I remember how to derive the formula for the cubic ( x^3 + a x^2 + bx + c = 0) by remembering these three steps: 1) Eliminate the quadratic term by a substitution x = y - a/3 (so that new cubic in y has no y^2 term). 2) Eliminate the fractions and then make a substitution absorbing the lead coefficient (it will be u = 3y). 3) Write u as the sum of two cube roots (so u = s^(1/3) + t^(1/3) ). (I don't really "remember" step 2 so much as I just "do it", because it's so natural; it's crying out to be done. But Step 1, and especially Step 3, I need to commit to memory. Step 3 is really the main one for me at least. I've derived this so many times that I now do remember the derivation without trouble.) With that setup, the solution is basically forced onto you by the algebra. You'll be forced into some algebra showing that you need to find s and t that simultaneously solve 2 equations: one for s+t in terms of the given coefficients, and another for st in terms of the given coefficients. But that means that s and t are the solutions to the quadratic: z^2 - (s+t) z + (st) = 0... and again, you know the values of s+t and st in terms of the original given a, b, and c. It works out like this: x^3 + a x^2 + bx + c = 0 Step 1: Let y = x + a/3, so that x = y - a/3. Then (y - a/3)^3 + a (y - a/3)^2 + b(y - a/3) + c = 0 [ y^3 - a y^2 + (a^2/3) y - a^3/27 ] + a [ y^2 - (2a/3) y + a^2/9 ] + b [ y - a/3 ] + c = 0 y^3 + ( a^2/3 - 2a^2/3 + b ) y + ( - a^3/27 + a^3/9 - ab/3 + c ) = 0 y^3 + ( - a^2/3 + b ) y + ( 2 a^3/27 - ab/3 + c ) = 0. Have a cubic in y with no quadratic term. Step 2: Eliminate the fractions and then make a substitution absorbing the lead coefficient (you can already see the coefficient expressions appearing in the Lagrange Resolvent) Multiply both sides by 27: 27 y^3 + ( - 9 a^2 + 27 b ) y + ( 2 a^3 - 9 ab + 27 c ) = 0 (3y)^3 - 3 ( a^2 - 3 b ) (3y) + ( 2 a^3 - 9 ab + 27 c ) = 0 u^3 - 3 ( a^2 - 3 b ) u + ( 2 a^3 - 9 ab + 27 c ) = 0, where u = 3y. Let P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c, so P and Q are known values determined by the original cubic equation. The original cubic equation in x is now the cubic equation in u given by: u^3 - 3 P u + Q = 0. Step 3: Suppose s and t are values (possibly complex) such that u = s^(1/3) + t^(1/3). Then u^3 = [ s^(1/3) + t^(1/3) ]^3 = s^(3/3) + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t^(3/3) = s + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t = s + 3 [ s^(2/3) t^(1/3) + s^(1/3) t^(2/3) ] + t = s + 3 s^(1/3) t^(1/3) [ s^(1/3) + t^(1/3) ] + t = s + 3 s^(1/3) t^(1/3) [ u ] + t = 3 [ s^(1/3) t^(1/3) ] u + [ s + t ] = 3 [ st ]^(1/3) u + [ s + t ]. Thus u = s^(1/3) + t^(1/3) implies that u^3 - 3 [ st ]^(1/3) u - [ s + t ] = 0. That's a universal formula, but now make it apply to the specific equation u^3 - 3 P u + Q = 0, where P and Q are known. That requires that - 3 [ st ]^(1/3) = - 3 P, and - [ s + t ] = Q, and so that requires: st = P^3 and s+t = -Q. Thus s and t are the two solutions to the quadratic: z^2 - (s+t) z + (st) = 0, so here meaning z^2 + Q z + P^3 = 0. And that's the desired formula. Tracing it back in terms of x, get: x = y - a/3 = u/3 - a/3 = [ - a + s^(1/3) + t^(1/3) ] / 3, where s and t are the two solutions to the quadratic: z^2 + Q z + P^3 = 0, where P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c.
The fact that you put on your effort to write such a long,tedious answer is absolutely mind bongling....damn your commitment towards maths is truly golden
@@lavanyajain2722 Thanks, that was nice. However, the truth is, it isn't as demanding as it seems. When you spend your life swimming in a topic, things that can seem impressive to others really aren't that big a deal. Writing this up took a little investment in time, but it's worthwhile (ya know, gotta spread math love to everyone in the world who wants it) and it's not really an effort. I re-derive it any time I need to solve a cubic (it's now quicker than looking it up), so it's second nature by now. If you'd spent as many hours of your life focused on math as I have, it would be the same for you. The effort in the write-up is almost entirely in thinking about the best way to give a presentation that others can benefit from, not really thinking about the math itself.
Any Asian parent: "Don't accept sweets from strangers, it might be drugs" This guy: "Don't accept quintic equations from strangers, it might be unfactorable"
@@hellopleychess3190 The quadratic formula always factors it. Essentially, if you have an integer quadratic, ax² + bx + c, with a,b,c all integers, and you want all integers in the factors, the discriminant, ∆ = (b² - 4ac) will always tell you whether that's possible. If ∆ is a perfect square, then yes; if not, no.
@@hellopleychess3190 Yes, well it seems you didn't describe the problem you were having. What actually IS the quadratic you're trying to factor? I.e., what are a, b, and c? And what exactly does your "factoring method" consist of? (There are many ways to find factors of a quadratic polynomial.) Fred
“Don’t try to factor someone’s random quintic equation. It’s not safe.” 😂😂😂 I love this problem and I enjoy your humor and smiles of satisfaction as you solve it! This teacher appreciates your enthusiasm for math!
I plugged the equation into Wolfram Alpha because I wanted to know the complex solutions, and here they are: x ≈ -0.1378 - 1.5273i x ≈ -0.1378 + 1.5273i
@@pauselab5569 There's the two real solutions that BPRP found first, and then another real solution from the cubic. The last two are imaginary. 5 total.
Since you asked how I'm doing. I've been a subscriber for a long time. Back when I started at community college years ago your calculus 2 videos were very helpful in my studying. After that I would watch your videos periodically. Six years after initially subscribing I'm about to graduate with a degree in computer science. I'll continue to watch your videos even though I don't need calculus 2 videos to help me study because I like the content you make. Keep up the good work.
just graduated with my B.A. in applied mathematics, lots of which was thanks to you and your brilliant videos about differential equations. but these are just the fun random videos I liked watching when im making a coffee, doing dishes, eating breakfast, etc. the beauty and joy of mathematics, and the humor throughout make this content better than the rest
in 2012 I finished my mathematics studies during my 2 years of intensive preparation in the so-called ``classes preparatoires`` in Morocco, which is a French education system that focuses on maths and physics for 2 successive years followed by a national competition to get a place in engineering schools. I graduated as an engineer and worked for many years and I am still enjoying your videos tackling some difficult calculus problems, and when you finish and find the solution it is like a stress-relieving process for me. Thank you for the great job, which is way better than watching the shitty social media content that has no point in our modern age.
My parents always told me never take quintic equations from strangers! lol thank you for these videos. I'm just watching for fun because it's calming to watch someone work through algebraic problems, especially after finishing a calc 3 course :)
For degree 3 the Lagrange resolvent isn't actually that bad to compute. It's all about symmetric polynomials of the roots, the key idea is to find quantities that remain fixed even if you switch around the roots. To do so let's call x_1, x_2 and x_3 the three roots of the cubic x^3+bx^2+cx+d, then we have by Vieta's formulas the three following polynomials: x_1 + x_2 +x_3 = -b x_1*x_2 + x_1*x_3 +x_2*x_3 = c x_1*x_2*x_3 = -d These polynomials are symmetric polynomials of the roots since clearly if you switch around the roots you still get the same three quantities. In fact they are the simplest symmetric polynomials of three variables and are known as the elementary symmetric polynomials of 3 variables and an important theorem is that any symmetric polynomial of n variables can be expressed using the elementary polynomials of n variables and therefore their values can be expressed using the coefficients of an equation of degree n. Knowing this Lagrange sets two quantities: y_1 = x_1 + j*x_2 + j^2*x_3 where j is any of the two roots of j^2+j+1 = 0 y_2 = x_1 + j^2*x_2 + j*x_3 Then you can show that the sum of the cubes of those two quantities stays the same even if you switch the roots around. This means the sum of the cubes is a symmetric polynomial of the roots which can then be expressed using the 3 elementary symmetric polynomials above and therefore y_1^3 + y_2^3 can be expressed using the coefficients b,c and d of the cubic to be solved. You can also show that the product of y_1 and y_2 is also invariant by switching around the roots so the product of the two quantities can also be expressed using the coefficients of the original cubic. In other words you can get this system of equations: y_1^3 + y_2^3 = something expressed with b, c and d y_1*y_2 = something else expressed with b, c and d Clearly if you take the cube of the second equation you get a system where you're looking for quantities for which you know the sum and the product, y_1^3 and y_2^3 will therefore be the roots of a quadratic equation and this equation will be the Lagrange resolvent. If you do your math right you will find y_1^3+y_2^3 = -2b^3 + 9bc - 27d and y_1*y_2 = b^2 - 3c. If you manage to do it you might try to apply the same reasoning for a quartic equation and find a resolvent that will be a cubic though this is noticeably longer to compute. Happy computation :)
I started watching your channel when I was in the 7th grade, that year I won the math Olympiad with 60/60 and it was so hard that the second highest score in my school was 47. All thanks to your videos. Also, I could not solve everything back then, but now I understand almost all of them.
My semester is indeed almost done :) I'm in grad school, and taking advanced algebra this semester. Next year I'll be starting my research for my master's thesis. But in my class we actually covered the insolvability of the general quintic a few weeks ago, so this video relates to what I'm learning about in my class!
@@roberttelarket4934 um... false? Abstract algebra is probably going to be a requirement for any kind of pure math masters/doctoral degree you can find, including the US which is where I'm going to school actually.
@@drfpslegend4149: You are apparently not in the U.S. Advanced Algebra is not the term used for an undergraduate Abstract Algebra or Modern Abstract Algebra course. Advanced Algebra was only used in my days in high school(usually as a senior) in the late 1960s and it definitely had nothing of the nature of Modern Abstract Algebra(groups, rings, etc.). Even today the latter is probably almost never taught in high school here! If accepted into a math grad program you must have as pre-requisite introductory Modern Abstract Algebra during your undergraduate years.
@@roberttelarket4934 a quick google search does indeed show that advanced algebra is a course that can be taken by senior undergraduate or master degree students at some universities in the u.s.
I am reaching 12th Grade in September so I am preparing for math and physics already,your videos are very enjoyable and informative,keep it up! Sending love from Greece
Will you record video about quintics solvable by radicals f.e when quinics are solvable by radicals and how tho solve them How good you are in Numerical methods Here polynomial roots can be approximated by QR method for eigenvalues Two of the companion matrices are already in upper Hessenberg form so reduction to Hessenberg form is not needed (matrix with coefficients in first row xor last column and ones just below main diagonal) To get QR decomposition you can use Householder reflections xor Givens rotations (rotations are easier to derive) Unshifted method is A_{1}=A A_{i}=Q_{i}R_{i} A_{i+1}=R_{i}Q_{i} Clever choice of shift should accelerate convergence but for repeated eigenvalues it is still slow Doubly implicit shift is tempting because matrix entries are real but eigenvalues may be complex Complex eigenvalues can be calculated from 2x2 blocks
The complex solutions are straightforward, though. In your formule 1/3 (-a + ³√z1 + ³√z2) you apply a small change: 1/3 (-a + ω ³√z1 + 1/ω ³√z2) and 1/3 (-a + 1/ω ³√z1 + ω ³√z2) where ω is a complex cubic root of unity (ω³ = 1, ω = -1/2 + i√3/2) It's really the same recipe for Cardano's formula.
The formula you gave is even more convenient if you use w2 (omega squared) instead of 1/w, as they're equivalent and both equal to the conjugate of w. w = -1/2 + i*sqrt(3)/2 w2 = -1/2 - i*sqrt(3)/2 Of course, it's the same thing. I just think the 1/w (reciprocal) notation makes the computation look harder than it really is.
The good "news" is that noone asks a mathematician "Yes, but why have you tried this (method/idea etc.) ?" as soon as it works. The winner is always right. Bravo for having created this working example !
I have found the two complex solutions, and they are much longer than the real solution to the cubic to write out (4 cube roots!) This quintic was quite the journey, and I plan to do more of these. Here are the complex solutions: cbrt = cube root sqrt = square root x = (1/3) - cbrt[(65+15*sqrt(21))/432] - cbrt[(65-15*sqrt(21))/432] ± {cbrt[(65*sqrt(3)+45*sqrt(7))/144] - cbrt[(65*sqrt(3)+45*sqrt(7))/144]}*i Note: I combined fractions into the cube roots because I think that they look ugly outside the cube roots. I have to give credit to @rslitman for showing the technique in the comments. Hopefully I didn't make any typos.
Well my answer is 1 real root and 2 complex roots from the cubic equation but a little bit different from yours. Real solution according to Cardano: X=(1/3)*[1+(1/cbrt(2))*(cbrt(65+15*sqrt(21))+cbrt(65-15*sqrt(21))]. The complex solutions are of course much longer.
My preferred method for extracting the other two roots is to simply observe that in the process of deriving the cubic formula, the two cube roots, p and q, multiply to equal a constant. So you can simply multiply one root by (-1+sqrt(3)i)/2 and the other by (-1-sqrt(3)i)/2, and vice versa, assuming the first pair of cube roots you took already produced the correct answer, of course...
Semester's done, watching for fun to help fix the c in business calculus. Thanks for your video, it's above my education but you explain things really well.
I finished my Numerical Analysis II final exam last week - your videos let me reminisce on the good ol' days of Calculus and Algebra. How I'm doing is very relieved to have finished that class.
For f(x) =x^5-5x+3, we find that: f(-1) =7, f(0) =3, f(1) =-1, f(2) =25, then x must be near to 1 or x=1+v and x=1-w, and v, w are small number. For x=1+v, using trap method, we have f(1,.5) =3.0938 and f(1.25) = -0.1928' then 1.25
How to arrive at the complex solutions: 1. Keep the 1/3 fractions in place. 2. Multiply the first cube root by (-1 + sqrt(-3))/2. 3. Multiply the second cube root by (-1 - sqrt(-3))/2. 4. The above steps give you the first complex solution. To get the other one, swap the placement of the two complex multipliers. You may recognize the complex numbers as containing the unit circle values of the cos and sin of 2pi/3 and 4pi/3 (or 120 degrees and 240 degrees). In fact, the multiplier of the real root is 1 + 0i, the unit circle values of the cos and sin of 0 (or 0 degrees).
If you already have a quadratic equation in the form 0 = x² + px + q (which of course could easily be achieved with any quadratic equasion by dividing ax² + bx + c = 0 by a), the quadratic formula reduces to x = -p/2 +- sqrt((p/2)-q) and thus is much less convoluted.
It’s fair to say the mechanics you explained are fantastic. For me. It’s substitution of trigonometric identity that is REQUIRED in order to come to a simplified expression. And that is where I stopped there’s a lot of thought in this video. Thank you it’s grand.
There is a trick that can be used in cases like this. If the polynomial has a factorization, reducing the polynomial modulo various primes will reduce the factors as well, and there are tools for factoring polynomials modulo primes that can speed up the process (Berlekamp's algorithm comes to mind). In this case, two obvious primes to reduce by are the coefficients 3 and 5. Reducing mod 5 and factoring produces x^5+3 = (x+3)^5 mod 5, which goes quickly since all five factors are linear. If a factorization of the original polynomial exists at all, there must be a quadratic factor and a cubic factor, and (x+3)^2 = x^2+6x+9 = x^2+x+4 mod 5 must be the quadratic factor. Now the constant term 4 does not divide the constant term of x^5-5x+3 over the integers, but x^2+x+4 = x^2+x-1 mod 5 and -1 divides +3. Testing x^2+x-1 shows that it is actually a factor of the original polynomial, and the other (cubic) factor comes from long division (and is equal to (x+3)^3 = x^3+9x^2+27x+27 modulo 5, as expected).
With a little linear algebra there is a general solution for a system of n quadratics in n unknowns. What you do is that for matrices A and B, you have x†Ax + Bx + C = 0 for a symmetric A where x is the vector of variables and x† is its transpose. Since A is symmetric you can choose a basis to diagonalize A, then (xS^(-1))† D (Sx) + BSx + C = 0. Then you can complete the square and take the square root. In finding the square root of D each eigenvalue has two square roots and so there are 2^n solutions.
At 13:43 You know what? You *don't* still have to check this! You used the fourth and second equations to derive the values for A, so those A values _must_ satisfy those two equations. The only one you need to check for consistency is the third one (and you already found it was consistent when A=1). This is guaranteed unless you think you've made a mistake, in which case it's a useful check of your accuracy. But it's most definitely not necessary.
Perhaps one of your best videos yet. For all practical purposes, though, an iterative formula gives real roots to any level of precision we choose: x₀ = ?; xₙ₊₁ = (4xₙ⁵ − 3)/(5xₙ⁴ − 5). x₀ = −2: xₙ → −1.61803398875... = −(√5 + 1)/2; x₀ = 0: xₙ → 0.61803398875... = (√5 − 1)/2; x₀ = 2: xₙ → 1.27568220364.... = complicated stuff with cube roots and square roots.
@@MichaelRothwell1 To be honest, I can't remember exactly. It's based on my own variation of the Newton-Raphson method, developed independently to solve specific problems. I think it's a little less versatile, but it seems to provide exactly the same rate of convergence.
x^5 -5x +3 converted is 1000(-5)3 in base x which is 99953 in base 10. I.e. 10^5 -5*10 +3 = 99953. Using prime factorisation we get that 99953 = 7 * 109 *131. If it can be broken into 2 factors it's either (7*109) * 131, (7*131) * 109 or 7 * (109*131). I.e. 763 * 131, 917 * 109 or 7 * 14279. Of these (we can try all 3 combinations if we want) 917 * 109 looks nicest. We can change them to (1000-100+20-3) * (100+10-1), or 1(-1)2(-3) * 11(-1) which converted from base x to polynomials gives (x^3-x^2+2x-3) * (x^2+x-1) , which we can quickly see is our answer. (x^3-x^2+2x-3) * (x^2+x-1) = x^5 -5x +3
I'm trying to follow your approach but I'm confused as to how you know 917*109 is the combination that works and not the others. Also, couldn't you write 917 as 1000-100+10+7, which would lead to a totally different polynomial...so how do you the correct form?
@@ryugasen I don't have a general solution/approach (perhaps one exists out there) but the rule of thumb is to have lower magnitude coefficients. 917 could be 1000-100+10+7 or 1000-100+20-3 opr perhaps something else, but I prefer 1000-100+20-3 since the coefficients are max 3 there, while the other has a coefficient of 7. Similarly I prefer 917*109 over the other combinations. But again, we can check all reasonable combinations if we want to since there are not that many.
I tried to prove langrange resolvent based on the construction of cubic formula and the three solutions would be (-a+(w^j)*³√z1+(w^(3-j))*³√z2)/3 Where j=0,1,2 and w=(1+i√3)/2
@@mathmathician8250 I thought it was the other way around since x³ - 1 = (x - 1)(x² + x + 1) and you get the -1 in the numerator due to the -b of the quadratic formula. I've seen some TH-cam vids use your version, but I also saw my version by just searching "cube roots of unity" in Google. I don't know what's going on
Very nice computations and explanations. Just a hint: the conjugate complex of a cubic are often given in lange math encyclpediae. I am used to the notation y1 = u + v y2,3 = - (u + v)/2 +- (u - v)2 * sqrt(3) * i where y denotes the variable of the reduced (depressed) cubic and u and v are the two cube roots. Often, the Tschirnhaus Transformation x = y - b/(3a) is used in order to transform ax^3 + bx^2 + cx + d = 0 to the reduced form y^3 + 3py + q = 0 So in the end we get x1 = -b/(3a) + ... etc. So thwre is no need to do the polynomialndivision to get y2 and y3 (it has already been done in the y2,y3 formulae above).
I just worked a cubic and a quartic using this method. The cubic worked out nicely because I was able to find one real factor for the binomial, so I only had to use one ax term in the quadratic. The quartic ended up factoring into two quadratics, both with a “square root type” middle term. Not so hard to factor the quartic, but a REAL pain to find the roots using the quadratic formula. This quintic, though…no, this is too much for me. Thank you for all the methodology. I’ve never used the cubic formula and now I know why 😝
thanks for asking how it's going :) I actually had a further math final today D: and I left a bunch of it empty, because I did some simplification error and couldn't get a vector equation for a 3d line 😔 but I do love your videos! you speak with such excitement, I can see how much you love math I too love math, but sometimes I feel like it doesn't love me as much
The last two complex solutions can be found by multiplying complex cubic roots of 1 with the cunic roots of Z1 and Z2, like this: Real solution: X=(-a+Z1^(1/3)+z2^(1/3))/3 First complex solution: X=(-a+t*Z1^(1/3)+s*z2^(1/3))/3 Second complex solution: X=(-a+s*Z1^(1/3)+t*z2^(1/3))/3 Where s and t are the complex cubic roots of 1. t=exp(2pi/3 i) s=exp(-2pi/3 i)
let f(x) = x^5 - 5x + 3. f(-2) = -32 + 10 + 3 = -19. f(0) = 0 - 0 + 3 = 3 f(1) = 1 - 5 + 3 = -1 f(2) = 32 - 1 + 3 = 34 Thus there is a real root between -2 and 0, a second real root between 0 and 1, and a third real root between 1 and 2. Let f'(x) = 5x^4 - 5 be the first derivative of f(x) and use x = x - f(x)/f'(x) to iteratively update estimates of roots of f(x) = 0. Set x = -1 or so to iteratively compute the real root between -2 and 0. Set x = 0.5 or so to iteratively compute the real root between 0 and 1. Set x = 1.5 or so to iteratively compute the real root between 1 and 2. With these three roots use synthetic division to obtain a remaining quadratic g(x) = 0. Use quadratic formula to get the remaining two roots.
It's not as precise or fun but you can also graph y=x^5 and y=5x-3 and find the intersection points. The x-coordinates of those points are the solutions.
Wow i‘m so amazed! I dont really have a clue about math, but i think its really great how many people can talk about this topic with you. Greeting from Germany🇩🇪
OK I used a root-finder. But the equation fits to phi=0.618... phi^2 = - phi + 1 phi^3 = 2 phi - 1 phi^4 = -3 phi + 2 phi^5 = 5 phi - 3 etc The coefficients are alternating, offset terms of the Fibonacci sequence. One to store in the memory bank EDIT: You could apply this idea to any equation of the type x^5 = Ax + B. I.e put x^2 = ax + b x^3 = ax^2 + bx = (a^2+b) x + ab ... x^5 = (a^4 + 3 a^2 b + b^2) x + ab(a^2 + 2 b) In this case A = a^4 + 3 a^2 b + b^2 = 5 [1] B = ab (a^2 + 2 b) = -3 [2] Solve [1] as a quadratic in a^2 a^2 = (-3 +/- sqrt(5b^2+20))/2 To get rational a^2 guess b^2=1 a^2 must be the positive root i.e. a^2=1 Solve [1] as a quadratic in b b = (-3*a^2 +/- sqrt(5 a^4+20))/2 b = (-3 +/- 5)/2 We must have b^2=1 so b = 1 Substitute in [2] 3a = -3 a = -1 The quadratic factor is x^2 - ax - b = x^2 + x - 1
A white van pulled up while I was walking and asked me to solve a quintic equation. Luckily BPRP taught me never to talk to strangers who ask quintic equations so I ran away.
Nice work. When calculating discriminant to solution of the Langrange resolver, you might want to take a short cut and factor before taking the square root of the discriminant, i.e b = 65 = 13*5, therefore b^2 = 13^2*5^2. 4*a*c = -4*1*125 = -4*5^2. You can factor out 5^2 so the discrimiannt is now 5^2*(13^2 + 5*4) = 5^2*(169 + 20) = 5^2*(189). 189 is divisible by 3 (sum of digits), and so is 189 / 3 = 69, 69 / 3 = 21 = 7*3. So you have (5^2)*(3^3)*7. You can now take the square root and you have 5*3*sqrt(7*3)
We got this lil thing in Italy called "Ruffini's Rule", we're taught this during algebra classes but everyone tends to forget it (cause it's tedious). I'm pretty sure that would've allowed us to find some real solutions from the beginning. What do you think about this? Cheers, Thomas! 💗
14:25 jaaay broh we got it 😂😂😂 i have only the 10 th class finished so that the only thing i undeestand bec. that only what i learn at 10th.class i just know x^2+2x+4=0😂
Hello! I am thinking about the "If a stranger gives you this quintic equsion....", and it gives giggles because it sounds like "Don't try this at home" xD. Regardless of me not being a student at any university, i still find your video entertaining
No need to solve for exact decimal to decimal complex solution But, you can numerically calculate the value of 'r' and then form the last quadratic which could be solved further. Will land up on an approximate complex solution.
Or try to find Delta A = 1 B =-5 C= 3 Delta = B²-4AC If delta more then 0 / positive it has 2 answers If its 0 it has 1 answer If its negative it doesnt have answers
Me deciding to follow up with my math teacher to actually understand a lesson for once; Him starting the lesson 0:45 I look down at my book for one second and then look back at him; Him literally the next second: 30:10
Please, please, please! Teach us everything about the Lagrange resolvent. Are there resolvents for polynomial equations of other powers? If there are, please give examples. How did Lagrange come up with this formula? Wikipedia gives a general formula for Lagrange resolvents (sum from i=0 to n-1 X sub i omega ^i), where omega is a primitive nth root of unity. How does that apply here, or in polynomial equations of other powers? I haven't seen anything like that on TH-cam. Thanks a lot!
You can get the complex roots easier if you just multiply the first cube root in the real root by one complex cube root of unity (-1/2 + sqrt(3)i/2) and the second one by the other complex cube root of unity ( (-1/2 - sqrt(3)i/2)). Then the other complex root is the conjugate of the first - or, you can reverse the 2 cube roots of unity and that will also give the second complex solution.
Another form by brute force, suppose that q(x) is a quadratic factor of f(x)=x^5-5*x+3, then f(0)=3, f(1)=-1. We can assume that q is monic, q(x)=x^2+b*x+c. Therefore q(0) divides 3 and q(1) divides -1. We take many finitely cases for q(x). This method proves that factoring polynomials (with integers coeficients) is equivalent to factor integer numbers. But this method is pure brute force.
Wiki states that if you multiply the cubic root parts you got by (-1+sqrt(3)i)/2, then you will get the 1st complex solution and 2nd complex solution if you multiply by (-1-sqrt(3)i)/2 instead. In other words, if formula for the real root is: x1 = 1/3 + 1/3 * cbrt(A) + 1/3 * cbrt(B) Then x2 = 1/3 + 1/3 * cbrt(A) * (-1+ sqrt(3)i)/2 + 1/3 * cbrt(B) * (-1+ sqrt(3)i)/2 x3 = 1/3 + 1/3 * cbrt(A) * (-1 - sqrt(3)i)/2 + 1/3 * cbrt(B) * (-1 - sqrt(3)i)/2 That makes the same sense as ± part of quadratic equation formula, please notice that: 1^2 = (-1)^2 = 1, 1^3 = [(-1+ sqrt(3)i)/2]^3 = [(-1- sqrt(3)i)/2]^3 = 1
This is not quite correct. You need to multiply one cube root by one non-real cube root of unity and the the other cube root by the other non-real cube root of unity, as explained in the comment by Risu0chan.
Well you see, suppose for integers a, b, p, and q, if x^2-ax-b divides x^5-px-q, then a divides q^2. Because of the small margins, I won't leave a proof.
Sheesh bro lemme sleep. Ngl, I really find your vids so satisfying to watch that it made me appreciate more the innate beauty of Math. God bless you man!!! Much love from ph!
There is an exam at my grad school which we have to take to start the research stage of our Ph.D. There is almost always a polynomial on there and they ask us if it can be factored using whole numbers. Usually there is a trick that makes it quick, but I was doing one of the older tests and one test writer gave one that had to be solved like this. It was very mean.
When you have x = 1/3(-a + cube root of something + cube root of the conjugate of something). Can't we get the complex solutions if we multiply the cube roots by the appropriate cube roots of unity?
Yes, you can, as explained by rslitman. Basically you multiply the first cube root by ω (where ω is a non-real cube root of 1) and the other cube root by ω² (the complex conjugate of ω). Then you do the same again, but multiply the cube roots by ω² and ω (i.e. in the opposite order).
I remember how to derive the formula for the cubic ( x^3 + a x^2 + bx + c = 0) by remembering these three steps:
1) Eliminate the quadratic term by a substitution x = y - a/3 (so that new cubic in y has no y^2 term).
2) Eliminate the fractions and then make a substitution absorbing the lead coefficient (it will be u = 3y).
3) Write u as the sum of two cube roots (so u = s^(1/3) + t^(1/3) ).
(I don't really "remember" step 2 so much as I just "do it", because it's so natural; it's crying out to be done. But Step 1, and especially Step 3, I need to commit to memory. Step 3 is really the main one for me at least. I've derived this so many times that I now do remember the derivation without trouble.)
With that setup, the solution is basically forced onto you by the algebra. You'll be forced into some algebra showing that you need to find s and t that simultaneously solve 2 equations: one for s+t in terms of the given coefficients, and another for st in terms of the given coefficients. But that means that s and t are the solutions to the quadratic:
z^2 - (s+t) z + (st) = 0... and again, you know the values of s+t and st in terms of the original given a, b, and c.
It works out like this:
x^3 + a x^2 + bx + c = 0
Step 1: Let y = x + a/3, so that x = y - a/3.
Then
(y - a/3)^3 + a (y - a/3)^2 + b(y - a/3) + c = 0
[ y^3 - a y^2 + (a^2/3) y - a^3/27 ] + a [ y^2 - (2a/3) y + a^2/9 ] + b [ y - a/3 ] + c = 0
y^3 + ( a^2/3 - 2a^2/3 + b ) y + ( - a^3/27 + a^3/9 - ab/3 + c ) = 0
y^3 + ( - a^2/3 + b ) y + ( 2 a^3/27 - ab/3 + c ) = 0.
Have a cubic in y with no quadratic term.
Step 2: Eliminate the fractions and then make a substitution absorbing the lead coefficient
(you can already see the coefficient expressions appearing in the Lagrange Resolvent)
Multiply both sides by 27:
27 y^3 + ( - 9 a^2 + 27 b ) y + ( 2 a^3 - 9 ab + 27 c ) = 0
(3y)^3 - 3 ( a^2 - 3 b ) (3y) + ( 2 a^3 - 9 ab + 27 c ) = 0
u^3 - 3 ( a^2 - 3 b ) u + ( 2 a^3 - 9 ab + 27 c ) = 0,
where u = 3y.
Let P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c, so P and Q are known values determined by the original cubic equation.
The original cubic equation in x is now the cubic equation in u given by:
u^3 - 3 P u + Q = 0.
Step 3: Suppose s and t are values (possibly complex) such that u = s^(1/3) + t^(1/3).
Then
u^3 = [ s^(1/3) + t^(1/3) ]^3
= s^(3/3) + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t^(3/3)
= s + 3 s^(2/3) t^(1/3) + 3 s^(1/3) t^(2/3) + t
= s + 3 [ s^(2/3) t^(1/3) + s^(1/3) t^(2/3) ] + t
= s + 3 s^(1/3) t^(1/3) [ s^(1/3) + t^(1/3) ] + t
= s + 3 s^(1/3) t^(1/3) [ u ] + t
= 3 [ s^(1/3) t^(1/3) ] u + [ s + t ]
= 3 [ st ]^(1/3) u + [ s + t ].
Thus u = s^(1/3) + t^(1/3) implies that
u^3 - 3 [ st ]^(1/3) u - [ s + t ] = 0.
That's a universal formula, but now make it apply to the specific equation u^3 - 3 P u + Q = 0, where P and Q are known.
That requires that
- 3 [ st ]^(1/3) = - 3 P, and
- [ s + t ] = Q,
and so that requires:
st = P^3 and s+t = -Q.
Thus s and t are the two solutions to the quadratic:
z^2 - (s+t) z + (st) = 0, so here meaning
z^2 + Q z + P^3 = 0.
And that's the desired formula. Tracing it back in terms of x, get:
x = y - a/3 = u/3 - a/3 = [ - a + s^(1/3) + t^(1/3) ] / 3, where s and t are the two solutions to the quadratic:
z^2 + Q z + P^3 = 0,
where P = a^2 - 3 b, and Q = 2 a^3 - 9 ab + 27 c.
This is purely amazing!
Woww.. I would prefer sudoku hard mode instead of solving this. Thanks for the video. I learned Lagrange resolvent 👍
Can you make on the derivation of parametric forms of pythagorean quadruples...
Like
m² + n² - p² - q²
2(mq + np)
2(nq - mp)
m² + n² + p² + q²
The fact that you put on your effort to write such a long,tedious answer is absolutely mind bongling....damn your commitment towards maths is truly golden
@@lavanyajain2722
Thanks, that was nice.
However, the truth is, it isn't as demanding as it seems. When you spend your life swimming in a topic, things that can seem impressive to others really aren't that big a deal. Writing this up took a little investment in time, but it's worthwhile (ya know, gotta spread math love to everyone in the world who wants it) and it's not really an effort. I re-derive it any time I need to solve a cubic (it's now quicker than looking it up), so it's second nature by now. If you'd spent as many hours of your life focused on math as I have, it would be the same for you. The effort in the write-up is almost entirely in thinking about the best way to give a presentation that others can benefit from, not really thinking about the math itself.
Any Asian parent: "Don't accept sweets from strangers, it might be drugs"
This guy: "Don't accept quintic equations from strangers, it might be unfactorable"
And not just Asian parents! My Italian mom told me (as a kid) that.
Fred
I'm trying to factor a quadratic equation right now but it has 11 and 17 and I can't figure out how to factor it
@@hellopleychess3190 The quadratic formula always factors it.
Essentially, if you have an integer quadratic, ax² + bx + c, with a,b,c all integers, and you want all integers in the factors, the discriminant, ∆ = (b² - 4ac) will always tell you whether that's possible.
If ∆ is a perfect square, then yes; if not, no.
@@ffggddss it seems you didnt understand the problem? I was trying to solve using the factoring method and 11 and 17 had no shared factor.
@@hellopleychess3190 Yes, well it seems you didn't describe the problem you were having. What actually IS the quadratic you're trying to factor?
I.e., what are a, b, and c? And what exactly does your "factoring method" consist of? (There are many ways to find factors of a quadratic polynomial.)
Fred
“Don’t try to factor someone’s random quintic equation. It’s not safe.” 😂😂😂 I love this problem and I enjoy your humor and smiles of satisfaction as you solve it! This teacher appreciates your enthusiasm for math!
Thank you!
It's dangerous to do it alone so you need to Bring radicals.
why couldnt you just do the rational root theorem
I plugged the equation into Wolfram Alpha because I wanted to know the complex solutions, and here they are:
x ≈ -0.1378 - 1.5273i
x ≈ -0.1378 + 1.5273i
@coomf any imaginary or irrational root of a polynomial must have a conjugate as a root (assuming the polynomial has rational, real coefficients)
@@nightytime only if the coefficients are real
@@ghotifish1838 yes
shouldn't there be 5 solutions though?
@@pauselab5569 There's the two real solutions that BPRP found first, and then another real solution from the cubic. The last two are imaginary. 5 total.
I hate that I have to keep turning down people that try to hand me quintics when I’m walking on the sidewalk. Those people are so annoying.
Since you asked how I'm doing. I've been a subscriber for a long time. Back when I started at community college years ago your calculus 2 videos were very helpful in my studying. After that I would watch your videos periodically. Six years after initially subscribing I'm about to graduate with a degree in computer science. I'll continue to watch your videos even though I don't need calculus 2 videos to help me study because I like the content you make. Keep up the good work.
Wow 6 years!! Thank you for the continuous support and congrats on graduating! I wish you the best in everything!
And remember kids : Do not take quintic equation from a stranger ! 😜
😆
😂😂
x^5-x+1
😂😂😂
😂unless there is obvious roots
just graduated with my B.A. in applied mathematics, lots of which was thanks to you and your brilliant videos about differential equations. but these are just the fun random videos I liked watching when im making a coffee, doing dishes, eating breakfast, etc. the beauty and joy of mathematics, and the humor throughout make this content better than the rest
in 2012 I finished my mathematics studies during my 2 years of intensive preparation in the so-called ``classes preparatoires`` in Morocco, which is a French education system that focuses on maths and physics for 2 successive years followed by a national competition to get a place in engineering schools. I graduated as an engineer and worked for many years and I am still enjoying your videos tackling some difficult calculus problems, and when you finish and find the solution it is like a stress-relieving process for me. Thank you for the great job, which is way better than watching the shitty social media content that has no point in our modern age.
My parents always told me never take quintic equations from strangers! lol thank you for these videos. I'm just watching for fun because it's calming to watch someone work through algebraic problems, especially after finishing a calc 3 course :)
For degree 3 the Lagrange resolvent isn't actually that bad to compute. It's all about symmetric polynomials of the roots, the key idea is to find quantities that remain fixed even if you switch around the roots.
To do so let's call x_1, x_2 and x_3 the three roots of the cubic x^3+bx^2+cx+d, then we have by Vieta's formulas the three following polynomials:
x_1 + x_2 +x_3 = -b
x_1*x_2 + x_1*x_3 +x_2*x_3 = c
x_1*x_2*x_3 = -d
These polynomials are symmetric polynomials of the roots since clearly if you switch around the roots you still get the same three quantities. In fact they are the simplest symmetric polynomials of three variables and are known as the elementary symmetric polynomials of 3 variables and an important theorem is that any symmetric polynomial of n variables can be expressed using the elementary polynomials of n variables and therefore their values can be expressed using the coefficients of an equation of degree n.
Knowing this Lagrange sets two quantities:
y_1 = x_1 + j*x_2 + j^2*x_3 where j is any of the two roots of j^2+j+1 = 0
y_2 = x_1 + j^2*x_2 + j*x_3
Then you can show that the sum of the cubes of those two quantities stays the same even if you switch the roots around. This means the sum of the cubes is a symmetric polynomial of the roots which can then be expressed using the 3 elementary symmetric polynomials above and therefore y_1^3 + y_2^3 can be expressed using the coefficients b,c and d of the cubic to be solved. You can also show that the product of y_1 and y_2 is also invariant by switching around the roots so the product of the two quantities can also be expressed using the coefficients of the original cubic. In other words you can get this system of equations:
y_1^3 + y_2^3 = something expressed with b, c and d
y_1*y_2 = something else expressed with b, c and d
Clearly if you take the cube of the second equation you get a system where you're looking for quantities for which you know the sum and the product, y_1^3 and y_2^3 will therefore be the roots of a quadratic equation and this equation will be the Lagrange resolvent.
If you do your math right you will find y_1^3+y_2^3 = -2b^3 + 9bc - 27d and y_1*y_2 = b^2 - 3c. If you manage to do it you might try to apply the same reasoning for a quartic equation and find a resolvent that will be a cubic though this is noticeably longer to compute. Happy computation :)
I aint reading all of the fnaf
I started watching your channel when I was in the 7th grade, that year I won the math Olympiad with 60/60 and it was so hard that the second highest score in my school was 47. All thanks to your videos.
Also, I could not solve everything back then, but now I understand almost all of them.
I wish i had something like this in my school.
Which Olympiad?
My semester is indeed almost done :) I'm in grad school, and taking advanced algebra this semester. Next year I'll be starting my research for my master's thesis. But in my class we actually covered the insolvability of the general quintic a few weeks ago, so this video relates to what I'm learning about in my class!
Yay! 😀 quintics are always fun
In U.S. grad schools you don’t take advanced algebra.
@@roberttelarket4934 um... false? Abstract algebra is probably going to be a requirement for any kind of pure math masters/doctoral degree you can find, including the US which is where I'm going to school actually.
@@drfpslegend4149: You are apparently not in the U.S. Advanced Algebra is not the term used for an undergraduate Abstract Algebra or Modern Abstract Algebra course. Advanced Algebra was only used in my days in high school(usually as a senior) in the late 1960s and it definitely had nothing of the nature of Modern Abstract Algebra(groups, rings, etc.). Even today the latter is probably almost never taught in high school here!
If accepted into a math grad program you must have as pre-requisite introductory Modern Abstract Algebra during your undergraduate years.
@@roberttelarket4934 a quick google search does indeed show that advanced algebra is a course that can be taken by senior undergraduate or master degree students at some universities in the u.s.
I am reaching 12th Grade in September so I am preparing for math and physics already,your videos are very enjoyable and informative,keep it up! Sending love from Greece
You know shit gets real when he has to pull out the blue pen
and the green one too
Try this quintic equation (ft cubic formula) next 👉 th-cam.com/video/YkEPMf1l2os/w-d-xo.html
Sure.
I love how you say "ft cubic formula"
4:27 who is the stranger that hands quintic equations to everyone? I want to know
Will you record video about quintics solvable by radicals
f.e when quinics are solvable by radicals and how tho solve them
How good you are in Numerical methods
Here polynomial roots can be approximated by QR method for eigenvalues
Two of the companion matrices are already in upper Hessenberg form so reduction to Hessenberg form is not needed
(matrix with coefficients in first row xor last column and ones just below main diagonal)
To get QR decomposition you can use Householder reflections xor Givens rotations
(rotations are easier to derive)
Unshifted method is
A_{1}=A
A_{i}=Q_{i}R_{i}
A_{i+1}=R_{i}Q_{i}
Clever choice of shift should accelerate convergence but
for repeated eigenvalues it is still slow
Doubly implicit shift is tempting because matrix entries are real but eigenvalues may be complex
Complex eigenvalues can be calculated from 2x2 blocks
I run my application for eigenvalues written in C# to solve this equation
It works but can be improved
The complex solutions are straightforward, though. In your formule 1/3 (-a + ³√z1 + ³√z2) you apply a small change:
1/3 (-a + ω ³√z1 + 1/ω ³√z2) and 1/3 (-a + 1/ω ³√z1 + ω ³√z2)
where ω is a complex cubic root of unity (ω³ = 1, ω = -1/2 + i√3/2)
It's really the same recipe for Cardano's formula.
Very nice, but there's a typo in your value for ω, which should be -1/2 + i√3/2.
@@MichaelRothwell1 Thank you, I edited it.
The formula you gave is even more convenient if you use w2 (omega squared) instead of 1/w, as they're equivalent and both equal to the conjugate of w.
w = -1/2 + i*sqrt(3)/2
w2 = -1/2 - i*sqrt(3)/2
Of course, it's the same thing. I just think the 1/w (reciprocal) notation makes the computation look harder than it really is.
The good "news" is that noone asks a mathematician "Yes, but why have you tried this (method/idea etc.) ?" as soon as it works. The winner is always right. Bravo for having created this working example !
I have found the two complex solutions, and they are much longer than the real solution to the cubic to write out (4 cube roots!) This quintic was quite the journey, and I plan to do more of these. Here are the complex solutions:
cbrt = cube root
sqrt = square root
x = (1/3) - cbrt[(65+15*sqrt(21))/432] - cbrt[(65-15*sqrt(21))/432] ± {cbrt[(65*sqrt(3)+45*sqrt(7))/144] - cbrt[(65*sqrt(3)+45*sqrt(7))/144]}*i
Note: I combined fractions into the cube roots because I think that they look ugly outside the cube roots.
I have to give credit to @rslitman for showing the technique in the comments.
Hopefully I didn't make any typos.
Well my answer is 1 real root and 2 complex roots from the cubic equation but a little bit different from yours.
Real solution according to Cardano:
X=(1/3)*[1+(1/cbrt(2))*(cbrt(65+15*sqrt(21))+cbrt(65-15*sqrt(21))].
The complex solutions are of course much longer.
My preferred method for extracting the other two roots is to simply observe that in the process of deriving the cubic formula, the two cube roots, p and q, multiply to equal a constant. So you can simply multiply one root by (-1+sqrt(3)i)/2 and the other by (-1-sqrt(3)i)/2, and vice versa, assuming the first pair of cube roots you took already produced the correct answer, of course...
Semester's done, watching for fun to help fix the c in business calculus. Thanks for your video, it's above my education but you explain things really well.
I finished my Numerical Analysis II final exam last week - your videos let me reminisce on the good ol' days of Calculus and Algebra. How I'm doing is very relieved to have finished that class.
For f(x) =x^5-5x+3, we find that:
f(-1) =7, f(0) =3, f(1) =-1, f(2) =25, then x must be near to 1 or x=1+v and x=1-w, and v, w are small number.
For x=1+v, using trap method, we have f(1,.5) =3.0938 and f(1.25) = -0.1928' then 1.25
How to arrive at the complex solutions:
1. Keep the 1/3 fractions in place.
2. Multiply the first cube root by (-1 + sqrt(-3))/2.
3. Multiply the second cube root by (-1 - sqrt(-3))/2.
4. The above steps give you the first complex solution. To get the other one, swap the placement of the two complex multipliers.
You may recognize the complex numbers as containing the unit circle values of the cos and sin of 2pi/3 and 4pi/3 (or 120 degrees and 240 degrees). In fact, the multiplier of the real root is 1 + 0i, the unit circle values of the cos and sin of 0 (or 0 degrees).
If you already have a quadratic equation in the form 0 = x² + px + q (which of course could easily be achieved with any quadratic equasion by dividing ax² + bx + c = 0 by a), the quadratic formula reduces to x = -p/2 +- sqrt((p/2)-q) and thus is much less convoluted.
It’s fair to say the mechanics you explained are fantastic. For me. It’s substitution of trigonometric identity that is REQUIRED in order to come to a simplified expression. And that is where I stopped there’s a lot of thought in this video. Thank you it’s grand.
台灣人路過
雖然大學不是念理工科系學到比較深的數學，但本身對代數和離散數學discrete math領域挺有興趣了解的
給個支持🎉
Why would I even attempt to solve a stranger's quintic equation? 🤔
😂😂
@user-yb5cn3np5q Regular kids are enticed by puppies, not polynomials. 😈
I can't even remember how I found your channel but I've been subscribed for quite a few years now. Love every video.
Thank you!!
The best part is 7:57, where he had to stop and think about -1+5, right after writing systems of equations to define 5 variables
There is a trick that can be used in cases like this. If the polynomial has a factorization, reducing the polynomial modulo various primes will reduce the factors as well, and there are tools for factoring polynomials modulo primes that can speed up the process (Berlekamp's algorithm comes to mind).
In this case, two obvious primes to reduce by are the coefficients 3 and 5. Reducing mod 5 and factoring produces x^5+3 = (x+3)^5 mod 5, which goes quickly since all five factors are linear. If a factorization of the original polynomial exists at all, there must be a quadratic factor and a cubic factor, and (x+3)^2 = x^2+6x+9 = x^2+x+4 mod 5 must be the quadratic factor. Now the constant term 4 does not divide the constant term of x^5-5x+3 over the integers, but x^2+x+4 = x^2+x-1 mod 5 and -1 divides +3. Testing x^2+x-1 shows that it is actually a factor of the original polynomial, and the other (cubic) factor comes from long division (and is equal to (x+3)^3 = x^3+9x^2+27x+27 modulo 5, as expected).
Very nice! Now I know what to do next time a stranger hands me a random quintic to factorise.
With a little linear algebra there is a general solution for a system of n quadratics in n unknowns. What you do is that for matrices A and B, you have x†Ax + Bx + C = 0 for a symmetric A where x is the vector of variables and x† is its transpose. Since A is symmetric you can choose a basis to diagonalize A, then (xS^(-1))† D (Sx) + BSx + C = 0. Then you can complete the square and take the square root. In finding the square root of D each eigenvalue has two square roots and so there are 2^n solutions.
Good explanation! 'e' is represented by more than 300 decimal places next to the whiteboard. That is also what I observed.
At 13:43 You know what? You *don't* still have to check this! You used the fourth and second equations to derive the values for A, so those A values _must_ satisfy those two equations. The only one you need to check for consistency is the third one (and you already found it was consistent when A=1). This is guaranteed unless you think you've made a mistake, in which case it's a useful check of your accuracy. But it's most definitely not necessary.
Perhaps one of your best videos yet. For all practical purposes, though, an iterative formula gives real roots to any level of precision we choose:
x₀ = ?; xₙ₊₁ = (4xₙ⁵ − 3)/(5xₙ⁴ − 5).
x₀ = −2: xₙ → −1.61803398875... = −(√5 + 1)/2;
x₀ = 0: xₙ → 0.61803398875... = (√5 − 1)/2;
x₀ = 2: xₙ → 1.27568220364.... = complicated stuff with cube roots and square roots.
How did you get this particular iterative formula?
@@MichaelRothwell1 To be honest, I can't remember exactly. It's based on my own variation of the Newton-Raphson method, developed independently to solve specific problems. I think it's a little less versatile, but it seems to provide exactly the same rate of convergence.
I’m a 16 year old kid in love with math, I love these videos, thank you so much!!!
Honestly, now I just want to wander round the streets handing out randomly generated quintic equations.
Solving degree 4 + equations through brute force is a real tedious task, good job.
x^5 -5x +3 converted is 1000(-5)3 in base x which is 99953 in base 10. I.e. 10^5 -5*10 +3 = 99953. Using prime factorisation we get that 99953 = 7 * 109 *131. If it can be broken into 2 factors it's either (7*109) * 131, (7*131) * 109 or 7 * (109*131). I.e. 763 * 131, 917 * 109 or 7 * 14279. Of these (we can try all 3 combinations if we want) 917 * 109 looks nicest. We can change them to (1000-100+20-3) * (100+10-1), or 1(-1)2(-3) * 11(-1) which converted from base x to polynomials gives (x^3-x^2+2x-3) * (x^2+x-1) , which we can quickly see is our answer.
(x^3-x^2+2x-3) * (x^2+x-1) = x^5 -5x +3
I'm trying to follow your approach but I'm confused as to how you know 917*109 is the combination that works and not the others. Also, couldn't you write 917 as 1000-100+10+7, which would lead to a totally different polynomial...so how do you the correct form?
@@ryugasen I don't have a general solution/approach (perhaps one exists out there) but the rule of thumb is to have lower magnitude coefficients. 917 could be 1000-100+10+7 or 1000-100+20-3 opr perhaps something else, but I prefer 1000-100+20-3 since the coefficients are max 3 there, while the other has a coefficient of 7. Similarly I prefer 917*109 over the other combinations. But again, we can check all reasonable combinations if we want to since there are not that many.
It took me 8 minutes 56 seconds to realize that he was holding a pokeball the whole time and that it is a microphone
I tried to prove langrange resolvent based on the construction of cubic formula and the three solutions would be
(-a+(w^j)*³√z1+(w^(3-j))*³√z2)/3
Where j=0,1,2 and w=(1+i√3)/2
Shouldn't w=(-1+i√3)/2? [With a minus sign in front of the 1]
Or, am I missing something?
My w satisfies w³=1 but your w satisfies w³=-1
@@mathmathician8250 I thought it was the other way around since x³ - 1 = (x - 1)(x² + x + 1) and you get the -1 in the numerator due to the -b of the quadratic formula. I've seen some TH-cam vids use your version, but I also saw my version by just searching "cube roots of unity" in Google. I don't know what's going on
Oh yeah just realized my mistake
Me encantó este problema; además, maravilloso el rostro de buena persona de este matemático. Gracias.
I enjoy your pain and frustrations - makes it more authentic. As a physicist, this is still stimulating 🤣
In a^3+b^3+c^3=∆ then ; √∆ = a+b+c
Where a,b,c are selected values
A:-1 b:-2 c:-3
≈ From ｉｎｄｉａｎ
I don't care whether you are an indian or not. Your math is wrong.
Very nice computations and explanations. Just a hint: the conjugate complex of a cubic are often given in lange math encyclpediae. I am used to the notation
y1 = u + v
y2,3 = - (u + v)/2 +- (u - v)2 * sqrt(3) * i
where y denotes the variable of the reduced (depressed) cubic and u and v are the two cube roots.
Often, the Tschirnhaus Transformation x = y - b/(3a) is used in order to transform
ax^3 + bx^2 + cx + d = 0
to the reduced form
y^3 + 3py + q = 0
So in the end we get
x1 = -b/(3a) + ...
etc.
So thwre is no need to do the polynomialndivision to get y2 and y3 (it has already been done in the y2,y3 formulae above).
Corrections:
y^3 + 3py + 2q = 0
y1 = u + v
y2,3 = -(u+v)/2 +- (u-v)/2 * sqrt3) * i
Shitty typos.
Please ignore the horizontal line. It must be
y2,3 = - (u + v)/2 +- (u - v)/2 * sqrt(3) * i
Finally!!!
I just worked a cubic and a quartic using this method. The cubic worked out nicely because I was able to find one real factor for the binomial, so I only had to use one ax term in the quadratic. The quartic ended up factoring into two quadratics, both with a “square root type” middle term. Not so hard to factor the quartic, but a REAL pain to find the roots using the quadratic formula. This quintic, though…no, this is too much for me. Thank you for all the methodology. I’ve never used the cubic formula and now I know why 😝
thanks for asking how it's going :)
I actually had a further math final today D:
and I left a bunch of it empty, because I did some simplification error and couldn't get a vector equation for a 3d line 😔
but I do love your videos! you speak with such excitement, I can see how much you love math
I too love math, but sometimes I feel like it doesn't love me as much
The last two complex solutions can be found by multiplying complex cubic roots of 1 with the cunic roots of Z1 and Z2, like this:
Real solution:
X=(-a+Z1^(1/3)+z2^(1/3))/3
First complex solution:
X=(-a+t*Z1^(1/3)+s*z2^(1/3))/3
Second complex solution:
X=(-a+s*Z1^(1/3)+t*z2^(1/3))/3
Where s and t are the complex cubic roots of 1.
t=exp(2pi/3 i)
s=exp(-2pi/3 i)
sometimes I do my math homework and just play one of your videos in the background, its very soothing
Rare footage of a kindergarten teacher at an asian school
With Newtons Method I got x equals about 0.6180,-1.6180 and 1.2757.
Btw. Love your Videos!
let f(x) = x^5 - 5x + 3.
f(-2) = -32 + 10 + 3 = -19.
f(0) = 0 - 0 + 3 = 3
f(1) = 1 - 5 + 3 = -1
f(2) = 32 - 1 + 3 = 34
Thus there is a real root between -2 and 0, a second real root between 0 and 1, and a third real root between 1 and 2.
Let f'(x) = 5x^4 - 5 be the first derivative of f(x) and use x = x - f(x)/f'(x) to iteratively update estimates of roots of f(x) = 0.
Set x = -1 or so to iteratively compute the real root between -2 and 0.
Set x = 0.5 or so to iteratively compute the real root between 0 and 1.
Set x = 1.5 or so to iteratively compute the real root between 1 and 2.
With these three roots use synthetic division to obtain a remaining quadratic g(x) = 0.
Use quadratic formula to get the remaining two roots.
It's not as precise or fun but you can also graph y=x^5 and y=5x-3 and find the intersection points. The x-coordinates of those points are the solutions.
This will only give you the three real solutions, not the two complex ones.
@@bjornfeuerbacher5514 graph it in the complex plane lol
Surprisingly the most mind blowing part of this for me was the 65^2 trick
or 60*60 + 5*60 + 5*60 + 25 = 3600 + 300 + 300 + 25 = 4225 (2 seconds)
timestamp?
22:50
@@szerednik.laszlo thanks :D
whats the idea behind that? just a coincidence?
15:32 Very good advice ! Now i'll know what to do the next time a stranger give me a quintic equation
15:15 I’m an old engineer who went to university 25 years ago. I’m watching for fun (of course) but also to keep my brain alert!!!
Not as EXTREME PATIENCE as the extreme algebra question, but still extreme patience.
In fact, I did a Chinese version first. That’s why this one went a lot more smoothly 😆
@@blackpenredpen Oh that's cool. I'm taking my AP Calc BC Exam this Monday so wish me luck! I believe you took yours in 2004?
Wow i‘m so amazed! I dont really have a clue about math, but i think its really great how many people can talk about this topic with you.
Greeting from Germany🇩🇪
Thank you!
OK I used a root-finder. But the equation fits to phi=0.618...
phi^2 = - phi + 1
phi^3 = 2 phi - 1
phi^4 = -3 phi + 2
phi^5 = 5 phi - 3
etc
The coefficients are alternating, offset terms of the Fibonacci sequence. One to store in the memory bank
EDIT: You could apply this idea to any equation of the type x^5 = Ax + B. I.e put
x^2 = ax + b
x^3 = ax^2 + bx
= (a^2+b) x + ab
...
x^5 = (a^4 + 3 a^2 b + b^2) x + ab(a^2 + 2 b)
In this case
A = a^4 + 3 a^2 b + b^2 = 5 [1]
B = ab (a^2 + 2 b) = -3 [2]
Solve [1] as a quadratic in a^2
a^2 = (-3 +/- sqrt(5b^2+20))/2
To get rational a^2 guess
b^2=1
a^2 must be the positive root i.e. a^2=1
Solve [1] as a quadratic in b
b = (-3*a^2 +/- sqrt(5 a^4+20))/2
b = (-3 +/- 5)/2
We must have b^2=1 so
b = 1
Substitute in [2]
3a = -3
a = -1
The quadratic factor is x^2 - ax - b = x^2 + x - 1
A white van pulled up while I was walking and asked me to solve a quintic equation. Luckily BPRP taught me never to talk to strangers who ask quintic equations so I ran away.
Nice work. When calculating discriminant to solution of the Langrange resolver, you might want to take a short cut and factor before taking the square root of the discriminant, i.e b = 65 = 13*5, therefore b^2 = 13^2*5^2. 4*a*c = -4*1*125 = -4*5^2. You can factor out 5^2 so the discrimiannt is now 5^2*(13^2 + 5*4) = 5^2*(169 + 20) = 5^2*(189). 189 is divisible by 3 (sum of digits), and so is 189 / 3 = 69, 69 / 3 = 21 = 7*3. So you have (5^2)*(3^3)*7. You can now take the square root and you have 5*3*sqrt(7*3)
189/3 is not 69
Oops, yes that should be 189 /3 = 63 = 7 * 3^2, allows you to factor out the 3 after the subtraction
4:28 i never accept quintic from stranger on the streets
We got this lil thing in Italy called "Ruffini's Rule", we're taught this during algebra classes but everyone tends to forget it (cause it's tedious). I'm pretty sure that would've allowed us to find some real solutions from the beginning. What do you think about this? Cheers, Thomas! 💗
There is also Horner's Method
Is it that the integer divisors of the constabt term ara possible sols?
@@Jj-gi1sg ara ara
Actually that's kinda the same thing with synthetic division, search about it (in English obviously). I'm Italian too anyway :)
14:25 jaaay broh we got it 😂😂😂 i have only the 10 th class finished so that the only thing i undeestand bec. that only what i learn at 10th.class i just know x^2+2x+4=0😂
Hello!
I am thinking about the "If a stranger gives you this quintic equsion....", and it gives giggles because it sounds like "Don't try this at home" xD. Regardless of me not being a student at any university, i still find your video entertaining
No need to solve for exact decimal to decimal complex solution
But, you can numerically calculate the value of 'r' and then form the last quadratic which could be solved further.
Will land up on an approximate complex solution.
Or try to find Delta
A = 1
B =-5
C= 3
Delta = B²-4AC
If delta more then 0 / positive it has 2 answers
If its 0 it has 1 answer
If its negative it doesnt have answers
Me deciding to follow up with my math teacher to actually understand a lesson for once;
Him starting the lesson 0:45
I look down at my book for one second and then look back at him;
Him literally the next second: 30:10
I like watching you cause I wanna go back to school eventually, don't want to forget everything in the mean time
Please, please, please! Teach us everything about the Lagrange resolvent. Are there resolvents for polynomial equations of other powers? If there are, please give examples. How did Lagrange come up with this formula?
Wikipedia gives a general formula for Lagrange resolvents (sum from i=0 to n-1 X sub i omega ^i), where omega is a primitive nth root of unity. How does that apply here, or in polynomial equations of other powers?
I haven't seen anything like that on TH-cam.
Thanks a lot!
Your explanation, as always, was very clear..
You can get the complex roots easier if you just multiply the first cube root in the real root by one complex cube root of unity (-1/2 + sqrt(3)i/2) and the second one by the other complex cube root of unity ( (-1/2 - sqrt(3)i/2)).
Then the other complex root is the conjugate of the first - or, you can reverse the 2 cube roots of unity and that will also give the second complex solution.
I just bought a hoodie from you it’s amazing I saw it and didn’t hesitate!
Thank you!! Which one did u get?
I bought the converge hoodie!
Eu gosto disso! Boa explicação detalhada!
Another form by brute force, suppose that q(x) is a quadratic factor of f(x)=x^5-5*x+3, then f(0)=3, f(1)=-1. We can assume that q is monic, q(x)=x^2+b*x+c. Therefore q(0) divides 3 and q(1) divides -1. We take many finitely cases for q(x).
This method proves that factoring polynomials (with integers coeficients) is equivalent to factor integer numbers. But this method is pure brute force.
Wiki states that if you multiply the cubic root parts you got by (-1+sqrt(3)i)/2, then you will get the 1st complex solution and 2nd complex solution if you multiply by (-1-sqrt(3)i)/2 instead.
In other words, if formula for the real root is:
x1 = 1/3 + 1/3 * cbrt(A) + 1/3 * cbrt(B)
Then
x2 = 1/3 + 1/3 * cbrt(A) * (-1+ sqrt(3)i)/2 + 1/3 * cbrt(B) * (-1+ sqrt(3)i)/2
x3 = 1/3 + 1/3 * cbrt(A) * (-1 - sqrt(3)i)/2 + 1/3 * cbrt(B) * (-1 - sqrt(3)i)/2
That makes the same sense as ± part of quadratic equation formula, please notice that: 1^2 = (-1)^2 = 1, 1^3 = [(-1+ sqrt(3)i)/2]^3 = [(-1- sqrt(3)i)/2]^3 = 1
This is not quite correct. You need to multiply one cube root by one non-real cube root of unity and the the other cube root by the other non-real cube root of unity, as explained in the comment by Risu0chan.
its also nice you can just use iteration / newton rapson to get a non exact answer in a few seconds :)
(x+phi)(x-1/phi)
(x^2+(phi-1/phi)x - 1)=0
phi^2-phi-1=0
phi - 1 - 1/phi = 0
phi - 1/phi = 1
(x^2 + x - 1) is a factor
كم هو رائع علم الجبر 😍!
11:44 That's what she said
I just watch your videos for fun, Steve! They are great! 👍
Thank you!
I watch math videos for fun, but this completely fried my brain
Very interesting. Greetings from Colombia.
Well you see, suppose for integers a, b, p, and q, if x^2-ax-b divides x^5-px-q, then a divides q^2. Because of the small margins, I won't leave a proof.
Sheesh bro lemme sleep. Ngl, I really find your vids so satisfying to watch that it made me appreciate more the innate beauty of Math. God bless you man!!! Much love from ph!
There is an exam at my grad school which we have to take to start the research stage of our Ph.D. There is almost always a polynomial on there and they ask us if it can be factored using whole numbers. Usually there is a trick that makes it quick, but I was doing one of the older tests and one test writer gave one that had to be solved like this. It was very mean.
Hello I'm fan from Cambodia.
Hello i am form india
Wow nice to meet you.
👍
By establishing max and min, one can see were are the 3 solutions: -1.x , 0.y , 1.z
hope you reach 1m subs soon with amazing vids like this
11:45
“Wow. D is big.”
-BPRP 2022
Well said 😂
"I dont know if you like extreme sports, but i like extreme math questions."
Story of my life
I'm struggling a bit with life right now, but watching your videos really helps. Thank you!
11:44 the best part
Saludos desde Perú BlackPenRedPen 💪📚
When you have x = 1/3(-a + cube root of something + cube root of the conjugate of something). Can't we get the complex solutions if we multiply the cube roots by the appropriate cube roots of unity?
Yes, you can, as explained by rslitman. Basically you multiply the first cube root by ω (where ω is a non-real cube root of 1) and the other cube root by ω² (the complex conjugate of ω). Then you do the same again, but multiply the cube roots by ω² and ω (i.e. in the opposite order).
@30:55 .....emotional..... (say it! say it! DAMAGE) aw... you missed a great chance for it😔
😆
If you want to try if you can devide through (X^2 + X -1) you an just try long division and find directly that (X^3 -X^2 + 2X -3) is the other factor.
Very nice and hardworking man.
Wow man... I loved your amazing video! I had never thought to solve like you did😮👌🏽👏🏽👏🏽
Thanks!
Bro you are just like quantum arrow... Fav maths guy😋😋😋