## Identity 2

This is prompted by another Catriona Shearer puzzle. Would you say it’s at all obvious that

$2 \arctan (\sqrt{3}/7) = \arccos (23/26)$ ?

Me neither. True, though. Look at this triangle:

By the law of cosines, $3 = 13 + 13 - 2\times13\cos(\theta)$ so $\theta = \arccos (23/26)$. But if you slice that triangle in half, you find its height is $7/2$ and so $\theta = 2\arctan(\sqrt{3}/7)$.

More generally, the half angle formula for the tangent is

$\tan(\theta/2) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}$

implying

$\cos\theta = \frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$.

Then if $\theta/2 = \arctan(a/b)$,

$2\arctan\left(\frac{a}{b}\right) = \arccos\left(\frac{b^2-a^2}{b^2+a^2}\right)$

and with $a=\sqrt{3}$, $b=7$ you get $2\arctan(\sqrt{3}/7) = \arccos(46/52) = \arccos(23/26)$.

## Identity

While struggling with a Catronia Shearer problem I stumbled across a proof of the trig identity

$\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$

which I hadn’t seen and probably wouldn’t have thought of if I’d been looking for it, though there’s nothing non-obvious about it in retrospect. Here it is. Draw a unit semicircle centered on O, with A and C the endpoints of the diameter. Draw chords AD and DC. (Angle ADC is a right angle.) Let angle CAD be $\theta$. By the Inscribed Angle Theorem, angle COD is $2\theta$. Drop a perpendicular from D to the diameter at B.

Now $\mathrm {AO} = 1$ and $\mathrm {OB} = \cos 2\theta$, so $\mathrm {AB} = 1+\cos 2\theta$. But $\mathrm {AC} = 2$ so $\mathrm {AD} = 2 \cos \theta$, so $\mathrm {AB} = 2 \cos^2 \theta$. Therefore $\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$.

## Deep thought required

The integer 10 can be written as the sum of three cubes: $10=2^3+1^3+1^3$. So can 6, although that isn’t obvious until you think of using negative integers: $6 = 2^3+(-1)^3+(-1)^3$. For that matter, there are easy solutions to the problem of finding such sums of three cubes for every integer up to 10 except for 4 and 5:

• $0 = 0^3+0^3+0^3$
• $1 = 1^3+0^3+0^3$
• $2 = 1^3+1^3+0^3$
• $3 = 1^3+1^3+1^3$
• $4 =$???
• $5 =$???
• $6 = 2^3+(-1)^3+(-1)^3$
• $7 = 2^3+(-1)^3+0^3$
• $8 = 2^3+0^3+0^3$
• $9 = 2^3+1^3+0^3$
• $10 = 2^3+1^3+1^3$

That prompts a few questions: Are there solutions for 4 and 5? Are there solutions for all integers? Or for an infinite number of integers?

The answers to the first two questions are, no and no. Start by observing that the cube of any multiple of 3, $(3n)^3$, is divisible by 9 (in fact 27). For numbers equal to 1 mod 3, $(3n+1)^3 = 27n^3+27n^2+9n+1$ so is equal to 1 mod 9, and for numbers equal to -1 mod 3, $(3n-1)^3 = 27n^3-27n^2+9n-1$ so is equal to -1 mod 9. So all cubes are equal to 0 or ±1 mod 9. From this you can figure out all sums of three cubes are equal to 0, 1, 2, or 3 mod 9. But 4 and 5 equal -4 and +4 mod 9. So they cannot be written as sums of three cubes; nor can 13, 14, 22, 23, 31, 32, ….

As for the third question, it’s conjectured all other integers can be expressed as sums of three cubes. But it’s not proved.

The integers up through 11 are easy. In fact you might realize 10 can be done another way: $10 = 4^3+(-3)^3+(-3)^3$. For that matter, $8 = 2^3+1^3+(-1)^3 = 2^3+2^3+(-2)^3$ and so on for an infinite number of solutions.

But 12 might take you a little longer: $12 = 10^3+7^3+(-11)^3$. Worse, $21 = 16^3+(-11)^3+(-14)^3$. Things are pretty easy up through 29, and then 30 is hard. Very hard.

Good luck coming up with the simplest known answer:

$30 = 2220422932^3+(-283059965)^3+(-2218888517)^3$.

Yikes!

So it goes. Lots of easy ones

$53 = 3^3+3^3+(-1)^3$

interspersed with slightly tougher ones

$52 = 60702901317^3 + 23961292454^3 + (-61922712865)^3$

But as of a few days ago, there were solutions known for all but two numbers less than 100 (other than the ±4 mod 9 impossibilities).

That was then. Andrew R. Booker has just published this result for the smallest hitherto-unsolved case:

$8866128975287528^3 + (-8778405442862239)^3 + (-2736111468807040)^3$

$= 696950821015779435648178972565490929714876221952$
$-676467453392982277424361019810585360331722557919$
$-20483367622797158223817952754905569383153664000$

$= 696950821015779435648178972565490929714876221952$
$-696950821015779435648178972565490929714876221919$

$= 33$

I hardly need add that this was not found with a Python script during Booker’s lunch hour. The paper goes into the number theoretic gymnastics involved in his computer search, which, he reports, “used approximately 15 core-years over three weeks of real time.”

So that’s 33 solved. We’re down to one remaining unsolved number under 100. And I’m ready to give it to you. Now.

Though I don’t think that you’re going to like it.

All right.

You’re really not going to like it.

All right. The smallest $n \ne \pm4\mod 9$ for which no expression as the sum of three cubes is known is…

is…

Forty-two.

## 2^n-gons

There’s nothing at all new about this, in fact it’s ancient, but maybe it’s new to you? So here goes.

Here’s a square inscribed in a unit circle:

Or to keep down the clutter, here’s just one quadrant of that:

One side of the square along with the radii at each endpoint forms a triangle. The length of one side of the square we’ll call $s_4$ and it is of course $\sqrt 2$. That means the perimeter of the square is $p_4 = 4\sqrt 2\approx 5.656854249$.

Now consider an inscribed octagon. Again, here’s just one quadrant.

One side of the square is labeled $s_8$; what’s its length? Well, drop a perpendicular from the vertex in the middle:

And now notice the right triangle with hypotenuse 1 and sides $x_8$ and $h_8$ is just half of a quadrant of an inscribed square, which means $h_8=s_4/2$. Then $x_8^2 = 1-h_8^2 = 1-s_4^2/4$. From that you can get $y_8=1-x_8$ and from that, $s_8^2=h_8^2+y_8^2$. The perimeter of the octagon is then $p_8 = 8s_8\approx 6.122934918$.

Well, that was so much fun, let’s do it again. Here’s a quadrant of a 16-gon:

The right triangle with hypotenuse 1 and sides $x_{16}$ and $h_{16}$ is just half of an eighth of an inscribed octagon, which means $h_{16}=s_8/2$. Then $x_{16}^2 = 1-h_{16}^2 = 1-s_8^2/4$, $y_{16}=1-x_{16}$, and $s_{16}^2=h_{16}^2+y_{16}^2$. The perimeter of the 16-gon is then $p_{16} = 16s_{16}\approx 6.242890305$.

You know the rest, right? From here you can do the 32-gon, 64-gon, 128-gon, and so on, getting $x_{2n}^2 = 1-h_{2n}^2 = 1-s_n^2/4$, $y_{2n}=1-x_{2n}$, $s_{2n}^2=h_{2n}^2+y_{2n}^2$, and perimeter $= p_{2n} = 2ns_{2n}$. For $n = 1024$, the perimeter is $p_{1024}\approx 6.283175451 = 2\times 3.1415877255$. As $n$ increases, the perimeter gets closer and closer to the circumference of the circle, so this gives a way to calculate $\pi$.

What about circumscribed polygons? If you look at the figures above, you can see the ratio of the size of a circumscribed square to an inscribed square is $1/x_8$. Likewise the ratio of the size of a circumscribed octagon to an inscribed octagon is $1/x_{16}$, and so on. So the perimeter of a circumscribed $n$-gon is $P_n = p_n/x_{2n}$. As $n$ increases, $P_n$ approaches $2\pi$ from above. And the average of $p_n$ and $P_n$ is a better approximation to $2\pi$ than either, though not by a lot.

 n h_n x_n y_n s_n p_n P_n average 4 1.41421356 5.65685425 8.00000000 6.82842712 8 0.70710678 0.70710678 0.29289322 0.76536686 6.12293492 6.62741700 6.37517596 16 0.38268343 0.92387953 0.07612047 0.39018064 6.24289030 6.36519576 6.30404303 32 0.19509032 0.98078528 0.01921472 0.19603428 6.27309698 6.30344981 6.28827340 64 0.09801714 0.99518473 0.00481527 0.09813535 6.28066231 6.28823677 6.28444954 128 0.04906767 0.99879546 0.00120454 0.04908246 6.28255450 6.28444726 6.28350088 256 0.02454123 0.99969882 0.00030118 0.02454308 6.28302760 6.28350074 6.28326417 512 0.01227154 0.99992470 0.00007530 0.01227177 6.28314588 6.28326416 6.28320502 1024 0.00613588 0.99998118 0.00001882 0.00613591 6.28317545 6.28320502 6.28319024

This is something like the way Archimedes calculated $\pi$ around 250 BC (I told you this was ancient), although he started with a hexagon rather than a square and only went up to a 96-gon. He didn’t have Google Sheets, though.

## Power trip (part 3)

On further thought the inequality

$\displaystyle a<\frac{b-a}{\ln(b/a)} for $a$ and $b>a$ positive real numbers

can be tightened up on the left side. I used

$\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{a}\right)(b-a)$

but in fact this can be improved to

$\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{b}\right)(b-a)+\frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)(b-a)$

$\displaystyle a\left(\frac{2b}{b+a}\right) < \frac{b-a}{\ln(b/a)} < b$.

For $b\gg a$, the left side is $\approx 2a$, while for $b = a(1+2\epsilon)$ where $\epsilon<<1$, the left side is $\approx a(1+\epsilon) = (a+b)/2$.

## Power trip (part 2)

That proof that $e^a>a^e$ for any positive $a\ne e$ relies on $\ln(e)=1$, so doesn’t generalize to anything nearly as simple relating $a^b$ to $b^a$. But let’s see what we can do.

Here’s an inequality which, if you’re like me, looks pretty mysterious:

$\displaystyle b>\frac{b-a}{\ln(b/a)}>a$ for $a$ and $b>a$ positive real numbers.

I mean, the difference between $a$ and $b$ doesn’t “know” anything about $a$ or $b$, right? And neither does their ratio. But the difference over the log of the ratio is always between $a$ and $b$? Weird!

(Okay, maybe you’re not so easily impressed, since if you know $b-a$ and $b/a$ you can get $a$ and $b$. Fine. I think it’s mysterious anyway.)

Well, let’s consider a function $f(x)$ with derivative $df(x)/dx = f'(x)$ monotonically decreasing (we could do increasing too, similarly). Let $a$ and $b>a$ be in the range where $f(x)$ and $f'(x)$ exist and $f'(x)>0$. Then

$\displaystyle \int_a^b f'(x)dx = f(b)-f(a)$

But that integral is the area under the curve from $a$ to $b$ and so

$\displaystyle f'(a)(b-a)>f(b)-f(a)>f'(b)(b-a)$.

Then $(b-a)>[f(b)-f(a)]/f'(a)$ and $[f(b)-f(a)]/f'(b)> (b-a)$ or

$\displaystyle \frac{f(b)-f(a)}{f'(b)}> (b-a)>\frac{f(b)-f(a)}{f'(a)}$.

But that makes sense given the definition of the derivative and the monotonic property of $f'(x)$; with a little thought you could’ve written this down directly. As obvious as it is, it takes a decidedly less obvious form if you specify $f(x) = \ln(x)$, $f'(x) = 1/x$, which gives

$\displaystyle b\ln(b/a)>(b-a)>a\ln(b/a)$

or the formerly mysterious

$\displaystyle b>\frac{b-a}{\ln(b/a)}>a$ QED.

I called the original post “Power trip” and this one “Power trip (part 2)”, and there haven’t been any powers here. Sorry. Here:

$\displaystyle a\ln(b/a) < b-a < b\ln(b/a)$

so

$\displaystyle (b/a)^a < e^{b-a} < (b/a)^b$.

That form is… neither obvious nor interesting, though. Unless $a=e$ in which case the left part is

$\displaystyle (b^e)/(e^e) < (e^b)/(e^e)$

or

$\displaystyle b^e

which is where we came in.

## Power trip

Here’s a mathematical paper of which I can smugly say I understood every word.

Ha ha. No, seriously, I like it. A visual proof that $e^\pi>\pi^e$.

But as Math with Bad Drawings pointed out someone else pointed out, the proof doesn’t depend on $\pi$ but can be generalized to any number $> e$, and as Math with Bad Drawings pointed out, a similar proof works for any positive number $< e$. That is, for all $a > 0$ and $a\ne e$, $e^a>a^e$.

Proof here if you don’t want to click through to MwBD:

If $a > e$,

$\displaystyle \frac{1}{e}(a-e) = \frac{a}{e}-1 > \int_e^a \frac{1}{x}dx = \ln(a)-\ln(e) = \ln(a)-1$

so

$\displaystyle \frac{a}{e} > \ln(a)$

or

$\displaystyle a > \ln(a^e)$

or

$\displaystyle e^a > a^e$

and similarly, if $b < e$,

$\displaystyle \frac{1}{e}(e-b) = 1-\frac{b}{e} < \int_b^e \frac{1}{x}dx = \ln(e)-\ln(b) = 1-\ln(b)$

so

$\displaystyle \frac{b}{e} > \ln(b)$

or

$\displaystyle b > \ln(b^e)$

or

$\displaystyle e^b > b^e$

QED