When the sequence a, b, c is geometric, ac - b^2 = 0. In this case, the same is true of the polynomial sequence, as we've seen. But what happens in general? Well, a(ax^2 + 2bx + c) = (ax + b)(ax) + a(bx + c), so a(ax^2 + 2bx + c) - (ax + b)^2 = a(bx + c) - b(ax + b).
-
Show this thread
-
And of course a(bx + c) - b(ax + b) = ac - b^2. That is, we conclude that, in general, a(ax^2 + 2bx + c) - (ax + b)^2 = ac - b^2. This is what is known as "completing the square." Since a(ax^2 + 2bx + c) = (ax + b)^2 + ac - b^2. If ac - b^2 > 0, then both sides are > 0.
1 reply 0 retweets 0 likesShow this thread -
It's hard to imagine making the algebra here any prettier, or more natural. And the result is very striking: a(ax^2 + 2bx + c) - (ax + b)^2 = ac - b^2 means that the polynomial analog of ac - b^2 turns out to have no x dependence -- and, indeed, is just ac - b^2 itself.
1 reply 0 retweets 0 likesShow this thread -
But what's most interesting about all this is that the same ideas apply all the way along the sequence of God's preferred polynomials. In particular, the next step is to compute a(ax^3 + 3bx^2 + 3cx + d) - (ax + b)(ax^2 + 2bx + c), which, it turns out, is equally lovely.
1 reply 0 retweets 0 likesShow this thread -
This computation leads to the so-called "depressed cubic" which is a cubic polynomial in ax + b with no quadratic term. And one can continue in this same way to arrive at "depressed" quartics, quintics, etc. They're all polynomials in ax + b, with no subleading term.
1 reply 0 retweets 0 likesShow this thread -
And since all these "depressed" polynomials are polynomials in ax + b, we know that when the coefficient sequence a, b, c, ... is geometric, they must take the form (ax + b)^n = 0 -- which means that their coefficients must all vanish in this case. Like ac - b^2, ad - bc, etc.
1 reply 0 retweets 0 likesShow this thread -
Indeed, one easily finds, for example, that a^2(ax^3 + 3bx^2 + 3cx + d) = (ax + b)^3 + 3(ac - b^2)(ax + b) + a(ad - bc) - 2b(ac - b^2). And this "depressed" cubic in ax + b has for its coefficients polynomials in ac - b^2 and ad - bc -- both zero in the geometric case.
1 reply 0 retweets 0 likesShow this thread -
I thought you might enjoy this string of ideas, which one can easily make into a "general theory of depression." It goes farther than this, too, as I imagine will not surprise you. Really quite lovely stuff, which puts "completing the square" into its proper algebraic context.
1 reply 0 retweets 1 likeShow this thread -
A final remark: in this context, the cubic formula takes on a peculiarly elegant and memorable form: if ax^3 + 3bx^2 + 3cx + d = 0, then ax + b = cbrt[(ac - b^2)(ar + b)] + cbrt[(ac - b^2)(as + b)], where r, s are the roots of (ac - b^2)x^2 + (ad - bc)x + (bd - c^2) = 0.
2 replies 0 retweets 1 likeShow this thread -
Replying to @MathPrinceps
@threadreaderapp God wants polynomials written a certain way, and I suspect he'd also like you to unroll the thread.2 replies 0 retweets 1 like
I'm embarrassed never to have learned how to do this. I suppose I'd better amend that situation.
-
-
Replying to @MathPrinceps @threadreaderapp
0h_0k Retweeted Thread Reader App
0h_0k added,
Thread Reader App @threadreaderappReplying to @80k_0k8Hi there is your unroll: Thread by@MathPrinceps: "@jamestanton God wants us to write our polynomials like this: 1ax^0 1ax^1 + 1bx^0 1ax^2 + 2bx^1 + 1cx^0 1ax^3 + 3bx^2 + […]" https://threadreaderapp.com/thread/1102050069228937216.html … Have a good day.
2 replies 0 retweets 1 like -
Replying to @80k_0k8 @threadreaderapp
Thank you very much. I am slowly becoming less ignorant.
1 reply 0 retweets 0 likes - 2 more replies
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.