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Chapter 2: Problem 3

Alter the program MonteCarlo to estimate the area of the circle of radius \(1 / 2\) with center at \((1 / 2,1 / 2)\) inside the unit square by choosing 1000 points at random. Compare your results with the true value of \(\pi / 4\). Use your results to estimate the value of \(\pi\). How accurate is your estimate?

Short Answer

Expert verified
The MonteCarlo estimates the circle's area and \( \pi \) using random points, with accuracy depending on the number of points. The more points used, the more accurate the estimate.

Step by step solution

01

Understanding the Problem

We need to estimate the area of a circle with radius \( \frac{1}{2} \) centered at \( \left( \frac{1}{2}, \frac{1}{2} \right) \) within a unit square, using a Monte Carlo method with 1000 random points.
02

Setting up the Monte Carlo Method

The Monte Carlo method involves randomly generating points within the unit square and determining what fraction of these points lie inside the circle. This fraction can be used to estimate the area of the circle.
03

Generating Random Points

Generate 1000 random points where each point \((x,y)\) has coordinates generated uniformly in the range \([0, 1]\), corresponding to the dimensions of the unit square.
04

Determining Points Inside the Circle

For a point \((x, y)\) to be inside the circle with radius \( \frac{1}{2} \), centered at \( \left( \frac{1}{2}, \frac{1}{2} \right) \), the condition \((x - 0.5)^2 + (y - 0.5)^2 \leq (0.5)^2 \) must be satisfied. Count the number of points that satisfy this condition.
05

Estimating the Circle's Area

Calculate the estimated area of the circle as the ratio of the number of points inside the circle to the total number of points (1000) multiplied by the area of the unit square (which is 1). Possible formula: \( \text{Estimated Area} = \frac{\text{Points inside circle}}{1000} \times 1 \).
06

Comparing with True Circle Area

The true area of the circle is \( \frac{\pi}{4} \). Compare this value with the estimated value from Step 5 by computing the difference between them.
07

Estimating Pi

Use the formula \( \text{Estimated } \pi = 4 \times \text{Estimated Area of Circle} \) to estimate \( \pi \).
08

Evaluating the Accuracy

Determine the accuracy by comparing the estimated \( \pi \) with the known value of \( \pi \approx 3.14159 \). Calculate the error as the absolute difference between the estimation and the actual \( \pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Area Estimation
In the Monte Carlo method, one exciting application is circle area estimation. We want to estimate the area of a circle that is perfectly fit inside a square. Imagine a circle with radius \( \frac{1}{2} \) inside a unit square. Here, the center of the circle is at \( \left( \frac{1}{2}, \frac{1}{2} \right) \). This specific setup helps us calculate the area of the circle using random simulation.The idea is straight forward:
  • First, envision how a circle would neatly fit within a square.
  • Picture tossing grains of rice randomly across the square. Let's pretend some grains fall inside the circle.
  • By counting how many grains land inside the circle, you get an idea of the circle's size relative to the square.
This simple activity captures the essence of Monte Carlo circle area estimation. By simulating this scenario, we not only approximate the circle's area but also link this estimation to pi.
Circle Within a Square
When we place a circle within a square, some cool geometry comes into play. Given our scenario, the square is a 'unit square', meaning each side is one unit long. A circle perfectly fitting into this square will be centered right in the middle, 鈥嶾((0.5, 0.5)鈥嶾), and its radius is usually half of the side of the square, which is \(0.5\).Here's why this is important:
  • The perimeter of the circle just touches the square at four points.
  • This perfect boundary condition is great for simulations. We easily see which random points land inside the circle.
  • It's a well-defined fit, making calculations neat and tidy.
Essentially, by looking at a circle within a square, we're creating a 'playground' for randomness, which Monte Carlo simulations utilize to great effect.
Estimating Pi
Estimating pi through this method is quite ingenious. Since the exact area of a quarter circle is \( \frac{\pi}{4} \) when related to the area of the unit square, we leverage this relationship to get to pi.This is how you can estimate pi:
  • Count the number of points that fall inside the circle.
  • Divide this by the total number of random points (1000 here).
  • This fraction gives you a direct estimate of \( \frac{\pi}{4} \).
  • Simply multiply the resulting fraction by 4 to approximate pi.
Thus, using a simple random event鈥攖he falling of points鈥攚e cleverly estimate a crucial constant like pi. It turns a mathematical concept into an interactive experiment!
Random Point Generation
Central to the Monte Carlo method is random point generation. Generating random points involves selecting random coordinates \((x, y)\) within a defined range鈥攈ere, both go from 0 to 1. Each coordinate is chosen independently and uniformly.What makes this powerful?
  • Randomness mimics the act of randomly tossing points onto the grid, simulating natural randomness.
  • Helps in covering the entire space inside the square. Burst of unbiased random points ensures no part is ignored.
  • This spread of points gives a reliable statistical sample that's crucial for estimating areas.
To summarize, each point represents an unbiased representation of a tiny piece of space. When we aggregate the outcomes (points inside or outside the circle), it supplies useful insights for calculations.

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Most popular questions from this chapter

At the Tunbridge World's Fair, a coin toss game works as follows. Quarters are tossed onto a checkerboard. The management keeps all the quarters, but for each quarter landing entirely within one square of the checkerboard the management pays a dollar. Assume that the edge of each square is twice the diameter of a quarter, and that the outcomes are described by coordinates chosen at random. Is this a fair game?

Write a program to carry out the following experiment. A coin is tossed 100 times and the number of heads that turn up is recorded. This experiment is then repeated 1000 times. Have your program plot a bar graph for the proportion of the 1000 experiments in which the number of heads is \(n,\) for each \(n\) in the interval \([35,65] .\) Does the bar graph look as though it can be fit with a normal curve?

For Buffon's needle problem, Laplace \(^{9}\) considered a grid with horizontal and vertical lines one unit apart. He showed that the probability that a needle of length \(L \leq 1\) crosses at least one line is $$ p=\frac{4 L-L^{2}}{\pi} $$ To simulate this experiment we choose at random an angle \(\theta\) between 0 and \(\pi / 2\) and independently two numbers \(d_{1}\) and \(d_{2}\) between 0 and \(L / 2 .\) (The two numbers represent the distance from the center of the needle to the nearest horizontal and vertical line.) The needle crosses a line if either \(d_{1} \leq(L / 2) \sin \theta\) or \(d_{2} \leq(L / 2) \cos \theta .\) We do this a large number of times and estimate \(\pi\) as $$ \bar{\pi}=\frac{4 L-L^{2}}{a} $$ where \(a\) is the proportion of times that the needle crosses at least one line. Write a program to estimate \(\pi\) by this method, run your program for 100 , 1000 , and 10,000 experiments, and compare your results with Buffon's method described in Exercise \(6 .\) (Take \(L=1 .)\)

Three points are chosen at random on a circle of unit circumference. What is the probability that the triangle defined by these points as vertices has three acute angles? Hint: One of the angles is obtuse if and only if all three points lie in the same semicircle. Take the circumference as the interval \([0,1] .\) Take one point at 0 and the others at \(B\) and \(C\).

Take a stick of unit length and break it into two pieces, choosing the break point at random. Now break the longer of the two pieces at a random point. What is the probability that the three pieces can be used to form a triangle?
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