I used to watch you a bunch in high school (2020-2022) when I was in pre calc (junior) and ap calc (senior). Although I could barely keep up with a lot of your videos because I hadn’t learned as much as I needed to understand it all, it was still super interesting and got me wanting to understand. I passed all of my classes with B-A’s, but quickly withdrew from math as a whole when I went into college mainly due to the difficultly with more new material and college taking more of my brain. I am studying psych and will get my bachelors in the winter of ‘24. However I have forgot most of calc as I didn’t need to take any more math classes either. However, I got Covid (again) and am near bed ridden till it goes away. And it sucks not being able to do anything productive. However, watching your videos makes me feel like I am doing something because I am trying to learn/re learn a lot of complex math. I mainly just wanted to say thankyou for making these videos. Even if I am in and out and not able (at the moment) to give your videos 100% effort for learning with trying the problems, they are still useful and interesting. I also appreciate you coming back to old videos with updates and more stuff for us to sink our teeth in. As well having these videos helped me a bit in my calc classes. Thankyou for making me feel a little bit smarter with each video and showing me the solutions and process to these seemingly impossible questions. Take care and continue being an absolute genius and wonderful person! ❤️ Happy holidays and a new years!
same but I loved maths when I was about 3-9 years old until my mom wanted me to "be better". Now I am form 1 and I rediscovered my interest towards maths because I standed out for myself.
Hey Ernest, I'm Ernest too, and I have always loved maths, but I've started to get seriously back into it over the last 10 years. I can't wait to retire (
@Chicken Leg No, it doesn't. He ain't a Japanese. You can find his biography and other credentials on the net. Check this one out: api-everipedia-org.cdn.ampproject.org/wp/s/api.everipedia.org/v2/wiki/amp/lang_en/blackpenredpen?usqp=mq331AQSKAFQApgBy46dyers4qcVsAEg
For those who want to know how to solve x^3 - x + 1 = 0, here is the method. First, note that (u + v)^3 = u^3 + 3vu^2 + 3uv^2 + v^3 = u^3 + v^3 + 3vu^2 + 3uv^2 = (u^3 + v^3) + 3uv(u + v). Therefore, let x = u + v. Hence x^3 = 3uvx + u^3 + v^3 = x - 1 implies 3uv = 1 and u^3 + v^3 = -1. 3uv = 1 implies v = 1/(3u), which implies u^3 + 1/(27u^3) = -1, which implies 27u^6 + 1 = -27u^3, which implies 27u^6 + 27u^3 + 1 = 0. Therefore, u^3 = [-27 + sqrt(621)]/54, or u^3 = [-27 - sqrt(621)]/54. From this, it can be shown rather trivially that v^3 is the conjugate of u^3. Therefore, without a loss of generality, u^3 = [-27 + sqrt(621)]/54 and v^3 = [-27 - sqrt(621)]/54, because addition is commutative. Therefore, u = -cbrt([sqrt(621) - 27]/54), and v = -cbrt([sqrt(621) + 27]/54), and as such, x = -[cbrt([sqrt(621) - 27]/54) + cbrt([sqrt(621) + 27]/54)] = -[cbrt([3·sqrt(69) - 27]/54) + cbrt([3·sqrt(69) + 27]/54)] = -[cbrt([sqrt(69) + 9]/18) + cbrt([sqrt(69) + 9]/18)]. This is the exact real answer, and you can show using some simple algebra that the result shown in the video actually simplifies to this. You don't need to actually know the cubic formula before hand. You only ever need to solve a system of equations to solve a cubic equation. In general, if you have a polynomial equation of the form y^3 = py + q, then by using the same method above, you can show that y = u + v, where 3uv = p, and u^3 + v^3 = q. Solving this system is always very easy as it reduces to solving a quadratic equation and taking the cube root of the solutions to that quadratic equation. Also, you can always convert a general polynomial equation x^3 + ax^2 + bx + c = 0 into the form above by letting x = y + a/3 and simplifying. Cubic equations are easier than they look. Quartic equations, on the other hand, are a genuine pain in the ass. Although they are always solvable, the method is significantly more tedious and annoying. EDIT: I made a dumb arithmetical mistake with my calculations, so I fixed it. EDIT 2: Split stuff into paragraphs, and fixed another dumb arithmetic mistake. See, this is why you don't do math after having pulled an all-nighter. *sigh*
Considering we ended up with an irrational value, a numeric solution would be as good. Any quintic with real coefficients has at least one real solution, so Newton-Raphson will give you a numeric solution. A quick inspection shows the expression changes sign between -1 and -2, so -1.5 would be a good starting point. I got -1.324717959 in four iterations. A numeric solution is not as "pure" as a formulaic solution, but the solution's the same and N-R is a _well-known_ _formula_ for solving these sort of problems.
Wow those expressions you added in red really makes a huge difference in factorization and it works so nicely. I like 6:35 the most. Def worth trying them out with hands for sure!
A quicker way to factor this problem is to use the zero property method at the beginning by adding and subtracting x^3 and then grouping (x^5+x^4+x^3) and (1-x^3).
Excelente lesson. That professor is perfect: he is smart and talented to teach, he sings, he plays multi color golf. What more? Very good!!! Greetings from Brasil. Thank you for your channel.
OMG! You actually did it twice! Soon, the whole world will know about that formula and the people behind them and the reason to teach the formula and the history instead of hiding it. Thank you so much!
@@blackpenredpen Wait, remember the video when you proved that sin(10°) is irrational? ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cN0vDlWjFiY.html
I approximated the solution right away using Newton's method to be -1.32472 (in only 4 iterations), but of course I liked your way of solving it using factoring and then cubic formula. Interesting video!
My pre-view attack: x⁵ + x⁴ + 1 would be -cyclotomic- monic, with all coefficients = 1, if three more terms (x³ + x² + x) were added. So let's do this: *[Note: Correction - "cyclotomic" has a more specialized meaning, which coincides with the one I intended, only when the degree is p-1 for some prime, p.]* x⁵ + x⁴ + 1 = x⁵ + x⁴ + x³ + x² + x + 1 - x(x² + x + 1) . . . multiply by (x-1), which will introduce the extraneous root, x=1; we will remove this later. (x-1)(x⁵ + x⁴ + 1) = x⁶-1 - x(x³ - 1) = (x³-1)(x³ - x + 1) = (x-1)(x² + x + 1)(x³ - x + 1) So: x⁵ + x⁴ + 1 = (x² + x + 1)(x³ - x + 1) The zeros of the first (quadratic) factor are the 2 complex cube-roots of 1. So the desired real zero of the quintic is the real zero of the cubic factor. And the general cubic equation has a solution formula - see Angel Mendez-Rivera's comment; also check out the video by Mathologer where he shows how to develop the solution, while complaining that the cubic formula should be taught in algebra classes, alongside the quadratic solution formula. Of course, if you don't care to find the closed-form answer, you can always use Newton's Method, or some other iterative method to compute it to any desired precision. Fred
Great ideia! Would just like to point out that the polynomial Σx^i (i going up to n) is only cyclotomic iff n+1 is prime therefore that polynomial isn’t cyclotomic.
@@bernardopicao267 Yes. I used the wrong term for the right polynomial; will correct that. BTW, what is the right name for the general polynomial, of degree (integer) n ≥ 0, that is simply p(x;n) = (xⁿ⁺¹ - 1)/(x - 1) = ∑₀ⁿ xᵏ ? @Davis John: I challenge you to point out an error, other than the already-identified error of terminology brought up by Bernardo. In particular, my conclusion that the real zero of x⁵ + x⁴ + 1 is the (unique) real zero of x³ - x + 1. I believe it is you who are in error. Fred
It seems solution lie between -1 and -2 since x=-1 gives the value of equation is +1 and when x = -2 value is -15 then I can use bisection method -1.5 etc till I get close to zero
I used a trick my teacher explained: If the exponents of a polynomial are each congruent to a different number mod 3, then it is divisible by x^2+x+1. Example: x^5+x+x^0 or x^8+x^7+x^3. It works for other number too. x^16 + x^11+x^6+x is divisible by x^3+x^2+x+1. This way I could factor the equation in a cubic and a quadratic equation. Unfortunately couldn't solve the cubic part, but at least I could find two solutions
@@ottoaberg5942 mod is short for modulus, the integer remainder of integer division. So 5 mod 3 = 2, 4 mod 3 = 1, and 0 mod 3 = 0. Since the answers differ the OP claims it is divisible by x^2+x+1.
@@ottoaberg5942 That is the assertion made by OP, and a quick check with Mathematica shows it seems to work. But I've not seen the proof for this rule myself...
Well, I actually applied Newton's formula on the integral of the expression just to figure out where the minimum is located (the actual problem). After analyzing the derivative of the problem, (second derivative in Newton's method), I picked -1 as initial value. After 7 iterations of next = xi - f'(xi)/f''(xi) arrived to x approx. -1.325.
x⁵ + x⁴ + 1 = 0 ⇒ x³(x² + x + 1) - x³ + 1 = 0 ⇒ x³(x² + x + 1) - (x³ - 1) = 0 ⇒ x³(x² + x + 1) - (x -1)(x² + x + 1) = 0 (because a³ - b³ = (a - b)(a² + ab + b²)); then we have (x² + x + 1)(x³ - x + 1) = 0 ⇒ x³ - x + 1 = 0, because x² + x + 1 = 0 does not have any real solution ( Δ = - 3 < 0) to be continued ... P.S: we need to note that if f(x) = x⁵ + x⁴ + 1 , then f(x) = 0 has only one real solution, and this solution is negative, between (-2) and (-1)); ... ... and this in using lim f(x) when x → - ∞; lim f(x) when x → + ∞; and f'(x) = 5x⁴ + 4x³ = x³(5x + 4) = x(5x + 4).x² x³ - x + 1 = 0 is easy to solve with Cardan's formula.
Me, being an engineer: Ok... google, what was the formula for Newton's method again? ...calculator goes brrr... Hmmm, something around -1.325. Lets put -1.5, it will do its job.
Hahaha i did the exact same thing took a couple seconds. Ahh it's around -1 , close enough. If we'd worked together we would have averaged it and nailed it.
I solved this by recognising almost immediately (by inspection since if we set x^3 = 1, we get x^2 + x + 1 =0 which also returns the conjugate complex roots of one) that that the two complex cube roots of one solved this. Which means linear factors of (x-omega) and (x-omega^2). Then by serial synthetic division and using the identities related to the complex cube roots of one, I got x^3 - x +1 =0. I solved this using the cubic solution I derived from first principles (recognising the reduced cubic form and putting x = z + 1/(3z) followed by z^3 = m and solving the quadratic). It was after this last step (when I'd already got the correct solution) that I watched your video hoping for more insight or a clever trick in how you solved that final cubic, but alas - you just regurgitated the cubic formula.
I just tested a bunch of numbers with two columns. One for x and one for f(x), and I got pretty close. I got to 1.32 which would be 0.02, which is quite close. I'm sure you'd get the answer eventually that way if you had enough patience.
Another interesting way to solve this equation is to use the symmetry of its solutions. Let z a solution of the equation Then to have z⁵+z⁴+1 z⁵ and z⁴ can be complex conjugate each other. Let ⃗z the conjugate of z. Then ⃗z⁴ will be conjugate of z⁴ then equal to z⁵ Then we have this awesome equation z⁵=⃗z⁴. Using exponential notation for complex numbers r⁵exp(5θ)=r⁴exp(-4θ) for r≠0 then r=1 to satisfy the equation and 5θ=-4θ+2πk we found that θ=2π/3 is a solution. then z=exp(±2π/3) are two solutions of the equation. Then since z are two cubic roots of the unit they satisfy the equation. z²+z+1=0 then x²+x+1 divide x⁵+x⁴+1.
I have a simpler way, you can derive the equation and find the extrim point to be x=0,x=-0.8 , by ploting it you see that at x=-0.8 you get y=0 ( a solution) at x=0 you get y=1, it is also noticible that the derivative of the function is 0 at this points and positive elsewere, that meens that this is a monotoic function and the only solution is at x=-0.8
I don't think you can go much further with that except compluter. In this process ultimately you have to solve this to find the intersection analytically
x⁵ + x⁴ + 1 = 0 ⇒ x³(x² + x + 1) - x³ + 1 = 0 ⇒ x³(x² + x + 1) - (x³ - 1) = 0 ⇒ x³(x² + x + 1) - (x -1)(x² + x + 1) = 0 (because a³ - b³ = (a - b)(a² + ab + b²)); then we have (x² + x + 1)(x³ - x + 1) = 0 ⇒ x³ - x + 1 = 0, because x² + x + 1 = 0 does not have any real solution ( Δ = - 3 < 0) to be continued ... P.S: we need to note that if f(x) = x⁵ + x⁴ + 1 , then f(x) = 0 has only one real solution, and this solution is negative, between (-2) and (-1)); ... ... and this in using lim f(x) when x → - ∞; lim f(x) when x → + ∞; and f'(x) = 5x⁴ + 4x³ = x³(5x + 4) = x(5x + 4).x² x³ - x + 1 = 0 is easy to solve with Cardan's formula.
In the beginning, I tried to write it as (x^4+1)(x+1), then I tried something with (x^3+1) and then think to something which make some terms simplifies and in the end find the answer. Nice, but Easy one.
Numberphile has a stimulating video on the plastic ratio, and the Wikipedia page on the plastic number (the same thing under a different name) is pretty good. Also, typing "plastic number" into Wolfram Alpha and then clicking on "More information" provides the start of an interesting mathematical tour.
It's a shame that I don't have anything to comment about before coming here, but I can't miss this golden opportunity to be this early lol.. I just want to say that I really like your videos and even though I still haven't officially learned calculus or trigonometry in school I watch every single one of your videos! :D
@@einsteingonzalez4336 That's what I like about these videos. Thanks to it, I think it will be easy learning complex numbers, since I already know the basics of what it is and how to use it etc, plus I see it everywhere and it becomes normal at this point! Same with trigonometry. I started to understand the small stuff that no one pays attention to, like the angles of pi related to the circle, euler's formula, comparing numbers, tricks to solving equations, and I can state many more.
Since we have to use a calculator anyway.. Why not just use Newton raphson's method? It gives the answer in only 4 iterations provided initial root is taken as x=-1.
I would have cranked the equation into an online polynomial solver, like Wolfram, it would give in two seconds the five solutions as stars in a nice complex plane. By the way, do you know that the family of polynomials equations: x-1=0, x^2+x-1=0, x^3+x^2+x-1=0, .... are related with the Fibonacci numbers, and of course with the Golden Ratio??
@@jimgraham6722 :: The number of iterations for convergence for Newton Rapson depends on the seed value you give in as Xo, and the precision you need. And also not to fall in a zone where the function has a valley, where the iterations will go up and down, oscillating.
@@powerdriller4124 Yes indeed, NM has some limitations, but to me it is fun to watch it resolve some otherwise difficult problems using a cheap programmable calculator.
What a nice guy! I converted the polynomial to a "depressed quintic" (sigh) y^5+y+1=0 with y=1/x. I guessed there would be roots y and y* with |y|=1. y=-1/2 +/- i sqrt(3)/2 turns out to work. Multiplying the terms for these gives the quadratic factor leaving the depressed cubic (double sigh). After that I Googled
in high school we were taught to factor any polynomial by dividing it by (x-b) where b is a factor of the 0-th term; that is 1 in your example. so we divided once by (x-1) and again by (x+1).
tha's what I did too. It was much faster. Unfortunately I got stuck because I did not know the cubic formula. I was hoping to see a solution that didn't require it. oh well.
Note that x^3=1 has three solutions: 1, w, w^2, we have w^5+w^4+1=w^2+w+1=(w^3-1)/(w-1)=0 and w^10+w^8+1=w+w^2+1=(w^3-1)/(w-1)=0. Therefore w and w^2 are the solutions of the function x^5+x^4+1=0. Thus the polynomial x^5+x^4+1 has factor x^2+x+1. One could simply conclude that x^5+x^4+1=(x^2+x+1)(x^3-x+1)
The solution is the negative of the plastic constant p, which solves the equation x^3 - x - 1 = 0 (equivalent to (x-1)(x)(x+1)=1 ), also solves the equation x^5 - x^4 - 1 = 0 (this can be obtained from bprp's equation by substituting u = -x) and is the limiting ratio of the Padovan and Perrin sequences which are analogous to the Fibonacci sequence and follow the rule F(n) = F(n-2) + F(n-3) = F(n-1) + F(n-5).
I could see that "w and w^2" are the solutions of the original equation, coz w^2 + w + 1 = 0 and w^3 = 1. So I got x^2 + x + 1 as a factor directly. Works in this particular case only though 😅
- 1,324718 My procédure was different, based on complex numbers. I differentiated once, twice, etc. and observed there was only one real root. So there are complex roots. Then I tried the assumption that two of the complex ones were on the unit circle, i.e. x=exp(+/- i thêta). In order for exp(i 5theta) +exp(i 4theta) to equ)al -1, aka exp(i π), we need sin(4theta) = - sin(5theta) and cos(4theta) + cos(5theta) = -1 Implying cos(4theta) = cos(5theta) = - 1/2 Which (a graphic is useful here) gives theta = +/- (2/3)π = +/- 120° or x = (1/2) +/- i (V3/2) or, combined, x²+x+1=0. The assumption proved correct. Dividing the original equiation by this factor leads to the cubic solving following your demonstration. Calling the two cube roots a and b, we have x0 = a+b and x1,x2 = (a+b)/2 +/- i |a-b|/2
You are very good and lucky to factorised the PENTIC=cubic*quadratic. Using remainder theorem f=pq+r .Within 5 seconds testing 0 r+ , -1 r+ , -2 r-. immediately the only Real solution between x=-1 and x=-2 .Since x=a+bi and x=c+di . Thus the best way is to use NewtonRaphson Method . follow by using scientisfic-Calculator or ComputerExcel .You can get answer-12digits or more Accuracy/Precision.
Although there is not an analytical expression for the solutions of quintic equations, there are conditions on the existence of closed expressions for certain coefficients. Those I think were developed by Abel, but I'm not quite sure.
I was able to reduce the problem to a cubic right away, just by looking for roots of unity. I spent about 20 minutes stumped looking for a trick to solve the cubic before thinking "Did this guy actually use the cubic formula?" Yep.
One solution is cube root of unity that is -1+√3i/2,-1-√3i/2 try putting them by x^3=-1 According to complex comcepts w,w^2 are roots of unity and w^3=1 and 1+w+w^2=0 Therefore the equation x^5+x^4+1=0 becomes x^2.x°3+x.x^3+1=0 now x^3=1 for x=w therefore equation becomes 1+w+w^2 HP
Your work is great but still solving the equation numerically is super hard and so I think that solving the equation graphically would be a lot easier You can just plot your equation and just note the intersection with the x axis
Your method is not precise enough. It depends upon using the human eye to produce a value that cannot be determined by eyeballing a graph. Determining the roots by purely analytical means is far more preferable and more precise.
Here is another trick to find the factorization of the polynomial. Let a be a primitive third root of unity. Hence a^3=1 and a^2+a+1=0. Substituting x=a in the given equation we get a^5+a^4+1=a^2+a+1=0. Hence both the primitive third roots must be roots of the given polynomial also. Hence x^2+x+1 must divide the given polynomial. The other factor is obtained easily by long division.
Equation with only 1, -1, and 0 : try complex roots of 1, say a , and 1/a must be solution so we get a new polynomial equation: 1+a+a^5=0. the difference yield the equation : a^4-a=0, a(a-1)(a^2+a+1)=0 : and try factoring the original equation by a^2+a+1.
There is something interesting in his way of looking at the numbers, mentioning several times the word "PEOPLE"! So it is not so abstract anymore from this point, the "people" are "working hard" on the right side of the table to give clear results presented on the left. :) The way of looking at math can be a reason why are the Chinese so good at this subject. :)
Oh. Actually, I got it wrong. He meant not general eqn but quintic of type "x**5 +p*x + q = 0". I only have thoughts that solutions of "z**5 +a* z**4 + b = 0" and eqn above are obviously related.
I just factored x^4(x+1)=-1, deduced that x must be negative (as x^4 is always positive) and larger in magnitude than -1, then set up a recurrence relation and let EXCEL solve it. Go the numerical answer right!
@@blackpenredpen No idea what Warrick used, but I used x(n+1) = (1+x(n)^(-4)+x(n))/2. Converges very rapidly from any initial x(0) 0. Of course, the answer should be negated [-x(n)]; I worked with positive numbers for ease of calc.
Note that both terms inside the cube roots are negative so if you are trying to calculate this numerically it might be useful to express them as the negative of the cube root of their absolute value otherwise you might end up picking one of the complex cube roots _i.e._ -((1/2 - sqrt(23/108))^(1/3)) - (1/2 + sqrt(23/108))^(1/3) ~ 1.32472
This is really genius. All is derived in a logical and understandable manner. I tried to figure out the polynomial multipöication but it got to complicated. Still I had the idea that maybe x^2 + 1 or x^2 + x + 1 could be the quadratic factor, since both of these end with 1 and have no real solution. But the grouping scheme was great and much easier!
Given that there is only one real non-repeated solution, this means it has to come from the cubic factor and not from the quadratic. This means that the cubic factor is a case where Cardano's formula works directly, because the cubic discriminant will be positive for the single non-repeated root. The quadratic factor could give us two real roots, or a real/repeated root, but not just one real/distinct root. If there were only one real root and it were repeated, it could come from the quadratic factor. This isn't possible with an odd degree polynomial, since a repeated root means there's a turning point at the root. It has to turn back around and cross the x-axis again. So any odd-degree polynomial with a real repeated root, will have another distinct root as well.
We can also use the concept of primitive root of unity. Suppose that w^3 = 1; w is the primitive cube root of unity =/ 1. So w^5 + w^4 + 1 = w^3 + w^2 + 1 => (x^2 + x + 1) is a factor.
