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by owyn
176 days ago
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The original post is written by AI so I will read it briefly, but your comment is fascinating. I got through undergrad math by brute force memorization and taking the C. Or sometimes the C-. The underlying concepts were never really clear to me. I did take a good online calculus class later that helped. However, I have questions:
"Turns out the quantity they needed exists, but couldn't be described in their notation" What is this about? Sounds interesting. "Statisticians just said "oh, that function" and gave it a new name." What is this? I never understood there is a relationship between quadratic equations and some kind of underlying mathematic geometric symmetry. Is there a good intro to this? I only memorized how to solve them. And the existential question. Is there a good way to teach this stuff? |
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In a polynomial equation, the coefficients can be written as symmetric functions of the roots: https://en.wikipedia.org/wiki/Vieta%27s_formulas - symmetric means it doesn't matter how you label the roots, because it would not make sense if you could say "r1 is 3, r2 is 7" and get a different set of coefficients compared to "r1 is 7, r2 is 3".
Since the coefficients are symmetric functions of the roots, that means that you can't write the roots as a function of the coefficients - there's no way to break that symmetry. This is where root extraction comes in - it's not a function. A function has to return 1 answer for a given input, but root extraction gives you N answers for the nth root of a given input. So that's how we're able to "choose" roots - consider the expression (r1 - r2) for a quadratic equation. That's not symmetric (the answer depends on which one we label as r1 and which we label as r2), so we can't write that expression as a function of the coefficients. But what about (r1 - r2)^2? That expression IS symmetric - you get the same answer regardless of how you label the roots. If we expand that out we get r1^2 - 2r1r2 + r2^2, which is symmetric, which means we can write it as a function of the coefficients. So we've come up with an expression whose square root depends on the way we've labeled the roots (using Vieta's formulas you can show it's b^2-4c, which you might recognize from the quadratic equation).
Galois theory is used to show that root extraction can only break certain types of symmetries, and that fifth degree polynomials can exhibit root symmetries that are not breakable by radicals.