This video is indeed original. I've once (about a year ago) been brute-forcing through all the math channels I could possibly find and got about 600. Then I went through them once by each and purged out all the trashy/too elementary ones. With that only 50 channels were left. What I can ensure: There has not been a single video of that kind! I can confirm this video's originality! Great video!
In f(X) = X^4+aX^3+bX^2+cX+d =0 we could check the coefficient and if 1+ a+ b+ c+ d=0 , X=1 that is one answer, then dividing f(X) by X-1 getting a Cubic equation to solve.. Likewise if 1- a+ b- c+ d=0 we have X= -1 that is one answer, and as it mentioned above, a Cubic equation needs to be solved for the other answers.
Sir Marvel. You are Captain Marvel of Maths, your are Thor of Maths, you are Martian Man Hunter of Math. Sir I have checked each and every equation.Each time I say , Sir you are great
Рік тому
When studying elliptic curves which can be represented by y^2 = x^4 + ax^3 + bx^2 + cx + d, for example, this approach follows naturally. Good work finding this if you haven't studied elliptic curves. Good even if you have.
This method has some similarities to the Ferrari method, but it is considerably simpler to describe and use. Consider the depressed quartic equation: y^4 + u*y^2 + v*y + w = 0 If v = 0, then the equation is quadratic in y^2 and easily solved. So assume v is non-zero for this method. Rearrange to y^4 = -u*y^2 - v*y - w
Add a term to both sides to make the left side a perfect square involving y^2 , and eventually, the right side a perfect square involving y : (y^2 + f)^2 = y^4 + 2*f*y^2 + f^2 = 2*f*y^2 + f^2 - u*y^2 - v*y - w (y^2 + f)^2 = (2*f - u)*y^2 - v*y + f^2 - w Complete the square on the right side : (y^2 + f)^2 = (2*f - u)*(y - v/2 / (2*f - u))^2 - (1/4)*v^2 / (2*f - u) + f^2 - w {1} (y^2 + f)^2 = (sqrt(2*f - u)*y - v/2 / sqrt(2*f - u))^2 where the last step requires that : - (1/4)*v^2 / (2*f - u) + f^2 - w = 0 which can be rearranged to a cubic polynomial in f : f^3 - u/2 *f^2 - w*f + u*w/2 - (1/8)*v^2 = 0
Note this can also be written (2*f - u)*(f^2 - w)/2 - (1/8)*v^2 = 0 , which shows that (2*f - u) cannot be 0 unless v = 0 , therefore non-zero v guarantees that (2*f - u) is nonzero. And if v = 0 , this method is not necessary since the original equation would be quadratic in y^2 and easily solved. Returning to {1} rearrange to the difference of two squares and factor: (y^2 + f)^2 - (sqrt(2*f - u)*y - v/2 / sqrt(2*f - u))^2 = 0 (y^2 + f - sqrt(2*f - u)*y + v/2 / sqrt(2*f - u)) * (y^2 + f + sqrt(2*f - u)*y - v/2 / sqrt(2*f - u)) = 0 (y^2 - g*y + f + 1/2 * v/g) * (y^2 + g*y + f - 1/2 * v/g) = 0 g = sqrt(2*f - u) The quartic has been factored into two quadratics which are easily solved.
If you get a rational root for the resolvent cubic, it simplifies calculations; the calculations are simplified further still if the root is a perfect square.
For the video, I found the animation to be distracting. I would have rather just seen each new line appear instantly, and then a pause for viewing, and then the next new line appear instantly. That is an interesting method to solve general quartics. I have not seen it before. All methods I have seen to solve general quartic equations involve solving a cubic equation. So when comparing quartic solution methods, it is essentially a question of how complicated or tedious it is to determine the cubic equation and then apply the root of the cubic to solve the quartic. I think the method in the video compares favorably with the Ferrari method, but perhaps is slightly behind the method I will outline below in another comment.
I tried to reduce general quartic to the biquadratic case and got equation of sixth degree Suppose we have equation a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x + a_{0} = 0 Lets substitute x = (pt+q)/(t+1) and multiply equation by (t+1)^4 You will get quartic in t Equate coefficients in terms t^3 and t to zero and you will get system of equation Solution to this system of equations leads us to polynomial equation of 10th degree but this 10th degree polynomial is divisible by polynomial a_{4}p^4+a_{3}p^3+a_{2}p^2+a_{1}p + a_{0} so there is polynomial equation of sixth degree left to solve
I tried to resolve the equation x^4-7x^2+11x-10=0 One of the real roots is x=2 The equation has another real root Can we find the exact value by using this brand new method?
This video is indeed original. I've once (about a year ago) been brute-forcing through all the math channels I could possibly find and got about 600. Then I went through them once by each and purged out all the trashy/too elementary ones. With that only 50 channels were left.
What I can ensure: There has not been a single video of that kind! I can confirm this video's originality!
Great video!
Excellent video. i wish videos like these got more recognition
Gutes Meisterwerk!👏👏👏
Great !
In f(X) = X^4+aX^3+bX^2+cX+d =0 we could check the coefficient and if 1+ a+ b+ c+ d=0 , X=1 that is one answer, then dividing f(X) by X-1 getting a Cubic equation to solve.. Likewise if 1- a+ b- c+ d=0 we have X= -1 that is one answer, and as it mentioned above, a Cubic equation needs to be solved for the other answers.
Sir Marvel. You are Captain Marvel of Maths, your are Thor of Maths, you are Martian Man Hunter of Math. Sir I have checked each and every equation.Each time I say , Sir you are great
When studying elliptic curves which can be represented by y^2 = x^4 + ax^3 + bx^2 + cx + d, for example, this approach follows naturally. Good work finding this if you haven't studied elliptic curves. Good even if you have.
Thank you.
I prefer this way to Ferrari's method. I think this way , has less calculations.
a good start . keep going
Is it that it has no root or no real roots, when the curve touches or cross the real line?
This method has some similarities to the Ferrari method, but it is considerably simpler to describe and use. Consider the depressed quartic equation:
y^4 + u*y^2 + v*y + w = 0
If v = 0, then the equation is quadratic in y^2 and easily solved. So assume v is non-zero for this method.
Rearrange to
y^4 = -u*y^2 - v*y - w
Add a term to both sides to make the left side a perfect square involving y^2 ,
and eventually, the right side a perfect square involving y :
(y^2 + f)^2 = y^4 + 2*f*y^2 + f^2 = 2*f*y^2 + f^2 - u*y^2 - v*y - w
(y^2 + f)^2 = (2*f - u)*y^2 - v*y + f^2 - w
Complete the square on the right side :
(y^2 + f)^2 = (2*f - u)*(y - v/2 / (2*f - u))^2 - (1/4)*v^2 / (2*f - u) + f^2 - w
{1} (y^2 + f)^2 = (sqrt(2*f - u)*y - v/2 / sqrt(2*f - u))^2
where the last step requires that :
- (1/4)*v^2 / (2*f - u) + f^2 - w = 0
which can be rearranged to a cubic polynomial in f :
f^3 - u/2 *f^2 - w*f + u*w/2 - (1/8)*v^2 = 0
Note this can also be written (2*f - u)*(f^2 - w)/2 - (1/8)*v^2 = 0 , which shows that (2*f - u) cannot be 0 unless v = 0 , therefore non-zero v guarantees that (2*f - u) is nonzero. And if v = 0 , this method is not necessary since the original equation would be quadratic in y^2 and easily solved.
Returning to {1} rearrange to the difference of two squares and factor:
(y^2 + f)^2 - (sqrt(2*f - u)*y - v/2 / sqrt(2*f - u))^2 = 0
(y^2 + f - sqrt(2*f - u)*y + v/2 / sqrt(2*f - u)) * (y^2 + f + sqrt(2*f - u)*y - v/2 / sqrt(2*f - u)) = 0
(y^2 - g*y + f + 1/2 * v/g) * (y^2 + g*y + f - 1/2 * v/g) = 0
g = sqrt(2*f - u)
The quartic has been factored into two quadratics which are easily solved.
If you get a rational root for the resolvent cubic, it simplifies calculations; the calculations are simplified further still if the root is a perfect square.
interesting
For the video, I found the animation to be distracting. I would have rather just seen each new line appear instantly, and then a pause for viewing, and then the next new line appear instantly.
That is an interesting method to solve general quartics. I have not seen it before. All methods I have seen to solve general quartic equations involve solving a cubic equation. So when comparing quartic solution methods, it is essentially a question of how complicated or tedious it is to determine the cubic equation and then apply the root of the cubic to solve the quartic. I think the method in the video compares favorably with the Ferrari method, but perhaps is slightly behind the method I will outline below in another comment.
I tried to reduce general quartic to the biquadratic case and got equation of sixth degree
Suppose we have equation a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x + a_{0} = 0
Lets substitute x = (pt+q)/(t+1) and multiply equation by (t+1)^4
You will get quartic in t
Equate coefficients in terms t^3 and t to zero
and you will get system of equation
Solution to this system of equations leads us to polynomial equation of 10th degree
but this 10th degree polynomial is divisible by polynomial a_{4}p^4+a_{3}p^3+a_{2}p^2+a_{1}p + a_{0}
so there is polynomial equation of sixth degree left to solve
Aurangzeb Alamgir the Great
I tried to resolve the equation
x^4-7x^2+11x-10=0
One of the real roots is x=2
The equation has another real root
Can we find the exact value by using this brand new method?