Author: John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

Meat-A-Morphosis: An Introduction to Functions

The video below was posted by my colleague Jason Ermer, founder of Collaborative Mathematics. From the YouTube description:

A cartoon about the proverbial mathematical “function machine”.

I was part of the creative team of mathematics teachers (including Patty Hill and Michael Word) who created this cartoon. It was originally (and still is) a component of the Honors Algebra 1 curriculum at Kealing Middle School in Austin, Texas.

I take full responsibility for the spelling error. (Can you find it? 🙂

Enjoy.

Another Reasoning Puzzle From ChefMongoose

I enjoyed this challenge.

Joseph Nebus's avatarnebusresearch

My friend ChefMongoose had another reasoning problem come to him, and I’m happy to share it further. It’s rather like that famous Singapore Birthday Problem that drove people crazy a couple of months ago. Here’s the problem:

I have a combination lock at work. There are three digits, all in the range 1 – 40; they’re all prime numbers. They’re X+Y, X+2Y, X+3Y — where X and Y are positive integers.

If I told you what X was but not Y, you wouldn’t be able to tell me the combination. If I told you what Y was but not X, you wouldn’t be able to tell me the combination. Now, what’s the combination?

I did work out the puzzle. It did make me notice a couple of strings of uniformly-spaced prime numbers I hadn’t done before, too, such as 3-13-23. (However, 3-13-23 isn’t one of the possible answers, because of…

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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.

 

 

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 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.