#quartic equation

[kuiperoid]

The History of the Cubic Equation, the Original Celebrity Feud

[also posted here]

[paper that I wrote for a class in undergrad that mostly served as an exercise in using LaTex]

Abstract

This document will describe the history and methods behind how the cubic equation was solved, specifically focusing on the work of specific sixteenth century Italian mathematicians. It will also touch upon the quartic and quintic equations.

1 The Life and Times of Tartaglia

Niccolo Fontana, known as “Tartaglia,” meaning “the stammerer,” was a Venetian mathematician during the early sixteenth century. His many accomplishments included engineering, working as a bookkeeper, and creating the first Italian translations of the works of Archimedes and Euclid, but he is most famous for his involvement in the 1535 Bologna University Mathematics Competition. Some years before, University of Bologna's Chair of Arithmetic and Geometry, Sciopione dal Ferro, solved one case of the cubic equation, a polynomial equation in which the highest variable exponent is three, a case only involving positive numbers. Though it is believed that dal Ferro solved this form around 1515, he kept it a secret until just before his death in 1526, at which point he shared this solution with his student and far inferior mathematician, Antonio Fior. Rumors began to spread that the cubic had been solved, eventually reaching Tartaglia. Tartaglia managed to find his own partial solution to the cubic in this time, leading the overconfident Fior to challenge him to a public math competition. According to the rules, each gave the other thirty problems to solve over the course of forty to fifty days. All of Fior's problems were set to the form x^3 + mx = n which dal Ferro had found, believing that Tartaglia would not be able to figure it out. However, Tartaglia had not only managed to solve all of the problems Fior had presented him with within the course of two hours, but he found a general method for solving all forms of the cubics before the contest was over. This ensured Tartaglia's victory [1].

Figure 1: A graph resulting from a cubic function

2 Enter Cardano

News of Tartaglia's win spread across Italy, eventually reaching Milan-based mathematician and general eccentric character Gerolamo Cardano. Though brilliant in mathematics and medicine, Cardano's illegitimate birth and generally abrasive personality had limited his opportunities, leading to a somewhat inconsistent professional life. Cardano managed to bring himself to prominence in the world of math with the publication of The Practice of Arithmetic and Simple Mensuration, the first of one hundred and thirty-one books he would write in his lifetime [2]. Cardano approached Tartaglia in 1539 with hopes of adding his solutions to the cubic to his next book, Practica Arithmeticae. Tartaglia was reluctant to share his method, having previously coded his method in a cryptic poem so that no one would figure it out, but did so after Cardano made an oath to not publish his ideas. Using this knowledge, Cardano spent the next six years working on the cubic and quartic equations, sometimes with the help of his assistant Lodovico Ferrari [1].

Ferrari was born in Bologna, Italy in 1522, a particularly politically tumultuous time for Northern Italy. Ferrari's father was killed in the army, meaning he had to move in with his uncle, Vincent. It so happened that Vincent's son Luke had run away to Milan for work and ended up briefly as a servant to Cardano. Luke ended up tiring of the work and going back home without telling Cardano. Cardano contacted Vincent to send his servant back, but Vincent thought that his fourteen-year-old nephew might make a better candidate. Upon Ferrari's arrival to Cardano's in November of 1936, Cardano learned that the teenager was literate and made him a secretary rather than a servant. He began to teach him mathematics, at which Ferrari proved to be so talented that he was able to teach by the age of eighteen and, by the age of twenty, he began working as a public lecturer in geometry. During this time, he and Cardano also worked on Tartaglia's cubic solutions in an attempt to solve the quartic as well. Between Tartaglia's findings and the work of another mathematician, Zuanne da Coi, Ferrari eventually found a solution for the quartic equation, polynomials for which the highest variable exponent is four. He wanted his work published, but it would be impossible to do so without revealing the work Tartaglia had produced, which Cardano had previously vowed to keep a secret. The two did further research and decided that because technically dal Ferro had solved a form of the cubic before Tartaglia, they could publish the findings about the cubic and the quartic without it being considered breaking Cardano’s oath. He featured these works in his book Ars Magna in 1545 [3].

Figure 2: A graph resulting from a quartic function

3 Ferrari Reigns Victorious

Needless to say, Tartaglia did not take well to the news of his findings being published, especially by someone who had previously promised not to. The alleged loophole counted for naught, as the portion about the cubic was very much Tartaglia's work. Ferrari responded to Tartaglia's anger by writing him an insulting letter in which he also challenged him to a public debate. Tartaglia saw no point in debating the relatively-unknown Ferrari, but did respond in an attempt to bring Cardano in. Tartaglia and Ferrari exchanged publicly read insulting letters for about a year. The rivalry was brought to its conclusion in 1548. Tartaglia was offered a job as a lecturer in Brescia and needed to participate in a public math competition with Ferrari to prove that he was worthy of the spot. Though less experienced in the ways of public debate than Tartaglia, Ferrari proved himself far more understanding of the concepts at hand, not only his own quartic equation, but the cubic as well. Tartaglia left that night in shame, leaving Ferrari as the winner [4].

4 The Failed Quintic

Though all proven mathematically brilliant, neither Ferrari, Cardano, nor Tartaglia ever even attempted to breach the quintic equation. It is thought that the first mathematician to attempt to actually solve for a specific quintic equation was French mathematician Joseph-Louis Lagrange in the eighteenth century. He tirelessly studied the previously developed methods for the quadratic, cubic, and quartic equations for clues. He focused on how rational numbers changed under permutations, but could not find an expression for powers above four. His successor came in the form of Italian mathematician Paolo Ruffini. Ruffini's work focused on the properties of permutations with the goal of proving that the quintic was not solvable. Though Ruffini did develop proofs, his work was considered vague for its focus on patterns instead of computation and was not well-accepted by mathematicians at the time. In the nineteenth century, Norwegian mathematician Niels Henrik Abel carried on with Ruffini’s work, publishing the completed versions of the proofs, and confirming that quintic functions cannot be solved with a single equation. This work is known to this day as Abel's Impossibility Theorem,or the Abel-Ruffini Theorem [5].

References

[1] Mastin, Luke. 16th Century Mathematicians. The Story of Mathematics, 2010.

http : //www.storyof mathematics.com/16thtartaglia.html

[2] JJ O’Connor and EF Robertson. Quadratic, Cubic, and Quartic Equations University of St. Andrews, February 1996.

http : //www−groups.dcs.st−and.ac.uk/history/HistT opics/Quadraticetcequations.html

[3] JJ OConnor and EF Robertson. Giralamo Cardano. School of Mathematics and Statistics at the University of Saint Andrews, June 1998

http : //www−history.mcs.st−andrews.ac.uk/Biographies/Cardan.html

[4] JJ OConnor and EF Robertson. Lodovico Ferrari. School of Mathematics and Statistics at the University of Saint Andrews, September 2005.

http : //www−groups.dcs.st−and.ac.uk/history/Biographies/F errari.html

[5] Fiona Brunk. Galois' Predecessors. School of Mathematics and Statistics at the University of Saint Andrews, January 2005.

http : //www−history.mcs.st−and.ac.uk/P rojects/Brunk/Chapters/Ch1.html

[enzoexposyto]

https://www.scribd.com/document/485003371/ESEMPIO-2-EQUAZIONI-e-DISEQUAZIONI-di-QUARTO-GRADO ESEMPIO 2 - EQUAZIONI e DISEQUAZIONI di 4° GRADO. CALCOLI, SOLUZIONI e GRAFICI PASSO PASSO - Equazione di una Quartica Passante per alcuni Punti - Sviluppi dell'Equazione - Controlli del Grafico - Punti di Flesso - Equazione e Disequazioni Associate all'Equazione della Quartica - Impostazione Generale delle Soluzioni - Soluzioni puntuali, con Grafico, dell'Equazione e delle Disequazioni - Sintesi delle Soluzioni #quartic #quartic_equations #biquadratic #biquadratic_function #equation #equations #function #functions #graph #biquadratic_equations #rational_function #biquadratiche #equazioni_biquadratiche #equazione #equazioni #equazioni_quartiche #funzioni_intere #funzioni_quartiche #funzioni_razionali #grafici #grafico #massimo #minimo #punto_di_flesso #quartica #funzioni_biquadratiche #funzione #funzioni https://www.youtube.com/user/enzoexposito http://www.enzoexposito.it/mobile/matematica.html http://www.enzoexposito.it/mobile/geom_analit_quartica.html https://www.slideshare.net/EnzoExposito1 https://www.linkedin.com/in/enzo-exposyto-aa970530/detail/recent-activity/shares/ https://www.facebook.com/enzoexposyto https://twitter.com/enzoexposyto https://www.tumblr.com/blog/enzoexposyto https://www.pinterest.it/enzoexposyto/_saved/ (presso World) https://www.instagram.com/p/ClJqWMELKt2/?igshid=NGJjMDIxMWI=

[musicalsiphonophore]

rare quartic equation W

[sabakos]

🔥matiyasevich's theorem/hilbert's tenth problem

Maybe something like matiyasevich's theorem was what Hilbert himself had in mind when he proposed the tenth problem, but I find "some of these are unsolvable" to be an entirely unsatisfying and incomplete theory of the solvability of diophantine equations!

For second degree diophantine equations in two variables (i.e. conic sections), we know how to determine the solvability of any given diophantine equation. I would like to see something similar like "For diophantine equations in m or fewer variables, of most nth degree..." which determines the upper bound on which diophantine equations can be solved. Similar to how we know that the quintic equation can't be solved in general in terms of radicals, but we know that the quartic can. Without that, I don't think we have a general theory of solvability yet.

[devotty]

I Solved A Quartic Equation in Two Ways

[bacanthony]

Can You Solve a Quartic Trigonometric Equation?

[truphysics]

Problem 8.1 - Griffith's Intro to QM

Problem 8.1 Use a gaussian trial function (Equation 8.2) to obtain the lowest upper bound you can on the ground state energy of (a) the linear potential: $V(x)=\alpha |x|$; (b) the quartic potential: $V(x)=\alpha x^4.$ Solution: Problem 8.1 Solution (Download) Find more Griffith’s solutions here. Do you prefer video lectures over reading a webpage? Follow us on YouTube to stay updated with…

[calebmatthew1]

Make use of the online root calculator to calculate the nth root

An Allcalculator.net's  root calculator aids in finding the roots of a given polynomial. You can quickly get the roots of any polynomial with the online roots calculator. Tools for using roots calculators speed up calculations and quickly display the roots or value of a variable.

What are the instructions for the root calculator?

Enter the number of the root in the root calculator you wish to calculate from in the "Number" field.

Enter the degree (default is 2) in the "Degree" field.

Change the calculation's precision by choosing the required number from the "Decimal" drop-down menu (the default is 20 decimal places).

In most circumstances, you do not need to click "Result" since the root calculator automatically discovers roots as you type letters or alter setting values.

Click "Reset" to return the calculator to its default settings.

How to find the root calculator?

Linear polynomials are those that have a degree of 1.

A quadratic polynomial has a degree of two.

A cubic polynomial has a degree of three.

A quartic polynomial has a degree of four.

A quintic polynomial has a degree of five.

An nth-degree polynomial is a polynomial with a degree (n) larger than 5.

Any degree of a polynomial reduces it to zero and identifies the roots of a given polynomial.

From the term "quad," which means square, comes the word "quadratic." In other terms, an "equation of degree 2" is a quadratic equation.

An equation of the form ax2 + bx + c = 0, where a is not equal to 0, is called a quadratic equation. a, b, and c are the coefficients of the quadratic equation. The discriminant formula to solve the quadratic equation is given:

X = -b ± √b2 – 4ac / 2a

What are the properties of square roots?

Some of the important properties of a square root calculator are as follows:

If two integers are multiplied in an equation using square roots, the entire equation may be written as a single sentence. 

                        √a x √b = √ (a x b) 

When two integers are split into their square roots in an equation, you can combine their square roots into a single one. 

                        √a / √b = √ (a / b)

You can divide the numbers into the root if a single number is the derivative of two different integers. 

                          √ (a x a) = a 

                          √9 = √ (3 x 3) = 3

If you want to express the square root exponentially, use the formula 1/2 

                          a = a1/2.    

The radicands (numbers inside the square root) can add or subtract two or more integers.

For instance, since 9 and 4 have identical radicals within, they may be added to or subtracted from. 

The square root of an equation becomes square if it is moved from the left to the right side or vice versa. 

                   √9 = 4 becomes

                   9 = 42

The square becomes a square root in the same way when it is shifted to the opposite side.  

                                         42= 9

                                          4= 92

If a square root referenced number is not a perfect derivative of any other number, it does not yield a clear result when multiplied. The solution will always be decimal or irrational.

For example, 

√26 = 5.09999.

The number's root will be illogical if it has zeros at the end. 

           √4000 = 63.24555.

An odd integer will always have an odd square root. 

For example, √9 = 3 and √121 = 11.

[designerbanana]

Mac os split screen horizontal

#MAC OS SPLIT SCREEN HORIZONTAL FULL#

#MAC OS SPLIT SCREEN HORIZONTAL PLUS#

#MAC OS SPLIT SCREEN HORIZONTAL FULL#

Business functions including Time-Value-of-Money (TVM), cash flows and amortization full screen interactive editor for solving TVM problems.Fifteen probability distribution functions including Normal, Student-t, Chi-square, Binomial and Poisson.Advanced statistics features including 10 hypothesis testing functions, seven confidence interval functions and one-way analysis of variance.Three statistical plot definitions for scatter plots, xy-line plots, histograms, regular and modified box-and-whisker plots and normal probability plots.List-based one- and two-variable statistical analysis, including logistic, sinusoidal, median-median, linear, logarithmic, exponential, power, quadratic polynomial, cubic polynomial and quartic polynomial regression models.Matrix operations including inverse, determinant, transpose, augment, reduced row echelon form and elementary row operations convert matrices to lists and vice-versa.Horizontal and vertical split-screen options.Eight different graph styles for differentiating the look of each graph drawn.Interactive analysis of function values, roots, maximums, minimums, integrals and derivative.Numeric evaluations given in table format for all graphing modes.User-defined list names lists store up to 999 elements.Sequence graphing mode shows time series plot, cobweb/stair-step plot and phase plots.Up to 10 graphing functions defined, saved, graphed and analyzed at one time.Graphs 10 rectangular functions, six parametric expressions, six polar expressions and three recursively-defined sequences.

#MAC OS SPLIT SCREEN HORIZONTAL PLUS#

Real and complex numbers calculated to 14-digit accuracy and displayed with 10 digits plus a two-digit exponent.Advanced functions accessed through pull-down display menus.Alphabetical CATALOG of all TI calculator operations in one menu.Change style and color of axes and grids on graphs.Displays up to four different representations, including graph, table, equation and data list screens.Emulates the core functionality of the TI-84 Plus CE graphing calculator.

[mathsodology]

MAT 2017 Q1 H

MAT 2017 Q1 H

Solution. We are given the following two conditions.

When

is divided by

, the remainder is 1, i.e.

The polynomial

has

as a factor, i.e.

Combining them together yields

From the equation (2), we find

Hence, the answer is (b).

Aside. When solving the simultaneous equations, one can of course directly substitute

into

and we arrive at the same equation.

[backslashdelta]

becky my beloved

props of nothing, what are the most top 5 math equations that are v satisfying and ✨just make sense✨ and what are the ones that are technically correct but don't pass the vibe check?

Myle my beloved!

This is a difficult question because I actually find that the ones that might "seem" incorrect are the most satisfying? Like, it's just so cool that something can look so complicated or weird (or so stupidly simply) and yet still be correct. I love that.

So instead, I'm going to kind of combine the two parts of your question, and give you five math things that I find really satisfying, but may not necessarily pass your vibe check.

c^2 = a^2 + b^2

We're starting out easy. The Pythagorean theorem. You have a right triangle and this is just always true regardless of the size and I just. Love it. So so neat.

x = [-b +/- sqrt(b^2 - 4*a*c)] / 2*a

The quadratic formula! This one is admittedly a bit more complicated, but really not that bad, and it's been drilled into my head since high school. I will never forget this formula and for that reason alone I feel like it needs to be included in this list. How cool is it that you can just always pop your a b and c into this formula and get all of the solutions to your quadratic equation? I think that's just so cool. Fun fact: there are also cubic and quartic formulas (they get very complicated), but it's been mathematically proven that you cannot have a formula for finding the solution of quintic polynomials or polynomials of any higher degree.

d (e^x) / dx = e^x

The derivative of e^x is itself. I just think that's so cool! Derivatives! If you were to continually take the derivative of any polynomial you would eventually get 0 but that never happens with e^x, you just e^x, forever, regardless of how many times you take it's derivative. What a fun property.

(A ⇒ B) ⇔ (¬B ⇒ ¬A)

The equivalence of contrapositives feels like cheating but it's not really an equation, but it's logic and it's an equivalence so I'm including it anyway. The way to read this in plain English is: "if 'if A then B' is true, then 'if not B then not A' is true, and vice versa". That probably didn't help. Let me give you an example instead.

Suppose A is the statement "The weather is warm," and B is the statement "I am wearing shorts." Then A ⇒ B says "If the weather is warm, then I am wearing shorts."

For the other side of the equivalence, using the statements we've chosen, ¬A is "The weather is not warm," and ¬B is "I am not wearing shorts." So ¬B ⇒ ¬A says "If I am not wearing shorts then the weather is not warm."

Now let's see if these statements are equivalent. If it's true that warm weather means I'm wearing shorts, then if I'm not wearing shorts it can't possibly be warm. And conversely, if me wearing something other than shorts means it's not warm, then if it is warm I must not not be wearing shorts, i.e. I am wearing shorts.

Maybe that didn't help either. Oops. I tried. Anyway, logic is cool as hell and I love it.

e^(pi*i) + 1 = 0

This is the equation that blows every math student's mind at some point because it's just. ridiculous. e, pi, and i are all ridiculous numbers. e and pi are both irrational and i isn't even a real number, and yet. And yet. You raise e to the power of pi*i and you add 1 and you get 0. It's just. Flabbergasting. It absolutely should not work but it does and it's so absolutely fantastic. She's beautiful I love her.

[enzoexposyto]

https://www.scribd.com/document/485003371/ESEMPIO-2-EQUAZIONI-e-DISEQUAZIONI-di-QUARTO-GRADO ESEMPIO 2 - EQUAZIONI e DISEQUAZIONI di 4° GRADO. CALCOLI, SOLUZIONI e GRAFICI PASSO PASSO - Equazione di una Quartica Passante per alcuni Punti - Sviluppi dell'Equazione - Controlli del Grafico - Punti di Flesso - Equazione e Disequazioni Associate all'Equazione della Quartica - Impostazione Generale delle Soluzioni - Soluzioni puntuali, con Grafico, dell'Equazione e delle Disequazioni - Sintesi delle Soluzioni #quartic #quartic_equations #biquadratic #biquadratic_function #equation #equations #function #functions #graph #biquadratic_equations #rational_function #biquadratiche #equazioni_biquadratiche #equazione #equazioni #equazioni_quartiche #funzioni_intere #funzioni_quartiche #funzioni_razionali #grafici #grafico #massimo #minimo #punto_di_flesso #quartica #funzioni_biquadratiche #funzione #funzioni https://www.youtube.com/user/enzoexposito http://www.enzoexposito.it/mobile/matematica.html http://www.enzoexposito.it/mobile/geom_analit_quartica.html https://www.slideshare.net/EnzoExposito1 https://www.linkedin.com/in/enzo-exposyto-aa970530/detail/recent-activity/shares/ https://www.facebook.com/enzoexposyto https://twitter.com/enzoexposyto https://www.tumblr.com/blog/enzoexposyto https://www.pinterest.it/enzoexposyto/_saved/ (presso World) https://www.instagram.com/p/ClJjj6hsUEK/?igshid=NGJjMDIxMWI=

[lthmath]

Ruffini

Paolo Ruffini, born on 22nd September 1765, was an Italian mathematician and philosopher. Ruffini's 1799 work marked a major development for group theory. Ruffini developed Joseph Louis Lagrange's work on permutation theory, following 29 years after Lagrange’s "Réflexions sur la théorie algébrique des equations" (1770–1771) which was largely ignored until Ruffini who established strong connections between permutations and the solvability of algebraic equations. Ruffini was the first to controversially assert the unsolvability by radicals of algebraic equations higher than quartics. This angered many members of the community such as Malfatti (1731–1807). Work in this area was later carried on by those such as Abel and Galois who succeeded in such a proof.

In mathematics, Ruffini's rule is a practical way for paper-and-pencil computation of the Euclidean division of a polynomial by a binomial of the form x – r. It was described by Paolo Ruffini in 1804. Ruffini's rule is a special case of synthetic division when the divisor is a linear factor.

[yellowmagicalgirl]

Juliet Dies in This Chapter Four: Behind the Scenes

No, this is not the real chapter four; I’m not going to break away from my “five stages of grief” format I’ve got going for me. You’ve got to wait probably a minimum of 12 hours for the chapter to post. However, tomorrow’s chapter is gonna involve a little bit of math. How I, the author, arrived at the solution will be irrelevant to the story and thus not included in the chapter. However, as someone who enjoys simple math problems like this one, I would like to share the way I found the solution. (I will be linking to this post when I post to tumblr and AO3 as an author’s note. Sorry, FFN people, I can’t do links while using that particular posting platform.)

Without further ado, let’s start.

(x^2 – 2x – 24)/(2x^4 – 5x^3 + 3x^2) ≥ 0

So this is our starting equation. A quadratic trinomial is being divided by a quartic trinomial, and this entire fraction must be greater than or equal to zero.

The trinomial in the numerator is very easily factored by the x-method. Please note that it has been years since I first learned how to factor, and I learned the method so that B is on top and A*C is on the bottom, so this may look a little wonky depending on how one was taught Algebra I.

[Image description: An “X” is drawn in red, with numbers in the four spaces created by the lines. The numbers written in the vertical compartments are red, and the numbers written in the horizontal compartments are blue. In the upper compartment is negative two, and in the lower compartment is negative twenty-four. In the left compartment is negative six, and in the right compartment is positive four.]

In the meantime, we can easily factor out an x squared from the denominator, giving us this equation:

((x - 6)(x + 4))/(x^2(2x^2 – 5x + 3)) ≥ 0

Once again, we’re going to use the x-method.

[Image description: Another “X”, with the same formatting as the previous one. In the upper compartment is negative five, and in the lower compartment is positive six. In the left compartment is negative two, and in the right compartment is negative three.]

However, we can’t just split (2x^2 – 5x + 3) into neat and simple factors of (x - 2) and (x - 3). Instead, we need to split our B term into

2x^2 – 5x + 3

2x^2 – 2x - 3x + 3

2x(x - 1) - 3(x - 1)

(2x - 3)(x - 1)

We can insert our factored expression back into the semi-factored equation that we got (2x^2 – 5x + 3) from.

((x - 6)(x + 4))/(x^2(2x - 3)(x - 1) ) ≥ 0

Do any terms cancel out? No, they don’t. Bummer. However, now that this equation has been completely factored, we know some key aspects.

x must be greater than or equal to positive six and negative four, but we can’t stop at the numerator. x must also never equal zero, positive three halves, and positive one.

Because this is an Algebra II/Pre-Calculus problem, I’m not going to solve for every single case manually. Instead, I will be using a scientific calculator. I’d use Excel, but I can’t copy and paste a graph. The results are as follows:

x < -4 ; positive

x = -4 ; zero

-4 < x < 0 ; negative

x = 0 ; undefined

0 < x < 1 ; negative

x = 1 ; undefined

1 < x < 3/2 ; positive

x = 3/2 ; undefined

x > 3/2 ; positive

Hence, x can be less than or equal to negative four, greater than positive one, and not equal to three halves. Or, to write the answer in a more elegant way,

x ≤ -4 U 1 ≤ x ≤ 3/2 U x ≥ 3/2

[devotty]

Quartic Equation Olympiad Tactics: Learn and Conquer

[arxt1]

The competition between the hydrodynamic instability from noise and magnetorotational instability in the Keplerian disks. (arXiv:2205.13230v1 [astro-ph.HE])

We venture for the comparison between growth rates for magnetorotational instability (MRI) and hydrodynamics instability in the presence of an extra force in the local Keplerian accretion flow. The underlying model is described by the Orr-Sommerfeld and Squire equations in the presence of rotation, magnetic field and an extra force, plausibly noise with a nonzero mean. We obtain MRI using Wentzel-Kramers-Brillouin (WKB) approximation without extra force for purely vertical magnetic field and vertical wavevector of the perturbations. Expectedly, MRI is active within a range of magnetic field, which changes depending on the perturbation wavevector magnitude. Next, to check the effect of noise on the growth rates, a quartic dispersion relation has been obtained. Among those four solutions for growth rate, the one that reduces to MRI growth rate at the limit of vanishing mean of noise in the MRI active region of the magnetic field, is mostly dominated by MRI. However, in MRI inactive region, in the presence of noise the solution turns out to be unstable, which are almost independent of the magnetic field. Another growth rate, which is almost complementary to the previous one, leads to stability at the limit of vanishing noise. The remaining two growth rates, which correspond to the hydrodynamical growth rates at the limit of the vanishing magnetic field, are completely different from the MRI growth rate. More interestingly, the latter growth rates are larger than that of the MRI. If we consider viscosity, the growth rates decrease depending on the Reynolds number.

from astro-ph.HE updates on arXiv.org https://ift.tt/vcFiSZN