/sci/ - Science & Math

[Anonymous No.15354584]

>"excuse me how do I solve a quartic polynomial?"
>"oh, its quite simple just use the quartic formula to find all 4 solutions."
>"great, what does it look like?"

Please simplify this eyesore for me.

[Anonymous No.15354595]

>>15354584
>he can't understand the quartic formula
NGMI, stick to studying engineering, brainlet.

[Anonymous No.15354607]

humblebragging for intellectual posers
>muh field is sooooooo smart, look at how big the formulas are
same retard will turn around and brag about the small ones being "elegant"
too narcissistic and desperate for attention to post it's math discussion in the /mg/, no genuine interest in the topic whatsoever, just another worthless poser

[Anonymous No.15354699]

I still don't understand why there is no formula for higher polynomials.

[Anonymous No.15354713]

>>15354607
We know anon even applied math was too hard for you, you can stop being angry now

[Anonymous No.15354729]

>>15354699
Study Galois theory

[Anonymous No.15354819]

>>15354729
so you can't give an answer
>solve polynomial with power of 1
>ok
>solve polynomial with power of 2
>ok
>solve polynomial with power of 3
>ok
>solve polynomial with power of 4
>ok
>try with 5
>NOOOOOOOOO YOU CAN'T JUST SOLVE A QUINTIC
why the fuck are mathfags like this??

[Anonymous No.15354825]

>>15354699
>>15354729
>>15354819
https://en.wikipedia.org/wiki/Bring_radical

[Anonymous No.15354829]

>>15354819
You're right, I can't give an answer. I did a semester of Galois theory at university 6 years ago, it all made perfect sense to me then but I've forgotten it now. Even if I did remember, I doubt a semester of Galois theory could be condensed into a single 4chan post.

[Anonymous No.15354876]

>>15354819
>so you can't give an answer
I can. https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem
Have no idea if that will be meaningful to you or not.
https://en.wikipedia.org/wiki/Galois_theory
The aforementioned galois theory is about why things are solvable, while abel-ruffini directly pertains to the absence of a formula reducible to radicals from its coefficients. That is, the absence of a general formula for finding a solution. Specific equations may of course nonetheless be solvable, just not in a generalizable form as with lower degrees.

[Anonymous No.15354888]

>>15354876
You're not giving an explanation, you're just giving him the name of the theorem. I'm pretty sure the guy was asking for an intuitive explanation of WHY quintic polynomials can't be solved with radicals. That can only be explained intuitively using Galois theory.

[Anonymous No.15354892]

>>15354829
The only thing you forgot is the definitions, taking a course doesn't makes you a genius, well however remembering the definitions is the only hardship

[Anonymous No.15354916]

Pointless to memorize Cardano or even Ferraris method. Just use some Polynomial divison or numerics like Newton-raphson, thats faster and easier

[Anonymous No.15354924]

>>15354819
It's a fact of nature that most things are not exactly solvable in the way that would appeal to 19-th century mathematicians. This is not so surprising.

[Anonymous No.15354927]

just transform it into a depressed quartic, loser.

[Anonymous No.15354948]

>>15354888
There are intuitive explanations on those pages. There is a reason I linked wikipedia for them both, as I believe anyone who understands enough to use words like "quintic" ought be able to at least understand the simple non-solvable quintic example therein for the galois theory page. Either way, fine,
>>15354819
The simplest possible intuition is actually from toplogy, as usually people are better able to understand visual demonstration. e.g. topological Galois theory. At a certain point the additional dimensions simply cannot be mapped onto a regular 2D plane, that is to say linear equations or functions with solutions in regular numbers, in the usual way you could with a 3D shape "collapsing" or "moving through" a 2D plane. As the solutions do not, at any point, "intersect" in the same way with this plane as do shapes produced in lower dimensions where its coefficients give definite points you can represent in 2D.

It would help to view what such a shape "moving through" a 2D plane ends up looking like. Usually, some criss-crossing lines such that any solution defining the shape or contact points would result in a violation of identity. Such as "1 = 2", but moreover also with larger series of contradictions at many points, perhaps "1 = 5 = 2 = 100". None of these are produced from any given example and are merely to represent the idea.
>>15354888
And if you've a problem with the attempted brevity given I am assuming I am thus explaining it to a high school student feel free to fucking do it yourself.

[Anonymous No.15355755]

>>15354948
Thanks dude. I appreciate your effort. Gonna read the wikipedia articles. I had a class on algebra but it sucked and didn't cover Galois theory.

[Anonymous No.15355775]

If you're so smart, solve quintics

[Anonymous No.15355790]

>>15355755
Most things, one way or other, can be shoved into simplifications one can at least intuitively understand by interactions of shapes. One simply has to get very creative in certain cases such that often times more than one shape is involved or even shapes across transformations or dimensions, or even multiple dimensions. As you might guess it stops being helpfully intuitive at a certain point and you devise yet further machines and intuitions.

Having wrong or only "partially correct" intuitions is fine so long as you understand their limitations. Think like that as one uses a tool, as one uses particular methods in solving increasingly complex problems as one learns from kindergarten to university. Develop your intuitions likewise and you'll be a lot better off.

[Anonymous No.15355809]

>>15354607
>same retard will turn around and brag about the small ones being "elegant"
they have no self-awareness because they're too low iq

[Anonymous No.15355825]

>>15355775
x^5 = 1
x = 1
too ez

[Anonymous No.15356314]

>>15355825
[math]x=e^{i2\pi n/5}[/math] for [math]n=0,1,2,3,4[/math]

[Anonymous No.15356407]

>>15354699
>>15354729
Important distinction here is that Galois theory does not imply that there can be no quintic formula in general, but that the formula can not be algebraic (that is consisting of only addition, subtraction, multiplication, division, integer powers and integer roots). Thus theoretically on might formulate a solution with transcendental functions, integrals, limits etc.

[Anonymous No.15357954]

>>15354584
to solve p(x)=0:
let x = (a*z+b)/(z+1), q(z) = p(x)*(z+1)^deg.
in the case deg=3, you want to solve for a,b to kill the z and z^2 coefficients then just cube root the thing.
In the case of deg = 4, you want to kill the z and z^3 coefficients and solve the quadratic (in z^2).

[Anonymous No.15358121]

>>15357954
>[math]x=\frac{(az+b)}{(z+1)}[/math]
>[math]p(z)=p(x)(z+1)^{deg}[/math]

formula names? or is it just Galois theory?

[Anonymous No.15358171]

>>15358121
Try it and you'll understand.
Start with degree 3.
You'll end up with the two equations:
3p(a)+p'(a)(b-a)=0
3p(b)+p'(b)(a-b)=0

Solve for a and b.
Then you'll get z = [-p(b)/p(a)]^(1/3)
Then get x from x = (a*z+b)/(z+1).

For degree 4 the equations are:
4p(a)+p'(a)(b-a)=0
4p(b)+p'(b)(a-b)=0
You'll need to use the cubic formula to solve this.

You get z^2 from solving the quadratic.

[Anonymous No.15358888]

>>15354948
Wikipedia is not intuitive

>>15354699
https://youtu.be/zCU9tZ2VkWc

[Anonymous No.15358952]

>>15355809
yep, its all about bragging and narcissism for them

[Anonymous No.15359468]

>>15358888
>Wikipedia is not intuitive
Intuition is relative to your degree of competence. An example to help construct a more accurate intuitive model differs, wildly, depending on if you're trying to help a middle-schooler, a high-schooler, an undergrad, and so on. Also their respective abilities for their ages.

[Anonymous No.15360206]

>>15354699
Basically, repeatedly taking the commutator subgroup of Sn fails to reach the trivial group when n>4 but does reach the trivial subgroup when n<=4.

The main insight is to impose an ordering on the set of roots of your polynomial.
This is useful for when you vary the coefficients of your polynomial over some path in C^n that starts and ends at the point corresponding to your polynomial. If you just consider the set of roots then you observe no change and you think no big deal.
With the ordered set of roots you observe the roots being permuted by letting the coefficients follow some loop in C^n.
Loops induce permutations.
Concatenating loops (doing one loop then another) corresponds to composition of permutations.
Basic complex analysis tells you the only way to go around a loop and get something different is if you go around a branch point.

Restricting our choice of loops to all compositions of commutators of two loops (corresponding to anything in the first commutator subgroup) guarantees no winding around zero for terms that are rational functions of the coefficients (w1 + w2 - w1 - w2 = 0). This means terms that have radical depth of 1 or less are unchanged.
Anything in the second commutator subgroup guarantees terms that have radical depth of 2 or less are unchanged.
...
Anything in the mth commutator subgroup guarantees terms that have radical depth of m or less are unchanged.

If the roots could not be written as functions of the coefficients with finite radical depth then no number of iterations would reach the trivial group.
S1 is trivial already so no radicals are required for degree 1.
S2 -> S1 takes 1 iteration so degree 2 requires at most radical depth 1.
S3 -> C3 -> S1 takes 2 iterations so degree 3 requires at most radical depth 2.
S4 -> A4 -> C2xC2 -> S1 takes 3 iterations so degree 4 requires at most radical depth 3.
S5 -> A5 -> A5 ... doesn't terminate so there could possibly be no finite radical depth solutions.