Tuesday, September 26, 2017

Tone It Down (How Cliches Weaken Your Case)

     Surely: the adverb of a man without an argument.

                —From Bad News, by Edward St. Aubyn

In politics, I think people are best served by plain writing that is dense with testable claims and publicly verifiable facts. Writers do try to "energize the base," and in campaigns there is a place for that. But few stylists are capable of elevated prose about live political issues. The more energetic a piece of political writing, the worse that piece of writing probably is.

And the less convincing it probably is. Clichés can make a writer sound like somebody on a rant—like a loud TV you just want to turn off. Extra words subtract gravity from what you're saying. Stock phrases cover up the fact that no argument is being made.

Enough preamble—let's see if the following examples speak for themselves. In every case, I think that striking words would have made for a more effective statement. Sources for the original statements are listed at the end.

Weaker:  I have no swastika or Third Reich related tattoos. PERIOD.

Stronger:  I have no swastika or Third Reich related tattoos.

Weaker:  But there wasn’t a shred of evidence that any insurer had “abused” the boy or his mom.

Stronger:  But there was no evidence that any insurer had abused the boy or his mom.

Weaker:  Jessie Ford, who’s like this great sociologist at NYU, …

Stronger:  Jessie Ford, a great sociologist at NYU, …

Weaker:  If we recognize men and women who identify with the genders they were assigned at birth (cisgender) and we recognize men and women who do not identify with their assigned gender (transgender), then surely we agree this difference is worth recording. 

Stronger:  If we recognize men and women who identify with the genders they were assigned at birth (cisgender) and we recognize men and women who do not identify with their assigned gender (transgender), then this difference is worth recording. 

I may gradually add more examples to the list over time. Here are the sources of the original statements, in order:

Sunday, September 17, 2017

Answers to Amy's Multiplication Puzzle

From an earlier post:

Amy multiplied an eight-syllable number by an eight-syllable number and obtained a four-syllable number. What could her numbers have been?

(These are whole numbers we're talking about.)

To solve this, I first reflected that an eight-syllable number is necessarily pretty large…therefore the four-syllable number must be pretty large…but what kinds of large numbers only have four syllables? Probably something like "twenty trillion." A number like that ends with a lot of zeros, which means that it contains a lot of factors of 10, or in other words a lot of 2s and 5s.

So, I tried putting powers of 2 against powers of 5, and pretty soon I hit upon the following fact:

128 × 125 = 16,000

This is (6-syllable) × (6-syllable) = (4-syllable). Not far from Amy's problem! To patch it up, I added "thousand" to both factors; doing so increases the syllable count in each factor by two, but doesn't change the syllable count of the answer (because "sixteen thousand" becomes "sixteen trillion").

128,000 × 125,000 = 16,000,000,000

Now we have (8-syllable) × (8-syllable) = (4-syllable), as desired. Amy's numbers could have been 128,000 and 125,000.


Using a computer, I also found some interesting cases:

1) The smallest instance of Amy's numbers that I could find was 1,120 × 6,250 = 7,000,000.

2) Some extreme versions of the puzzle, with answers.

Amy multiplied a 22-syllable number by a 22-syllable number and obtained a 4-syllable number. What could her numbers have been?

6,103,515,625 × 1,474,560,000,000,000,000,000 = 9 nonillion

Amy multiplied a 26-syllable number by a 69-syllable number and obtained a 4-syllable number. What could her numbers have been?

2,147,483,648 × 1,396,983,861,923,217,773,437,500 = 3 decillion

3) The four-syllable number 9 septillion can be written as a product of two 13-syllable numbers in three different ways:

9 septillion = 156,250 × 57 quintillion, 600 quadrillion

9 septillion = 7,812,500 × 1 quintillion, 152 quadrillion

9 septillion = 39,062,500 × 230 quadrillion, 400 trillion

4) The four-syllable number 1,000,000,002 can be written as a product of two 11-syllable numbers in two different ways:

1,000,000,002 = 11,829 × 84,538

1,000,000,002 = 23,658 × 42,269

That's it for Amy's puzzle! In a later post, I'll share my code for generating number names. Once you have the number name, counting syllables is easy; but converting a digit string into a string of words is harder. UPDATE 9/21/2017: Instead of sharing my code, here is a webpage with number-naming code in many programming languages.

Saturday, September 16, 2017

Answer to Eyeballing Rates Of Change

Watch carefully and you'll see the straight lines both glowing green at the moment when the blue area leads the red area by the greatest amount. Were you in the ballpark?

Here is a video in MP4 format. And here is one-page walkthrough of the Calculus for the general case where ¼ is replaced by a general parameter λ.

Tuesday, September 12, 2017

Eyeballing Rates Of Change

At about which point in this movie would you say that the numerical difference between the blue and red areas is greatest?

A video file that you can play, pause, rewind, and download is here. One way to share an answer would be to pause the video at the moment of your best guess, take a screenshot, and email it to me at zimblogzimblog@gmail.com.

I sometimes wonder if studying Calculus or working in a Calculus-heavy field makes people any better at perception tasks like this. I do believe that teaching kinematics for so many years improved my direct perception of acceleration; being able to see a couple of derivatives down has helped me to avoid some erratic drivers over the years.

In case any Calculus-trained readers would like to analyze the problem symbolically, I'll give the formulas I used to make the animation:

  • For any value of m ≥ 0, the blue region is defined by x ≥ 0, y ≥ 0, ym x, and yx ex
  • For any value of m ≥ 0, the red region is defined by x ≤ 0, y ≤ 0, y ≥ ¼ m x, and y ≥ x ex.

You should find a simple answer for mbest, the value of m that maximizes the difference blue areared area.

For some extra symbol pushing, replace y = ¼ m x with y = λ m x, where 0 < λ < 1. You should find \(m_{\rm best} = \lambda^{{\sqrt{\lambda}}/(1-\sqrt{\lambda})}\). The limit of mbest as λ approaches 1 is interesting.

Sunday, September 10, 2017

Amy's Multiplication Puzzle

You may recall this puzzle from an earlier post on this blog (also in Word Games 3):

Amy subtracted a three-syllable number from a three-syllable number and obtained a thirty-seven syllable number. What could her numbers have been?

(Answer here.)

Today it occurred to me that we could make a version of this puzzle using multiplication in place of subtraction. Here's a puzzle along those lines:

Amy multiplied an eight-syllable number by an eight-syllable number and obtained a four-syllable number. What could her numbers have been?

(These are whole numbers we're talking about.) I found one solution by hand, which I'll share in my next post along with any other solutions I receive.

Amy's (8, 8, 4) scenario isn't the only interesting case to look at, so feel free to share any "near misses" too, such as

(7-syllable number) × (5-syllable number) = (3-syllable number),

(6-syllable number) × (6-syllable number) = (4-syllable number),


Saturday, September 9, 2017

Free Speech Among The Deplorables: Where Are They Now?

An update on the principals from my December 2016 post "Free Speech Among The Deplorables."

Carl Paladino. Paladino's latest controversy began when he wrote an email with racist and obscene jokes and accidentally sent it to a newspaper reporter instead of to his friends. This set in motion a chain of events resulting last month in the state Commissioner of Education removing Paladino on other grounds. Paladino is appealing the dismissal to the state Supreme Court, where he appears to have a small but appreciable chance of winning his case.

Pamela Ramsey Taylor. Taylor wrote a racist comment on her Facebook page that led to a one-month suspension from her job as director of the nonprofit Clay County (West Virginia) Development Corporation. I originally reported that Taylor had been fired, but that wasn't accurate. She is apparently back at work.

Beverly Whaling. One of Taylor's Facebook friends, and Mayor of Clay at the time, Beverly Whaling resigned in the aftermath of the controversy. There is no recent news about her, and the incident is now part of the history of Clay, West Virginia.

Milo Yiannopoulos. Yiannopoulus lost his book deal with Simon & Schuster (and his job at Breitbart.com) after talking positively about sex between adults and underage boys. Yiannopolous self-published the book and is currently suing Simon & Schuster for breach of contract. He continues to be involved in free speech controversies.

Friday, September 8, 2017

All The Triangles With Area 1 and Perimeter 8

I just felt like looking at them.

There is a phase of the animation where the base and height both appear to remain constant over time, but that is only approximately the case. (I think if you had a rigorously constant base, then because the perimeter is constant, the third vertex would have to trace part of an ellipse, not a horizontal line. So the base and height can't both remain constant over time.)

Sunday, September 3, 2017

On DIV-ing And Dividing

Here's a problem that came home in my daughter's backpack:

Ebenezer has 97 lunch trays. He will make four stacks of lunch trays. He will put the same number of trays in each stack. He will put as many trays as possible in each stack. If Ebenezer does this, how many trays will be in each stack? How many trays will be left over?

Let's consider the first question first. How many trays will be in each stack?

I found the answer 24 by starting down the road of calculating 97 ÷ 4 far enough to see that result would be "24 and some stuff." (Too late, I remembered that I actually knew 4 × 24 = 96 offhand—it was still a good check on my answer.)

Thus my approach was to pursue, just far enough, a strategy of "divide and round down." If you work with computers, then you probably know that the operation "divide and round down" equals a computer command known as "DIV." For example, a computer will tell you that 16 DIV 3 = 5, and one way to check this answer would be to calculate 16 ÷ 3 = 5⅓ and round the result down.

I doubt this is how a computer actually calculates the value of 16 DIV 3, but "divide and round down" gives the right answer and shows the relationship between DIV and ÷, which is what I want to write about today.

A few more examples just to clarify how DIV works:

     32 DIV 7 = 4

     54 DIV 6 = 9

     3 DIV 4 = 0.

A friendly term for the DIV operation might be "arraying." Ebenezer is trying to array 97 things in a 4-by-something array. It's to be understood that when you are arraying things, the equal groups are to be as large as possible, and there might be some articles left over. I think the word "arraying" has a discrete connotation that helps convey its meaning. We array pieces of silverware when we set the table, but we don't "array" the milk when we pour some into each child's cup. (Update 9/4/17: Maybe you could think of DIV as a "packaging up" operation. That phrase avoids the spatial connotations of arraying. For example, if an elevator fits 6 people and there are 20 people who need to get to the roof, then it is a packaging up operation more than an arraying operation. Anyway.)

One nice thing DIV enables us to do is to write down a formula for remainders. For example, if we know that 97 DIV 4 = 24, then we can multiply the 24 by 4 (result: 96) and then what's left over is the remainder; in symbols,

remainder = 97 − [4 × (97 DIV 4)].

More generally, we have the following formula for remainders in a whole-number problem:

remainder = m − [× (m DIV n)].

There is a second well known computer operation, called MOD, which finds the remainder; so we could write

remainder = m MOD n

or using the previous expression,

m MOD n = m − [× (m DIV n)].

Rearranging this, we can say for any integer m and any nonzero integer n

m = × (m DIV n) + m MOD n.

This is the whole-number analogue of the much simpler relationship for rational numbers r and s ≠ 0,

r = s × (r ÷ s).

Notice there's no remainder here; "division with remainder" is a contradiction in terms.

Of course, the relationship r = s × (r ÷ s) is simpler only if you understand rational numbers (or at least fractions). The curriculum in late elementary school often doesn't do enough in that respect, leaving students in middle school stuck with cumbersome (or even faulty) whole-number ideas about operations.

Naturally there are many word problems in which arraying (rather than division) is called for—because a cup of flour can be partitioned any which way, but lunch trays, marbles, wheelchairs, and teachers can't be partitioned and remain lunch trays, marbles, wheelchairs, or teachers. So in a word problem, the context determines the best answer to give—whether this be the exact quotient of two given numbers, or the whole-number part of the quotient, the quotient rounded to the next greatest whole number, or even the remainder. Getting the right answer to such problems is better thought of as a modeling skill than as a calculation skill.

In fact, I suspect that the existing curriculum spends too much time having kids practice bare whole-number division problems with remainders: that confuses the subject of arithmetic by giving division two incompatible meanings. Of course, I would give students plenty of distributed/repetitive practice with multi-digit long division; I'd just set things up so that in all those practice problems, either the divisor divides the dividend evenly, or else the answer is to be expressed correctly as a fraction or decimal. (So that, in both cases, the true relationship r = s × (r ÷ s) holds. You should always be able to check a quotient by simply multiplying.) Meanwhile, situations with remainders would be prevalent in the word problems students work on.

Perhaps that approach is simplistic, but certainly I'll say that I was appalled once to see my daughter working on a giant drill worksheet with dozens problems like this:

  5 ÷ 4 = 1 R 1.

My daughter was terribly confused at first. "Isn't 5 divided by 4 equal to five-fourths?" she asked. I answered her by saying, "Yes it is. You are right. But for tonight, just write what they want." I didn't know what else to say. That worksheet penalizes students who have correctly learned basic fraction concepts and division concepts—concepts which, unlike whole-number concepts, are essential for the transition to algebra.

What a terrible and unnecessary conflict between ideas. There is a need for practicing calculations like 8,995 ÷ 7 = 1,285 so by all means, provide distributed/repetitive practice with those kinds of problems. But it can't be educational for students to rehearse statements like "5 ÷ 4 = 1 R 1," which lie about the division symbol in ways that grade-level students can detect.

I'll close with some better news, specifically in reference to this related earlier post about division. On a recent Saturday School I was elated to see, in a series of workbooks written in 2001 by my colleague Marsha Stanton, this page:

This is the best case I've seen so far of using a true equation to represent objects arrayed in equal groups with some objects left over. This would be worth doing as a way to represent word problems, as I wrote in the earlier post (one-page PDF at tinyurl.com/notdivision).

Saturday, September 2, 2017

What If The Mad King Had More Than Two Kids?

In the Riddle of the Mad King, we shared approximately 68% of the king's fortune with one son and shared the remainder with the other son, so as to simultaneously (1) preserve the king's bragging rights and (2) anger one of the sons.

What if there were three sons? Or many sons?

This raises the question of how to break 1 into parts x > y > z > … so that the least of the ratios 1/x, x/y, y/z, … is as large as possible.

For the case of two sons, we maximized the lesser of the two ratios by making both ratios equal. Let's assume that circumstance generalizes. If we give n sons the respective amounts x1x2, …, xn−1, 1 − x1x2 − … − xn−1, then some algebra shows that we'll have equality of the ratios

1/x1x1/x2 = x2/x3 = … = xn−2/xn−1 = xn−1/(1 − x1 − … − xn−1

if x1 is the unique root of the equation rn + 1 − 2r + 1 = 0 in the range 0 < r < 1. Once you know x1, you can easily find the rest of the shares by repeatedly applying the same ratio 1/x1 that the king uses.

For example, if there are three sons, then solve r4 − 2r + 1 = 0 for r ≈ 0.544, which is the first son's share. The second son's share is the square of this number, and the third son's share is the cube. Numerically the shares come out to be, approximately,

x1:  54%
x2:  30%
x3:  16%

Here is a graph for the case of three sons that generalizes the graph of the objective function f(x) from last time; here the shading is darker where Min{1/x, x/y, y/(1 − xy)} is greater—except that the small white dot shows the solution x1 = 0.5436…, x2 = 0.2955….

What if there are infinitely many sons? You don't need any algebra to make all the ratios equal in this case, as long as you happen to know that

1/2  +  1/4  +  1/8  +  1/16  +  … = 1.

We can take advantage of this fact to solve the problem in the case of infinitely many sons. Let's line everybody up: first the king, followed by an infinite lineup of his sons. The king starts by giving all of his money to the son on his left—who then gives half to the brother on his left, who then gives half of that to the brother on his left, and so on. In the limit, each son has twice as much as his neighbor, and the king's original fortune is twice what the richest son has.