Lessons from teaching gifted elementary students (Part 8b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received, in the students’ original handwriting. They wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

Here’s how I started the problem, using a trick that I use in my mathematical magic show. Suppose that there are only six numbers instead of twelve, and let the six numbers be $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ . Then here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

In other words, the top number can be obtained by using the numbers on the fifth row of Pascal’s triangle (recall that the fifth row of Pascal’s triangle has six numbers on it). Specifically, if I multiply the bottom numbers by the corresponding number in a row of Pascal’s triangle and add them up, I’ll get the number on top without having to compute all of the intermediate steps.

For the problem my students gave me, the bottom row has 12 numbers, which means I’ll need to use the 11th row of Pascal’s triangle. Also, as we’ll see, I was fortunate that my students gave me a simple pattern of consecutive squares for the numbers on the bottom row. Since the numbering in Pascal’s triangle starts on zero, the numbers in the bottom row are $(k+1)^2$ as $k$ varies from 0 to 11.

Putting all this together, I can conclude that

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$ .

Beginning with tomorrow’s post, I’ll discuss how I computed this sum without a calculator.

Lessons from teaching gifted elementary school students (Part 8a)

Here’s a question I once received, in the students’ original handwriting:

Here’s the explanation that my students told me (but didn’t write down): they wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. For example, the lower-left portion of the triangle would build like this (since 1+4=5, 4+9=13, 9+16=25, etc.):

18 38

5 13 25

1 4 9 16

Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

And I produced the answer in less than five minutes.

I’ll reveal how I got the answer so quickly in this series. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

Top 100 Math Blogs for Students and Teachers

Now this was an unexpected surprise. I recently received the following message:

Hi John,

My name is Anuj Agarwal. I’m Founder of Feedspot.

I would like to personally congratulate you as your blog Mean Green Math has been selected by our panelist as one of the Top 100 Math Blogs on the web.

http://blog.feedspot.com/math_blogs/

I personally give you a high-five and want to thank you for your contribution to this world. This is the most comprehensive list of Top 100 Math Blog on the internet and I’m honored to have you as part of this!

Also, you have the honor of displaying the following badge on your blog. Use the below code to display this badge proudly on your blog.

Best,
Anuj

I’m not gonna lie: it’s really, really flattering to see my blog listed on the same page (at #76) with some of the genuine heavy hitters out there in mathematical blogland.

This seems like as good an occasion as any to thank my former students — many of whom are now secondary teachers of mathematics somewhere in the Dallas-Fort Worth metroplex — for suggesting that I create this blog in the first place as a way of keeping touch with them after they graduated. I do hope that this blog has been a help.

Thoughts on Infinity: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students.

Part 1: Different types of countable sets

Part 2a: Divergence of the harmonic series.

Part 2b: Convergence of the Kempner series.

Part 3a: Conditional convergent series or products shouldn’t be rearranged.

Part 3b: Definition of the Euler-Mascheroni constant $\gamma$ .

Part 3c: Evaluation of the conditionally convergent series $\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} \dots$

Part 3d: Confirmation of this evaluation using technology.

Part 3e: Evaluation of a rearrangement of this conditionally convergent series.

Part 3f: Confirmation of this different evaluation using technology.

Part 3g: Closing thoughts.

Abbott and Costello and division

See also the similar post by Ma and Pa Kettle.

A Visual Proof of a Remarkable Trig Identity

Strange but true (try it on a calculator):

$\displaystyle \cos \left( \frac{\pi}{9} \right) \cos \left( \frac{2\pi}{9} \right) \cos \left( \frac{4\pi}{9} \right) = \displaystyle \frac{1}{8}$ .

Richard Feynman learned this from a friend when he was young, and it stuck with him his whole life.

Recently, the American Mathematical Monthly published a visual proof of this identity using a regular 9-gon:

Source: https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/1045091252206525/?type=3&theater

This same argument would work for any $2^n+1$ -gon. For example, a regular pentagon can be used to show that

$\displaystyle \cos \left( \frac{\pi}{5} \right) \cos \left( \frac{2\pi}{5} \right) = \displaystyle \frac{1}{4}$ ,

and a regular 17-gon can be used to show that

$\displaystyle \cos \left( \frac{\pi}{17} \right) \cos \left( \frac{2\pi}{17} \right) \cos \left( \frac{4\pi}{17} \right) \cos \left( \frac{8\pi}{17} \right) = \displaystyle \frac{1}{16}$ .

Katherine Johnson: NASA Pioneer and “Computer”

I recently stumbled on the article “This Black NASA Mathematician Was the Reason Many Astronauts Came Home — Their Life Depended on Her Calculations” featuring Katherine Johnson, one of the first African-Americans to work at NASA. The article features an interview with her; I recommend the whole thing highly.

Computing e to Any Power (Part 5)

In this series, I’m exploring the following ancedote from the book Surely You’re Joking, Mr. Feynman!, which I read and re-read when I was young until I almost had the book memorized.

One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for e^x, which is 1 + x + x^2/2! + x^3/3! Each term you get by multiplying the preceding term by x and dividing by the next number. For example, to get the next term after x^4/4! you multiply that term by x and divide by 5. It’s very simple.

When I was a kid I was excited by series, and had played with this thing. I had computed e using that series, and had seen how quickly the new terms became very small.

I mumbled something about how it was easy to calculate e to any power using that series (you just substitute the power for x).

“Oh yeah?” they said. “Well, then what’s e to the 3.3?” said some joker—I think it was Tukey.

I say, “That’s easy. It’s 27.11.”

Tukey knows it isn’t so easy to compute all that in your head. “Hey! How’d you do that?”

Another guy says, “You know Feynman, he’s just faking it. It’s not really right.”

They go to get a table, and while they’re doing that, I put on a few more figures.: “27.1126,” I say.

They find it in the table. “It’s right! But how’d you do it!”

“I just summed the series.”

“Nobody can sum the series that fast. You must just happen to know that one. How about e to the 3?”

“Look,” I say. “It’s hard work! Only one a day!”

“Hah! It’s a fake!” they say, happily.

“All right,” I say, “It’s 20.085.”

They look in the book as I put a few more figures on. They’re all excited now, because I got another one right.

Here are these great mathematicians of the day, puzzled at how I can compute e to any power! One of them says, “He just can’t be substituting and summing—it’s too hard. There’s some trick. You couldn’t do just any old number like e to the 1.4.”

I say, “It’s hard work, but for you, OK. It’s 4.05.”

As they’re looking it up, I put on a few more digits and say, “And that’s the last one for the day!” and walk out.

What happened was this: I happened to know three numbers—the logarithm of 10 to the base e (needed to convert numbers from base 10 to base e), which is 2.3026 (so I knew that e to the 2.3 is very close to 10), and because of radioactivity (mean-life and half-life), I knew the log of 2 to the base e, which is.69315 (so I also knew that e to the.7 is nearly equal to 2). I also knew e (to the 1), which is 2. 71828.

The first number they gave me was e to the 3.3, which is e to the 2.3—ten—times e, or 27.18. While they were sweating about how I was doing it, I was correcting for the extra.0026—2.3026 is a little high.

I knew I couldn’t do another one; that was sheer luck. But then the guy said e to the 3: that’s e to the 2.3 times e to the.7, or ten times two. So I knew it was 20. something, and while they were worrying how I did it, I adjusted for the .693.

Now I was sure I couldn’t do another one, because the last one was again by sheer luck. But the guy said e to the 1.4, which is e to the.7 times itself. So all I had to do is fix up 4 a little bit!

They never did figure out how I did it.

My students invariably love this story; let’s take a look at the third calculation.

Feynman knew that $e^{0.69315} \approx 2$ , so that

$e^{0.69315} e^{0.69315} = e^{1.3863} \approx 2 \times 2 = 4$ .

Therefore, again using the Taylor series expansion:

$e^{1.4} = e^{1.3863} e^{0.0137} = 4 e^{0.0137}$

$\approx 4 \times (1 + 0.0137)$

$= 4 + 4 \times 0.0137$

$\approx 4.05$ .

Again, I have no idea how he put on a few more digits in his head (other than his sheer brilliance), as this would require knowing the value of $\ln 2$ to six or seven digits as well as computing the next term in the Taylor series expansion.

Computing e to Any Power (Part 4)

In this series, I’m exploring the following ancedote from the book Surely You’re Joking, Mr. Feynman!, which I read and re-read when I was young until I almost had the book memorized.

One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for e^x, which is 1 + x + x^2/2! + x^3/3! Each term you get by multiplying the preceding term by x and dividing by the next number. For example, to get the next term after x^4/4! you multiply that term by x and divide by 5. It’s very simple.

When I was a kid I was excited by series, and had played with this thing. I had computed e using that series, and had seen how quickly the new terms became very small.

I mumbled something about how it was easy to calculate e to any power using that series (you just substitute the power for x).

“Oh yeah?” they said. “Well, then what’s e to the 3.3?” said some joker—I think it was Tukey.

I say, “That’s easy. It’s 27.11.”

Tukey knows it isn’t so easy to compute all that in your head. “Hey! How’d you do that?”

Another guy says, “You know Feynman, he’s just faking it. It’s not really right.”

They go to get a table, and while they’re doing that, I put on a few more figures.: “27.1126,” I say.

They find it in the table. “It’s right! But how’d you do it!”

“I just summed the series.”

“Nobody can sum the series that fast. You must just happen to know that one. How about e to the 3?”

“Look,” I say. “It’s hard work! Only one a day!”

“Hah! It’s a fake!” they say, happily.

“All right,” I say, “It’s 20.085.”

They look in the book as I put a few more figures on. They’re all excited now, because I got another one right.

Here are these great mathematicians of the day, puzzled at how I can compute e to any power! One of them says, “He just can’t be substituting and summing—it’s too hard. There’s some trick. You couldn’t do just any old number like e to the 1.4.”

I say, “It’s hard work, but for you, OK. It’s 4.05.”

As they’re looking it up, I put on a few more digits and say, “And that’s the last one for the day!” and walk out.

What happened was this: I happened to know three numbers—the logarithm of 10 to the base e (needed to convert numbers from base 10 to base e), which is 2.3026 (so I knew that e to the 2.3 is very close to 10), and because of radioactivity (mean-life and half-life), I knew the log of 2 to the base e, which is.69315 (so I also knew that e to the.7 is nearly equal to 2). I also knew e (to the 1), which is 2. 71828.

The first number they gave me was e to the 3.3, which is e to the 2.3—ten—times e, or 27.18. While they were sweating about how I was doing it, I was correcting for the extra.0026—2.3026 is a little high.

I knew I couldn’t do another one; that was sheer luck. But then the guy said e to the 3: that’s e to the 2.3 times e to the.7, or ten times two. So I knew it was 20. something, and while they were worrying how I did it, I adjusted for the .693.

Now I was sure I couldn’t do another one, because the last one was again by sheer luck. But the guy said e to the 1.4, which is e to the.7 times itself. So all I had to do is fix up 4 a little bit!

They never did figure out how I did it.

My students invariably love this story; let’s take a look at the second calculation.

Feynman knew that $e^{2.3026} \approx 10$ and $e^{0.69315} \approx 2$ , so that

$e^{2.3026} e^{0.69315} = e^{2.99575} \approx 10 \times 2 = 20$ .

Therefore, again using the Taylor series expansion:

$e^3 = e^{2.99575} e^{0.00425} = 20 e^{0.00425}$

$\approx 20 \times (1 + 0.00425)$

$= 20 + 20 \times 0.00425$

$= 20.085$ .

Again, I have no idea how he put on a few more digits in his head (other than his sheer brilliance), as this would require knowing the values of $\ln 10$ and $\ln 2$ to six or seven digits as well as computing the next term in the Taylor series expansion:

$e^3 = e^{\ln 20} e^{3 - \ln 20}$

$\approx 20 (1 +e^{ 0.0042677})$

$\approx 20 \times \left(1 + 0.0042677 + \frac{0.0042677^2}{2!} \right)$

$\approx 20.0855361\dots$

This compares favorably with the actual answer, $e^3 \approx 20.0855392\dots$ .

Computing e to Any Power (Part 3)

In this series, I’m exploring the following ancedote from the book Surely You’re Joking, Mr. Feynman!, which I read and re-read when I was young until I almost had the book memorized.

One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for e^x, which is 1 + x + x^2/2! + x^3/3! Each term you get by multiplying the preceding term by x and dividing by the next number. For example, to get the next term after x^4/4! you multiply that term by x and divide by 5. It’s very simple.

When I was a kid I was excited by series, and had played with this thing. I had computed e using that series, and had seen how quickly the new terms became very small.

I mumbled something about how it was easy to calculate e to any power using that series (you just substitute the power for x).

“Oh yeah?” they said. “Well, then what’s e to the 3.3?” said some joker—I think it was Tukey.

I say, “That’s easy. It’s 27.11.”

Tukey knows it isn’t so easy to compute all that in your head. “Hey! How’d you do that?”

Another guy says, “You know Feynman, he’s just faking it. It’s not really right.”

They go to get a table, and while they’re doing that, I put on a few more figures.: “27.1126,” I say.

They find it in the table. “It’s right! But how’d you do it!”

“I just summed the series.”

“Nobody can sum the series that fast. You must just happen to know that one. How about e to the 3?”

“Look,” I say. “It’s hard work! Only one a day!”

“Hah! It’s a fake!” they say, happily.

“All right,” I say, “It’s 20.085.”

They look in the book as I put a few more figures on. They’re all excited now, because I got another one right.

Here are these great mathematicians of the day, puzzled at how I can compute e to any power! One of them says, “He just can’t be substituting and summing—it’s too hard. There’s some trick. You couldn’t do just any old number like e to the 1.4.”

I say, “It’s hard work, but for you, OK. It’s 4.05.”

As they’re looking it up, I put on a few more digits and say, “And that’s the last one for the day!” and walk out.

What happened was this: I happened to know three numbers—the logarithm of 10 to the base e (needed to convert numbers from base 10 to base e), which is 2.3026 (so I knew that e to the 2.3 is very close to 10), and because of radioactivity (mean-life and half-life), I knew the log of 2 to the base e, which is.69315 (so I also knew that e to the.7 is nearly equal to 2). I also knew e (to the 1), which is 2. 71828.

The first number they gave me was e to the 3.3, which is e to the 2.3—ten—times e, or 27.18. While they were sweating about how I was doing it, I was correcting for the extra.0026—2.3026 is a little high.

I knew I couldn’t do another one; that was sheer luck. But then the guy said e to the 3: that’s e to the 2.3 times e to the.7, or ten times two. So I knew it was 20. something, and while they were worrying how I did it, I adjusted for the .693.

Now I was sure I couldn’t do another one, because the last one was again by sheer luck. But the guy said e to the 1.4, which is e to the.7 times itself. So all I had to do is fix up 4 a little bit!

They never did figure out how I did it.

My students invariably love this story; let’s take a look at the first calculation.

Feynman knew that $e^{2.3026} \approx 10$ and $e^1 \approx 2.71828$ , so that

$e^{3.3026} = e^{2.3026} \times e^1 \approx 10 \times 2.71828 = 27.1828$ .

That’s all well and good, but how did Feynman give an initial answer of $27.11$ in his head? The book doesn’t say, but we can guess that he used the Taylor series approximation:

$e^x = 1 + x - \displaystyle \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

$e^{-0.0026} \approx 1 - 0.0026$

Therefore,

$e^{3.3} = e^{3.3026} e^{-0.0026} \approx 27.1828 (1 - 0.0026)$

$\approx 27.18 - 27 \times 26 \times 0.0001$

$\approx 27.18 - 700 \times 0.0001$

$= 27.18 - 0.07$

$= 27.11$ .

However, I’m not going to lie… I have no idea (other than his sheer genius with numbers) how he was able to mentally tack on two extra digits. The above calculation, to four decimal places, yields $e^{3.3} \approx 27.1121$ , which is incorrect in the fourth decimal place. Indeed, if we use the more precise value of $\ln 10 = 2.3025851\dots$ , we would obtain

$e^{3.3} = e^{3.3025851\dots} e^{-0.0025851\dots} \approx 27.1828 (1 - 0.0025851\dots)$

$e^{3.3} \approx 27.112548\dots$ ,

which is still incorrect in the fourth decimal place. In other words, Feynman could not have just used the first two terms of the Taylor expansion of $e^x$ to obtain his answer of $27.1126$ . Therefore, Feynman must have used the next term in the Taylor series expansion as well as at least five decimal places of $\ln 10$ to make this calculation:

$e^{3.3} = e^{3.3025851\dots} e^{-0.0025851\dots} \approx 27.1828 \displaystyle \left(1 - 0.0025851\dots + \frac{(0.0025851\dots)^2}{2!} \right)$

And I have no idea how he could possibly have done this in his head in only a minute or two.