Bella: You gotta give me some answers.

Edward: “Yes”; “No”; “To get to the other side”; “1.77245…”

Bella (interrupting): I don’t want to know what the square root of pi is.

Edward (surprised): You knew that?

— Twilight

March 14 (month 3, day 14 of each year) is the day math-nerds celebrate the number *π* (3.14…), and you might be one of them. But if you’re getting tired of your *π* served plain, why not spice things up by combining the world’s favorite nerdy number with the world’s favorite nerdy operation?

The square root of *π* has attracted attention for almost as long as *π* itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius *r*=1, then its area is *πr*^{2} = *π*, so a square with side-length *s* has the same area as your circle if *s*^{2 } = *π*, that is, if *s* = sqrt(*π*). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(*π*) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of *π* crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

**A NONSENSICAL FACTORIAL**

Factorials are a handy shortcut in many counting problems. If someone asks you how many different ways there are to order the 52 cards of a standard deck, the answer is “52 factorial” (meaning 52 × 51 × 50 × … × 3 × 2 × 1, often written as “52!”); this answer takes a lot less time to say than “eighty unvigintillion, six hundred and fifty-eight vigintillion, …, eight hundred and twenty-four trillion”. But how should we define *n*! if *n* is not a counting number? What if, say, *n* is negative one-half?

A practical-minded person might gibe that someone who wants to know how many ways there are to order a stack consisting of negative one-half cards isn’t playing with a full deck. But this kind of craziness is often surprisingly fruitful; the symbolisms mathematicians come up with sometimes take on a life of their own, and living things want to grow. Or, as the great mathematician Leonhard Euler wrote (unless he didn’t^{1}), “My pen is a better mathematician than I am,” meaning that notations sometimes precede understanding. By trying to extend the factorial function to numbers like −1/2 that aren’t counting numbers, Euler was led to invent^{2} the gamma function, which is important not just in pure mathematics but in applications.

Specifically, Euler showed^{3} that when *n* is a positive integer, *n*! is equal to the integral

For those of you who know coordinate geometry but don’t know calculus, this expression represents the area bounded by the vertical axis *x* = 0, the horizontal axis *y* = 0, and the graph of the curve *y* = *x*^{n}*e*^{−}^{x}. Here *e* = 2.718… is the other famous nerdy number of math, dominating the world of exponentials and logarithms the way *π* rules the world of sines and cosines. The cool thing about that expression is that it makes sense when you replace *n* by −1/2 (corresponding to the area under the curve *y* = *x*^{−1/2}*e*^{−}^{x } in the picture below), and gives the value sqrt(*π*).

What’s more, if you find other integrals that coincide with *n*! when *n* is a positive integer, they’ll give you sqrt(*π*) when you replace *n* by −1/2. Metaphorically you could say that even though −1/2 times −3/2 times −5/2 times … times 3 times 2 times 1 isn’t equal to any number, the number it’s *trying* to equal is sqrt(*π*).

**A NONSENSICAL SUM**

Let’s consider a simpler example of an expression that’s “trying to equal” something. If *r* is a positive number that’s less than 1, then the infinite sum 1−*r*+*r*^{2}−*r*^{3}+… (an alternating geometric series) converges to 1/(1+*r*) in the sense that the partial sums 1, 1−*r*, 1−*r*+*r*^{2}, 1−*r*+*r*^{2}−*r*^{3},… get ever-closer to 1/(1+*r*) as the number of terms gets ever-bigger.^{4} Now, if *r* is 1, then 1/(1+*r*) makes sense and equals 1/2, but the alternating sum 1−1+1−1+… does not get ever-closer to 1/2 or to any other number. The partial sums 1, 1−1, 1−1+1, 1−1+1−1, … just vacillate between two values instead of converging: 1, 0, 1, 0, …

But what if we change the definition of convergence? Mathematician Ernesto Cesàro proposed a more permissive definition in which we take a sequence of numbers that annoyingly refuses to converge and replace it by a new, more tractable sequence whose *n*th term is the *average* of the first *n* terms of the old sequence.^{5}

If we apply Cesàro’s procedure to the sequence 1,0,1,0,… we get 1/1, (1+0)/2, (1+0+1)/3, (1+0+1+0)/4, … which sure enough converges to 1/2. And remember, 1/2 is exactly what we got from 1/(1+*r*) when we replaced *r* by 1.

So I’ve shown you two different shady methods of assigning a value to the non-converging sum 1−1+1−1+… . One method considers the more general sum 1−*r*+*r*^{2}−*r*^{3}+…, finds an expression for it that’s valid for all positive *r*<1, and then substitutes *r*=1; the method is called Abel summation. The other method replaces the ordinary notion of summation taught in calculus classes by Cesàro summation. And here’s an even simpler shady proof: Let *x* = 1 − 1 + 1 − 1 + … . Then

1 − *x* = 1 − (1 − 1 + 1 − 1 + …) = 1 − 1 + 1 − 1 + 1 − … = *x*,

so 1 − *x* = *x*, which yields *x* = 1/2.

What’s interesting is that even though these shady methods are shady in different ways, they all give the same value. This phenomenon − different shady methods arriving at the same answer − crops up a lot at the frontiers of math, and it often points to mathematical concepts that have not yet come into view. When we finally climb the hill that hid those concepts from view, we find that there’s nothing illegitimate about those concepts; they’re just more mathematically technical than, and not as attention-grabbing as, a formula like “1−1+1−1+…=1/2”. Such a formula, presented out of context, with all the sense-making scaffolding yanked away, is sort of like Lewis Carroll’s Cheshire cat: all that’s left is a (possibly infuriating) smirk.

An infamous example of mathematical YouTube sensationalism is the formula 1+2+3+… = −1/12 (and its less-celebrated sibling 1+1+1+… = −1/2). When equations like these are hyped outside of their proper context, math-fans sometimes respond with anger: “Those bastards are changing the rules again!” What often isn’t explained (because it’s fairly technical) is what the new rules are. What can and should be explained is that this rules-change is analogous to what you’d see if you were an American watching American football on TV and then changed the channel to watch Australian football. Well-educated sports fans don’t fume “They’re playing it wrong! Why, even my *gym teacher* knows more about football than these players do!”; they recognize that the Australians are playing a different sport. Same here, except that the name of the sport is “zeta-function regularization”^{6}, it’s played with numbers, and the players are analytic number theorists, only some of whom are Australian.

**LORD KELVIN’S DEFINITION OF A MATHEMATICIAN**

The physicist William Thomson, better known as Lord Kelvin, was a big fan of mathematics, calling it “the etherealization of common sense” and “the only good metaphysics”. According to an anecdote recounted by his biographer S. P. Thompson, Kelvin was a bit of an awestruck fanboy when it came to mathematicians themselves:

*Once when lecturing to a class he [Lord Kelvin] used the word “mathematician,” and then interrupting himself asked his class: “Do you know what a mathematician is?” Stepping to the blackboard he wrote upon it: *

*Then putting his finger on what he had written, he turned to his class and said: “A mathematician is one to whom that is as obvious as that twice two makes four is to you.”*

This formula (usually attributed to Gauss) is important and fairly well known, but I don’t think I know a single mathematician who regards the formula as obvious, so I don’t think Kelvin’s definition of a mathematician is a good one. Here’s an alternative definition for your consideration: A mathematician is one who recognizes the difference between what is obvious and what is merely familiar. Or: A mathematician is one who recognizes the difference between what is obvious and what one has come to understand in stages, by means of a nontrivial chain of trivial steps.

This seems like an ungrateful way to treat a scientist who, when all is said and done, was just trying to praise my kind. But a mathematician is one to whom flattery is annoying if it is inaccurate. (See, I can overgeneralize too!)

The expression

represents the area under the curve *y* = *e*^{−}^{x}^{2}.

The ∫ notation, which we saw earlier, is due to Leibniz, who chose a stylized version of the letter “*s*” to commemorate the fact that the way we can compute such areas is by first approximating them as __s__ums, and by then atoning for the error of the approximation by using ever-better approximations and seeing what the approximations converge to.

**SOLVING A PROBLEM BY MAKING IT SIMPLER**

Let’s talk about sums for a minute. Specifically, consider the problem of adding together all the numbers in the first four rows and first four columns of the multiplication table:

1 2 3 4

2 4 6 8

3 6 9 12

4 8 12 16

A trick for computing the sum is to consider that the product (1 + 2 + 3 + 4) × (1 + 2 + 3 + 4), if expanded out by the general distributive law, would have the sixteen numbers 1×1, 1×2, …, 4×3, and 4×4 as its constituent terms, so that the sum of the sixteen numbers must be 10 × 10, or 100. (“A mathematician is someone who works hard at being lazy,” said Murray Klamkin, riffing on George Pólya.)

You might want to apply the same idea to sum the numbers in the infinite array

1 1/2 1/4 1/8 …

1/2 1/4 1/8 1/16 …

1/4 1/8 1/16 1/32 …

1/8 1/16 1/32 1/64 …

…

(see Endnote #7).

This way of summing the numbers in an array is a cute trick, and it’s the sort of trick you see quite often in mathematics, where you reduce a two-dimensional problem to a one-dimensional problem, or more broadly solve a problem by reducing it to something that looks easier. But how often do you solve a problem by “reducing” it to something that looks harder? That counterintuitive tactic provides the nicest way I know to prove the famous formula of Gauss that Lord Kelvin quoted.

**SOLVING A PROBLEM BY MAKING IT MORE COMPLICATED**

Let *A* be the area between the *x*-axis and the curve *y* = *e*^{−}^{x}* ^{x}*, represented by Gauss’ integral. Then multivariate calculus can be used to show that

*A*

^{2}is the volume between the

*x*,

*y*-plane and the surface

*z*=

*e*

^{−}

^{x}

^{x}*e*

^{−}

^{y}

^{y}. (Note that I’m writing

*xx*and

*yy*in those exponents instead of the more common

*x*

^{2}and

*y*

^{2}because this version of the WordPress editor doesn’t do well with exponents-within-exponents.)

Using the laws of exponents, we can rewrite that right-hand side as *e*^{−(}^{x}^{x}^{ +}^{y}^{y}^{)}. This crucial step reveals that the surface has a surprising symmetry: it’s rotationally symmetric around the *z*-axis.^{8} That fact enables calculus students to treat the space between the *x*,*y*-plane and the surface *z* = *e*^{−(}^{x}^{x}^{ +}^{y}^{y}^{)} as a solid of revolution, and to compute its volume using the method of cylindrical shells. One gets *A*^{2} = *π*, from which it follows that *A* = sqrt(*π*): the formula on Lord Kelvin’s blackboard. (For more details, see John Cook’s writeup.)

I would modify Kelvin’s adage to say that a mathematician is someone who, having learned the preceding derivation, may not be able to remember the details, but remembers that circles play a role, and that the value of the integral therefore involves *π*. Or perhaps a mathematician is someone who finds the proof beautiful, irrespective of the beauty of the formula itself.

Gauss’ formula isn’t just a mathematical curiosity: the expression *e*^{−}^{x}* ^{x}* is closely related to the famous bell-shaped curve of statistics, and the fact that the area beneath it is sqrt(

*π*) explains where the

*π*in basic statistical formulas comes from.

To go back to the classic problem with which this essay began: You can’t *square* the *circle*, but you can do something much more important, namely, you can prove

by interpreting the *square* of the left hand side as a volume and then computing that volume using *circles*.

To give Lord Kelvin his due, let me credit him for praising mathematicians for conceptual understanding, even if his praise strikes me as excessive. His adage is a lot better than something along the lines of “A mathematician is someone who knows the first half-dozen digits of the square root of *π*.”

*Thanks to Henry Baker, Michael Collins, David Feldman, Sandi Gubin, David Jacobi, Kerry Mitchell, Cris Moore, Ben Orlin, Evan Romer, James Tanton, and Glen Whitney.*

Next month: Air from Archimedes.

**ENDNOTES**

#1. Can anyone provide a source, and/or the original non-English version of the aphorism?

#2. Euler defined Γ(*z*) as

which leads to a slightly annoying off-by-one issue: instead of Γ(*n*) = *n*! we have Γ(*n*) = (*n*−1)!. Another issue worth mentioning is that Euler’s definition only works for positive values of *z*; if *z* ≤ 0 the area under the curve becomes infinite. Fortunately a technique called analytic continuation can be used to figure out what expressions like Γ(−1/2) are “trying” to equal, but discussing that would take me too far from my theme.

Actually, it’s not too hard to guess what we might want Γ(−1/2) to equal: the formula *n*! = *n* (*n*−1)! corresponds to the formula Γ(*z*+1) = *z* Γ(*z*), which (with *z* = −1/2) tells us that Γ(−1/2) should be Γ(1/2)/(−1/2) = sqrt(*π*)/(−1/2) = −2 sqrt(*π*).

Here’s a plot of the gamma function (notice that it is undefined when *z* is zero or a negative integer):

The values of Γ(*z*) for *z* = 1/2, 3/2, etc. play a role in the general formula for the volume of an *n*-dimensional ball of radius *R*. This volume is given by

You might enjoy using the formula Γ(*z*+1) = *z* Γ(*z*) to compute Γ(3/2) and Γ(5/2) (given that Γ(1/2) = sqrt(pi)) and then use the values Γ(1) = 1, Γ(3/2) = sqrt(*π*) / 2, Γ(2) = 1, and Γ(5/2) = (3 sqrt(*π*)) / 4 to compute the volume of an *n*-dimensional ball for *n* = 0, 1, 2, and 3. Curiously, plugging in *n* = −1 tells us that the “volume of a (−1)-dimensional ball” of radius *R* is trying to equal 1/*πR*, but I have no idea what this might mean!

#3. Some facts about the gamma function are hard to prove, but one that’s doable by first-year calculus methods is the formula Γ(1) = 1 (using improper integrals). It’s only a little bit harder to prove the formula Γ(*z*+1) = *z* Γ(*z*) using integration by parts. With these two formulas, we can go on to prove

Γ(2) = 1 Γ(1) = 1,

Γ(3) = 2 Γ(2) = 2 × 1,

Γ(4) = 3 Γ(3) = 3 × 2 × 1,

etc. This shows that Γ(*n*) is indeed equal to (*n*−1)!.

#4. When *r* is between 0 and 1, the partial sum 1−*r*+*r*^{2}−…±*r** ^{n}* times 1+

*r*equals 1±

*r*

^{n}^{+1}, which converges to 1 as

*n*goes to infinity (since

*r*

^{n}^{+1}converges to 0 as

*n*goes to infinity); therefore 1−

*r*+

*r*

^{2}−…±

*r*

*converges to 1/(1+*

^{n}*r*).

#5. Cesaro’s definition is a conservative extension of the original limit concept. If you have a sequence that converges to some limit, then its “Cesàro-ization” converges to the same limit. The value of Cesàro’s method lies in the fact that that Cesàro-ization of a divergent (i.e., non-convergent) sequence is sometimes convergent. But the averaging trick isn’t a panacea; lots of divergent sequences remain divergent after you take those running averages. For instance, the series 1 − 2 + 4 − 8 + … converges to 1/3 by Abel summation, but no matter how many times you applying Cesàro’s trick to the sequence of partials sums, you get a divergent sequence. And neither Cesàro’s trick nor Abel’s lets you assign a meaningful finite number to a sum like 1+1+1+… or 1+2+3+…

#6. The black magic of zeta-regularization can also be used to show that “infinity factorial” (i.e., the product of all the positive integers) is trying to equal sqrt(2*π*) and that the product of all the primes is trying to equal 4*π*^{2}. See the article by E. Muñoz García and R. Pérez Marco listed in the References.

#7. The sum arises as the expansion of 1+1/2+1/4+1/8+… times 1+1/2+1/4+1/8+…, and since 1+1/2+1/4+1/8+… equals 2 (in the sense that the partial sums converge to 2), the sum of all those numbers equals 2 times 2, and as Lord Kelvin said, “twice two makes four”. On the other hand, we can take that table of numbers and group it by diagonals to get 1×1 + 2×1/2 + 3×1/4 + 4×1/8 + … So we’ve shown that the infinite sum 1/1 + 2/2 + 3/4 + 4/8 + 5/16 + … (where the numerators increase by adding 1 and the denominators increase by doubling) converges to 4. A mathematician is someone who thinks this is cute. Not “obvious” — just cute.

#8. Suppose the points (*x*,*y*) and (*x*‘,*y*‘) are at equal distance from the origin; call that distance *r*. Then we have *x*^{2}+*y*^{2} = *r*^{2} = (*x’*)^{2}+(*y’*)^{2}, so *e*^{−(}^{x}^{x}^{ +}^{y}^{y}^{)} equals *e*^{−((x’)(x’)}^{ +(y’)(y’)}^{)}, which tells us that the three-dimensional graph of the function *e*^{−(}^{x}^{x}^{ +}^{yy}^{)} has rotational symmetry. One thing I love about the proof of Gauss’ formula is the way it brings together so much mathematics: *e*, *π*, and even the Pythagorean Theorem!

**REFERENCES**

E. Lamb, Does 1+2+3… Really Equal −1/12?, Roots of Unity (blog hosted by Scientific American), 2014.

E. Muñoz García and R. Pérez Marco, “The Product Over All Primes is 4*π*^{2} “, Commun. Math. Phys. 277, 69–81 (2008).

David Richeson, Tales of Impossibility: The 2000-Year Quest to Solve the Mathematical Problems of Antiquity, Princeton University Press (2019).

Videos about “−(1/2)!”:

Presh Talkwalkar, “What is the Factorial of 1/2?”

blackandredpen, “The Gamma function & the Pi function”

blackandredpen, “But can you do negative factorial?”

Euler’s Academy, “Gamma Function: (−1/2)!”

Videos about 1+2+3+… = −1/12:

Numberphile, ASTOUNDING: 1+2+3+4+5+… = −1/12

Mathologer, Numberphile v. Math: the truth about 1+2+3+…=−1/12

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D.O.You didn’t mention a fun fact that the 2 integrals in the essay are related by a simple substitution. Would be tricky to explain to people who do not know calc.

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Yagub AliyevI have a photo of square root of pi appearing accidentally inside a fruit. I know it sound unbelievable but it is true. If you want to put into your nice article then inform me. I can give for free.

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jamesproppPost authorYes please! Email it to my gmail account.

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jamesproppPost authorYou can find the image here: https://faculty.uml.edu//jpropp/mathenchant/059-peach.jpg

Yagub writes: “My daughter Khadija spotted this and told me. While I was taking this photo she ate the other half where the remaining formula for the square root of pi could potentially be. A second later this part was also eaten. I found and read your paper because of this.”

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jonathanjcrabtreeFrom a given circle, the square root of pi is easy to draw.

Simply drop a length of string measuring a quarter of the way around a circle down through the centre.

Imagine dropping it from the 12 o’clock position on a clock face down the centre towards, yet not reaching 6 o’clock.

Then draw a perpendicular line from the end of this line to the circumference of the circle.

From the point where the perpendicular line from the vertical meets the circumference, draw a line to the 12 o’clock position.

The length of this hypothesis line equals the square root of pi.

If you’d like to see this, have a look at this video, with proofs in the description.

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jamesproppPost authorThanks!

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jamesproppPost authorBob Baillie writes:

The title reminds me of a tongue-in-cheek story, “First-Born Smarter Than Their

Siblings? We’ll See”, that I heard on NPR in 2007. I give the link below. The

reporter on the street was deliberately asking people questions that NPR knew

they wouldn’t be able to answer.

The first question to each person was, “What is the square root of pi?” Their

answers were hilarious! One guy knew the first couple of digits of pi, but that

was the closest anybody got.

I remember thinking at the time, “Oh, if only they had asked me!” I could have

given them the first 10 decimal places. I would have had 2 minutes of fame, but

better yet, I would loved to have seen the look on their faces when I answered a

question they clearly regarded as unanswerable.

Here’s the link. Enjoy!

https://www.npr.org/templates/story/story.php?storyId=11291639

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