I remember how I was happy when I find the integration of the Gaussian function in my high school years. It was an instantaneous spark which make me realize that increasing the integration dimension may bring a solution. Moreover, one can find a series expansion for Pi number after evaluating the integral.
Gaussian Integral
What is the Gaussian integral: Finding the area under the Gaussian function. I = \int_{-\infty}^{+\infty}\, e^{-\alpha x^2}\,dx
Figure 1: Gaussian(x) = exp(-x2)
There is no known function whose derivative is equal to Gaussian . Therefore, we need something different to get an answer for this integration.
Let’s define another integral in 2-D:
\hat I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\alpha (x^2+y^2)}\,dx\,dy
Because the integration domain is all real space in 2D and exponential function is separable, x and y may be integrated independently.
\hat I = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\alpha (x^2+y^2)}\,dx\,dy = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-\alpha x^2} e^{-\alpha y^2} \,dx\,dy = \bigg(\int_{-\infty}^{+\infty}e^{-\alpha x^2} \,dx\bigg) \bigg(\int_{-\infty}^{+\infty}e^{-\alpha y^2} \,dy\bigg)
This equation says that \hat I = I^2 . If we can find a method to calculate \hat I, we can solve our problem. \hat I simply calculates volume under the {e^{-\alpha (x^2+y^2)}} function. Alternatively, we can calculate same volume with cylindrical coordinates by summing infinitesimal cylindrical volume elements f(r{,}\theta)rd\theta dr where {r^2=x^2+y^2} , {\theta=\arctan(\frac{y}{x})}. What this actually means is that you can calculate the volume under the lovely Gaussian blanket by summing either infinitesimal rectangular boxes or infinitesimal cylindrical boxes.
Figure 2: Gaussian(x,y) = exp(-x2-y2)
\hat I =\int_{0}^{2\pi} \int_{0}^{+\infty} e^{-\alpha r^2}\,dr\,rd\theta
Now, we can integrate this:
\hat I = {\int_{0}^{2\pi}d\theta} {\int_{0}^{+\infty} e^{-\alpha r^2}\, \frac{d(r^2)}{2}}
All integrals can be done with known functions:
\hat I = (\theta |_{0}^{2\pi})(\frac{e^{-\alpha r^2}}{2\alpha}\bigg|_{0}^{+\infty} ) = \frac{\pi}{\alpha}
Then, the beautiful answer for our Gaussian integral which connects e to \pi:
{\int_{-\infty}^{+\infty} e^{-\alpha x^2}\,dx} = \sqrt[]{\frac{\pi}{\alpha} }
Pi Approximator
If \alpha = 1 , we have: \sqrt[]{\pi} = {\int_{-\infty}^{+\infty} e^{-x^2}\,dx}We can expand exponential to its Taylor series: e^{u} = 1 + u + \frac{u^2}{2} + ... = \sum\limits_{n=0}^{\infty} {\frac{u^n}{n!}}\Rightarrow e^{-x^2}=1 - x^2 + \frac{x^4}{4} + ... = \sum\limits_{n=0}^{\infty} {({-1})^n \frac{x^{2n}}{n!}}
Thus, we can evaluate the integral with this expansion:
{\int_{-\infty}^{+\infty} e^{-x^2}\,dx}={\int_{-\infty}^{+\infty}\sum\limits_{n=0}^{\infty} {({-1})^n \frac{x^{2n}}{n!}}\,dx}={\sum\limits_{n=0}^{\infty} {({-1})^n \frac{x^{2n+1}}{(2n+1)n!}}}\bigg|_{-\infty}^{+\infty}Finally we have exact expression for \pi: \pi = \bigg(\sum\limits_{n=0}^{\infty} {({-1})^n \frac{x^{2n+1}}{(2n+1)n!}}\bigg|_{-\infty}^{+\infty}\bigg)^2
To numerically calculate this, we should approximate this expression. This is equal to assigning all infinities to a finite number:
\pi\approx\bigg(\sum\limits_{n=0}^{k} {({-1})^n \frac{x^{2n+1}}{(2n+1)n!}}\bigg|_{-a}^{+a}\bigg)^2One thing to notice is if we increase a, we should increase k much faster in order to get convergence. Increasing a and k will cause more floating point error in computation. Therefore, we need to find ideal a, then a large enough k. In Figure 1, we can see that exponential is very small after |x|>3. With these observations, I tried some a and k. Eventually, I got a pi approximation with 8 decimal point accuracy. You can try it with below code:
function my_pi(a=4.5,k=70.0)
total = 0.0
for n=0:k
total = total + (-1)^n * a^(2n+1)/((2n+1)*factorial(n))
end
return (2*total)^2 #total includes only +a part of the integration. Thus, we doubled it here.
end
julia> my_pi(a=4.5,k=70.0) 3.141592659154459 julia> pi π = 3.1415926535897...
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