Short answer: I did. I had a friend who was programming a massive computer to instruct a massive laser cutting machine to cut pieces of shoe leather before stiching them together into the final shoe. For the curved cuts, they approximated by cubic equations and hence needed their intersection points to know when to stop the cuts. Therefore they needed to equate the 2 cubic equations and then find the roots of the single cubic in order to know when to stop cutting. They were using was Newton's Method to find the roots. Problem was, in being able to find a reasonable x0 to start the iteration. As they had no picture, they really had no idea how to pick a resonable x0. Remember they were only dealing with small portions of each curved cut and as such needed to find only 1 specific real root. Newtion's Method was often generating the wrong real root or if x0 happened to be close to a stationary point then the computations went arwy. Enter the cubic equation. I wrote a C program which solved the cubic (using cases on d) but ran into a different intermittent problem - the same problem that Cardano encountered. That is, when d = b^2/4 + a^3/27 was negative ===> a is negative and sometimes produced a negative value for d. So when d < 0 I went over to the triple angle cosine formula, as you did, and my problem was solved. Now I probably could have used some features of the C programming language to do Complex Number Arithmetic but this seemed like an easy fix so I tried it. As arc_cosine in C returns the k = 0 value, just add 2pi/3 and 4pi/3 to the angle to get all solutions from 0 to 2pi. Then back substitute. Also remember, they were not interested in exact solutions but floating point values to pretty much any degree of accuracy desired. Anyway the whole thing worked like a charm. I'll leave out (1) the details of single, double and triple roots and how to sort out the different cases and (2) how to deal with d = 0, 0 in floating point arithmetic, but suffice it to say the whole process made my day. Who would have thought??? So cool. I guess I could also add that (1) d 0 yields 1 single root. And that a cubic can ONLY have (1) 3 real - 0 complex or (2) 1 real - 2 complex roots.
Interestingly, the cosine substitution works precisely when the usual cubic formula "fails" (involving cube roots of complex numbers). If you write the numbers in the cubic formula as complex exponentials, then the cosines fall out of that.
And a fun thing, since you know one of these is 4+0i, you can get an exact value of the inverse cosine of the expression based on choosing the root corresponding to x=4+0i
The issue with the cubic formula is that, besides giving complicated expressions that turn out to be equal to simple values, there are other methods that can sometimes find said values much faster. The rational root theorem (or simple inspection) would lead to the value x=2 in the original problem; the polynomial could then be factored into (x-2)(x²+2x-24), which can then be further factored into (x-2)(x-4)(x+6), yielding all three solutions.
you don't. if the argument is not a real number between -1 and 1, you will get a complex solution for the inverse cos and in at least one case for n, the cos will cancel all the irrationality and give a real result.
@kpaasial. That is a good question. I don't have the space here, but the arugument of the inverse cosine consists of a ratio involving b^2/4 and a^3/27 which are the 2 terms in the cubic discriminant d = b^2/4 + a^3/27. I know that's not much to go on but I think you can get to your -1 to 1 interval from there. I'll see what I can come up with.
@kpaasial. Got it. Take a look at the 2 cases for d = b^2/4 +a^3/27. i.e. when b^2/4 + a^3/27 0 Now check the argument of the inverse cosine in each case. You will find it trapped bewteen -1 and +1.
Very good! This solution method is also given in the "CRC Standard Math Tables", page 10 in the 27th edition. I've never used it, and can't, off hand, think of when it would be better than the method described on page 9. Perhaps in the days before computers, when it would be easier to find tables of cosines than tables of cube roots?
You didn't say (but it matters) use this when there are three real roots. And the reason it is useful is that Cardano's Method is quite a mess unless your cubic has only one real root. Writing the (depressed) cubic in Vieta's Form x³ = 3Ax + 2B allows you to dispense with fractions in both Cardano and this form. Now you might wonder "what cubic is solved if you use the sines for this angle?". The cubic which is cognate to the one using cosines, namely x³ = 3Ax - 2√(A³ - B²). For your exemplar this works out to x³ = 3(28/3)x + 2(80/(3√3)) if I escaped blunder. But (error or not) the point is the trig substitutions give a lot back for your effort (getting to theta).
In case of three real roots, the standard cubic formula (with its two cube roots) gives exactly the same answer as the trig version. Problem is that the real numbers are, so to speak, "encrypted" in the corners of several overlapping triangles inside of the complex plane. The approach shown may therefore be interpreted as a shortcut for the direct access of those purely real numbers.
All these depressed cubics with 0 as the coefficient of x^2 have roots of the form -2m, m+ n and m - n. You get the expression x^3 - (3m^2 - n^2)x + 2m(m^2 - n^2).
Our 2nd sem will start next week. I already watched your 50 integrals video and currently watching the 100 integrals part 1. I wont watch this video for now, I just want to say, thank you very much!!!!
There isn’t really any way to simplify the complex expressions from the original video. According to galois theory, if all three roots to a cubic equation are real but irrational, the roots cannot be expressed exactly using any finite radical expression with real numbers only. So you can only write the roots with complex numbers, or an infinite radical. If you want to actually calculate/approximate the root numerically, just use wolframalpha or any calculator that can do roots of complex numbers.
I have 2 questions about the Lambert W function; can you please answer them? 1. Why is the Lambert W function only in base e? Why can’t we have, say, W base 3 or base 10 or base pi (or something like that? 2. Logartithms have special properties like the product property (log(a*b) = log(a) + log(b)), expontential property (log(a^b) = b*log(a)), etc. Does the Lambert W function have similar properties?
Im not him, but I will try to answer it, perhaps there is more that I dont know 1. I think this is for definition, the person who created it, had the objective of creating an inverse function, the inverse of x*e^x, but later was generalised to all the functions who had the same structure 2. I dont think so
"Why is the Lambert W function only in base e? Why can’t we have, say, W base 3 or base 10 or base pi (or something like that?" Well, it's quite easy to solve exponential equations with other bases using W, so why bother using additional functions for other bases? "Does the Lambert W function have similar properties?" There are the quite complicated formula W(x) + W(y) = W(xy (1/W(x) + 1/W(y)), which holds for x, y > 0, and W(nx^n/W(x)^(n-1)) = nW(x) for n, x > 0.
It's still possible for have positive p and ending up with a real solution. Since any cubic equation f(x)=x^3+px+q has at least one real root due to the Intermediate Value Theorem.
@@blackpenredpen Yes I understand that. I meant for this formula specifically shown in the video, does it only work with a negative "p" value? For example if the equation was something like f(x)=x^3+4x+5, then p = 4 and when you insert p=4 into the formula used, you will have sqrt(-4/3) and also sqrt(-3/4) which will have no real solutions.
What happens when the absolute value of that stuff inside cos^-1 expression is greater than 1? This formula only makes sense for specific values of p and q.
I was trying to solve a cubic equation, then I got curious about the inverse of the middle part. Wolfram Alpha magically gave the sin-arcsin result. I tried to search for it to learn more but found nothing. Seems like I searched for the wrong thing all along. Should've just read the Wikipedia article for the solving method.
I was watching your video "th-cam.com/video/-EDdRZcgLa4/w-d-xo.html - solving equations but they get increasingly awesome" and wondered if it would be possible to solve the equation x^6+x^5+x^4+x^3+x^2+x+1=0. I came up with this solution: x^3+x^2+x+1+1/x+1/x^2+1/x^3 (/x^3) (x+1/x)^3+(x+1/x)^2+(x+1/x)-3x-3/x-2+1=0 (x+1/x)^3+(x+1/x)^2+(x+1/x)-3(x+1/x)-1=0 (x+1/x)^3+(x+1/x)^2-2(x+1/x)-1=0. Then substitution (x+1/x)=y and y^3+y^2-2y-1=0 - cubic equation and I don't know what to do with it (maybe cubic formula). Maybe this solution is wrong, but if you find one, please put it on TH-cam.
![แค่บอก - Atom ชนกันต์ [OFFICIAL MV]](http://i.ytimg.com/vi/kgwhtg8bAxY/mqdefault.jpg)
So you want to prove the cubic formula: th-cam.com/video/ULsPhWmqhyc/w-d-xo.html
proof for the quartic when?
@@bain8renn
I have a proof for it but it's too big to fit in my mind.
When you've forgotten the formula but need the score and creat new formula
Lmao i do this all the time. It wastes soo much of my exam time tho
Short answer: I did. I had a friend who was programming a massive computer to instruct a massive laser cutting machine to cut pieces of shoe leather before stiching them together into the final shoe. For the curved cuts, they approximated by cubic equations and hence needed their intersection points to know when to stop the cuts. Therefore they needed to equate the 2 cubic equations and then find the roots of the single cubic in order to know when to stop cutting. They were using was Newton's Method to find the roots. Problem was, in being able to find a reasonable x0 to start the iteration. As they had no picture, they really had no idea how to pick a resonable x0. Remember they were only dealing with small portions of each curved cut and as such needed to find only 1 specific real root. Newtion's Method was often generating the wrong real root or if x0 happened to be close to a stationary point then the computations went arwy.
Enter the cubic equation. I wrote a C program which solved the cubic (using cases on d) but ran into a different intermittent problem - the same problem that Cardano encountered. That is, when d = b^2/4 + a^3/27 was negative ===> a is negative and sometimes produced a negative value for d. So when d < 0 I went over to the triple angle cosine formula, as you did, and my problem was solved. Now I probably could have used some features of the C programming language to do Complex Number Arithmetic but this seemed like an easy fix so I tried it. As arc_cosine in C returns the k = 0 value, just add 2pi/3 and 4pi/3 to the angle to get all solutions from 0 to 2pi. Then back substitute.
Also remember, they were not interested in exact solutions but floating point values to pretty much any degree of accuracy desired. Anyway the whole thing worked like a charm. I'll leave out (1) the details of single, double and triple roots and how to sort out the different cases and (2) how to deal with d = 0, 0 in floating point arithmetic, but suffice it to say the whole process made my day. Who would have thought??? So cool.
I guess I could also add that (1) d 0 yields 1 single root. And that a cubic can ONLY have (1) 3 real - 0 complex or (2) 1 real - 2 complex roots.
Wow, that is really cool. The more I learn about cubic equations, the more fascinated I get.
that's crazy stuff man , im really interested in it!! do u have it on github?
Interestingly, the cosine substitution works precisely when the usual cubic formula "fails" (involving cube roots of complex numbers). If you write the numbers in the cubic formula as complex exponentials, then the cosines fall out of that.
In this situation, the cubic equation has 3 distinct real roots.
I have used this for programming a solver for cubic equations of state in chemical engineering thermodynamics
Really nice way to solve it!
It's good to see different kinds of solutions.
I love the t-shirt (الجبر it's an arabic word means Algebra)
Math with bprp is love❤❤
And a fun thing, since you know one of these is 4+0i, you can get an exact value of the inverse cosine of the expression based on choosing the root corresponding to x=4+0i
The issue with the cubic formula is that, besides giving complicated expressions that turn out to be equal to simple values, there are other methods that can sometimes find said values much faster. The rational root theorem (or simple inspection) would lead to the value x=2 in the original problem; the polynomial could then be factored into (x-2)(x²+2x-24), which can then be further factored into (x-2)(x-4)(x+6), yielding all three solutions.
How does this guarantee that the expression inside the inverse cos is between -1 and 1?
you don't. if the argument is not a real number between -1 and 1, you will get a complex solution for the inverse cos and in at least one case for n, the cos will cancel all the irrationality and give a real result.
@kpaasial.
That is a good question. I don't have the space here, but the arugument of the inverse cosine consists of a ratio involving b^2/4 and a^3/27 which are the 2 terms in the cubic discriminant d = b^2/4 + a^3/27. I know that's not much to go on but I think you can get to your -1 to 1 interval from there. I'll see what I can come up with.
@kpaasial.
Got it. Take a look at the 2 cases for d = b^2/4 +a^3/27.
i.e. when b^2/4 + a^3/27 0
Now check the argument of the inverse cosine in each case. You will find it trapped bewteen -1 and +1.
Very good! This solution method is also given in the "CRC Standard Math Tables", page 10 in the 27th edition. I've never used it, and can't, off hand, think of when it would be better than the method described on page 9. Perhaps in the days before computers, when it would be easier to find tables of cosines than tables of cube roots?
You didn't say (but it matters) use this when there are three real roots. And the reason it is useful is that Cardano's Method is quite a mess unless your cubic has only one real root. Writing the (depressed) cubic in Vieta's Form x³ = 3Ax + 2B allows you to dispense with fractions in both Cardano and this form. Now you might wonder "what cubic is solved if you use the sines for this angle?". The cubic which is cognate to the one using cosines, namely x³ = 3Ax - 2√(A³ - B²). For your exemplar this works out to x³ = 3(28/3)x + 2(80/(3√3)) if I escaped blunder. But (error or not) the point is the trig substitutions give a lot back for your effort (getting to theta).
In case of three real roots, the standard cubic formula (with its two cube roots) gives exactly the same answer as the trig version. Problem is that the real numbers are, so to speak, "encrypted" in the corners of several overlapping triangles inside of the complex plane. The approach shown may therefore be interpreted as a shortcut for the direct access of those purely real numbers.
from Morocco thank you very much
simply amazing!!
Maybe add an "o" in trigonmetry. No, I shouldn't be like that. Nice video anyway.
Thanks tho. I just fixed the title.
All these depressed cubics with 0 as the coefficient of x^2 have roots of the form -2m, m+ n and m - n. You get the expression x^3 - (3m^2 - n^2)x + 2m(m^2 - n^2).
Our 2nd sem will start next week. I already watched your 50 integrals video and currently watching the 100 integrals part 1.
I wont watch this video for now, I just want to say, thank you very much!!!!
Could also use the Rational Root Theorem finding factors of p/q or 48 / 1. Roots -6, 4 and 2 work!
Beautiful problem 🚀👌
Finally it's blackpenredpen and not blackpixelredpixel
please make a video on how to find sin(10)
in your old video you did not tell us how to approximate the complex expression
There isn’t really any way to simplify the complex expressions from the original video.
According to galois theory, if all three roots to a cubic equation are real but irrational, the roots cannot be expressed exactly using any finite radical expression with real numbers only.
So you can only write the roots with complex numbers, or an infinite radical.
If you want to actually calculate/approximate the root numerically, just use wolframalpha or any calculator that can do roots of complex numbers.
I don't see here cube roots. Weather it mean that cube equations can be solved with compass and ruler? May be angle trisection too?😁
I have 2 questions about the Lambert W function; can you please answer them?
1. Why is the Lambert W function only in base e? Why can’t we have, say, W base 3 or base 10 or base pi (or something like that?
2. Logartithms have special properties like the product property (log(a*b) = log(a) + log(b)), expontential property (log(a^b) = b*log(a)), etc. Does the Lambert W function have similar properties?
Im not him, but I will try to answer it, perhaps there is more that I dont know
1. I think this is for definition, the person who created it, had the objective of creating an inverse function, the inverse of x*e^x, but later was generalised to all the functions who had the same structure
2. I dont think so
"Why is the Lambert W function only in base e? Why can’t we have, say, W base 3 or base 10 or base pi (or something like that?"
Well, it's quite easy to solve exponential equations with other bases using W, so why bother using additional functions for other bases?
"Does the Lambert W function have similar properties?"
There are the quite complicated formula W(x) + W(y) = W(xy (1/W(x) + 1/W(y)), which holds for x, y > 0, and W(nx^n/W(x)^(n-1)) = nW(x) for n, x > 0.
For real solutions, does this mean that "p" needs to be negative to work since there are two square roots with a negative sign and "p" in it?
It's still possible for have positive p and ending up with a real solution. Since any cubic equation f(x)=x^3+px+q has at least one real root due to the Intermediate Value Theorem.
@@blackpenredpen Yes I understand that. I meant for this formula specifically shown in the video, does it only work with a negative "p" value? For example if the equation was something like f(x)=x^3+4x+5, then p = 4 and when you insert p=4 into the formula used, you will have sqrt(-4/3) and also sqrt(-3/4) which will have no real solutions.
What happens when the absolute value of that stuff inside cos^-1 expression is greater than 1? This formula only makes sense for specific values of p and q.
For a test woildnt you need to prove the thing for the 3 cos thingy too)?
I was trying to solve a cubic equation, then I got curious about the inverse of the middle part. Wolfram Alpha magically gave the sin-arcsin result. I tried to search for it to learn more but found nothing. Seems like I searched for the wrong thing all along. Should've just read the Wikipedia article for the solving method.
If we want to stay in real numbers this formula solves casus irreducibilis
I use it when I would like to avoid complex while solving cubic equations
how about we take x = usin(thetha)???
you can do that too, then you just need to use the triple angle formula for sin: sin(3θ)=3sin(θ)-4sin³(θ)
I literally watched the cubic formula videos before this video was uploaded
What's the catch?
I got x={-6,2,4} for my solution set using RRT and long division.
How about if p>0...
Thats ALOT OF PENS!!! Money well spent sir
read Tranter's "Techniques of Mathematical Analysis", pp.132-133.
what a name in your t shirt !!!!
algebra in arabic
الجبر shirt ♥♥♥♥
Asiáticos são muito inteligentes mesmo.
Cheese
I didn't even need to use Wikipedia.
Do with the quartic formula or the quintic
there exists no quintic formula and this is impossible to do with the quartic
@ggaattyy2 ok i forgot that the quintic formula doesnt exist
This is the best that I can offer: th-cam.com/video/L8W684pCHyc/w-d-xo.html
I also discovered it earlier.
I was watching your video "th-cam.com/video/-EDdRZcgLa4/w-d-xo.html - solving equations but they get increasingly awesome" and wondered if it would be possible to solve the equation x^6+x^5+x^4+x^3+x^2+x+1=0. I came up with this solution:
x^3+x^2+x+1+1/x+1/x^2+1/x^3 (/x^3)
(x+1/x)^3+(x+1/x)^2+(x+1/x)-3x-3/x-2+1=0
(x+1/x)^3+(x+1/x)^2+(x+1/x)-3(x+1/x)-1=0
(x+1/x)^3+(x+1/x)^2-2(x+1/x)-1=0. Then substitution (x+1/x)=y and
y^3+y^2-2y-1=0 - cubic equation and I don't know what to do with it (maybe cubic formula). Maybe this solution is wrong, but if you find one, please put it on TH-cam.