Evaluating the Sum of a Geometric Sequence

This is a short blog post that shows you an easy and intuitive way to derive the formula for the summation of an infinite geometric sequence. Let \( 0 \leq p < 1 \), and let \( a \) be some constant; then we wish to find the value of \( x \) such that $$ x = \sum_{k=0}^\infty a p^k $$ Writing out the first few terms of the summation, we get: $$ x = a + a p + a p^2 + a p^3 + \dots $$ Rewriting the equation by factoring out a \( p \) from every term except the first, we get: $$ x = a + p (a + a p + a p^2 + \dots) $$ Notice that the expression in parenthesis is exactly how \( x \) is defined. Replacing the expression with \( x \) leaves us with: $$ x = a + p x $$ Solving the equation for \( x \) yields $$ x = \frac{a}{1-p} $$ Just remember this simple derivation and you will never have to look up the formula for evaluating the sum of an infinite geometric sequence ever again!

Comments

Post a Comment

Popular posts from this blog

Efficiently Remove Duplicate Rows from a 2D Numpy Array

Suppose you have a 2d numpy array and you want to remove duplicate rows (or columns).  In this blog post, I'll show you a trick you can use to do this more efficiently than using np.unique(A, axis=0).  This algorithm has time complexity $ O(\max(n \log{n}, n m)) $ for an $ n \times m $ matrix, and works almost surely.  By "almost surely" I mean that it is a randomized algorithm that works correctly with probability $1$.  To find the unique rows of a matrix $A$, the algorithm works by generating a random vector $x$ of real numbers, computing the dot product $ y = A x $, then analyzing the unique elements of $y$.  The indices of unique elements in $y$ is the same as the unique row indices in $A$ with probability $1$.  Below is an example to demonstrate how it works:

> import numpy as np > A = np.random.randint(low=0, high=3, size=(10, 2)) > A array([[2, 1], [1, 2], [0, 0], [2, 2], [1, 2], [0, 0], [0, 2], [2, 1], …
Keep reading

Multi-Core Programming with Java

Processors speeds are no longer growing at the rate we've grown accustomed to. Moore's law states that computers double in speed every 18 months. Moore's law is still alive and well, but we just need to find alternate ways to speed up computers. That's where multi-core programming comes into play. The idea behind it is that 2 heads are better than one and in the right situation 2 processors can work on the same problem to solve it twice as fast. Parallel programming is becoming increasingly important these days and there are many different flavors of it including distributed computing, shared memory computing, and GPU computing. Today I am going to focus on shared memory computing, or using multi-core computers rather than many single core computers. This blog post is intended to serve as a beginners guide to parallel programming with Java. It is recommended that the reader have some java experience and is familiar with some of the features of Java 8 (namely lambda exp…
Keep reading

Beat the Streak: Day Three

Image
In order to maximize your probability of beating the streak, you should (1) predict the probability that a batter will get a hit in a given game given the game parameters and (2) determine if it's worthwhile to risk your current streak in order to possibly improve it by 1 or 2 games. In this blog post, I outline my solution to (2). In previous blog posts, I've hinted at what I do to solve (1) and will continue that discussion in a later blog post.

Motivation When your current streak is short, the optimal strategy is to pick the best player every day, regardless of how likely their probability of getting a hit is (to a certain extent). However, as your streak grows, you have an important decision to consider: is it better to pick the best player today and possibly lose your current streak, or skip picking a player and instead maintain your current streak. Naturally, this decision should be guided by the players probability of getting a hit, as well as the distribution of these …
Keep reading