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.

 

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:

Feynman identity

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}.

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.

Computing e to Any Power (Part 2)

In this series, I’m looking at a wonderful anecdote from Nobel Prize-winning physicist Richard P. Feynman from his book Surely You’re Joking, Mr. Feynman!. This story concerns a time that he computed e^x mentally for a few values of x, much to the astonishment of his companions.

Part of this story directly ties to calculus.

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.

As noted, this refers to the Taylor series expansion of e^x, which is can be used to compute e to any power. The terms get very small very quickly because of the factorials in the denominator, thus lending itself to the computation of e^x. Indeed, this series is used by modern calculators (with a few tricks to accelerate convergence). In other words, the series from calculus explains how the mysterious “black box” of a graphing calculator actually works.

Continuing the story…

“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!”

For now, I’m going to ignore how Feynman did this computation in his head and instead discuss “the table.” The setting for this story was approximately 1940, long before the advent of handheld calculators. I’ll often ask my students, “The Brooklyn Bridge got built. So how did people compute e^x before calculators were invented?” The answer is by Taylor series, which were used to produce tables of values of e^x. So, if someone wanted to find e^{3.3}, they just had a book on the shelf.

For example, the following page comes from the book Marks’ Mechanical Engineers’ Handbook, 6th edition, which was published in 1958 and which I happen to keep on my bookshelf at home.

ExponentTable

Look down the fifth and sixth columns of this table, we see that e^{3.3} \approx 27.11. Somebody had computed all of these things (and plenty more) using the Taylor series, and they were compiled into a book and sold to mathematicians, scientists, and engineers.

But what if we needed an approximation better more accurate than four significant digits? Back in those days, there were only two options: do the Taylor series yourself, or buy a bigger book with more accurate tables.

Computing e to Any Power (Part 1)

Whenever I teach natural logarithms, I always share the following anecdote 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.

In this series, I’d like to take a deeper look at this wonderful anecdote.