Different Ways of Solving a Contest Problem: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on different ways of solving the contest problem “If $3 \sin \theta = \cos \theta$ , what is $\sin \theta \cos \theta$ ?”

Part 1: Drawing the angle $\theta$

Part 2: A first attempt using a Pythagorean identity.

Part 3: A second attempt using a Pythagorean identity and the original hypothesis for $\theta$ .

Zeno’s Hairodox

Source: http://joegp.com/halfcut/

Useless Numerology for 2016: Part 5

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

$2016 = 2^{11} - 2^5$

Establishing that these two expressions are equal takes a little bit of work. We begin by dividing the sum into even and odd values of $n$ . The odd values of $n$ from $n = 1$ to $n = 63$ have the form $n = 2k+1$ , where $k$ varies between $k = 0$ and $k = 31$ . The even values of $n$ from $n = 0$ to $n = 62$ have the form $n = 2k$ , where again $k$ varies between $k = 0$ and $k = 31$ .

$\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2 = \displaystyle \sum_{n=0 \atop n =2k+1}^{63} (-1)^{n+1} n^2 + \displaystyle \sum_{n=0 \atop n =2k}^{63} (-1)^{n+1} n^2$

$= \displaystyle \sum_{k=0}^{31} (-1)^{(2k+1)+1} (2k+1)^2 + \displaystyle \sum_{k=0}^{31} (-1)^{2k+1} (2k)^2$

$= \displaystyle \sum_{k=0}^{31} \left[ (-1)^{2k+2} (2k+1)^2 + (-1)^{2k+1} (2k)^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[(2k+1)^2 - (2k)^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[4k^2 + 4k + 1 - 4k^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[4k + 1 \right]$

This last sum is an arithmetic series. The first ( $k = 0$ ) term is $1$ , the last term is $4(31) + 1$ , and there are $32$ terms. Using the formula for an arithmetic series, we find

$\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2= \displaystyle \frac{32 [1 + 4(31) + 1]}{2}$

$= \displaystyle \frac{32[4(31) + 2]}{2}$

$= 32[2(31) + 1]$

$= 32[2(32-1) + 1]$

$= 32[2(32) - 2 + 1]$

$= 32[64 - 1]$

$= 2^5[2^6 -1]$

$= 2^{11} - 2^5$ .

Useless Numerology for 2016: Part 4

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

$2016 = (1+2+...+8+9)^2 - (1+2)^2$

To evaluate the top series, we use the formula

$1^3 + 2^3 + \dots + n^3 = \displaystyle \frac{n^2 (n+1)^2}{4}$

$= \displaystyle \left( \frac{n(n+1)}{2} \right)^2$

$= (1 + 2 + \dots + n)^2$ ,

where we used the formula for an arithmetic series. Therefore,

$3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 = (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3) - (1^3 + 2^3)$

$= (1 + 2 + \dots + 9)^2 - (1 + 2)^2$ .

Useless Numerology for 2016: Part 3

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = 1+2+3 + \dots + 62 + 63$

$2016 = 2^{11} - 2^5$

In this post, we’ll explore why these two expressions have to be equal.

The sum $1 + 2 + 3 + \dots + 62 + 63$ is an arithmetic series. The first term is $1$ , the last term is $63$ , and there are $63$ terms in the series. Using the formula for an arithmetic series, we find

$1 + 2 + 3 + \dots + 62 + 63 = \displaystyle \frac{(63)(1 + 63)}{2}$

$= \displaystyle \frac{63 \times 64}{2}$

$= 63 \times 32$

$= (64-1) \times 32$

$= (2^6 - 1) \times 2^5$

$= 2^{11} - 2^5$ .

Useless Numerology for 2016: Part 2

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5$

$2016 = 2^{11} - 2^5$

Not surprisingly, there’s a natural reason why these two expressions are equal. (However, there isn’t a natural reason why the answer happens to match the current year other than coincidence.)

To begin, $2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10}$ is a finite geometric series. The first term is $2^5 = 32$ , the common ratio is $2$ , and there are $6$ terms in the series. Using the formula for a finite geometric series,

$2^5 + 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} = \displaystyle \frac{2^5 (1-2^6)}{1-2} = \frac{2^5-2^{11}}{-1} = 2^{11} - 2^5$ ,

thus establishing that these two expressions are equal.

Useless Numerology for 2016: Part 1

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

$2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5$

$2016 = 1+2+3 + \dots + 62 + 63$

$2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

$2016 = (1+2+...+8+9)^2 - (1+2)^2$

$(2 + 0 + 1)! = 6$

$2016 = 2^{11} - 2^5$

$2016 = 2016 \times 1$

The last tongue-in-check equation is my favorite.

In this series, I’ll explain why these different expressions for $2016$ have to be equal to each other. I’ll begin with tomorrow’s post.

How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction

Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by $\sec^2 x$ , and the substitution $u = \tan x$ .

Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution $u = \tan \theta/2$ .

Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle

Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and partial fractions.

Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and contour integration using the real line and an expanding semicircle.

Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

The Inspection Paradox

From http://allendowney.blogspot.com/2015/08/the-inspection-paradox-is-everywhere.html:

Airlines complain that they are losing money because too many flights are nearly empty. At the same time passengers complain that flying is miserable because planes are too full. They could both be right. When a flight is nearly empty, only a few passengers enjoy the extra space. But when a flight is full, many passengers feel the crunch.

Once you notice the inspection paradox, you see it everywhere. Does it seem like you can never get a taxi when you need one? Part of the problem is that when there is a surplus of taxis, only a few customers enjoy it. When there is a shortage, many people feel the pain.

This article gives multiple examples (including computations) of the Inspection Paradox (also known as the Friendship Paradox), meaning that even the simple concept of “average” can be a little elusive. See also http://www.technologyreview.com/view/523566/how-the-friendship-paradox-makes-your-friends-better-than-you-are/

The Wrong Door, Or Why Math Gets a Bad Rap

I really enjoyed this news article on how to motivate young students to enjoy mathematics. Surprise, surprise: it isn’t by doing a whole bunch of rote arithmetic or algebra problems.

http://www.maa.org/news/the-wrong-door-or-why-math-gets-a-bad-rap