Once you solved for u at 19:45, there is a simpler way of solving for v. In the video, you noted that you could have done the same process for v and so obtained that v could be written in the same form as u and then recognized that y=u+v could be expressed in three cases but that the only valid case was where the square roots under the cube root were opposite signs for u and v. There is a faster way to figure this out without resorting to plugging in all three cases into the original depressed cubic. Instead, we can recall that one of the equations in our original system was u^3+v^3=-q which was from around 16:15, meaning v^3=-u^3-q. Because u^3=-q/2 (+ or - ) sqrt[(q^2/4)+(p^3/27)], we can substitute this expression for u^3 in the above equation. Thus, v^3=-[-q/2 (+ or - ) sqrt[(q^2/4)+(p^3/27)]]-q= q/2-q (- or +) sqrt[(q^2/4)+(p^3/27)]=-q/2 (- or +) sqrt[(q^2/4)+(p^3/27)] (Keep in mind that the (+ or - ) in front of the square root turned into (- or +) because of the negative in front of u^3 in the equation v^3=-u^3-q. Although both (+ or - ) and (- or +) represent simply that the square root is either positive or negative, switching from (+ or - ) to (- or +) simply clarifies that if we know the original sign of the square root, then the negative will turn it into the opposite sign. For example, if we knew that this square root expression was negative, then the (+ or - ) in front of the square root resolves to simply (-), but the negative in front of u^3 in this equation causes the square root to being positive so (- or +) is represented by (+)). This implies v= cbrt[-q/2 (- or +) sqrt[(q^2/4)+(p^3/27)]], which represents the third case because given that u has a (+ or - ) for the square root expression under the cube root, the fact that the square root expression under the cube root for v has a (- or +) indicates that the square root expressions for u and v have opposite signs. Thus, if we pick one to be the positive case (say let u be the case where the square root expression is positive) then the other will be the negative case and the choice does not matter since y=u+v and addition is commutative. Solving for v in this way after solving for u means that we automatically eliminate the other two cases for y=u+v that you listed (where the square root expressions for u and v have the same sign ) without having to plug in these cases into the depressed cubic.
@angelosterizakis7635
Жыл бұрын
I cannot thank you enough. Clear, effortless, step by step method. I now understand the cubic formula thanks to your video. I find the cubic formula especially useful when finding exact trigonometric ratios of trisected angles. Huge respect for you.
@holyshit922
2 жыл бұрын
This idea can be generalized to quartic Now I show differences between this method for cubic and for quartc If you want to use this idea for quartic 1st step substitution y=x+b/(4a) after this step we should get equation y^4 + py^2 + qy + r = 0 2nd step substitution y=u+v+w We try to expand (u+v+w)^4 to be able to compare coefficients with equation y^4 + py^2 + qy + r = 0 (u+v+w)^4=2(u^2+v^2+w^2)(u+v+w)^2+8uvw(u+v+w)+4(u^2v^2+u^2w^2+v^2w^2)-(u^2+v^2+w^2)^2 In my opinion although extending method for cubics to solve quartics is possible a lot less computation we will have if we solve quartics by factoring into two quadratics first
@em_zon2643
2 жыл бұрын
Very interesting approach! Thank you! Merry Christmas ☃️!
@andreben6224
2 жыл бұрын
For the depressed cubic via the calculus shortcut, you can simply use the Taylor expansion of f and then you can see that the x shift is also the amount which cancels f ' ' So you can derive it a second time the y=x+b/3a in a second way :) N.B. this can all be done in any positive caracteristic field (provided we allow ourselves to add cubic roots of unity, so splitting fields), but I'm not sure how to generalise it to arbitrary fields (I suppose only fields of characteristic 3 would pose a "threat" ^_^;; )
@duncanzhu490
8 ай бұрын
This video is absolutely amazing! Thank you so much!
@ees996
9 ай бұрын
At 39min50s, you mention this Discriminant equals 0, but it doesn't. D = q^2/4 + p^3/27 .What formula did you use to calculate p and q??
@MichaelMaths_
9 ай бұрын
Maybe double check your calculation, it should be D = (-686/27)²/4 + (-49/3)³/27 = (2⋅7³)²/(4⋅27²) - (7²)³/(27⋅3³) = (4⋅7⁶)/(4⋅27²) - 7⁶/27² = 7⁶/27² - 7⁶/27² = 0.
@ees996
9 ай бұрын
@@MichaelMaths_ how do u find the q= -686/27 and p= -49/3 ?
@MichaelMaths_
9 ай бұрын
@@ees996Either you can get them through the substitution x = y - b/(3a) like at the start of the video or use the calculus shortcuts (p = df/dx and q = f(x) both at x = -b/(3a) where f is the cubic function) which basically comes from Taylor expansion about x = -b/(3a).
@ees996
9 ай бұрын
@@MichaelMaths_ p= b - a^2/3 = 16 - 2^2/3 = 14.6666. Q = c + (2a^3-9ab)/27 = 10 + (2*2^3 - 9*2*16)/27 = -0.07. I also tried with a=1,b=8,c=5,d=-50 and still doesnt work
@MichaelMaths_
9 ай бұрын
@@ees996 It seems like the formulas you have for p and q are for a monic cubic equation of the form x³ + ax² + bx + c, but in the video example you need to first divide both sides by the leading coefficient, which is 2 in this case. After doing this, we will have a = 8, b = 5, and c = -50 according to this form and then you will correctly find p = -49/3 and q = -686/27.
@pmccarthy001
2 жыл бұрын
I found an error in the reduction from the general cubic equation to the depressed cubic equation at about 11:05 in the video. I don't seem to be able to post a screenshot or image here. Trying to post MathML appears to freeze the whole webpage. I can try posting the TeX, but I doubt it'll convert it to TeX... $$ \left( y^3-\frac{by^2}{a}+\frac{b^2y}{3a^2}-\frac{b^3}{27a^3} ight) +\frac{b}{a}\left( y^2-\frac{2by}{3a}+\frac{b^2}{9a^2} ight) +\frac{c}{a}\left( y-\frac{b}{a} ight) +\frac{d}{a}=0 \\ y^3\cancel{-\frac{by^2}{a}}+\frac{b^2y}{3a^2}-\frac{b^3}{27a^3}\cancel{+\frac{by^2}{a}}+\frac{3b^3}{27a^3}+\frac{cy}{a}-\frac{bc}{3a^2}+\frac{d}{a}=0 \\ y^3+\left( \frac{cy}{a}+\frac{b^2y}{3a^2} ight) +\left( \frac{-b^3}{27a^3}+\frac{3b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a} ight) =0 $$ If that doesn't work, this is the best text rendition I could get of it. I couldn't get bold, italic, or underline to post here. (3) ((y^3-(by^2)/a+(b^2 y)/(3a^2 )-b^3/(27a^3 ))+b/a (y^2-2by/3a+b^2/(9a^2 ))+c/a (y-b/a)+d/a=0 (4) y^3-(by^2)/a+(b^2 y)/(3a^2 )-b^3/(27a^3 )+(by^2)/a (-2by/3a)+(3b^3)/(27a^3 )+cy/a-bc/(3a^2 )+d/a=0 (5) y^3+(cy/a+(b^2 y)/(3a^2 ) (-(2b^2 y)/(3a^2 )))+((-b^3)/(27a^3 )+(3b^3)/(27a^3 )-bc/(3a^2 )+d/a)=0) In (4), (-2by/3a) s/b (-(2b^2 y)/(3a^2 ))). That term s/b the same in (4), (5). It's almost as if you were copying from notes and the notes had this error.
@MichaelMaths_
2 жыл бұрын
Oh I see, yeah that should have been written as -(2b^2 y)/(3a^2) distributing the b/a at (4). I probably misread my notes for the video while recording, though the end result for the depressed cubic is still verified to be correct.
@pmccarthy001
2 жыл бұрын
@@MichaelMaths_ Yes, that's why I thought you were probably reading from notes. Although I've never seen the 1st derivative trick to help reduce the general cubic to the depressed equation. With that, it almost makes the purely algebraic reduction almost simply of theoretical interest. Do you think that 1st derivative trick makes Cardano's method competitive with other methods of root-finding for a wider class of equations?
@MichaelMaths_
2 жыл бұрын
@@pmccarthy001 Yeah that was just something I stumbled upon while messing around with depressed cubics in a graphing calculator, and comparing coefficients showed that it really works. I do think using that trick brings Cardano's method to a more competitive level, close to how the quadratic formula performs when solving for the roots of quadratics, but you just need a little differentiation before plugging in values. It can be worth it to use this over something like Newton's method for cubics, but probably not for analytical solutions of solvable degree 4 polynomials and higher.
@robertjennings4236
Жыл бұрын
I'm having difficulty with the 3rd example in particular with the cos(pi/12). I have used the half-angle formulas but i still cannot match any of the given real solutions. Can someone help? Thanks
@MichaelMaths_
Жыл бұрын
You could use the half angle formula, but here are some steps I used with only angle addition/subtraction and phase shifts by π. For n = 0: cos(π/12) = cos((3 - 2)π/12) = cos(π/4 - π/6) (and use cos(x - y) formula) For n = 1: cos(π/12 + 2π/3) = cos(3π/4) = cos(π - π/4) = -cos(π/4) (and use special value) For n = 2: cos(π/12 + 4π/3) = cos(17π/12) = cos((12 + 5)π/12) = cos(π + 5π/12) = -cos(5π/12) = -cos((2 + 3)π/12) = -cos(π/6 + π/4) (and use cos(x + y) formula)
@robertjennings4236
Жыл бұрын
Thanks to your help, I finished example 3 this morning. Yay! Thanks very much!!
@MichaelMaths_
5 ай бұрын
@@Podcast690 No restrictions were made on the coefficients so yes, it can
@polarbear_195
11 ай бұрын
Explain it's derivatative ( sorry wrong spells ) how did you get 686 in 40:53 ?
@MichaelMaths_
11 ай бұрын
y(x) is just the function on the left side of the equation. Essentially this calculus shortcut to get the depressed equation coefficients in terms of derivatives comes from a Taylor expansion of y(x) at x = -b/(3a). For the value of y(-b/3a) = y(-8/3), we have (-8/3)³ + 8(-8/3)² + 5(-8/3) - 50 = -512/27 + 512/9 - 40/3 - 50 = (-512 + 1536 - 360 -1350)/27 = -686/27.
@TerryFerrellmathematics
3 ай бұрын
Wow!!! Love it.
@shibam4182
Жыл бұрын
52:32 where did you get 2πn/3 in the bracket and where did you find n=0,1,2 and can we solve D
@MichaelMaths_
Жыл бұрын
Cosine is 2π periodic so +2πn doesn’t add any contribution, but if we take that part divided by 3 from the root, it now gives a contribution and you only need n = 0,1,2 since larger n gives repeated values. I think you can take cube roots of complex numbers without using polar form, but it’s longer and using DeMoivre’s theorem is much easier, or avoid trig by not fully simplifying lol.
@shibam4182
Жыл бұрын
Can you plz solve x^3-3x+1=0 find all the three solutions
@shibam4182
Жыл бұрын
@@MichaelMaths_But how what is the formula for taking the cube roots of complex number without using polar form
@@shibam4182 Not sure if there is any simple formula, but you can probably set the cube root equal to some arbitrary complex number a + bi, cube both sides, and make real and imaginary parts equal to get a system of equations to solve for a and b.
@arkdevil1475
8 ай бұрын
What app do you use to write ?
@MichaelMaths_
8 ай бұрын
I use Milton infinite canvas but might be switching soon
@vengeancegaming5389
Жыл бұрын
How do you get 63? 38:20
@MichaelMaths_
Жыл бұрын
The 63 in the cube root comes from a part of the formula, namely -q/2
@JoseAntonio-ng5yu
2 ай бұрын
The calculus shortcut ain't really a shortcut, it's pretty much the same as the direct formulas por p and q, however it's MUCH easier to memorize
@em_zon2643
Жыл бұрын
Good One!
@AsBi1
11 ай бұрын
very helpful
@lazarusunkwon6
2 жыл бұрын
Why is the inflection point the center?
@MichaelMaths_
2 жыл бұрын
Since a cubic is odd, it has half-turn symmetry about the inflection point and that can be treated as its center. However, unlike higher degree odd polynomials, there will always be exactly one inflection point and this keeps it centered about that point no matter the terms/coefficients in the equation.
@lazarusunkwon6
2 жыл бұрын
@@MichaelMaths_ Thank you. Great video by the way.
@notesfromundergroundenjoyer
Жыл бұрын
hey i have a question i know when the imaginarys are negative and plus they cancel and give an answer but how do you get the other 2 to be real since it doesn't make sense sorry its just that the ending did not explain enough for me@@MichaelMaths_
@ees996
9 ай бұрын
you mention r is magnitude, what is that??
@MichaelMaths_
9 ай бұрын
Basically magnitude of u (or v since complex conjugate doesn’t affect it). I realize now that this can be more concretely written as 2 |u| cos(arg(u) + 2πn/3) for D < 0.
@ees996
9 ай бұрын
@@MichaelMaths_ did u forget |u| has exp 1/3
@MichaelMaths_
9 ай бұрын
@@ees996 The expression for u is in a cube root so we could say |u³|^(⅓) but that’s the same as |u|.
Пікірлер: 52