41–50 of 50 posts
It's fortunate that we so seldom need to solve cubic or higher order polynomial equations. Quadratic equations come up in physics all the time, because it's very common to have a constant acceleration. But when does one ever come across a constant jerk?
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As others pointed out, this is just a fallacy. Physics (as well as Math, CS, and all other disciplines) is full of non-linear, chaotic problems that are way too hard to solve, so academia doesn't like to teach them to new students and even experts aren't delighted to deal with them, because it's frustrating that we can't solve them (in a "beautiful" and "elegant" way). My own hot take is that Physics is stuck on a th…
>My own hot take is that Physics is stuck on a theory of everything for the last 100 years, because we are stuck on some fundamental limitations what Math is able to model. Your previous assumption reads more like math being able to model it, but scientists preferring elegant, aesthetic and simplistic solutions. Looking at my PhD physics neighbor in student housing a few years back, it doesn't seem true though. He wa…
It's fortunate that we so seldom need to solve cubic or higher order polynomial equations. Quadratic equations come up in physics all the time, because it's very common to have a constant acceleration. But when does one ever come across a constant jerk?
I always wondered why that is for a long time (in basic physics that at least). It wasn't until grad school that I understood that position and velocity are enough to uniquely determine a trajectory on a manifold.
Nice write-up! Incidentally I ran all the exercises through ChatGPT which seems to have completely crushed them, with work shown. Forget about assigning grades via math homework, it only took a few minutes (though I should go through and check it by hand I think). [edit looking at the answers in more detail it seems to go down some strange rabbit holes that probably were not what the author intended] Also, if one wan…
Does this work for more advanced maths? It's a shame the amount you can spend per book for self-study, without access to the exercise answers needed to confirm your comprehension.
I feel like this article is a bit backwards - solving quadratic equations is only easy if you have access to the square root function, which by definition is a solver of quadratic expressions. Without it, one needs to resort to iterative root-finding, which works for polynomials of any order.
Computing (or computing with ) a square root is much easier than solving a quadratic equation! - A square root’s result is always nonnegative (or in the complex case, has nonnegative real part) and unique, whereas the quadratic equation has two solutions. - You can manipulate the root function symbolically much better than the roots of an equation. sqrt(2)^2 is 2, but figuring out that `solution(x^2-2=0)^2` reduces t…
I don’t know what definition you’re using, but if we take the square root of Y to be a number X such that X*X = Y, then the above statement isn’t true.
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Computing (or computing with ) a square root is much easier than solving a quadratic equation! - A square root’s result is always nonnegative (or in the complex case, has nonnegative real part) and unique, whereas the quadratic equation has two solutions. - You can manipulate the root function symbolically much better than the roots of an equation. sqrt(2)^2 is 2, but figuring out that `solution(x^2-2=0)^2` reduces t…
>A square root’s result is always nonnegative (or in the complex case, has nonnegative real part) I don’t know what definition you’re using, but if we take the square root of Y to be a number X such that X*X = Y, then the above statement isn’t true.
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>A square root’s result is always nonnegative (or in the complex case, has nonnegative real part) I don’t know what definition you’re using, but if we take the square root of Y to be a number X such that X*X = Y, then the above statement isn’t true.
The principal square root (ie, the one with positive part) is commonly referred to as the square root, especially since the radical symbol is explicitly defined to produce the principal root.
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That’s not right. For example, if you have access to any sort of radical you still can’t solve the quintic.
But if you have access to an extended set of operations (ultraradicals), in particular an operation that solves a parameterized quintic, you can solve all quintics. https://en.m.wikipedia.org/wiki/Bring_radical