Risk-o-phile: Royal Navy and Silicon Valley

Opening lines of Patrick O’Brian’s very fine book “H.M.S Surprise”:

But I put it to you, my lord, that prize-money is of essential importance to the Navy. The possibility, however remote, of making a fortune by some brilliant stroke is an unparalleled spur to the diligence, the activity, and the unremitting attention of every man afloat.

They’re talking about the British Navy, during Admiral Lord Nelson’s heyday around 1800. Ship’s officers and crew were rewarded, sometime substantially, when they captured enemy war and merchant ships. Capturing a prize required skill and luck, and was a rare and random event, but a real motivator.

Just like starting a software company.

The Royal Navy was a very important institution at the time, projecting power and keeping Napoleon at bay. Luckily for England it was also very effective, despite all the political maneuvering, and the inefficiencies imposed by an hereditary aristocracy.

The work was hard, the hours were long, and the women were scarce.

Sound familiar?

The cubic formula

My son and I were trying to use the cubic formula the other day to find the roots of a cubic equation. It’s not as simple as the quadradic formula:

    \[ \begin{array} {*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}\]

We used the method where you first convert the cubic into a depressed cubic. A depressed cubic is missing the x^2 term. It looks like this:

    \[ \begin{array} {*{20}c} {ax^3 + cx + d = 0} \\ \end{array}\]

You convert a normal cubic into a depressed cubic with a substitution. Starting with a general cubic polynomial:

    \[ \begin{array} {*{20}c} {ax^3 + bx^2 + cx + d = 0} \\ \end{array}\]

Make the substitution:

    \[ \begin{array} {*{20}c} {x = y - \frac{b}{3a}} \\ \end{array}\]

And you end up with the depressed cubic:

    \[ \begin{array} {*{20}c} {ay^3 + (c - \frac{b^2}{3a})y + (d + \frac{2b^3}{27a^2} - \frac{bc}{3a}) = 0} \\ \end{array}\]

Finally you find the roots for the depressed cubic to solve for y and substitute back to find the roots for the original equation. Solving for the depressed cubic is a multi-step operation, but it’s not too difficult. The above link explains it nicely.

To make it easier, we decided to try to solve a cubic that was already depressed. We picked the following because one of the three roots (x = 1) is obvious.

    \[ \begin{array} {*{20}c} {x^3 + x - 2 = 0} \\ \end{array}\]

The other two roots are easy to discover. Since (x = 1) is a root, we know {x^3 + x - 2 = 0} can be expressed {(x - 1)P = 0} where P is a quadratic expression. Long division yields {(x^3 + x - 2) = (x - 1)(x^2 + x + 2)} which tells us the other two roots are {x = \frac{-1 \pm \sqrt{-7}}{2}}.

So now we know what to expect.

When you solve this depressed quadratic you get:

    \[ \begin{array} {*{20}c} {x_1 = \sqrt[3]{\sqrt{\frac{28}{27}} + 1} - \sqrt[3]{\sqrt{\frac{28}{27}} - 1}}\\ \end{array}\]

If you put it thru a calculator, you’ll find the above expression indeed equals 1. I was a little surprised to find two irrationals separated by an integer in a non-obvious way. I also found it unsatisfying, since I could not simplify this expression all the way to 1 using algebra.

Let’s look at another cubic solution. Wikipedia tells us the following formula will give us the real root:

    \[ \begin{array} {rl} {x_1 = } & {- \frac{b}{3a}} \\ {} & {- {\frac{1}{3a}}\sqrt[3]{\frac{1}{2}\left(Q + \sqrt{Q^2 - 4(b^2 - 3ac)^3}\right)}} \\ {} & {- {\frac{1}{3a}}\sqrt[3]{\frac{1}{2}\left(Q - \sqrt{Q^2 - 4(b^2 - 3ac)^3}\right)}} \\ \end{array}\]


    \[ \begin{array} {rl} {Q =} & {2b^3 - 9abc + 27a^2d} \\ \end{array}\]

To solve {x^3 + x - 2} = 0 we plug (a = 1), (b = 0), (c = 1), and (d = -2) into the above to get:

    \[ \begin{array} {rl} {x_1 =} & {- \frac{1}{3}\sqrt[3]{\frac{1}{2}\left[-54 + \sqrt{(54^2 + 4*27)}\right]}} \\ {} & {- \frac{1}{3}\sqrt[3]{\frac{1}{2}\left[-54 - \sqrt{(54^2 + 4*27)}\right]}} \\ \end{array}\]

which simplifies to

    \[ \begin{array} {*{20}c} {x_1 = - \frac{1}{3}\left[\sqrt[3]{-27 + 6\sqrt{21}} + \sqrt[3]{-27 - 6\sqrt{21}}\right]} \\ \end{array}\]

and finally evaluates (with a calculator) to:

    \[ \begin{array} {*{20}c} {x_1 = 1}\\ \end{array}\]

Although that’s interesting, it’s no more satisfying than the first solution. I feel I’m missing something since I can’t take it all the way to (x=1) algebraically. It was still a fun exercise though. I’ve occasionally wondered about the methods for solving cubic equations, and now I’ve finally seen it done.

I can see why they don’t teach this in high school. It’s too complicated to ask people to memorize, so they wouldn’t want it on a test. And it gives funny looking results, full of square- and cube-roots, that are difficult to manipulate.

Hmmm, now I’m wondering about a general method for finding the roots of a 4th-order equation. Or even a for a 5th-order equation.

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