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13 / 2 / 13 / ?

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  No. 163
>>475 >>479 >>682
What the fuck is a Polynomial and how do I solve one?
>>
  No. 164 >>165 >>175
A polynomial is an expression of the form
∑k=0nan−kxk,a≠0\sum_{k=0}^n a_{n-k}x^k,a\neq0
with coefficients defined over a field (such as
R\mathbb{R}
,
C\mathbb{C}
or
GF(pk)GF(p^k)
). Polynomial algebra itself forms a ring. In a nutshell, it's a sum of powers of
xx
where every power has is multiplied by a number coefficient

P=a0xn+a1xn−1+⋯+an−1x+anP = a_0 x^n + a_1 x^{n-1} +\cdots + a_{n-1}x + a_n


Solving polynomial equations means finding values of
xx
such that when the value is substituted the polynomial evaluates to zero
∑k=0nan−kxk=0\sum_{k=0}^n a_{n-k}x^k = 0


Second degree polynomial equation
a0x2+a1x+a2=0a_0x^2+a_1x + a_2=0
is called a quadratic equation and can be solved using a quadratic formula

x1,2=−a1±a12−4a0a22a0x_{1,2} = \frac{-a_1\pm\sqrt{a_1^2 - 4a_0a_2}}{2a_0}

Cubic and quartic formulas also exist but they are much more involved while formulas for degree five or higher have been proven not to exist by the Abel-Ruffini theorem.

Solutions to a polynomial equation are called roots, which may or may not exist depending on the polynomial equation and field the polynomial is defined over, but according to the fundamental theorem of algebra solutions always exist for polynomial equations over the field of complex numbers
C\mathbb{C}
. Knowing roots allows us to factor a polynomial:
∑k=0nan−kxk=a0∏k=0n(x−xk)\sum_{k=0}^n a_{n-k}x^k = a_0\prod_{k=0}^n(x-x_k)

Polynomial coefficients and roots are also linked via Vieta's formulas.

Generally finding solutions to or factoring a polynomial is an NP-hard problem (hard even for computers) but, of course, numeric approximation methods exist (like Newton's method).

While finding solutions to a polynomial equation is hard, evaluating a polynomial for any given
xx
is extremely easy (it's just multiplication and addition). Many functions can be represented by a polynomial via Taylor series (or a special case called Maclaurin series) and thus be computed easily by a computer or by hand with as much precision as you'd like. For example:
sin⁡x=∑k=0∞(−1)kx2k+1(2k+1)!=x−x33⋅2⋅1+x55⋅4⋅3⋅2⋅1−…\sin x = \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k + 1)!} = x - \frac{x^3}{3\cdot 2\cdot 1} + \frac{x^5}{5\cdot 4 \cdot 3\cdot 2\cdot 1}-\dots

For small
xx
,
sin⁡x≈x\sin x\approx x
is a good approximation.


https://en.wikipedia.org/wiki/Polynomial
https://en.m.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem
https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra
https://en.wikipedia.org/wiki/Vieta%27s_formulas
https://en.wikipedia.org/wiki/Root-finding_algorithms
https://mathworld.wolfram.com/MaclaurinSeries.html
>>
  No. 165

File: 8A8CFA56-78E0-4DD4-BE5D-7F3E7E00F2FC.jpeg ( 309.75 KB , 650x650 , 1650548632700.jpeg )

Image
>>164
Thanks bubs
>>
  No. 175
>>164
The ring of power series R[[x]] is not the ring of polynomials R[x]
>>
  No. 179
ggggggggggggggg
>>
  No. 180 >>473
Try using a premade computer algorithm to solve your polynomials. It uses the highly advanced techniques of guessing
>>
  No. 472
Is possible extend Ruffini's regola to irrational equation with small edit?
e.g. The result of equation
−x2+6x+22x+2+24=0 -x^2+6x+22 \sqrt{x+2}+24=0
is x=14
using Ruffini
−1+6+2216+2414−14−112−24−1−8−24140 \begin{array}{c|c c c|c} & -1 & +6 & +22\sqrt{16} & +24 \\ 14 & &-14 & -112 & -24 \\ \hline &-1 & -8 & -\frac{24}{14} & 0\\ \end{array}

but the quoto is
−x−8+22x+2+4 -x-8+\frac{22}{\sqrt{x+2}+4}
>>
  No. 473
>>180
do any of them use machine learning to narrow down where to guess? have we gotten that far?
>>
  No. 475
>>163
polynomial is when you have multiple wives
>>
  No. 479 >>494
>>163
I once knew how to solve polynomials but nowadays anything above degree 2 polynomials is too difficult to me
>>
  No. 494
>>479
Me too, I solve 2-degree equation like this:
x2+ax+b=0x=p+iq x^2+ax+b=0\\ x=p+iq

where
i2=−1i^2=-1

(p+iq)2+a(p+iq)+b=0{p2−q2+ap+b=02iqp+aiq=0{q=±a24+aa2+bp=−a2 \\(p+iq)^2+a(p+iq)+b=0\\ \begin{cases} p^2-q^2+ap+b=0\\ 2iqp+aiq=0 \end{cases} \\ \begin{cases} q=\pm \sqrt{\frac{a^2}{4}+a\frac{a}{2}+b} \\p=-\frac{a}{2} \end{cases}

because i don't remember formula.
>>
  No. 681
waow
>>
  No. 682 >>683
>>163
A poly is the sum of plus mono, until quartic eq. there are the formulas; from 5-degree eq. on... there is a general process, that is still secret
>>
  No. 683
>>682
Test

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