# Richard Feynman’s Integral Trick

“I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] showed how to differentiate parameters under the integral sign — it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. [If] guys at MIT or Princeton had trouble doing a certain integral, [then] I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” (Surely you’re Joking, Mr. Feynman!)

I read Surely You’re Joking, Mr. Feynman! dozens of times when I was a teenager, and I was always curious about exactly what this integration technique actually was. So I enjoyed reading this article about the Leibniz Integration Rule: https://medium.com/dialogue-and-discourse/richard-feynmans-integral-trick-e7afae85e25c

# Computing e to Any Power: 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 examining one of Richard Feynman’s anecdotes about mentally computing $e^x$ for three different values of $x$.

Part 1: Feynman’s anecdote.

Part 2: Logarithm and antilogarithm tables from the 1940s.

Part 3: A closer look at Feynman’s computation of $e^{3.3}$.

Part 4: A closer look at Feynman’s computation of $e^{3}$.

Part 5: A closer look at Feynman’s computation of $e^{1.4}$.

# 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:

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

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

# Lessons from teaching gifted elementary students (Part 6b)

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 surprising depth of mathematical knowledge.

Here’s a question I once received:

255/256 to what power is equal to 1/2? And please don’t use a calculator.

Here’s how I answered this question without using a calculator… in fact, I answered it without writing anything down at all. I thought of the question as

$\displaystyle \left( 1 - \epsilon \right)^x = \displaystyle \frac{1}{2}$.

$\displaystyle x \ln (1 - \epsilon) = \ln \displaystyle \frac{1}{2}$

$\displaystyle x \ln (1 - \epsilon) = -\ln 2$

I was fortunate that my class chose 1/2, as I had memorized (from reading and re-reading Surely You’re Joking, Mr. Feynman! when I was young) that $\ln 2 \approx 0.693$. Therefore, we have

$x \ln (1 - \epsilon) \approx -0.693$.

Next, I used the Taylor series expansion

$\ln(1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} \dots$

to reduce this to

$-x \epsilon \approx -0.693$,

or

$x \approx \displaystyle \frac{0.693}{\epsilon}$.

For my students’ problem, I had $\epsilon = \frac{1}{256}$, and so

$x \approx 256(0.693)$.

So all I had left was the small matter of multiplying these two numbers. I thought of this as

$x \approx 256(0.7 - 0.007)$.

Multiplying $256$ and $7$ in my head took a minute or two:

$256 \times 7 = 250 \times 7 + 6 \times 7$

$= 250 \times (8-1) + 42$

$= 250 \times 8 - 250 + 42$

$= 2000 - 250 + 42$

$= 1750 + 42$

$= 1792$.

Therefore, $256 \times 0.7 = 179.2$ and $256 \times 0.007 = 1.792 \approx 1.8$. Therefore, I had the answer of

$x \approx 179.2 - 1.8 = 177.4 \approx 177$.

So, after a couple minutes’ thought, I gave the answer of 177. I knew this would be close, but I had no idea it would be so close to the right answer, as

$x = \displaystyle \frac{\displaystyle \ln \frac{1}{2} }{\displaystyle \ln \frac{255}{256}} \approx 177.0988786\dots$

# Square roots and Logarithms Without a Calculator: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do 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 computing square roots and logarithms without a calculator.

Part 1: Method #1: Trial and error.

Part 2: Method #2: An algorithm comparable to long division.

Part 3: Method #3: Introduction to logarithmic tables. At the time of this writing, this is the most viewed page on my blog.

Part 4: Finding antilogarithms with a table.

Part 5: Pedagogical and historical thoughts on log tables.

Part 6: Computation of square roots using a log table.

Part 7: Method #4: Slide rules

Part 8: Method #5: By hand, using a couple of known logarithms base 10, the change of base formula, and the Taylor approximation $ln(1+x) \approx x$.

Part 9: An in-class activity for getting students comfortable with logarithms when seen for the first time.

Part 10: Method #6: Mentally… anecdotes from Nobel Prize-winning physicist Richard P. Feynman and me.

Part 11: Method #7: Newton’s Method.

# New Math, by Tom Lehrer

To add a little levity to the recent posts about the Common Core, here’s the song “New Math” by Tom Lehrer that poked fun at the New Math curriculum of the 1960s.

For some context, here’s a succinct summary of the new math fad by Richard Feynman, who won the Nobel Prize in Physics and served on a commission for choosing math textbooks for California in 1964:

I understood what they were trying to do. Many [Americans] thought we were behind the Russians after Sputnik, and some mathematicians were asked to give advice on how to teach math by using some of the rather interesting modern concepts of mathematics. The purpose was to enhance mathematics for the children who found it dull.

I’ll give you an example: They would talk about different bases of numbers — five, six, and so on — to show the possibilities. That would be interesting for a kid who could understand base ten — something to entertain his mind. But what they turned it into, in these books, was that every child had to learn another base! And then the usual horror would come: “Translate these numbers, which are written in base seven, to base five.” Translating from one base to another is an utterly useless thing. If you can do it, maybe it’s entertaining; if you can’t do it, forget it. There’s no point to it.

From the chapter “Judging Books by their Covers” in Surely You’re Joking, Mr. Feynman!

# Teaching for understanding and teaching procedures

Many critics of the current state of mathematics education take issue with asking students to explain their reasoning. They’d rather students just apply an algorithm and get the answer.

The following is quoted from QED: The Strange Theory of Light and Matter, where Richard Feynman describes how he’s going to explain for a lay audience the techniques behind quantum mechanics that earned him a Nobel Prize. (By the way, I highly recommend this book.)

How am I going to explain to you the things I don’t explain to my students until they are third-year graduate students? Let me explain it by analogy.

The Maya Indians were interested in the rising and setting of Venus as a morning “star” and as an evening “star” – they were very interested in when it would appear. After some years of observation, they noted that five cycles of Venus were very nearly equal to eight of their “nominal years” of 365 days (they were aware that the true year of seasons was different and they made calculations of that also). To make calculations, the Maya had invented a system of bars and dots to represent numbers (including zero), and had rules by which to calculate and predict not only the risings and settings of Venus, but other celestial phenomena, such as lunar eclipses.

In those days, only a few Maya priests could do such elaborate calculations. Now, suppose we were to ask one of them how to do just one step in the process of predicting when Venus will next rise as a morning star – subtracting two numbers. And let’s assume that, unlike today, we had not gone to school and did not know how to subtract. How would the priest explain to us what subtraction is?

He could either teach us the numbers represented by the bars and dots and the rules for “subtracting” them, or he could tell us what he was really doing: “Suppose we want to subtract 236 from 584. First, count out 584 beans and put them in a pot. Then take out 236 beans and put them to one side. Finally, count the beans left in the pot. That number is the result of subtracting 236 from 584.”

You might say, “My Quetzalcoatl! What tedium, counting beans, putting them in, taking them out – what a job!”

To which the priest would reply, “That’s why we have the rules for the bars and dots. The rules are tricky, but they are a much more efficient way of getting the answer than by counting beans. The important thing is, it makes no difference as far as the answer is concerned: we can predict the appearance of Venus by counting beans (which is slow, but easy to understand) or by using the tricky rules (which is much faster, but you must spend years in school to learn them).”

To understand how subtraction works – as long as you don’t have to actually carry it out – is really not so difficult.

That’s my position: I’m going to explain to you what the physicists are doing when they are predicting how Nature will behave, but I’m not going to teach you any tricks so you can do it efficiently. You will discover that in order to make any reasonable predictions with this new scheme of quantum electrodynamics, you would have to make an awful lot of little arrows on a piece of paper. It takes seven years – four undergraduate and three graduate to train our physics students to do that in a tricky, efficient way. That’s where we are going to skip seven years of education in physics: By explaining quantum electrodynamics to you in terms of what we are really doing, I hope you will be able to understand it better than do some of the students!

In the same way, I want students in 2nd and 3rd grades to understand what they are really doing when they subtract, and not just mindlessly follow a procedure to get an answer that they do not really understand.

Where I tend to agree with most critics of the Common Core is that students are asked to write miniature essays to explain their reasoning, and that’s probably a bad idea. Even though I want students to understand why subtraction works, 2nd and 3rd graders are still learning how to write complete sentences and can get easily frustrated with explaining their reasoning in paragraph form. I think there are better ways (like drawing pictures) of assessing whether young children really understand subtraction that is more developmentally appropriate.

Throughout grades K-10, students are slowly introduced to the concept of angles. They are told that there are $90$ degrees in a right angle, $180$ degrees in a straight angle, and a circle has $60$ degrees. They are introduced to $30-60-90$ and $45-45-90$ right triangles. Fans of snowboarding even know the multiples of $180$ degrees up to $1440$ or even $1620$ degrees.

Then, in Precalculus, we make students get comfortable with $\pi$, $\displaystyle \frac{\pi}{2}$, $\displaystyle \frac{\pi}{3}$, $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{\pi}{6}$, and multiples thereof.

We tell students that radians and degrees are just two ways of measuring angles, just like inches and centimeters are two ways of measuring the length of a line segment.

Still, students are extremely comfortable with measuring angles in degrees. They can easily visualize an angle of $75^o$, but to visualize an angle of $2$ radians, they inevitably need to convert to degrees first. In his book Surely You’re Joking, Mr. Feynman!, Nobel-Prize laureate Richard P. Feynman described himself as a boy:

I was never any good in sports. I was always terrified if a tennis ball would come over the fence and land near me, because I never could get it over the fence – it usually went about a radian off of where it was supposed to go.

Naturally, students wonder why we make them get comfortable with measuring angles with radians.

The short answer, appropriate for Precalculus students: Certain formulas are a little easier to write with radians as opposed to degrees, which in turn make certain formulas in calculus a lot easier.

The longer answer, which Precalculus students would not appreciate, is that radian measure is needed to make the derivatives of $\sin x$ and $\cos x$ look palatable.

Source: http://mathworld.wolfram.com/CircularSector.html

1. In Precalculus, the length of a circle arc with central angle $\theta$ in a circle with radius $r$ is

$s = r\theta$

Also, the area of a circular sector with central angle $\theta$ in a circle with radius $r$ is

$A = \displaystyle \frac{1}{2} r^2 \theta$

In both of these formulas, the angle $\theta$ must be measured in radians.

Students may complain that it’d be easy to make a formula of $\theta$ is measured in degrees, and they’d be right:

$s = \displaystyle \frac{180 r \theta}{\pi}$ and $A = \displaystyle \frac{180}{\pi} r^2 \theta$

However, getting rid of the $180/\pi$ makes the following computations from calculus a lot easier.

2a. Early in calculus, the limit

$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

is derived using the Sandwich Theorem (or Pinching Theorem or Squeeze Theorem). I won’t reinvent the wheel by writing out the proof, but it can be found here. The first step of the proof uses the formula for the above formula for the area of a circular sector.

2b. Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$, we replace $x$ by $\theta/2$ to find

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \frac{2\sin^2 \displaystyle \left( \frac{\theta}{2} \right)}{ \theta}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \sin \left( \frac{\theta}{2} \right) \cdot \frac{\sin \displaystyle \left( \frac{\theta}{2} \right)}{ \displaystyle \frac{\theta}{2}}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0 \cdot 1$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0$

3. Both of the above limits — as well as the formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ — are needed to prove that $\displaystyle \frac{d}{dx} \sin x = \cos x$ and $\displaystyle \frac{d}{dx} \cos x = -\sin x$. Again, I won’t reinvent the wheel, but the proofs can be found here.

So, to make a long story short, radians are used to make the derivatives $y = \sin x$ and $y = \cos x$ easier to remember. It is logically possible to differentiate these functions using degrees instead of radians — see http://www.math.ubc.ca/~feldman/m100/sinUnits.pdf. However, possible is not the same thing as preferable, as calculus is a whole lot easier without these extra factors of $\pi/180$ floating around.

# Square roots and logarithms without a calculator (Part 10)

This is the fifth in a series of posts about calculating roots without a calculator, with special consideration to how these tales can engage students more deeply with the secondary mathematics curriculum. As most students today have a hard time believing that square roots can be computed without a calculator, hopefully giving them some appreciation for their elders.

Today’s story takes us back to a time before the advent of cheap pocket calculators: 1949.

The following story comes from the chapter “Lucky Numbers” of Surely You’re Joking, Mr. Feynman!, a collection of tales by the late Nobel Prize winning physicist, Richard P. Feynman. Feynman was arguably the greatest American-born physicist — the subject of the excellent biography Genius: The Life and Science of Richard Feynman — and he had a tendency to one-up anyone who tried to one-up him. (He was also a serial philanderer, but that’s another story.) Here’s a story involving how, in the summer of 1949, he calculated $\sqrt[3]{1729.03}$ without a calculator.

The first time I was in Brazil I was eating a noon meal at I don’t know what time — I was always in the restaurants at the wrong time — and I was the only customer in the place. I was eating rice with steak (which I loved), and there were about four waiters standing around.

A Japanese man came into the restaurant. I had seen him before, wandering around; he was trying to sell abacuses. (Note: At the time of this story, before the advent of pocket calculators, the abacus was arguably the world’s most powerful hand-held computational device.) He started to talk to the waiters, and challenged them: He said he could add numbers faster than any of them could do.

The waiters didn’t want to lose face, so they said, “Yeah, yeah. Why don’t you go over and challenge the customer over there?”

The man came over. I protested, “But I don’t speak Portuguese well!”

The waiters laughed. “The numbers are easy,” they said.

They brought me a paper and pencil.

The man asked a waiter to call out some numbers to add. He beat me hollow, because while I was writing the numbers down, he was already adding them as he went along.

I suggested that the waiter write down two identical lists of numbers and hand them to us at the same time. It didn’t make much difference. He still beat me by quite a bit.

However, the man got a little bit excited: he wanted to prove himself some more. “Multiplição!” he said.

Somebody wrote down a problem. He beat me again, but not by much, because I’m pretty good at products.

The man then made a mistake: he proposed we go on to division. What he didn’t realize was, the harder the problem, the better chance I had.

We both did a long division problem. It was a tie.

This bothered the hell out of the Japanese man, because he was apparently well trained on the abacus, and here he was almost beaten by this customer in a restaurant.

Raios cubicos!” he says with a vengeance. Cube roots! He wants to do cube roots by arithmetic. It’s hard to find a more difficult fundamental problem in arithmetic. It must have been his topnotch exercise in abacus-land.

He writes down a number on some paper— any old number— and I still remember it: $1729.03$. He starts working on it, mumbling and grumbling: “Mmmmmmagmmmmbrrr”— he’s working like a demon! He’s poring away, doing this cube root.

Meanwhile I’m just sitting there.

One of the waiters says, “What are you doing?”.

I point to my head. “Thinking!” I say. I write down $12$ on the paper. After a little while I’ve got $12.002$.

The man with the abacus wipes the sweat off his forehead: “Twelve!” he says.

“Oh, no!” I say. “More digits! More digits!” I know that in taking a cube root by arithmetic, each new digit is even more work that the one before. It’s a hard job.

He buries himself again, grunting “Rrrrgrrrrmmmmmm …,” while I add on two more digits. He finally lifts his head to say, “$12.0$!”

The waiter are all excited and happy. They tell the man, “Look! He does it only by thinking, and you need an abacus! He’s got more digits!”

He was completely washed out, and left, humiliated. The waiters congratulated each other.

How did the customer beat the abacus?

The number was $1729.03$. I happened to know that a cubic foot contains $1728$ cubic inches, so the answer is a tiny bit more than $12$. The excess, $1.03$, is only one part in nearly $2000$, and I had learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. So all I had to do is find the fraction $1/1728$, and multiply by $4$ (divide by $3$ and multiply by $12$). So I was able to pull out a whole lot of digits that way.

A few weeks later, the man came into the cocktail lounge of the hotel I was staying at. He recognized me and came over. “Tell me,” he said, “how were you able to do that cube-root problem so fast?”

I started to explain that it was an approximate method, and had to do with the percentage of error. “Suppose you had given me $28$. Now the cube root of $27$ is $3$ …”

He picks up his abacus: zzzzzzzzzzzzzzz— “Oh yes,” he says.

I realized something: he doesn’t know numbers. With the abacus, you don’t have to memorize a lot of arithmetic combinations; all you have to do is to learn to push the little beads up and down. You don’t have to memorize 9+7=16; you just know that when you add 9, you push a ten’s bead up and pull a one’s bead down. So we’re slower at basic arithmetic, but we know numbers.

Furthermore, the whole idea of an approximate method was beyond him, even though a cubic root often cannot be computed exactly by any method. So I never could teach him how I did cube roots or explain how lucky I was that he happened to choose $1729.03$.

The key part of the story, “for small fractions, the cube root’s excess is one-third of the number’s excess,” deserves some elaboration, especially since this computational trick isn’t often taught in those terms anymore. If $f(x) = (1+x)^n$, then $f'(x) = n (1+x)^{n-1}$, so that $f'(0) = n$. Since $f(0) = 1$, the equation of the tangent line to $f(x)$ at $x = 0$ is

$L(x) = f(0) + f'(0) \cdot (x-0) = 1 + nx$.

The key observation is that, for $x \approx 0$, the graph of $L(x)$ will be very close indeed to the graph of $f(x)$. In Calculus I, this is sometimes called the linearization of $f$ at $x =a$. In Calculus II, we observe that these are the first two terms in the Taylor series expansion of $f$ about $x = a$.

For Feynman’s problem, $n =\frac{1}{3}$, so that $\sqrt[3]{1+x} \approx 1 + \frac{1}{3} x$ if $x \approx 0$. Then $\latex \sqrt[3]{1729.03}$ can be rewritten as

$\sqrt[3]{1729.03} = \sqrt[3]{1728} \sqrt[3]{ \displaystyle \frac{1729.03}{1728} }$

$\sqrt[3]{1729.03} = 12 \sqrt[3]{\displaystyle 1 + \frac{1.03}{1728}}$

$\sqrt[3]{1729.03} \approx 12 \left( 1 + \displaystyle \frac{1}{3} \times \frac{1.03}{1728} \right)$

$\sqrt[3]{1729.03} \approx 12 + 4 \times \displaystyle \frac{1.03}{1728}$

This last equation explains the line “all I had to do is find the fraction $1/1728$, and multiply by $4$.” With enough patience, the first few digits of the correction can be mentally computed since

$\displaystyle \frac{1.03}{500} < \frac{1.03}{432} = 4 \times \frac{1.03}{1728} < \frac{1.03}{400}$

$\displaystyle \frac{1.03 \times 2}{1000} < 4 \times \frac{1.03}{1728} < \frac{1.03 \times 25}{10000}$

$0.00206 < 4 \times \displaystyle \frac{1.03}{1728} < 0.0025075$

So Feynman could determine quickly that the answer was $12.002\hbox{something}$.

By the way,

$\sqrt[3]{1729.03} \approx 12.00238378\dots$

$\hbox{Estimate} \approx 12.00238426\dots$

So the linearization provides an estimate accurate to eight significant digits. Additional digits could be obtained by using the next term in the Taylor series.

I have a similar story to tell. Back in 1996 or 1997, when I first moved to Texas and was making new friends, I quickly discovered that one way to get odd facial expressions out of strangers was by mentioning that I was a math professor. Occasionally, however, someone would test me to see if I really was a math professor. One guy (who is now a good friend; later, we played in the infield together on our church-league softball team) asked me to figure out $\sqrt{97}$ without a calculator — before someone could walk to the next room and return with the calculator. After two seconds of panic, I realized that I was really lucky that he happened to pick a number close to $100$. Using the same logic as above,

$\sqrt{97} = \sqrt{100} \sqrt{1 - 0.03} \approx 10 \left(1 - \displaystyle \frac{0.03}{2}\right) = 9.85$.

Knowing that this came from a linearization and that the tangent line to $y = \sqrt{1+x}$ lies above the curve, I knew that this estimate was too high. But I didn’t have time to work out a correction (besides, I couldn’t remember the full Taylor series off the top of my head), so I answered/guessed $9.849$, hoping that I did the arithmetic correctly. You can imagine the amazement when someone punched into the calculator to get $9.84886\dots$

# Square roots and logarithms without a calculator (Part 8)

I’m in the middle of a series of posts concerning the elementary operation of computing a root. This is such an elementary operation because nearly every calculator has a $\sqrt{~~}$ button, and so students today are accustomed to quickly getting an answer without giving much thought to (1) what the answer means or (2) what magic the calculator uses to find roots. I like to show my future secondary teachers a brief history on this topic… partially to deepen their knowledge about what they likely think is a simple concept, but also to give them a little appreciation for their elders.

To begin, let’s again go back to a time before the advent of pocket calculators… say, 1952.

This story doesn’t go back to 1952 but to Boxing Day 2012 (the day after Christmas). For some reason, my daughter — out of the blue — asked me to compute $\sqrt[19]{25727}$ without a calculator. As my daughter adores the ground I walk on — and I want to maintain this state of mind for as long as humanly possible — I had no choice but to comply. So I might as well have been back in 1952.

In the past few posts, I discussed how log tables and slide rules were used by previous generations to perform this calculation. The problem was that all of these tools were in my office and not at home, and hence were not of immediate use.

The good news is that I had a few logarithms memorized:

$\log_{10} 2 \approx 0.301$, $\log_{10} 3 \approx 0.477$, $\log_{10} 7 = 0.845$,

and $\ln 10 = 2.3$.

I had the first two logs memorized when I was a child; the third I memorized later. As I’ll describe, the first three logarithms can be used with the laws of logarithms to closely approximate the base-10 logarithm of nearly any number. The last logarithm was important in previous generations for using the change-of-base formula from $\log_{10}$ to $\ln$. It was also prominently mentioned in the chapter “Lucky Numbers” from a favorite book of my childhood, Surely You’re Joking Mr. Feynman, so I had that memorized as well.

I also knew that $\ln(1+x) \approx x$ for $x = 0$ from the Taylor expansion of $\ln(1+x)$.

To begin, I first noticed that $25727 \approx 25600$, and I knew I could get $\log_{10} 25600$ since $25600 = 2^8 \times 100$. So I started with

$\log_{10} 25727 = \log_{10} \left(100 \times 256 \times \displaystyle \frac{257.27}{256} \right)$

$\log_{10} 25727 \approx \log_{10} 100 + 8 \log_{10} 2 + \log_{10} 1.005$

$\log_{10} 25727 \approx 2 + 8(0.301) + \displaystyle \frac{\ln 1.005}{\ln 10}$

$\log_{10} 25727 \approx 4.408 + \displaystyle \frac{0.005}{2.3}$

$\log_{10} 25727 \approx 4.408 + 0.002$

$\log_{10} 25727 \approx 4.410$

I did all of the above calculations by hand, cutting off after three decimal places (since I had those base-10 logarithms memorized to only three decimal places). Therefore,

$\log_{10} 25727^(1/19) = \displaystyle \frac{1}{19} \log_{10} 25727 \approx \displaystyle \frac{4.410}{19} \approx 0.232$

So, to complete the calculation, I had to find the value of $x$ so that $\log_{10} x = 0.232$. This was by far the hardest step, since it could only be done by trial and error. I forget exactly what steps I tried, but here’s a sample:

$\log_{10} 2 \approx 0.301$. Too big.

$\log_{10} 1.5 = \log_{10} \displaystyle \frac{3}{2} = \log_{10} 3 - \log_{10} 2 \approx 0.477 - 0.301 = 0.176$. Too small.

$\log_{10} 1.6 = \log_{10} \displaystyle \frac{2^4}{10} = 4\log_{10} 2 - \log_{10} 10 \approx 4(0.301) - 1 = 0.203$. Too small.

$\log_{10} 1.8 = \log_{10} \displaystyle \frac{2 \cdot 3^2}{10} \approx 0.301 + 2(0.477) - 1 = 0.255$. Too big.

Eventually, I got to

$\log_{10} 1.71 = \log_{10} \displaystyle \frac{3^2 \cdot 19}{100}$

$\log_{10} 1.71 = 2\log_{10} 3 + \log_{10} 19 - \log_{10} 100$

$\log_{10} 1.71 \approx 2(0.477) + \displaystyle \frac{\log_{10} 18 + \log_{10} 20}{2} - 2$

$\log_{10} 1.71 \approx -1.046 + \frac{1}{2} (\log_{10} 2 + 2 \log_{10} 3 + \log_{10} 2 + \log_{10} 10)$

$\log_{10} 1.71 \approx -1.046 + \frac{1}{2}(2.556)$

$\log_{10} 1.71 \approx 0.232$

So, after a hour or two of arithmetic, I told her my answer: $\sqrt[19]{25727} \approx 1.71$. You can imagine my sheer delight when we checked my answer with a calculator:

In Part 9, I’ll discuss my opinion about whether or not these kinds of calculations have any pedagogical value for students learning logarithms.