#polynomial math

[inkskinned]

i love you vaccines i love you research i love you reading the book instead of having chatgpt summarize it i love you critically thinking rather than reacting to a headline i love you investigating the source material i love you science i love you math even though you are personally my enemy (math/yn slowburn) i love you writing even though you try to stab me a lot i love you Experts in Your Field i love you Using The Brain

[birdssong]

save me notes app cecil . notes app cecil save me

[tyra-altavilla]

Pre-Calculus Meme with my dog

[astrozxmbie]

this is how I feel when I enjoy and like the topic we are learning about in class

[maybeinanotherworld]

the other day, i said i was tired of seeing legendre polynomials everywhere and a friend of mine said, with no explanation whatsoever, "le gender polynomials" and i just think that's the funniest thing i have ever heard

[bubbloquacious]

Jacobson's Basic algebra 1 gives such a nice argument for why algebraic numbers have minimum polymials, making it a corollary of the fact that polynomial rings over fields are principal ideal domains. Consider a field extension E/F (i.e. a field E that contains a field F). For any u ∈ E we have in E the subring F[u] and the subfield F(u). These are defined as the smallest subring and subfield of E respectively that contain both F and u. If E = F(u) for some u ∈ E, then E/F is called a simple field extension.

Let F(u)/F be an arbitrary simple field extension. Consider the polynomial ring F[X]. By the universal property of polynomial rings (which is essentially what one means when they say that X is an indeterminate), there is a unique ring homomorphism F[X] -> F(u) that sends any a ∈ F to itself and that sends X to u (you can think of this homomorphism as evaluation of a polynomial at u). If the kernel of this homomorphism is the zero ideal, then F[u] is isomorphic to F[X], so u is transcendental over F. If the kernel is non-zero, then by definition there are non-zero polynomials p ∈ F[X] such that p(u) = 0 in F(u), so u is algebraic over F. Because F[X] is a principal ideal domain, there is a polynomial p that generates the kernel. In other words, p divides all polynomials q such that q(u) = 0 in F(u). Two polynomials (over a field) generate the same ideal if and only if they differ by a constant factor, so there is a unique monic minimum polynomial in F[X] for u.

[yandereshingeki]

I’m at the point where I see the quadratic equation and I recoil i am so tired of the daily math i’ve been doing May’s gonna have me feeling like this if I actually pass this class

[er-cryptid]

Synthetic Division [Ex. 2]

Patreon

[nablavii]

So today was the second course day into algebraic geometry. We've studied how projective geometry involves non-measurable stuff. I mean, once you step into real analysis or some algebra, you find things like metric, norm, and so on, but sometimes you gotta step back a little, where things lose measure and you only have a ruler and something to make circumferences. Then, you can ask: Are there any relations between geometrical symmetries and polynomial roots? Well... You've gotta find out.

ESFM IPN - Dr. César Lozano Huerta.

[uncertain-noob]

Continued Fractions of Polynomial Roots

Imagine you've got a polynomial, and you want the continued fraction representation of one of its roots. First things first, if you count the sign changes in the coefficients as you go from highest to lowest, you've got an upper bound on the number of roots that poly has that are positive. If the number of changes is 1 or 0, the bound is exact. This is due to Descartes. Much more here https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots Now, imagine the root you care about is the largest positive root. We want to find its integer part to start our continued fraction journey. If we shift the poly by mapping x to x+1, we move that root one step towards the negative ie we subtract one from the value of its highest root. If we keep doing this, Descartes test tells us that eventually all the coefficients of the poly will have one sign, and the root will have changed from positive to negative. If we backtrack one step, we know the largest integer shift which still has one positive root. The root of that shifted poly will be the fractional part of the root we cared about, and the amount we shifted by is the first number in our continued fraction. Okay, but that's only the first term. Yeah, but all you have to do now is take your poly, substitute x -> 1/x and then multiply by x^n, where n is the degree, to get back a poly in x, whose roots are all the inverses of the roots before we transformed. And the root we care about, the inverse of the fractional part of the root we first cared about, is the only, and largest, root of the poly greater than 1. The other roots, after a few iterations, all end up in the unit disc with negative real part. This is stuff i read here https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.135.107&rep=rep1&type=pdf but i think its pretty well a 'folk theorem' ie, so many dudes worked it out more or less at the same time, and it just sort of moved into the general conscience. But I could be wrong. Also, for anyone implementing with a machine, some caveats. The coefficients of the poly are not well controlled. To get more than a few dozen partial quotients in the cf, you'll need arbitrary precision integers, like scala's BigInt. I shifted by increasing powers of 2, tested how many roots i had, and then used a binary search after i had an upper bound. Mostly partial quotients are low, so there just stepping through might be smarter. See the Gauss Kuzmin distribution for further guidance. https://mathworld.wolfram.com/Gauss-KuzminDistribution.html

[kneadingmaths]

There's a nifty trick to solve self-reciprocal polynomials! Did you know it?

[cryptablog]

how do you factor trinomials.

[cherryribbcns]

shimi shimi coco pop shimi shimi pop pop. suffering from exams? we'll suffer together, love

-👾 anon

UGH I HATEHATEHATE EXAMS SO FRICKIN MUCH &&& IM WORRIED AS HELL TOO LIKE NOOOOOO

[roblingoblin285]

this is literally insane but whumpee who's forced to do math problems under the threat of torture. like whumper goes "rewrite that polynomial function as the product of linear factors or i'm going to lash you with a hot poker"

[existinglimits]

A particularly fun equation

(x^2-5x+5)^(x^2-11x+30)=1

What I rather enjoy about this problem is that Wolfram Alpha doesn't get all of the solutions. While algorithms are powerful, they are not omnipotent.

And I don't mean just computer algorithms - I missed the same two solutions as Wolfram because I only used the standard approach - logarithms. (Luckily, getting something wrong makes it stick in my head better than getting something right, so this problem has lived in my brain rent free for over a month.)

Sometimes the thoughtful/creative/thorough application of more "basic" concepts is the way to go.

Hint:

What are three different cases where a^b = 1?

[roboyomo]

Good luck yomomo 💥

JUST CAME OUT OF THE ALGEBRA FINAL

EVEN THOUGH IT WAS THE ONE I WAS LEAST WORRIED FOR, I STILL THOUGHT "what the Fuck do i do there"