Try this quintic equation next: kzbin.info/www/bejne/faCqpImCo8triM0
@taektok31522 жыл бұрын
Damn..!!!!!!, i cant hold my mother finger to not subscribe this channel.. So i follow you sir.. Im so sorry.. 🥺🥺🥺🥺🥺
@debashishkumarbehera5012 жыл бұрын
Can u find the solution of the equation given by sinx+e^x=0
@tetraktys27862 жыл бұрын
Please, can you give us dificult equations, you makes me feel good at math So can you give us the solution of this equation? w(ln(10/x))=-w(-ln(10*(10^(1/x)))) Just kidding, only I want the equation, please
@leonardobarrera28162 жыл бұрын
@@tetraktys2786 Tetraktys, I was solving, and I didn't get the answer, thanks
@mumujibirb2 жыл бұрын
I also found one for a calculator, but it uses newton's method
@kummer452 жыл бұрын
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 KZbin must be benevolent with you. Teachers makes the difference. You are one of them.
@hoogreen2 жыл бұрын
^^
@oreos2653 Жыл бұрын
bro wtf i thought this stopped at quadratics bro where the 3rd and 4th and 5th exponent come from
@chuashanganluciennhps9992 Жыл бұрын
@@oreos2653 there is more to math than you think :)
@Mystery_Biscuits2 жыл бұрын
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.
@willie333b2 жыл бұрын
Hmm imaginary numbers?
@Mystery_Biscuits2 жыл бұрын
@@willie333b well, yeah of course this won’t find all the solutions but this video was just interested in finding the real solution to certain quintics…
@78anurag2 жыл бұрын
Niels Henrik Abel crying rn
@rohanmehta85365 ай бұрын
He proved the unsolvability of the general quintic,not for all of them.
@rohanganapathy82 жыл бұрын
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!!
@johnchessant30122 жыл бұрын
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.
@obinator90652 жыл бұрын
Now do the quartic equation, but with the formula for general solutions.
@fantiscious2 жыл бұрын
I don't think mathologer provides the formula, but he does provide the method
@kummer452 жыл бұрын
@@fantiscious The topic BEGS for the long derivation. However the quadratic and cubic should be done first and the roots of unity for z^3=1 should be discussed first. There is some complex number algebra needed in such adventure. Binomial theorem plays a big role too completing the square, the cube and the quartic. Many books evade this because the algebra goes heavy on the details, moreover when Galois Theory enters when we confront the quantic The whole thing becomes a tour the force in Abstract Algebra. Besides the theory is extremely beautiful.
@NintendoGamer789 Жыл бұрын
Wiki page covers it pretty well, I was just too lazy to look for the details when I first searched this up in like 2021 lol
@shibam4182 Жыл бұрын
ax⁴+bx³+cx²+dx+e=0(a≠0) Substitute x=y-b/4a It will convert to y⁴+py²+qy+r=0(after dividing by a) Then y=(±√p+2λ±√-3p-2λ±2q/√p+2λ)/2 (where p+2λ≠0) And for the value of λ we have to solve a cubic equation in λ term 8λ³+20pλ²+(16p²-8r)λ+(4p³-4pr-q²)=0 Also if q is positive then {+()+√()-()}/2 {+()-√()-()}/2 {-()+√()+()}/2 {-()-√()+()}/2 & If q is negative then {+()+√()+()}/2 {+()-√()+()}/2 {-()+√()-()}/2 {-()-√()-()}/2 In this way the equation gives 4 roots
@DefenderTerrarian2 жыл бұрын
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.
@afuyeas99142 жыл бұрын
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.
@acelm8437Ай бұрын
It's nice that b^2-5ac resembles the discriminant for quadratic equations!
@ntth742 жыл бұрын
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)
@letstalksciencewithshashwa95272 жыл бұрын
Bprp : quintic equation Yt captions : green tea equation
@rihankota20212 жыл бұрын
literally Awesome ur one of the best calculus teacher u have seen ur explanation is awesome👍🏻🥰...
@Peter_19862 жыл бұрын
No matter how much I practise math, blackpenredpen always manages to bring up new challenges.
@LuigiElettrico Жыл бұрын
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 aaaaand..." 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!
@trollme.trollmehard.95242 жыл бұрын
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.
@wetwillyis_18812 жыл бұрын
I think someone needs to make you and intro, with music, that makes you seem like the final boss of a math game.
@thereaper37452 жыл бұрын
I have a question : Can we solve this equation ? a^x + bx + c = 0 Please share it
@aayushkc58242 жыл бұрын
yes x = (-b W((a^(-c/b) log(a))/b) - c log(a))/(b log(a))
@shophaune22982 жыл бұрын
a^x + bx + c = 0 e^(x ln a) + bx + c = 0 e^(x ln a) = -bx - c (-bx -c)e^(-x ln a) = 1 (-bx -c)e^(-x ln a)a^(-c/b) = a^(-c/b) (-bx -c)e^(-x ln a)e^(-c/b ln a) = a^(-c/b) (-bx -c)e^(-x ln a -c/b ln a) = a^(-c/b) (-x -c/b)e^(ln a (-x -c/b)) = a^(-c/b)/b ln a (-x -c/b)e^(ln a (-x -c/b)) = a^(-c/b) ln a /b ln a (-x -c/b) = W(a^(-c/b) ln a /b) where W is the product logarithm (-x -c/b) = W(a^(-c/b) ln a /b)/ ln a x + c/b = -W(a^(-c/b) ln a /b)/ ln a bx + c = -bW(a^(-c/b) ln a /b)/ ln a bx = (-bW(a^(-c/b) ln a /b) - c ln a)/ ln a x = (-bW(a^(-c/b) ln a /b) - c ln a)/ (b ln a) when a = e, this simplifies to: x = -W( e^(-c/b)/b) - c / b
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?
@astecheee15192 жыл бұрын
Just use "inspect graph" on Desmos. Duh.
@shapshooter77692 жыл бұрын
@@astecheee1519 But it's not Lagrange, it's most likely Newton-Rhapson
@chaiotic2 жыл бұрын
This is so good! There's something about polynomial equations that's just so neat, like cool puzzles
@Migeters2 жыл бұрын
My guy I love the videos best calculus channel on yt
@scottleung95872 жыл бұрын
Wow, I wish this worked for all cases of quintic functions - but this is a good start!
@pepebriguglio61258 ай бұрын
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 🙏
@tobybartels84262 жыл бұрын
Once you have one real solution, the other solutions are trivial. Because you can factor out x−r (where r is the root that you found) from the original polynomial to get a quartic polynomial, then apply the quartic formula to that.
@ErezMarom Жыл бұрын
9:47 impressing people is the best reason for inventing a formula.
@Chill----2 жыл бұрын
A very smooth solution indeed.(for the general case)
@kamalrihani96092 жыл бұрын
The coefficients are numbers that have a precise relationship, so this solution is for some specific five-degree equations and not all equations. De Galois is right again. Thans for you.
@harshrajsinhsarvaiya30242 жыл бұрын
Nice T-shirt
@DavidvanDeijk2 жыл бұрын
Nice how you structured the video, with first the plain example and next the advanced with every thing matching the location on the board, very visual like blackpenredpen is famous for.
@vishalmishra30462 жыл бұрын
Any general Cubic equation can be changed to x^3 + 3 m x = 2 n. Then, x = (n + √D)^(1/3) + (n - √D)^(1/3) where cubic discriminant D = n^2 + m^3. To solve Quintic equation, let's change x to be sum of 5th root (instead of 3rd root) and change D to quintic discriminant D = n^2 + m^5. Therefore, IF { x = (n + √D)^(1/5) + (n - √D)^(1/5) } THEN { x^5 + 5 m x (x^2 + m) = 2 n } So, transform your Quintic equation to the form, x^5 + 5 m x (x^2 + m) = 2 n to get x = (n + √D)^(1/5) + (n - √D)^(1/5) where quintic discriminant D = n^2 + m^5.
@johnedwards4394Ай бұрын
Great topic. Rarely discussed.
@dharunrahul17002 жыл бұрын
Plot twist : THE QUINTIC EQUATION FORMULA WAS FOUND BY STEVE ( bprp)
@fabiosantucci66282 жыл бұрын
6:51 look the bottom on the right
@Abacadaba7022 жыл бұрын
Merch idea: cat and indeterminate forms as a hoodie?????
@samehwaleed487211 ай бұрын
I can with ax^5 +bx^4 +cx^3 +dx^2 +ex+f=0 But f =retaion ship between a, b, c, d, e So can you put 5 exchange no exchange a, b, c only
@PennyLapin2 жыл бұрын
The hilarity of seeing what the English captions interpret "quintic" as each time
@trollme.trollmehard.95242 жыл бұрын
Ahh! They're everywhere! :)
@nanamacapagal8342 Жыл бұрын
Just realized that's how the cubic formula was derived x^3 + 3px + 2q = 0 becomes a little more obvious it resembles the form if p = b/a and q = c/(2a) we can also solve more general cubics: for Ax^3 + Bx^2 + Cx + D = 0, make the substitution x = u - B/(3A), then after simplifying, divide through by A. You're back to u^3 + 3pu + 2q = 0 The basic cubic formula carries the rest from there, and the only thing left to do is to undo the substitution
@dav1gamer28710 ай бұрын
Please do this for the other formulas
@SuperYoonHo2 жыл бұрын
Awesome!!! I uploaded the link:)
@tonyhaddad1394 Жыл бұрын
Great strategie
@tomctutor2 жыл бұрын
Just thinking out loud here! What if we have the general quintic(omial) with integer coefficients: f(x)= ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0 then make a linear transform x=mt+n such that f(t) = a(m,n)t^5 + (0)t^4 + c(m,n)t^3 + d(m,n)t^2 + e(m,n)t + f(m,n) = 0 (choose m,n to make b(m,n)=0) Now make another linear transform for t= ps + q f(s) = a(p,q)s^5 + c(p,q)s^3 +(0)s^2 + e(p,q)s +f(p,q) = 0 (choose p,q to make d(p,q) = 0) Now make a final linear transform for s =ju + v f(u) = a(j,k)u^5 + c(j,k)u^3 + e(j,k)u + f(j,k) = 0 (choosing j,k so that e(j,k)/c(j,k) are in the correct ratio to match your fornula?) By the Fundamental Theorem any quintic must have a real solution, so this procedure is guaranteed to get that solution at least? I am not 100% sure this would work but maybe some programmer out there can write a program to check out my idea! 😁
@bjornfeuerbacher5514 Жыл бұрын
"Now make another linear transform for t= ps + q f(s) = a(p,q)s^5 + c(p,q)s^3 +(0)s^2 + e(p,q)s +f(p,q) = 0 (choose p,q to make d(p,q) = 0)" The problem is already in this step: You have to insert t = ps + q into a(m,n)t^5. If you choose p,q to make d(p,q) = 0, this will result in a non-vanishing term with s^4.
@tomctutor Жыл бұрын
@@bjornfeuerbacher5514 I see what you mean, if you t-transform out one s^n term then another spurious t^m term will resurface again! Seems to be a no win scenario, ah at least you can always get rid of one of the lower order terms! As I said just an idea. But thanks for pointing out the logical flaw.🤔
@iyziejane2 жыл бұрын
What's the minimal set of special functions to add to arithmetic + radicals to allow for a quintic formula?
@gigantopithecus82549 ай бұрын
elliptic functions
@ardizzle062 жыл бұрын
i too want the quintic in my life
@Eric-uf7dw2 жыл бұрын
Next do the general quartic formula :)
@brian124632 жыл бұрын
28:37 look at top middle
@shobhitnegi6756 Жыл бұрын
it work on my pc thx bro vеry much
@niltondasilva1645 Жыл бұрын
Maravilha! Eu gosto desses cálculos DIFÍCEIS da equação de grau 5. 👏👏👏👏👏👏
@ahmadmazbouh2 жыл бұрын
5:55 i would use (-1)
@Eli-kz3bw2 жыл бұрын
That c/a at 16:27 seems more sinister than at first glance. Since the coefficients are guaranteed to be rational numbers, the c/a seems to force you into having a rational constant as well. To me though it seems that c only needs to be a real constant. Do I have all that right?
@angelespinosa9062 жыл бұрын
Muy interesante!
@Dan_Tyme2 жыл бұрын
Love the videos! Where did you get that identities poster behind you?
@PrudentialViews2 жыл бұрын
YAY!
@ruud97672 жыл бұрын
I can't believe I was here at this historic moment!
@charliearmour16282 жыл бұрын
Great, very entertaining. I loved it.
@yamanktail65732 жыл бұрын
Chines guys are too powerful ☠️
@AliHassan-hb1bn Жыл бұрын
I would say integral of secx is ln of its derivative. Thank you
@blackpenredpen Жыл бұрын
?
@ferre-jv3rp2 жыл бұрын
Can u solve this equation please? Ln (sin(x)) = log(cos(x)) Greetings from Argentina
@tsurutajunichiro62502 жыл бұрын
hello, I want to present to you a problem from the moroccan math olympiad, here it is find all solutions for x and y in: x^4 + y^4 + 2 = 4xy hope you can solve it, Good Luck
@trix6098 ай бұрын
Can you do the ferrari's fouth formula?
@Monster-nn.0072 жыл бұрын
U did it man👌✨
@abu-karz2 жыл бұрын
hi bprp Pls take a close pic of that information chart of that unit circle behind you and give it in the community tab
@Ninja207042 жыл бұрын
Is there a formula for a general quartic equation? I know that for quintic and higher degree polynomials we can’t always get exact answers, but I forgot if its true for quartics
@d.l.74162 жыл бұрын
There is one but its very long
@Ninja207042 жыл бұрын
@@d.l.7416 Thank you. Even the general cubic formula already looks incredibly long and confusing for me
@himalperera30072 жыл бұрын
Why integration of 1/(x.lnx) by parts gives 0=1
@militantpacifist40872 жыл бұрын
4:33 Fly wanted to be in the video.
@chrisleon27 Жыл бұрын
I shall make more contribution in the future if I keep my word
@antoniogeronimocabral1262 жыл бұрын
how about solving this equation?: 64+log[2](x)=x
@dad-ms8mz Жыл бұрын
I guess ur the most productive person .
@infinity-yy3qn2 жыл бұрын
find the range of f(x) f(x) = cosx[sinx + √(sin²x + sin²z)] here z is constant value sir please solve this problem
@evionlast2 жыл бұрын
I'm gonna check that Chinese Wikipedia often, I don't know Chinese so... wish me luck 🤞
@blackpenredpen2 жыл бұрын
I actually didn’t really read the description. I just read the formula 😆
@giuseppemalaguti4352 жыл бұрын
al min 23 circa manca la radice 5 a denominatore(red)
@абдулазизмаримбоев2 жыл бұрын
Hello thank you very much for it. Can you also try 100 trigonometric equations? It will be great, if you decide to solve it
@kinshuksinghania42892 жыл бұрын
This is great even though it’s just a special case
@VinTheFox2 жыл бұрын
Seems like c/a is expressed as a ratio so that 0 can't be allowed. Otherwise the formula would be incomplete, because the equation can have multiple real solutions.
@awildstevey2 жыл бұрын
Y’all know the know the meme where newton and like hawking are holding back Einstein? We need one but with Abel and Galois
@blackpenredpen2 жыл бұрын
😂
@Manluigi Жыл бұрын
I can do this for polinomials of degree bigger
@stevemonkey66662 жыл бұрын
👍 this is a great mathematical journey
@Mono_Autophobic2 жыл бұрын
He's obsessed with quintic equations
@blackpenredpen2 жыл бұрын
Who isn’t? 😆
@sendai-shimin8 ай бұрын
Please challenge also the equation "型式2" on this Wikipedia.
@yaleng45972 жыл бұрын
5:23 What about PhD degree?
@blackpenredpen2 жыл бұрын
😆
@chessematics2 жыл бұрын
I became very very very upset after learning that angle trisection is impossible.
@blackpenredpen2 жыл бұрын
Doubling a cube is also impossible… 😔
@chessematics2 жыл бұрын
@@blackpenredpen yeah. But i was 7th grade and literally working my bowels out to find a way of trisecting an angle. Just for the record, i had no idea how formal constructions or proofs work. I just wanted to brute force it. That was 2019 February. In 202 i found a way of pretty closely approximating ANY rational multiple (i.e. multiplying the angle with ANY rational number). But it was still just a close approximation and of course, i must admit with all my grief, an angle in general can't be trisected. I got to know the doubling of cube and squaring of circle at the at the same time.
@schizoframia48742 жыл бұрын
The quintish formula 🙃
@spencerswaggington2 жыл бұрын
i love your videos!
@kiwithemaniaguy Жыл бұрын
Playing Rhythm Hell, don't send a quintic plus cubic equation
@kaddaderrer61522 жыл бұрын
Fantastic
@chrisleon27 Жыл бұрын
Hello from China
@blackpenredpen Жыл бұрын
Greetings! Thank you for your super chat!
@-rachmaninoff2 жыл бұрын
Can you solve e^3x + 5x^2 - e^2x = 2 ? :)
@Magic738052 жыл бұрын
Sir, I humbly request you to start a new series for mathematics. There’s an exam of maths in india. This exam is conducted by UPSC. Actually this exam is also known as maths optional in UPSC. I’m an aspirant and preparing for this maths optional exam. If possible please start this series so that we can get some help from you sir.
@bruh____7842 жыл бұрын
√36 (possibly)
@sabasmoreno67052 жыл бұрын
Congratulations
@phenixorbitall3917 Жыл бұрын
Magic
@minimath58822 жыл бұрын
Math finds a way!
@mansoormohamedali2 жыл бұрын
Hello Mr. BpRp. I have a question that has been keeping me all night these past few days. Hope you can help. Question: Let have any area on earth ( a city or a region). That we can calculate or know its area. We have n (a natural number) number of fixed spot we can put on the area.How can we calculate the minimal distance we can place n number of fixed spot such that we have the same distance between any point on the edge of the boundary to the next fixed spot or from fixed spot to fixed spot. For example: if we have n =1, so the only logical place to put the fixed point is at the center. From the the center ( fixed spot) to any point in the boundary are the same distance. What is the reasoning in the case n = 2,or n = 3, … n = inf?
@Ceratops172 жыл бұрын
Okay so the first question is do you want your set (region) do be convex, this means every connection between points is also in the set itself. Also these points you mentioned just have the same distance to the center if your set is a ball. (If I misunderstood you, please correct me.) Consider a ellipses for example, we need a closed set M to also have the points on the boundary B, you can’t find a point x where the distance from any point on the boundary is the the same for all of them so to write it mathematically, no x (element) M: d(x,b)=d(x,c) for b,c (element) B and x (not element) bc (I mean the straight line connecting b and c here)
@social63322 жыл бұрын
wow good.
@ahmedalhattab962 жыл бұрын
[√(x^2+5) - √(x+5)]÷(x-1)=3-√(7) Can you solve this equation for me plz
@identityelement77292 жыл бұрын
Could someone solve any quintic equation if this person would have an countable infinite amount of such special case formulaes? One formulae is not enough I know but more than One?
@amirhamzakhan2198 Жыл бұрын
Who has patience to invent this formula for the first time?
@prodromoskonstandas1552 жыл бұрын
Shouldn't t have 10 roots ? Because it's a tenth degree ?
@That_One_Guy...2 жыл бұрын
As BPRP said in the video, the number of solution in t variable is actually only 1/2 of the degree of t because the other half is the conjugate of the other one (so there won't actually be 10 instead it will just be 5), and the other 4 required solution is hidden away and to find them all i suspect you either : 1) Probably gonna need to use trig function to discover it (just like in cubic function). 2) Multiply the first solution by the fifth root of unity (correct me if i'm wrong on this one because i just recently read it from wikipedia about cardano's formula). These root of unity are all of the solution of x^5 = 1.