared errors:

import math
import numpy as np
import scipy.optimize as optimize


def tahn(xs, a):
    return [math.tanh(math.sqrt(2 / math.pi) * (x + a * x**3)) for x in xs]


def sigmoid(xs, a):
    return [2 * (1 / (1 + math.exp(-a * x)) - 0.5) for x in xs]


print_points = 0
np.random.seed(123)
# xs = [-2, -1, -.9, -.7, 0.6, -.5, -.4, -.3, -0.2, -.1, 0,
#       .1, 0.2, .3, .4, .5, 0.6, .7, .9, 2]
# xs = np.concatenate((np.arange(-1, 1, 0.2), np.arange(-4, 4, 0.8)))
# xs = np.concatenate((np.arange(-2, 2, 0.5), np.arange(-8, 8, 1.6)))
xs = np.arange(-10, 10, 0.001)
erfs = np.array([math.erf(x/math.sqrt(2)) for x in xs])
ys = np.array([0.5 * x * (1 + math.erf(x/math.sqrt(2))) for x in xs])

# Fit tanh and sigmoid curves to erf points
tanh_popt, _ = optimize.curve_fit(tahn, xs, erfs)
print('Tanh fit: a=%5.5f' % tuple(tanh_popt))

sig_popt, _ = optimize.curve_fit(sigmoid, xs, erfs)
print('Sigmoid fit: a=%5.5f' % tuple(sig_popt))

# curves used in https://mycurvefit.com:
# 1. sinh(sqrt(2/3.141593)*(x+a*x^2+b*x^3+c*x^4+d*x^5))/cosh(sqrt(2/3.141593)*(x+a*x^2+b*x^3+c*x^4+d*x^5))
# 2. sinh(sqrt(2/3.141593)*(x+b*x^3))/cosh(sqrt(2/3.141593)*(x+b*x^3))
y_paper_tanh = np.array([0.5 * x * (1 + math.tanh(math.sqrt(2/math.pi)*(x + 0.044715 * x**3))) for x in xs])
tanh_error_paper = (np.square(ys - y_paper_tanh)).mean()
y_alt_tanh = np.array([0.5 * x * (1 + math.tanh(math.sqrt(2/math.pi)*(x + tanh_popt[0] * x**3))) for x in xs])
tanh_error_alt = (np.square(ys - y_alt_tanh)).mean()

# curve used in https://mycurvefit.com:
# 1. 2*(1/(1+2.718281828459^(-(a*x))) - 0.5)
y_paper_sigmoid = np.array([x * (1 / (1 + math.exp(-1.702 * x))) for x in xs])
sigmoid_error_paper = (np.square(ys - y_paper_sigmoid)).mean()
y_alt_sigmoid = np.array([x * (1 / (1 + math.exp(-sig_popt[0] * x))) for x in xs])
sigmoid_error_alt = (np.square(ys - y_alt_sigmoid)).mean()

print('Paper tanh error:', tanh_error_paper)
print('Alternative tanh error:', tanh_error_alt)
print('Paper sigmoid error:', sigmoid_error_paper)
print('Alternative sigmoid error:', sigmoid_error_alt)

if print_points == 1:
    print(len(xs))
    for x, erf in zip(xs, erfs):
        print(x, erf)

Output:

Tanh fit: a=0.04485
Sigmoid fit: a=1.70099
Paper tanh error: 2.4329173471294176e-08
Alternative tanh error: 2.698034519269613e-08
Paper sigmoid error: 5.6479106346814546e-05
Alternative sigmoid error: 5.704246564663601e-05
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    $\begingroup$ Why is the approximation needed? Couldn't they just use erf function? $\endgroup$
    – SebiSebi
    May 3, 2019 at 16:51
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First note that $$\Phi(x) = \frac12 \mathrm{erfc}\left(-\frac{x}{\sqrt{2}}\right) = \frac12 \left(1 + \mathrm{erf}\left(\frac{x}{\sqrt2}\right)\right)$$ by parity of $\mathrm{erf}$. We need to show that $$\mathrm{erf}\left(\frac x {\sqrt2}\right) \approx \tanh\left(\sqrt{\frac2\pi} \left(x + a x^3\right)\right)$$ for $a \approx 0.044715$.

For large values of $x$, both functions are bounded in $[-1, 1]$. For small $x$, the respective Taylor series read $$\tanh(x) = x - \frac{x^3}{3} + o(x^3)$$ and $$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \left(x - \frac{x^3}{3}\right) + o(x^3).$$ Substituting, we get that $$ \tanh\left(\sqrt{\frac2\pi} \left(x + a x^3\right)\right) = \sqrt\frac{2}{\pi} \left(x + \left(a-\frac{2}{3\pi}\right)x^3\right) + o(x^3) $$ and $$ \mathrm{erf}\left(\frac x {\sqrt2}\right) = \sqrt\frac2\pi \left(x - \frac{x^3}{6}\right) + o(x^3). $$ Equating coefficient for $x^3$, we find $$ a \approx 0.04553992412 $$ close to the paper's $0.044715$.

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  • $\begingroup$ So why the paper does not use $0.04553992412$? Any idea? $\endgroup$
    – Wei Zhong
    Aug 5, 2021 at 21:22
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    $\begingroup$ You have to remember that the o(x^3) remainders hide an infinite number of small terms, and these terms are different for each of the functions (small-o notation). Numerical methods would probably take into account some of what gets lost in the higher-order terms. $\endgroup$ Aug 11, 2021 at 21:19
  • $\begingroup$ Thanks, that makes sense. I am interested to know if the value is approaching theirs if we expand to the o(x^5). $\endgroup$
    – Wei Zhong
    Aug 11, 2021 at 22:39

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