Let's Solve A Special Type of Quartic
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I used the quartic formula and got that one of the roots is (1/2)(sqrt(2)+sqrt(1+i*sqrt(31))-sqrt(1-i*sqrt(31))). You have shown several quartics with cubic and quadratic terms missing. So how about x^4+8x=-12?
By inspection, I know that this equation has at least two real roots. One real root is just a bit smaller than 1 and the other real root is somewhat smaller than -2. Fixed-point iteration x←(7-x^4)/8. For x=1, iterate to x=0.81881... For x=-2.2, iterate x←-∜(7-8x) to x=-2.233025...
Note the two different forms of fixed-point iteration. One form or the other may be needed to converge for different values of x.
Generally a particular fixed-point iteration equation will converge if the absolute value of its derivative at x is less than 1.
Is less than one for all the domain right? Or for a specific neighborhood with its center at the solution?
Solution according to Ludovico Ferrari (student of Geronimo Cardano),
who invented it as the first method to solve the general quartic in around 1540
(published by Cardano, together with the solution to the general cubic, in "Ars Magna" in 1545):
x^4 + 8x = 7
x^4 = -8x + 7
Ansatz: (x^2 + z)^2 = x^4 + 2zx^2 + z^2 for any value of z.
So add 2zx^2 + z^2 to both sides:
x^4 + 2zx^2 + z^2 = 2zx^2 - 8x + (z^2 + 7)
The right side is a perfect square iff. its discriminant is zero.
Here we have
a = 2z
b = -8
c = z^2 + 7
Discriminant D = b^2 - 4ac = 0 or, to put it simpler, b^2 = 4ac:
(-8)^2 = 4*(2z)*(z^2 + 7)
64 = 8*z*(z^2 + 7)
Divide by 8:
8 = z*(z^2 + 7)
8 = z^3 + 7z
z^3 + 7z - 8 = 0
This cubic resolvent has a solution z = 1, as 8 = 1 + 7.
Plug z = 1 into the quartic above:
x^4 + 2x^2 + 1 = 2x^2 - 8x + 8
x^4 + 2x^2 + 1 = 2*(x^2 - 4x + 4)
Both sides can now be written as perect squares:
(x^2 + 1)^2 = (sqrt(2)*((x - 2))^2
Two possibilities by taking the square root:
First quadratic:
x^2 + 1 = sqrt(2)*(x - 2)
x^2 + 1 = sqrt(2)*x - 2*sqrt(2)
x^2 - sqrt(2)*x + (1 + 2*sqrt(2)) = 0
Discriminant D
= 2 - 4*(1 + 2*sqrt(2))
= 2 - 4 - 8*sqrt(2)
= -2 - 8*sqrt(2)
= (-2)*(1 + 4*sqrt(2))
Two complex solutions:
x1, x2
= ( sqrt(2) +- sqrt(2)*sqrt(1 + 4*sqrt(2))*i ) / 2
= ( 1 +- sqrt(1 + 4*sqrt(2))*i ) / sqrt(2)
Second quadratic:
x^2 + 1 = - sqrt(2)*(x - 2) = 0
x^2 + 1 = - sqrt(2)*x + 2*sqrt(2)
x^2 + sqrt(2)*x + (1 - 2*sqrt(2)) = 0
Discriminant D
= 2 - 4*(1 - 2*sqrt(2))
= 2 - 4 + 8*sqrt(2)
= -2 + 8*sqrt(2)
= 2*(4*sqrt(2) - 1)
Two real solutions:
x3, x4
= ( - sqrt(2) +- sqrt(2)*sqrt(4*sqrt(2) - 1) ) / 2
= ( - 1 +- sqrt(4*sqrt(2) - 1) ) / sqrt(2)
Are you sure you are correct
You missed sign after moving terms to the other side of equation
but I like your approach
@@holyshit922 Yes, I had some wrong signs in the quadratics because I was typing them into my smartphone in the rain at a bus stop, which was a bit tedious :-) I have corrected and expanded the solution. It now matches the solution of the video.
perfect 😍
wow!
Typically in a problem like this I spit the constant in to two numbers so one of them being a positive complete square on LHS. The number 7 has two basic options 16-9 and 8-1. Only the 8-1 worked in this case. Then added and subtracted the middle term 2x^2. It produces the same result as yours. This is method 2b ( or not to be. lol. )
Nice!
Thanks!
Good one. Hey, consider this though it's maybe too easy.
Take the product of four consecutive integers, add one, take the square root. For example, sqrt(3*4*5*6 + 1) = 19. Prove that this always yields an integer.
You mean the product, not the sum. Interesting observation, straightforward proof.
@@altosrule6387 You are right, thanks I will edit. It's not hard to get to the quartic but factoring it was a puzzle for me!
This is easy. Let n be the smallest of the four consecutive integers, then the other three integers are n + 1, n + 2 and n + 3 and we need to prove that
n(n + 1)(n + 2)(n + 3) + 1
is the square of an integer. First we multiply the outer two factors and the inner two factors of the product of the four consecutive integers which gives
(n² + 3n)(n² + 3n + 2) + 1
Now, the average of the two factors n² + 3n and n² + 3n + 2 is n² + 3n + 1 and the difference of n² + 3n + 2 and n² + 3n is 2, which means that we can rewrite n² + 3n as n² + 3n + 1 − 1 and n² + 3n + 2 as n² + 3n + 1 + 1 so we have
(n² + 3n + 1 − 1)(n² + 3n + 1 + 1) + 1
We now have a product of the difference and the sum of n² + 3n + 1 and 1 and applying the difference of two squares identity (a − b)(a + b) = a² − b² we can rewrite this as
(n² + 3n + 1)² − 1² + 1
which is
(n² + 3n + 1)²
and this is indeed the square of an integer for any integer n, which completes the proof.
@@NadiehFanTold you I thought it was easy. But happy you found it fun! :D
Good problem
heavy!
Your claim at 7:35 that factoring a quartic with real coefficients by converting this into a difference of two squares with real coefficients is not always possible is *false* and it is easy to prove that this is in fact always possible.
Since any quartic equation can be written as a monic quartic equation, we can without loss of generality assume we have a monic quartic equation
x⁴ + ax³ + bx² + cx + d = 0
where all coefficients a, b, c, d are assumed to be real. Note that this includes all monic quartic equations which lack one or more terms, such as a cubic term or a quadratic term, or both, since each of the coefficients can be zero.
Now, it is well known (and easy to prove) that _complex_ solutions of algebraic equations with _real_ coefficients always occur as _conjugate pairs_ of complex solutions. That is, if x = x₀ is a complex zero of a polynomial with real coefficients, then x = x̄₀ is also a zero of the same polynomial. This is a simple consequence of the fact that the conjugate of any sum or product of complex numbers is equal to the sum or product of the conjugates of those numbers as well as the fact that any complex number which happens to be a real number is its own conjugate.
Also, according to the factor theorem, a polynomial contains a factor (x − x₀) _if and only if_ x₀ is a zero of that polynomial. If a polynomial with _real_ coefficients has a complex zero x = x₀ and therefore contains a factor (x − x₀) it will also have a conjugate zero x = x̄₀ and therefore contain a factor (x − x̄₀). But if a polynomial contains both a factor (x − x₀) and a factor (x − x̄₀) this means that this polynomial contains a quadratic factor
(x − x₀)(x − x̄₀) = x² − (x₀ + x̄₀)x + x₀x̄₀
where both the coefficients −(x₀ + x̄₀) and x₀x̄₀ are _real_ because both the sum and the product of two conjugate complex numbers are real.
Since complex zeros of polynomials with real coefficients always occur as conjugate pairs, this means that a quartic polynomial with real coefficients can have either 0 or 2 or 4 complex zeros, that is, it can either have 0 or 1 or 2 conjugate pairs of complex zeros, together with either 4 or 2 or 0 real zeros. Since any conjugate pair of complex zeros results in a quadratic factor with real coefficients and since obviously any pair of real zeros also gives a quadratic factor with real coefficients, it follows that _any monic quartic polynomial with real coefficients can be factored into two monic quadratic polynomials with real coefficients_ regardless of the presense or absence of complex zeros.
So, for any monic quartic equation
x⁴ + ax³ + bx² + cx + d = 0
where all coefficients a, b, c, d are real there exist _real_ numbers p, q, r, s such that this equation can be written as
(x² + px + q)(x² + rx + s) = 0
Now, if we consider the _identities_
x² + px + q = x² + ½(p + r)x + ½(q + s) + ½(p − r)x + ½(q − s)
x² + rx + s = x² + ½(p + r)x + ½(q + s) − ½(p − r)x + ½(q − s)
we can write the equation as
(x² + ½(p + r)x + ½(q + s) + ½(p − r)x + ½(q − s))(x² + ½(p + r)x + ½(q + s) − ½(p − r)x + ½(q − s)) = 0
and applying the difference of two squares identity this can be written as
(x² + ½(p + r)x + ½(q + s))² − (½(p − r)x + ½(q − s))² = 0
Since p, q, r, s are real, all coefficients ½(p + r), ½(q + s), ½(p − r), ½(q − s) are also real. So, we have proved that _any monic quartic polynomial with real coefficients can be written as a difference of two squares where the first term is the square of a monic quadratic polynomial with real coefficients and the second term is the square of either a linear polynomial with real coefficients or the square of a real constant_
Q.E.D.
I have studied the quartic intensely and agree. Furthermore, I could prove the logic assumption that every of the three solutions to the cubic resolvent leads to one of the three possible combinations of the four linear factors if you group them on pairsnto the two quadratics.
It follows that if the cubic resolvent has only one real solution, there exists only one factorization of the quartic into two quadratics with REAL coefficients, and that we get two real and two complex solutions of the quartic.
The other two cases are a bit more difficult to describe.
If the cunic resolvent has three real solutions, any of them leads to two quadratics with real coefficients. Either both quadratics have real solutions, then the quartic has four real solutions. Or both quadratics have complex solutions, then the quartic has four complex solutions.
It cannot happen that the quartic has two real and two complex solutions but its cubic resolvent only one real solution, because if that were the case, there would exist two factorizations of the into two quadratics with complex coefficients, leasing to four complex solutions.
All said of course assumes that the original quartic has real coefficients. And I have omitted the special case of multiple roots (zero discriminant of both the cubic resolvent and the original quartic).
Wow! Nice 😍
The 2nd method looks simpler...
(x^2 - sqrt(2)*x+1+2*sqrt(2)) is a factor
I used Bairstow's method with pseudo random initial guess which I wrote in Pascal
This method still neds to be improved
(maybe by implementing correct way to choose initial guess or maybe by replacing Newton's method for system of two nonlinear equations created by equating remainder of division polynomial by quadratic to zero)
Pozdrawiam serdecznie i życzę miłego dnia
Thank you! Likewise 🥰
Wow
You didn't demonstrate that complex values of *a* lead to complex values of *x* before discarding them.
It is not pleasant, but it doesn't actually matter which of the three solutions for a you choose, in the end you will always find the exact same four roots of the quartic equation. This is so because each of the three values for a (and the value for b that goes with it) corresponds to one of the three possible factorizations of the quartic into two quadratics.
Good point.
(x - x₁)(x - x₂) and (x - x₃)(x - x₄)
(x - x₁)(x - x₃) and (x - x₂)(x - x₄)
(x - x₁)(x - x₄) and (x - x₂)(x - x₃)
I'd start by noticing that x=-1 is a root, factor out (x+1) and then move on to the cubic.
The test for x=-1 is almost as simple as the test for x=1.
No, x = −1 is not a root.
I believe x = - e^((Pi)i) also works
Nice!
Thanks!