I’m back! Croatia, Greece, Turkey… all behind me. In the meantime, I’ve fallen even more in love with math.stackexchange and have ended up as a temporary moderator for philsophy.stackexchange (check them out). To announce my return, a little fun:

#### from Loers Hey everyone thanx for the amazing effort that u provide us with , over here . just gotta a simple questions why cannot we differentiate |x|when x = 0 ? or let’s say |x+2| when x = -2 this is really annoying me I cannot see a proper reason for it thanx again

We strive to develop our humor. The Chaz (quite the internet sensation, if you haven’t run across him) writes:

Yes, the absolute value looks like a “V”. My point (pun intended) is that at the tip of the “V”, you can place a line that only touches once, but there are infinitely many such lines.

Or you can go a little overboard. I wrote:

It’s all pointy-like there, and derivatives don’t like sharp objects. They’re timid creatures, tortured even longer than students on this forum (I know – hard to believe!) by math teachers and old enough to remember not only when teachers were allowed to hit or throw chalk at their students, but were also allowed to stab them.

To be honest (in the sense that completely false stories can still convey a sense of truth), derivatives once thought to themselves, “Why shouldn’t we differentiate absolute values?” This, of course, made absolute value very happy, because the differentiables were one of the most exclusive clubs in all functiondon (not counting those smooth functions – they’re pricks… albeit very smooth ones). So they let him in. But to their dismay, they realized that a new function arrived in functiondom, calling himself absolute value’s derivative. And he was bipolar! Sometimes he was as positive as can be, and everyone loved him. He was very consistent. But then, without any notice or change, he was suddenly incredibly negative! It was terrible! Worse, there was a moment in the middle where they couldn’t tell what he was! So they quickly tossed absolute value out of the club, and he’s felt very ostracized ever since – you see, he’s even continuous, and these guys always feel like they deserve more than they do.

I end by nothing that it is completely ordinary to anthropomorphize derivatives.

#### Prioryofxd writes: … * … * … = 27 How do i find out whats behind the empty spaces they must be the same number?

To this, the great CRGreathouse responds:

(algebra)
You’re trying to solve x * x * x = 27, that is, x^3 = 27. Start by transposing to get x^3 – 27 = 0 and then factor as (x – 3)(x^2 + 3x + 9) = 0. The roots are the zeros of the first factor together with those of the second factor. For the first x = 3 works by inspection, and there are no further solutions by the Fundamental Theorem of Algebra. The second can be factored with the quadratic formula, giving
$\dfrac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 9} } {2} = \frac{-3}{2} \pm 3 i \sqrt{3/4}$

(complex analysis)
Alternately, notice that the roots are evenly spaced about the circle about the origin with magnitude 3. In particular the solutions are each 3 times a third root of unity:

$3e^{2 \pi i k / 3}$ for k = 0, 1, or 2. You can expand this using Euler’s formula to get $3 \cos{ 2 \pi k / 3} + i \sin{ 2 \pi k /3}$.
which are equivalent to the above three solutions. You can equate these with the relevant terms from the quadratic formula to derive trig identities if desired…(geometry)
You might reasonably despair, knowing that constructing the general solution to this problem is equivalent to the Delian problem of doubling the cube, long known to be impossible with ruler and straightedge. But never fear! A linkage or marked ruler provide solutions to cubic equations like this. Another powerful approach: use origami, following Huzita’s axioms. These are related to the Pierpont primes, Sloane’s A005109.

(numerical analysis)
But perhaps you aren’t willing to just guess answers until you find one. You might be tempted to use Newton’s method on x^3 – 27 using its derivative 3x^2, but it’s well-known that if you choose a bad starting point you may never converge on the answer. Bisection is slow, and further you need to be able to bracket the interval, which puts you back in the position of guessing again! I recommend the Schonhage-Gourdon method, as implemented (e.g.) in gp (see my .sig). Try

Code: Select all
polroots(x^3-27)

(Diophantine equations)
Like geometry, the outlook seems bleak at first: I don’t even know if cubic Diophantine equations are known to be decidable in general. Quartic Diophantine equations in sufficiently many (58) variables are known to be unsolvable in the general case, as they can encode universal problems. But fortunately univariate cubics are solvable due to a method of del Ferro, Tartaglia, and Cardano. Noteworthy: this method typically requires working with imaginary numbers, even if all roots are real!

(algebraic number theory)
It is well-known that the Eisenstein integers are a unique factorization domain, and it’s not hard to check that $27 = -(1 + 2\omega)^6$ with $\omega = e^{2 \pi i /3}$. Thus the equation is of the form $x^3 + y^6 = 0$ and so the solutions can be found through usual factorization techniques.

(Galois theory)
Fortunately the equation is not of degree five or higher, or by Abel’s theorem there would be no general closed-form solution using root extraction, multiplication, and addition. For lower degrees such methods do exist: in particular Cardano’s method, mentioned above. But even degree five is attackable: the Glashan-Runge-Young criteria show if a particular equation is solvable with radicals, and if not Hermite showed a method using elliptic integrals. Sixth degree is expected to be harder…

That’s really quite magical in a way. Along the same lines, but not as artfully presented is my answer to the question of how to find the line between two points, but that’s for another forum. But for now, I’m back to blogging!