The Bell Curve, Gaussian Integral, & Error Function#
Revised
02 Apr 2023
Bell Curve, the Gaussian Function#
\( \begin{aligned} f:\mathbb{R}\to\mathbb{R} \\ \text{domain}\,\,\,D&=\mathbb{R}=(-\infty,\infty) \\ \text{image}\,\,\,I&=\mathbb{R}^+\impliedby e^{-x^2}\gt0 \\ \text{even function}\,\,\,e^{-x^2}&=f(x)=f(-x)=e^{-(-x)^2} \\ \text{horizontal asymptote}\,\,\,0&=\lim_{x\to-\infty}e^{-x^2}=\lim_{x\to\infty}e^{-x^2} \\ \text{vertical asymptote}\,\,\,&\text{none, since the denominator is never equal to zero} \end{aligned} \)
First Derivative
\( \begin{aligned} f'(x)&=\frac{d}{dx}e^{-x^2} \\ u&=-x^2 \\ \frac{d}{dx}u&=-2x \\ f'(x)&=\frac{d}{dx}e^u =e^u\frac{d}{dx}u =-2xe^{-x^2} &&\text{by chain rule} \end{aligned} \)
Critical Points
\( \begin{aligned} f'(x)&=-2xe^{-x^2}=0 \implies -2x=0 \implies x=0 \end{aligned} \)
\(f\) is a global maximum at \(x=0\).
First Derivative Test
\( \begin{aligned} x&\lt0 \implies f'(-1) =-2(-1)e^{-(-1)^2} =2e^{-1} &&\gt0 \\ x&\gt0 \implies f'(1) =-2(1)e^{-(1)^2} =-2e^{-1} &&\lt0 \end{aligned} \)
\(f\) increases on the interval \((-\infty,0)\) and decreases on the interval \((0,\infty)\).
Second Derivative
\( \begin{aligned} f''(x) &=\frac{d}{dx}\left(-2xe^{-x^2}\right) \\ &=-2\frac{d}{dx}\left(xe^{-x^2}\right) \\ &=-2\left[x\frac{d}{dx}\left(e^{-x^2}\right)+\left(\frac{d}{dx}x\right)e^{-x^2}\right] &&\text{by product rule} \\ &=-2\left(-2x^2e^{-x^2}+e^{-x^2}\right) \\ &=2e^{-x^2}\left(2x^2-1\right) \end{aligned} \)
Inflection Points
\( \begin{aligned} f''(x) &=2e^{-x^2}\left(2x^2-1\right) =0 \implies 4x^2-2=0 \implies x^2=\frac{1}{2} \implies x=\pm\frac{1}{\sqrt{2}} \end{aligned} \)
Second Derivative Test
\( \begin{aligned} x&\lt-\frac{1}{\sqrt{2}} &&\implies f''(-1) =2e^{-(-1)^2}\left(2(-1)^2-1\right) =2e^{-1}(2-1) =2e^{-1} &&\gt0 \\ -\frac{1}{\sqrt{2}}\lt x&\lt\frac{1}{\sqrt{2}} &&\implies f''(0) =2e^{-(0)^2}\left(2(0)^2-1\right) =2(0-1) =-2 &&\lt0 \\ \frac{1}{\sqrt{2}}&\lt x &&\implies f''(1) =2e^{-(1)^2}\left(2(1)^2-1\right) =2e^{-1}(2-1) =2e^{-1} &&\gt0 \end{aligned} \)
\(f\) is concave up on the intervals \(\left(-\infty,-\frac{1}{\sqrt{2}}\right)\cup\left(\frac{1}{\sqrt{2}},\infty\right)\) and concave down on the interval \(\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\).
Gaussian Integral#
The total area under the curve represents the sumd of all probabilities and is equal to unity.
Thus \( \begin{aligned} \frac{1}{\sqrt{\pi}} \end{aligned} \) is a common normalization factor.
An indefinite integral must be rewritten as a limit
\( \begin{aligned} \int_{-\infty}^\infty e^{-x^2}\,dx =\lim_{A\to\infty}\int_{-A}^A e^{-x^2}\,dx \end{aligned} \)
but, impossibility of \(u\)-substitution
\( \begin{aligned} \text{I}&\overset{\text{def}}{=}\int_{-\infty}^\infty e^{-x^2}\,dx =\int_{-\infty}^\infty e^{-y^2}\,dx \\ &\implies \\ \text{I}^2&=\int_{-\infty}^\infty e^{-x^2}\,dx\int_{-\infty}^\infty e^{-y^2}\,dx \\ &=\underset{\mathbb{R}^2}{\iint}e^{-x^2}e^{-y^2}\,dx\,dy \\ &=\underset{\mathbb{R}^2}{\iint}e^{-(x^2+y^2)}\,dA \\ &=\lim_{R\to\infty}\int_0^{2\pi}\int_0^R e^{-r^2}\,r\,dr\,d\theta &&\text{by polar transformation equations} \\ &=\lim_{R\to\infty}\int_0^{2\pi}d\theta\int_0^R e^{-r^2}\,r\,dr \\ &=2\pi\lim_{R\to\infty}\int_0^R e^{-r^2}\,r\,dr \\ u&=-r^2 \implies r=0\to u=0,r=R\to u=-R^2 \\ du&=-2r\,dr \implies -\frac{1}{2}\,du=r\,dr \\ \text{I}^2 &=2\pi\lim_{R\to\infty}\int_0^{-R^2} e^u\left(-\frac{1}{2}\right)\,du \\ &=-\pi\lim_{R\to\infty}\int_0^{-R^2} e^u\,du \\ &=\pi\lim_{R\to\infty}\int_{-R^2}^0 e^u\,du \\ &=\pi\lim_{R\to\infty}e^u\Big|_{-R^2}^0 \\ &=\pi\lim_{R\to\infty}\left(1-e^{-R^2}\right) \\ &=\pi\left(\lim_{R\to\infty}1-\lim_{R\to\infty}e^{-R^2}\right) \\ &=\pi(1-0) \\ &=\pi \\ \text{I}&=\sqrt{\pi} \end{aligned} \)
Resources#
[Y] blackpenredpen. (24 Oct 2022). “how Laplace solved the Gaussian integral”. YouTube.
[Y] blackpenredpen. (27 Sep 2018). “The Bell Curve”. YouTube.
[Y] blackpenredpen. (26 Sep 2018). “the impossible integral of e^(-x^2) & the error function”. YouTube.
[Y] MIT OpenCourseWare. (04 Jan 2011). “Integral of exp(-x^2) | MIT 18.02SC Multivariable Calculus, Fall 2010”. YouTube.