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
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 UA-cam must be benevolent with you. Teachers makes the difference. You are one of them.
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.
@@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…
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!!
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.
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.
@@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.
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
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.
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)
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?
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.
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 🙏
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!
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
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.
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.
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
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! 😁
"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.
@@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.🤔
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.
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.
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
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?
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.
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
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?
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)
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.
both the general form of the equation and the final formula could be made simpler by using only 2 parameters instead of 3 take u = b/a and 2v = c/a; that way neither of the expressions will have fractions the substitution, in that case, would be x = t - u/t
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.
Would that mean that this formula can be generalized to any quintic that can be transformed into the quintic form mentioned at the start? If so, what properties does this particular class of quintics obey?
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?
@@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.
Try this quintic equation next:
ua-cam.com/video/GoGsVLnf8Rk/v-deo.html
Damn..!!!!!!, i cant hold my mother finger to not subscribe this channel..
So i follow you sir..
Im so sorry..
🥺🥺🥺🥺🥺
Can u find the solution of the equation given by sinx+e^x=0
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
@@tetraktys2786 Tetraktys, I was solving, and I didn't get the answer, thanks
I also found one for a calculator, but it uses newton's method
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 UA-cam must be benevolent with you.
Teachers makes the difference. You are one of them.
^^
bro wtf i thought this stopped at quadratics bro where the 3rd and 4th and 5th exponent come from
@@oreos2653 there is more to math than you think :)
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.
Hmm imaginary numbers?
@@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…
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!!
Niels Henrik Abel crying rn
He proved the unsolvability of the general quintic,not for all of them.
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.
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.
It's nice that b^2-5ac resembles the discriminant for quadratic equations!
Now do the quartic equation, but with the formula for general solutions.
I don't think mathologer provides the formula, but he does provide the method
@@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.
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
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
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.
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)
literally Awesome ur one of the best calculus teacher u have seen ur explanation is awesome👍🏻🥰...
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?
Just use "inspect graph" on Desmos. Duh.
@@astecheee1519 But it's not Lagrange, it's most likely Newton-Rhapson
No matter how much I practise math, blackpenredpen always manages to bring up new challenges.
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.
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 🙏
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!
I have a question :
Can we solve this equation ?
a^x + bx + c = 0
Please share it
yes
x = (-b W((a^(-c/b) log(a))/b) - c log(a))/(b log(a))
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
It’s solvable with the W function
@@artophile7777 Yes.
That's solvable using lambert W function :
a^x = -bx-c
a^x/b = -x - c/b
1/b = -(x + c/b) * a^(-x)
a^(-c/b) / b = -(x + c/b) * a^( -(x + c/b))
ln(a) * a^(-c/b) / b = -ln(a) * (x + c/b) * e^(-ln(a) * (x + c/b))
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.
This is so good! There's something about polynomial equations that's just so neat, like cool puzzles
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.
Wow, I wish this worked for all cases of quintic functions - but this is a good start!
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
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! 😁
"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.
@@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.🤔
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.
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.
Great strategie
Bprp : quintic equation
Yt captions : green tea equation
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
A very smooth solution indeed.(for the general case)
Love the videos! Where did you get that identities poster behind you?
My guy I love the videos best calculus channel on yt
6:51 look the bottom on the right
9:47 impressing people is the best reason for inventing a formula.
Please do this for the other formulas
Nice T-shirt
Great topic. Rarely discussed.
28:37 look at top middle
What's the minimal set of special functions to add to arithmetic + radicals to allow for a quintic formula?
elliptic functions
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
There is one but its very long
@@d.l.7416 Thank you. Even the general cubic formula already looks incredibly long and confusing for me
Maravilha! Eu gosto desses cálculos DIFÍCEIS da equação de grau 5. 👏👏👏👏👏👏
Next do the general quartic formula :)
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?
Merch idea: cat and indeterminate forms as a hoodie?????
hi bprp Pls take a close pic of that information chart of that unit circle behind you and give it in the community tab
Great, very entertaining. I loved it.
Can u solve this equation please?
Ln (sin(x)) = log(cos(x))
Greetings from Argentina
Awesome!!! I uploaded the link:)
i too want the quintic in my life
it work on my pc thx bro vеry much
Muy interesante!
5:55 i would use (-1)
4:33 Fly wanted to be in the video.
Plot twist : THE QUINTIC EQUATION FORMULA WAS FOUND BY STEVE ( bprp)
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.
I would say integral of secx is ln of its derivative. Thank you
?
The hilarity of seeing what the English captions interpret "quintic" as each time
Ahh! They're everywhere! :)
find the range of f(x)
f(x) = cosx[sinx + √(sin²x + sin²z)]
here z is constant value
sir please solve this problem
U did it man👌✨
This is great even though it’s just a special case
how about solving this equation?: 64+log[2](x)=x
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
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?
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)
Why integration of 1/(x.lnx) by parts gives 0=1
Can you do the ferrari's fouth formula?
Hello thank you very much for it. Can you also try 100 trigonometric equations? It will be great, if you decide to solve it
I shall make more contribution in the future if I keep my word
👍 this is a great mathematical journey
I can't believe I was here at this historic moment!
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.
Please challenge also the equation "型式2" on this Wikipedia.
I guess ur the most productive person .
What is this algebra? What problem are you trying to solve here?
Can you solve e^3x + 5x^2 - e^2x = 2 ? :)
al min 23 circa manca la radice 5 a denominatore(red)
Hello from China
Greetings! Thank you for your super chat!
He's obsessed with quintic equations
Who isn’t? 😆
i love your videos!
I can do this for polinomials of degree bigger
√36 (possibly)
I think someone needs to make you and intro, with music, that makes you seem like the final boss of a math game.
both the general form of the equation and the final formula could be made simpler by using only 2 parameters instead of 3
take u = b/a and 2v = c/a; that way neither of the expressions will have fractions
the substitution, in that case, would be x = t - u/t
Fantastic
Y’all know the know the meme where newton and like hawking are holding back Einstein? We need one but with Abel and Galois
😂
Chines guys are too powerful ☠️
I'm gonna check that Chinese Wikipedia often, I don't know Chinese so... wish me luck 🤞
I actually didn’t really read the description. I just read the formula 😆
Shouldn't t have 10 roots ? Because it's a tenth degree ?
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.
Would that mean that this formula can be generalized to any quintic that can be transformed into the quintic form mentioned at the start?
If so, what properties does this particular class of quintics obey?
YAY!
Math finds a way!
The quintish formula 🙃
Playing Rhythm Hell, don't send a quintic plus cubic equation
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?
I became very very very upset after learning that angle trisection is impossible.
Doubling a cube is also impossible… 😔
@@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.
[√(x^2+5) - √(x+5)]÷(x-1)=3-√(7)
Can you solve this equation for me plz
The equation looks like the cubic formula.
Congratulations
Magic
Who has patience to invent this formula for the first time?
Awesome
wow good.