Well, I divided this essay into two parts because the first was getting too long, and now I think the second is twice as long. But that just means it’s twice as interesting. Then the report on the amazing incompetence of Sam Bankman at running FTX came out, and that was topical, so I wanted to publish my comments on that immediately. But I think I must publish this as hot on Bankman’s heels as his creditors because I want to hand it out in class tomorrow to talk about next week. After all, we need to get to Conway’s theorems too. So here’s the rest.
In Part I, I talked about equations both the simple and quadratical, and the Binomial Theorem about which the Major-General was teeming with a lot of news. Now we come to many cheerful facts about the square of the hypotenuse.
The hypotenuse is the long diagonal side of a right triangle, where a right triangle is any triangle that has a right angle, which is a 90-degree angle. Figure 3 shows an example of a right triangle with sides of length a, b, and c, where c is the length of the long side, the hypotenuse. That’s the triangle you see repeated 8 times there.
The binomial quadratic expression (x+y)2 relates to how we proved the Pythagorean Theorem in class, which says that the square of the hypotenuse equals the sum of the squares of the other two sides of any right triangle. See Figure 3. The Pythagorean Theorem says that c2 = a2 + b2. That can be proved because in the left panel of Figure 3, (a+b)2 = a2 + b2 + 2ab, as we found in Part I, and ab/2 is the area T of the right triangle the Theorem talks about, but in the right panel (a+b)2 = c2 + 4T. Thus, a2 + b2 + 4T = c2 + 4T, so a2 + b2 = c2 , so we’ve proved the Theorem.
One student then pointed out that it also came up in another proof we had in class two months ago, the Odd Numbers Squared Minus One Are Divisible by Eight Theorem.
Theorem: Suppose a square's side is an odd number. Then the area of the square minus one is divisible by eight.
Example: Suppose a square’s side is 7 feet long. The area of that square is 49 square feet (7 times 7). Subtract 1, and you 48, which is divisible by 8.
Proof. Let the side of the square have length X. Since X is odd, we can write it as X = 2N+1 for some number N. See Figure 4.
What we must show is that (2N+1)2 -1 is divisible by 8.
\( \begin{eqnarray} (2N+1)^2 -1 & = & (2N+1)(2N+1) -1\\ && \nonumber\\ & = &(2N+1) (2N) + (2N+1) (1) -1 \nonumber\\ && \nonumber\\ & = &2N \cdot 2N + 2N\cdot 1+ 2N\cdot 1 + 1\cdot 1 -1\\ && \nonumber\\ & = & 4N^2 + 2N + 2N + 1 -1 \\ && \nonumber\\ & = & 4N^2 + 4N\\ && \nonumber\\ & = &4 (N^2+N)\\ && \nonumber\\ & = & 4 N (N+1) \end{eqnarray} \)
Clearly 4N(N+1) is divisible by 4. But we can show that N(N+1) is even (a lemma), so N(N+1) is divisible by 2 and the whole thing is divisible by (4) (2) = 8. The reason the lemma “N(N+1) is even” is true is that either N is even or (N+1) is even, so N(N+1) is an even number multiplied by an odd number. But anything multiplied by an even number is even. So N(N+1) is even, so it is divisible by 2 and 4N(N+1) is divisible by 8. Quod erat demonstrandum.
So much for verse 1 of the Major-General’s Song. That’s most of the math in Pirates of Penzance, except for the calculations that indicate to the noble Frederick “That though you’ve lived twenty-one years, yet, if we go by birthdays, you’re only five and a little bit over!”, the key mathematico-legal idea of the entire opera.
But we do have three more lines of math in The Major-General’s Song:
“I'm very good at integral and differential calculus;
I know the scientific names of beings animalculous:…”
and
“In conics I can floor peculiarities parabolous;”
Integral calculus is about computing the area under curves. Differential calculus is about computing the slope of curves. Consider the equation y = 6x - x2 shown in Figure 5.
If x = 1 then y = 6(1) - (1)(1) = 5. The slope at (1,5) is 4, the same as the slope of the line y = 1 + 4x that goes from (0, 1) to (1, 5) and beyond and which just barely touches the curve. The shaded area under the curve between x = 4 and x = 5 and above the x-axis equals 6 2/3. Calculus got me those numbers using the formulas and calculations below.
\(\begin{eqnarray} If \; y = x^n &\; then\;& \frac{dy}{dx} = n x^{n-1},\; so && \\ && \\ \frac{dy}{dx} &= &6 - 2x = 6 - 2(1) = 4. \\ &&\\ Also, \; &\int x^ndx = \frac{x^{n+1}}{n+1},&\; so \\ &&\\ \int_4^5 (6x - x^2)dx& = & \left|^5_4 3x^2 - \frac{x^3}{3} \right. \\ &&\\ & = & \left( 3(5)^2 - \frac{5^3}{3} \right) - \left( 3(4)^2 - \frac{4^3}{3} \right) \\ &&\\ & = & \left( 3(25) - \frac{125}{3} \right) - \left( 3(16) - \frac{64}{3} \right) \\ &&\\ & = & 75 - \frac{125}{3} - 48 + \frac{64}{3} \\ &&\\ & = & 27 - \frac{61}{3} \\ &&\\ & = & \frac{81}{3} - \frac{61}{3} \\ &&\\ & = & \frac{20}{3} \\ &&\\ & = & 6 \frac{2}{3} \\ \end{eqnarray}\)
If you take calculus, or look around on the Web, you can find out why those formulas are true. They’re half of what you need to know for practical calculus, since you can approximate any function over a finite range by combinations of polynomials like y = xn.
As for parabolas, the Greeks found that you could get the most important shapes of curves—circles, ellipses, parabolas, and hyperbolas— by slicing a cone four different ways. A parabola has an equation like y = 6x - x2; it’s a symmetric hill that rises gently to a highest point and then declines. It is also the shape of the path a ball makes when you throw it and it first rises, then falls under the quadratical pull of gravity. Thus, the curve in Figure 5 is a parabola. Figure 6 shows how all the conic sections are sliced out. The Greeks hadn’t discovered algebra (an Arabic word), so they had to do everything geometrically.
Having read all this, you are now well-equipped for battle, at least so long as the enemy doesn’t use the full-caliber Binomial Theorem against you or ask you tricky calculus questions like “What is the slope of eix? (see the last part of my “(a+b^n)/n = x”. Therefore, God Exists”). Do you think the People’s Liberation Army has it in their arsenal, or are my fears imaginary?
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