de Moivre quintic formula

We will explore the de Moivre quintic formula. This is just a formula for a special case of the quintic equation in the form of x^5+5ax^3+5a^2x+b=0. By Galois theory, there's no formula that solves all the quintic equations. But don't be upset, because here's a nice formula for a special quintic equation.

0:00 a quintic formula? finally!
9:54 how to get that quintic form
28:46 other insane quintic cases

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[kummer]

I am a mathematician. I can safely say that my respect to you increases exponentially.

We need a detailed derivation for the Bashkara formula or quadratic, the cubic and the quartic polynomial equation with real or complex coefficients. Many people and lots of books don't give the steps for the derivation of these formulas. I found cubic and quartic in ordinary differential equations. THEY DO HAPPEN THERE and of course with those weird Lagrangians and Hamiltonians.

I've seen your channel lately seeing an increment in difficulty tackling hard problems. Please continue doing so. You deserve getting the Patreon and and more subscribers. The algorithm of YOUTUBE must be benevolent with you.

Teachers makes the difference. You are one of them.

[Mystery_Biscuits]

11:03 I think the minus being there has a more important role in general. When looking for solutions, we need to make sure that the range of the substitution still completely covers the desired range of answers. For example, the substitution x= t+1/t has the range (-infty, -2] union [2, infty). This means, if the solution to the equation lies inside (-2, 2), there does not exist a t that can find it. This is why the minus version works better. x = t -1/t has the whole real line as it's range so the problem of potentially missing a solution is removed.

[rohanganapathy]

Well nowadays since there are so many ways to find solutions to these big equations without actually solving them but with computers, no one has ever tried anything such a way. This is actually really fun! This is why I love mathematics, we do the impossible!!

[johnchessant]

That's really cool! The x = t + 1/t trick works in reverse when you're trying to find the 5th roots of unity.
x^4 + x^3 + x^2 + x + 1 = 0 -> x^2 + x + 1 + 1/x + 1/x^2 = 0
and now we set t = x + 1/x to get t^2 + t - 1 = 0. Solve for t using quadratic formula, then use the result to solve for x using quadratic formula again.

[obinator]

Now do the quartic equation, but with the formula for general solutions.

[LuigiElettrico]

At 9:51 he says: "Keep all this in mind, because this was a very nice demonstration and now of course I will have to erase the board
21:07 - "Extra B (?) Tada (?), that's very nice eh? And now I just have to erase the board"
More hidden messages please :D And of course best math material as always!

[chrisleon]

I shall make more contribution in the future if I keep my word

[DefenderTerrarian]

I just started Calc 1, and your videos have let me get a leg up on the competition, so to speak. Starting college two years early (I graduated out of tenth grade) I felt that I would be, how to say..., a little dumber than the other students in Calc 1, but thanks to you I am more familiar with Calculus. Keep up the good work.

[ntth]

8:48 the imaginary roots can be find by mutiply the t value with 5 fifth-root of 1: e^0/5 (which is the value used in the video); e^i(2pi/5); e^i(-2pi/5); e^i(4pi/5); e^i(-4pi/5)

[rihankota]

literally Awesome ur one of the best calculus teacher u have seen ur explanation is awesome👍🏻🥰...

[Peter_]

No matter how much I practise math, blackpenredpen always manages to bring up new challenges.

[afuyeas]

If the equation is of the form ax^5 + bx^3 + cx + d = 0 it will be solvable using this formula if and only if b^2-5*a*c = 0; the quintics that satisfy this condition are known as "De Moivre's quintics" and notice how the form of the quintic in the video satisfy this relationship.

The reason why the formula ends up looking very much like the cubic formula is that it can be derived using the same idea. Let x = u + v, if the quintic satisfies the initial condition stated above then you can impose a condition on the product u*v so that you're only left with u^5+v^5 = - d. Take the fifth power of the product and you get a quadratic that is then easily solved.

It has to be noted there are clear limitations with the nature of the roots of De Moivre's quintics. If what is inside the square root (let's call it the discriminant) is a real number then you only get 1 real root and 2 pairs of complex conjugates, if the discrimant is negative then you always add two complex conjugates to get your five roots so you end up with 5 real roots. Finally if the discrimant is 0 then you can show that the equation has 1 simple root and 2 distinct double roots, all real if the coefficients are real. In conclusion such equations will never have 3 real roots and a pair of conjugate real roots because the formula given is incapable of expressing such a case, nor will you ever get 1 double root, 1 simple root and 1 pair of complex conjugate roots with this formula even though there are quintics that have such roots.

I really like De Moivre's quintics because even though there are very specific cases of quintics they still give you some hints about the general quintic not being solvable in radicals because the limitations of radical expressions are very clear when solving them.

[shashwatsharma]

Bprp : quintic equation
Yt captions : green tea equation

[pepebriguglio]

I see why you chose the third formula. First and second options are very complicated. The first one is fascinating though! All powers of x are allowed to have non-zero coefficients, and still there is only restrictions on two of those coefficients: those of x and x². That is indeed impressive 🙏

[quanticansh]

instead of using general solutions, can I make an infinite series inverse function using the Lagrange's inversion theoram, to get the accurate roots of any degree polynomial equation I guess?

[Migeters]

My guy I love the videos best calculus channel on yt

[chaiotic]

This is so good! There's something about polynomial equations that's just so neat, like cool puzzles

[ErezMarom]

9:47 impressing people is the best reason for inventing a formula.

[trollme.trollmehard.]

I honestly watch these discoveries with a little apprehension. I would love to drop that and just focus on helping where my own skills / research lies, so would appreciate knowing how people manage this better.

[thereaper]

I have a question :

Can we solve this equation ?

a^x + bx + c = 0

Please share it