/sci/ - Science & Math

Plug in random numbers and record on a number line where its positive and negative.
Home in on your zeros
If they're not integers or easily recognizable roots, you're fucked.
If you find a zero or two, synthetic division to make the rest of the zeros easier.
Or you could always just use the quartic formula.
Pic related

>>7970816
Also, you're screwed on this one cause any integer plugged into this equation outputs an odd number, and zero is even.

These are the answers Wolfram gives me

>>7970805

>>7970825
>4 solutions
>the complex one is not a complex number
Uhh

It's not first grader shit. There's a binomial theorem that you learn about in introductory discrete math.
I'm too drunk to actually check if this is right, but it looks like your cubed root is

4^(1/3)x + 3^(1/3)

>>7970829
Thanks for the formula, it's rather usefull, to be honest, I've never seen this one, or at least I don't remember.

>those related queries

>>7970927
>try it
>how would a dog see a plot of (4 x^3-12 x^2-4 x+3)-0

>>7970805
You can always try the Newton-Raphson approximation method; I don't know Latex so I can't type out the formula, but it's a quick google search away!

>>7970927
So you need to take 192mg per kg to die of caffeine. Neat.

>>7970805
Just proof there are no "nice" roots, with a criterion of your choice. Eisenstein or that other one I forgot the name.

>>7970805
If you wanted to do it pen and paper then I'd suggest using Newton's method (which will give you a rational approximation to the real root, in this case taking [math] x _{0} = \frac {1} {2} [/math] gives [math] x \approx \frac {10} {26} [/math]) or use the cubic formula to find all three roots.

>>7970805
I usually plug in [-3;3], if something works polynomial division. Your function is fucked though, I think you made a mistake somewhere.

Ruffini's mehtod

>>7971551
Isn't ruffini only for integer or rational roots?

>>7970854

This is a version of the general cubic formula, which can be used to solve (your) OP's polynomimal equation. I have derived it in the past and verified that (given real/complex caveats which I forget) it can in fact be simplified to this; this is about as "nice" a complete version, or algebraic solution, as can be expressed.

There is an alternative form of the formula where the hairy stuff is moved into denominators, and wiki purports that that form is "better" for some reason that I haven't appreciated. see cubic equation on wiki for more.

I am going to test this form versus the OP's equation via hand calculation, because I could use the practice and I'm curious.

>>7971961

Consider OP's >>7970805 polynomial equation [math] ax^3 + bx^2 + cx + d = 0, [/math] , where [math] a = 4, \;\;\; b = -12, \;\;\; c = -4, \;\;\; d = 3 [/math] .

If memory serves, these coefficients a-d can then be plugged and chugged "one-to-one" in the formulae >>7970829 without any special thought or replacement of terms. I've worked with and test-driven this version of the cubic formula before, but I haven't actually inspected the below yet so if I run into a goof we'll soon know.

The relevant bits of the formula >>7970829 simplify as follows:

[math] \displaystyle - \frac{b}{3a} = 1 , \;\;\; - \frac{1}{3a} = - \frac{1}{12} , \;\;\; \frac{1 \pm i \sqrt{3} }{6a} = \frac{1 \pm i \sqrt{3} }{24} \;\;\; aka \;\; A^+, A^-, \;\;\; ... = \sqrt[3]{-1944 \pm 24 i \sqrt{5727} } \;\; aka \;\; C^+, C^- [/math]

This gives the roots

[math] \displaystyle x_1 = 1 - \frac{1}{12} C^+ - \frac{1}{12} C^- , \\ x_2 = 1 + A^+ C^+ + A^- C^- , \\ x_3 = 1 + A^- C^+ + A^+ C^- . [/math]

; this treatment and analogous "nice" forms of same are often described as /Cardano forms/ of the roots. Notice that the thing under that third root radical is in both cases a strictly complex number (not purely real, not purely imaginary), having two components that are, in the grand scheme of things, pretty close together (that other 5727 radical thing with 24 is about 1816-something, close to being "square" with 1944).

Since OP's cubic has real coefficients, we may expect at least one real root, or else something has gone fucky. It is also highly probable that I've made at least one arithmetic mistake somewhere. I would imagine that the next step is to apply De Moivre's formula to get a nicer form for those third root-things, but I think I'll eat lunch first.

>>7972063

Awesome, please keep updating your progress

OP are you familiar with Galois Theory from algebra? You have to use the Galois groups to derive the formula for a triatic equation. Or just look it up online. Then just plug in the values into the formula. It's simple.

>>7972063

Although it will become onerous, we wish to express the third-root form from this post itself as a complex number with real and imaginary parts, or as a choice of one of the three nth (third) roots implied by taking the cube root of a complex number. This obliges us to employ De Moivre's formula, which in turn obliges us to express the radicand in polar form. We will take the C^+ form first by way of example, though both must be completed, per the above.

In the case of C+, suffice to say that the radicand is a complex number in the second quadrant, not far from the line "y=-x", denoting not a functional notation so much as a relationship between real and imaginary parts of numbers on that line in the plane. Where r is the radius (modulus) and theta the angle or argument (and the latter expressions are a simple does-the-answer-make-sense check, we find that

[math] \displaystyle r = 1536 \sqrt{3} , \;\;\; \theta = \tan^{-1} \bigg( - \frac{24 \sqrt{5257} }{1944} \bigg) + \pi \;\; "=" \;\; 2.39 > \frac{3}{4} \pi [/math] .

Which leads to the polar/Euler/De-Moivre-y expression of such-and-such, and the extraction of the third roots of this can be put thus (the third root of r part simplifies down rather nicely):

[math] \displaystyle C^+ = \cos \Bigg( \frac{1}{3} \bigg( \tan^{-1} \bigg( - \frac{24 \sqrt{5257} }{1944} + \pi \bigg) + 2 \pi k \bigg) \Bigg) + i \sin \Bigg( \frac{1}{3} \bigg( \tan^{-1} \bigg( - \frac{24 \sqrt{5257} }{1944} + \pi \bigg) + 2 \pi k \bigg) \Bigg) . [/math]

where k = 0, 1 or 2. We have plugged-and-chugged using https://en.wikipedia.org/wiki/De_Moivre's_formula#Roots_of_complex_numbers

Recall that C- is a distinct value (the conjugate), appearing in another part of the roots we wish to express and check. The above must therefore be repeated for C-.

Now is as good a time as any to re-state (my) goal: give and verify 'closed-form' (if not algebraic, inverse trig) solutions to OP's polynomial. I have gone out on some limbs, however.

>>7970805
Newton-Raphson or Bisection solver algorithms if a numerical solution is good enough

>>7970805
Her's an algorithm from algebra 2, basically generate some numbers using the factors of the last and first number, then when you get those number try synthetic division till one works.

>>7972285
forgot the link:
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml

factor theorem and remainder theorem.

>>7972063

Although it will become onerous we wish to express the third-root form from this post as a complex number with real and imaginary parts, or as a choice of one of the three nth (third) roots implied by taking the cube root of a complex number. This obliges us to employ De Moivre's formula, which in turn obliges us to express the radicand in polar form. We will take the C^+ form first by way of example, though both must be completed, per the above.

In the case of C+, suffice to say that the radicand is a complex number in the second quadrant, not far from the line "y=-x", denoting not a functional notation so much as a relationship between real and imaginary parts of numbers on that line in the plane. Where r is the radius (modulus) and theta the angle or argument (and the latter expressions are a simple does-the-answer-make-sense check, we find that

[math] \displaystyle r = 1536 \sqrt{3} , \;\;\; \theta = \tan^{-1} \bigg( - \frac{24 \sqrt{5257} }{1944} \bigg) + \pi \;\; "=" \;\; 2.39 > \frac{3}{4} \pi , [/math]

Which leads to the polar/Euler/De-Moivre-y expression of etc, and the extraction of third roots of this can be put thus (the third root of r part simplifies down rather nicely):

[math] \displaystyle C^+ = 8 \sqrt{3} \Bigg( \cos \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( - \frac{24 \sqrt{5257} }{1944} \big) + \pi \bigg) + 2 \pi k \bigg) + i \sin \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( - \frac{24 \sqrt{5257} }{1944} \big) + \pi \bigg) + 2 \pi k \bigg) \Bigg) . [/math]

where k = 0, 1 or 2. We have plugged-and-chugged using https://en.wikipedia.org/wiki/De_Moivre's_formula#Roots_of_complex_numbers

Recall that C- is a distinct value (the conjugate), appearing in another part of the roots we wish to express and check. The above must therefore be repeated.

Now is as good a time as any to re-state (my) goal: give and verify 'closed-form' (if not algebraic, inverse trig) solutions to OP's polynomial. I have gone out on some limbs.

>>7972292

Toward the end of this post I should more properly have said that the /radicands/ of the third root are the things that are conjugages, not C+ and C- themselves for my present purposes (though I would expect the six roots to maybe have conjugate-forms but never mind that right now).

Brief reconsideration of C-'s radicand, however, leads us to state its analogous form (since we've already bashed out the code, just need pm tweaks):

[math] \displaystyle C^+ = 8 \sqrt{3} \Bigg( \cos \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( \frac{24 \sqrt{5257} }{1944} \big) - \pi \bigg) + 2 \pi k \bigg) + i \sin \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( \frac{24 \sqrt{5257} }{1944} \big) - \pi \bigg) + 2 \pi k \bigg) \Bigg) . [/math]

>>7972292

I should more properly have said toward the end of this post that the /radicands/ of C+ and C- are the things that are conjugates, not C+ and C- themselves (though that might be the case, never mind that now).

Since I've already bashed out the code for C+, I might as well give C- (just have to tweak some pm's for a very similar arctan argument form):

[math] \displaystyle C^- = 8 \sqrt{3} \Bigg( \cos \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( \frac{24 \sqrt{5257} }{1944} \big) - \pi \bigg) + 2 \pi k \bigg) + i \sin \bigg( \frac{1}{3} \bigg( \bigg( \tan^{-1} \big( \frac{24 \sqrt{5257} }{1944} \big) - \pi \bigg) + 2 \pi k \bigg) \Bigg) . [/math]

>>7972063

So, with choices of k for /two/ distinct third-root forms (or what I might better have written as k' just now in the case of C-), and with reference to this post, what's left are multiplications and additions of "x+iy" complex-number-forms, to arrive at our roots for OP's polynomial in closed form. We would then be obliged to, y'know, actually check by plugging in.

And I thought of doing this earlier, but before I carry this silliness any further, a simple graph of OP's polynomial in functional form reveals three distinct roots, /all real/, and agreeing with the general feel of this post >>7970825 I should therefore expect to collapse my silliness down to three distinct real numbers when I'm done, and have them all actually work, or else what's the point.

>>7972361

And, while inspecting x_1 when we choose k = k' = 0, happily I am satisfied that the imaginary parts do indeed cancel (in this case), as expected. However I find my first serious problem: this route's decimal approximation does not match any of >>7970825 , which I do concur is what wolfram spits out when you have it solve the original polynomial.

So, so many places where I could have made a goof. Can you see a mistake that I made? Although doubtful in this case, one also wonders if there's any diffrences in how the CAS interprets the two approaches.

>>7972503

Happily, my mistake is easily identified upon review of the above. Between two posts, I started consistently replacing the correct "5727" term from >>7972063 , with the incorrect "5257", which was then reproduced consistently throughout >>7972292
>>7972338 and >>7972503 . Upon double checking, the rest of what I did still looks right to me. Especially since I can now plug in the corrected C+ and C- (replace per the above), and end up with this, which agrees with the earlier "solver". The imaginary parts in this case still cancel nicely.

So I'm satisfied that I have used the formula >>7970829 , together with appropriate choices of ks in the De Moivre representation earlier, to arrive at a closed-form representation one of OP's three roots (pic related). Specifically, the single negative one among its three distinct real roots.

I think we learned this in precalc. Use Descarte's law of signs and the rational zero theorem to narrow down possible roots and then plug in random roots in synthetic division. Do it twice to get a quadratic and then factor or use formula.

Actually inspecting the products A(+-)C(+-) in the other roots, while still holding k = k' = 0 as is our wont, yields up numerical approximations in WA which again agree with >>7970805 , once I cheat again and punt back to it. Inspecting these products one-by-one shows that they associated forms /are/ complex conjugates, and consequently their imaginary parts vanish in their respective sums. So since what I have by this point is a nice product of complex numbers (in various signing-cases) of the form (a+bi)(c+di), where I can simply ignore the imaginary part, I instead move to compute the respective real parts, adding 1 in each case, per our forms in >>7972063 .

I should also mention in passing that WA's straight-solver appears to give complicated stuff in the denominators of certain terms, which is at-a-distance from this "Cardano form" that I've used all along. This goes directly back to my missing information about the substantive difference between the two forms which are mentioned on wiki's cubic equation page. It has to do with choice of roots, something that I've glossed over in this exercise.

>>7972841

the "agree with" clause here is rather addressed to the picture in >>7970825 .

>>7972854

Furthermore, the two positive real roots, although not of course complex conjugates, yet have for their expressions nearly identical forms, but for some judicious changes of sign. They can be expressed in joint form as follows:

[math] \displaystyle x_{2,3} = 1 + \frac{2 \sqrt{3}}{3} \cos \bigg( \frac{1}{3} \bigg( \tan^{-1} \big( \pm_{2} \frac{24 \sqrt{5727} }{1944} \big) \pm_{3} \pi \bigg) \bigg) \pm_{1} 2 \sin \bigg( \frac{1}{3} \bigg( \tan^{-1} \big( \pm_{2} \frac{24 \sqrt{5727} }{1944} \big) - \pm_{3} \pi \bigg) \bigg) [/math]

The indices of x and the pm signs are somewhat unrelated formalities; the central pm_1 option affords one choice, but pm_2 and pm_3 must be of opposite signs, I just haven't edited these. They check with the above. These are the closed forms of the above two remaining positive real roots. It is tempting to check their arithmetic mean, which is about 1.8072. Note also that the above schema affords the opposite-signing route of arriving at the same two real roots.

Pretty much the last thing to do would be to actually plug these closed forms into OP's polynomial.

>>7970805
You have a computer, so put that badboy into Excel.