I love how you explain more than one method for the same problem
@SyberMath
3 жыл бұрын
Oh, thanks!
@holyshit922
3 жыл бұрын
This can be solved using trigonometry using triple angle formula for cosine or sine cos(3t)=4cos^3t-3cost
@SyberMath
3 жыл бұрын
Absolutely!
@leif1075
2 жыл бұрын
@@SyberMath why did you question whether it's a polynomial when it obviously is?
@leif1075
2 жыл бұрын
@@SyberMath woukdnt you just make it 2x cubed minus 1 for a and b since both are cubes instead kf making y just 2x?
@pmccarthy001
3 жыл бұрын
Wow! I don't think I've ever seen someone approach the cubic from quite that perspective before. You display quite a bit of depth of understanding of this material.
@SyberMath
3 жыл бұрын
Wow, thanks!
@snejpu2508
3 жыл бұрын
Random, not connected observation: if you start from 1 and then add 1 and reverse the fraction, you will get 1/2. If you continue adding 1 and reversing a fraction, you will generate Fibonacci numbers in the denominators. That surprised me, because I've never known this.
@Qermaq
3 жыл бұрын
Not only that, upon iteration of x = 1/(x+1), the fraction itself tends to (sqrt(5)-1)/2, which is the limit of dividing F(n) by F(n+1). You can determine sqrt(5) slowly but accurately by simply doing what you were doing for a very long time.
@user-mx8sj1nc6v
3 жыл бұрын
From the subject "relationship between a function and its derivetives" I know that a three degree polinom f(x)=ax^3 .... if a ia positive, the graph is increasing-decreasing-increasing or only increasing. In this case it has two values that zero the derivetive. So -2 is minimum and 2 is max. So the roots are above 2 , below -2 and between 0 and 1. But since the roots are not "nice" numbers, the suggested way of solving is very good.
@SyberMath
3 жыл бұрын
Thank you!
@n_digit_flow_9621
3 жыл бұрын
Awesome sir, I really impressed by this channel
@SyberMath
3 жыл бұрын
Thank you!
@hsjkdsgd
3 жыл бұрын
I, by my intuition, proceeded with the 2nd method. Finished it in style!
@SyberMath
3 жыл бұрын
Nice!
@timetraveller2818
3 жыл бұрын
funny how i just learned cardano's method this morning
@SyberMath
3 жыл бұрын
No way! 😁
@leif1075
2 жыл бұрын
@@SyberMath but if someone didnt know cardanos formula i dont see how or why anyone would or could derive it even Cardano even me..don't you agree?
@danielsittner7570
3 жыл бұрын
I've got a cool question you (and anyone who reads this for that matter) can solve. There is an equilateral triangle with side lengths of 10 cm. On each vertex there is a turtle such that turtle 1 always looks at turtle 2, turtle 2 always looks at 3 and so on. They all start to move simultaneously in a speed of 10 cm per second while maintaining their "target" turtle in their sight. The turtles are dots. 1. How long will it take for them to meet and how long will they have to walk? 2. How many rounds will they go around the center of the equilateral triangle they started on?
@SyberMath
3 жыл бұрын
I'm sharing this problem in the community tab. You can follow up there: kzitem.info/rock/W4czokv40JYR-w7u6aXZ3gcommunity
@happiness2477
3 жыл бұрын
i just found your channel yesterday and i am loving the content :D By the way one question for you *Find the sum of all integers n, such that 1 ≤ n ≤ 1998 and* *that 60 divides* *n³ - 30n² + 100n*
@gaussianPsycho
3 жыл бұрын
2nd method is just beautiful
@SyberMath
3 жыл бұрын
Thank you so much 😊
@UttamKumar-kt4jp
3 жыл бұрын
Both the methods were equally good .just loved it.
@SyberMath
3 жыл бұрын
Glad you liked it!
@usernamehere94
3 жыл бұрын
Very interesting, where did you discover this method?
@vanshr.sachan264
3 жыл бұрын
For the second method don't we have to go with the assumption that the value of x at which the function 8x³ - 6x -1 is zero is between -1 and 1 (as cos fucntions range is from -1 to 1) Am I missing something... Pls let me know...
@angelmendez-rivera351
3 жыл бұрын
You do not need to make the assumption, since it is a provable fact that cos(3·α) = 4·cos(α)^3 - 3·cos(α) for every real α is true. This is just the triple-angle formula, which can be proven from the sum-to-product formula of the cosine function, which, if you will, is the functional equation that defines the cosine function to begin with.
@MarcoMate87
3 жыл бұрын
I think you are right, this fact has to be checked before continuing if we are looking for real solutions. One has to check that at least one real solution belongs to the interval [-1,1]. In fact, if for example, we had 8x³ - 6x - 4 = 0 instead of 8x³ - 6x -1 = 0, the triple angle technique doesn't work apparently. In that case, we have 4x³ - 3x = 2; if we use cos(3·α) = 4·cos(α)^3 - 3·cos(α) and x = cos(α) we obtain cos(3·α) = 2 which is impossible over the real numbers. 8x³ - 6x - 4 = 0 has one real solution and two complex conjugate solutions, and the real solution is greater than 1. In our case we can apply the zeros theorem for continuous functions for f(x) = 8x³ - 6x -1. For example f(0) = -1, f(1) = 1, so the function has (at least, in this case exactly) one real zero in the interval (0, 1).
@NadiehFan
2 жыл бұрын
This video seems to have given rise to some misunderstandings regarding the trigonometric solution of cubic equations. It is definitely NOT true that one or all of the real solutions of the cubic must be on the interval [−1, 1] in order for the method to work. ANY cubic which has only real solutions can be solved trigonometrically. In fact, the method named after Cardano (but first discovered by Del Ferro) to solve a reduced cubic (that is, a cubic that lacks the quadratic term) will - paradoxically - result in cube roots of *complex* numbers if and only if all three solutions of the cubic are *real* and distinct. This is not properly explained in the video. Any attempt to find these cube roots of complex numbers algebraically leads to - guess what - a cubic equation which is equivalent to the original equation you were trying to solve in the first place. So, it seemed to early algebraists they were running around in circles (no trigonometric pun intended). That's why they named this case the casus irreducibilis (the irreducible case). Somewhat later, towards the end of the 16th century, Vieta came up with a method to solve cubics with only real roots trigonometrically. Here's a modern presentation of his method. Let's assume we have a reduced cubic z³ + pz + q = 0 where p and q are real numbers. The method named after Cardano generates cube roots of complex numbers whenever (q/2)² + (p/3)³ < 0 and this is precisely the condition for the equation to have three distinct real roots. So, how do we use the triple angle identity cos 3φ = 4∙cos³φ − 3∙cos φ to solve a reduced cubic equation? The idea is to first rewrite our cubic equation as z³ + pz = −q and then substitute z = r∙cos φ This will give us r³∙cos³φ + p∙r∙cos φ = −q Now obviously we want to get 4∙cos³φ rather than r³∙cos³φ at the left hand side so we divide both members by r³ and multiply both members by 4 to get 4∙cos³φ + (4p/r²)∙cos φ = −4q/r³ If we compare this with 4∙cos³φ − 3∙cos φ we can see that all we need to do now to get the desired expression at the left hand side is to choose r in such a way that 4p/r² = −3 and therefore we can choose r = 2·√(−p/3) Note that this value of r is real and positive, since (q/2)² + (p/3)³ < 0 implies p < 0. With this choice of r, we have 4∙cos³φ − 3∙cos φ = −½q/√((−p/3)³) and therefore cos 3φ = −½q/√((−p/3)³) From this, we can get a value of 3φ on the interval [0, π] and therefore a value of φ on the interval [0, ⅓π] which satisfies this equation. But we can add any integer multiple of 2π to 3φ and therefore any integer multiple of 2π/3 to φ, and since z = r∙cos φ with r = 2·√(−p/3) the roots of the cubic equation are therefore given by z = 2·√(−p/3)·cos(φ + 2kπ/3) where we can take any three consecutive integer values of k (e.g. k = 0, 1, 2) to obtain all three roots. Note that the condition (q/2)² + (p/3)³ < 0 implies that the absolute value of −½q/√((−p/3)³) is smaller than 1, which means that the trigonometric solution of the reduced cubic equation z³ + pz + q = 0 is always possible if this equation has three distinct real roots.
@LOL-gn7kv
3 жыл бұрын
Great video ! Another way could be using cardan's method of solving cubics as it is already in the form of a reduced cubic
@SyberMath
3 жыл бұрын
Wasn't this Cardano's method? 🤔
@angelmendez-rivera351
3 жыл бұрын
Method #1 is literally Cardano's method.
@SyberMath
3 жыл бұрын
Thanks for the confirmation, Angel!
@CaradhrasAiguo49
Жыл бұрын
9:25 or note that the angles from 5 * pi/9 + n * 2pi/3 are 2*pi-compliments of the 3 angles found from pi/9 + n*2*pi/3, e.g. 5 * pi / 9 + 13 * pi / 9 = 2 * pi and so on, and cos(2 * pi - theta) = cos(theta) for all real theta
@tonyhaddad1394
3 жыл бұрын
Good job !!! Very helpful video thanks !!!
@SyberMath
3 жыл бұрын
Glad it was helpful!
@caeoranger
3 жыл бұрын
Hi at 2:38 you gave a+b=2x, if that is so then a^3 +b^3 should be in terms of x as well (should not equal 1)
@SyberMath
3 жыл бұрын
The goal is to get a reduced cubic in a+b
@caeoranger
3 жыл бұрын
@@SyberMath but the substitution is incorrect
@vishalmishra3046
3 жыл бұрын
*General Formula* Transform equation to x^3 = 3mx + 2n. If m > 0 and m^3 > n^2 then x = 2√m Cos T/3 where Cos T = n/m^(3/2). Let's apply the general formula to this problem. x^3 = 3(1/4)x + 2(1/16). So, m = 1/4 and n = 1/16. Applying pre-condition, m > 0 and m^3 = 1/64 > 1/256 = n^2, so the formula applies in this equation. Cos T = n / m^(3/2) = (1/16) / (1/8) = 1/2 = Cos 60. Therefore, x = 2√m Cos T/3 = 2 √(1/4) Cos 60/3 = 1 x Cos 20 = Cos 20. Add/Subtract 120 for the other 2 real but negative solutions, -Cos 40 and -Cos 80, or simply x = Sin (70, -10, -50).
@SyberMath
3 жыл бұрын
Interesting
@GenZClub
3 жыл бұрын
May I ask what app you use to draw?
@SyberMath
3 жыл бұрын
Of course! I use Notability
@user-mx8sj1nc6v
3 жыл бұрын
Obviously, if it had rational roots there were 1 , -1 , 1/8 , -1/8 , 1/4 , -1/4 , 1/2 , -1/2
@SyberMath
3 жыл бұрын
Absolutely!
@user-ow8hy8cv6b
3 жыл бұрын
I actually did it with the second method you showed and seconds before the end of the video I was ready to comment about the other solutions from the trigonometric equation. It's important to wait till the end and listen carefully 😂
@SyberMath
3 жыл бұрын
Great job!
@That_One_Guy...
3 жыл бұрын
1) Is the first method just to get one imaginary root or maybe 2-3 ?? 2) is second method applicable to all cubic ??
@angelmendez-rivera351
3 жыл бұрын
The first method does not give any imaginary or complex roots. All the solutions are real numbers, even though they are expressed using complex numbers in radical symbols.
@SyberMath
3 жыл бұрын
Should be applicable
@That_One_Guy...
3 жыл бұрын
@@angelmendez-rivera351 that's weird, does that mean we somehow only take the real part to get the real root ? I don't know how you convert complex root into real root
@angelmendez-rivera351
3 жыл бұрын
@@That_One_Guy... You take the principal cube root, and if you do the mathematics correctly, you do just get a real number.
@user-mx8sj1nc6v
3 жыл бұрын
Please let me say, about your first solution, that there is a CARDON's Method where you put x=u+v , and you build a quadratic equation that its roots are u^3 and v^3 , and then you can find 3 values for u and 3 for v (including complex) and then you add them in pairs to find 3 values of x. Here is a link to an example. ( He devides polinoms, but this is not nacesery ). kzitem.info/news/bejne/zqmEq4muqYWAbKw
@SyberMath
3 жыл бұрын
Yess!
@agnibeshbasu3089
3 жыл бұрын
Nice video, but can you bring a series on USAMO problems?
@SyberMath
3 жыл бұрын
They would be beyond my reach! Sorry!
@agnibeshbasu3089
3 жыл бұрын
@@SyberMath ok no problem...
@HasanGun
3 жыл бұрын
Good solution!👏
@SyberMath
3 жыл бұрын
Thank you!
@christopherrice4360
3 жыл бұрын
WHOA i have never seen a polynomial solved using methods of substitution before! Very intriguing to use advanced methods on simpler math problems🧐 hmmm... can this "method" of math tools grow up and go beyond polynomials or is there a limit on how much substitution can apply to in mathematics?
@SyberMath
3 жыл бұрын
Glad you enjoyed it!
@sahilsinghbhandari444
3 жыл бұрын
Nice question and nice method for solving through trigonometry👍👍
@SyberMath
3 жыл бұрын
Thank you!
@tahmidrahaman772
3 жыл бұрын
Can you plz solve this in your next video: (x-5)(x-6)=25/(24)² Anyway great video!
@SyberMath
3 жыл бұрын
Thank you!
@aryanmukherjee5679
3 жыл бұрын
Very Nice Video
@SyberMath
3 жыл бұрын
Thank you!
@georget8008
2 жыл бұрын
in the 2nd method, by setting x=Cosa, don't we restrict our solutions to the domain [-1,1]? Don't we ignore any other solutions outside this interval?
@SyberMath
2 жыл бұрын
We don't because this cubic has three solutions and we find all of them. If there were no real solutions for x on the interval [-1,1], then cos(3alpha) would not give us a solution on that interval. Consider 8x^3-6x-2=0. The real solution will be greater than 1!
@romeisbig6485
2 жыл бұрын
9:10 I didn't understand this part. How did 5pi/3 come there?
@SyberMath
2 жыл бұрын
cos(5pi/3)=cos(pi/3)
@CengTolga
3 жыл бұрын
How do we know which n values to pick for the cubic roots? Why can it not be n = 5, n = -2, etc?
@SyberMath
3 жыл бұрын
What is n?
@aptilious2774
3 жыл бұрын
Great content 😁
@SyberMath
3 жыл бұрын
Glad you enjoyed it!
@gauravsmatharu
3 жыл бұрын
Nice video 👍
@SyberMath
3 жыл бұрын
Thanks 👍
@damiennortier8942
2 жыл бұрын
I have all my time to see yours videos ;-)
@SyberMath
2 жыл бұрын
💖
@damiennortier8942
2 жыл бұрын
@@SyberMath but just wen I enjoy the videos 😅❤️
@adityaghosh2844
3 жыл бұрын
The second method is great.....👍🙏.....
@SyberMath
3 жыл бұрын
Glad you think so!
@saralsubodhgyanbysubodhsir
3 жыл бұрын
Can this second method work for any cubic equation or is it some kind of exception or something. Please can anybody tell me that🙏🙏🙏
@SyberMath
3 жыл бұрын
It does! It's Cardano's method
@mariomestre7490
3 жыл бұрын
Genial, gràcies
@SyberMath
3 жыл бұрын
You're welcome! Thanks!
@tmacchant
2 жыл бұрын
My question is case of 1/2=cos(5pi/3).
@SyberMath
2 жыл бұрын
cos(5pi/3)=cos(pi/3)
@SnapSnippetsHQ
3 жыл бұрын
What's your age narrator??
@SyberMath
3 жыл бұрын
You need to make a guess but I cannot answer your question since it's personal info! 😁
@SnapSnippetsHQ
3 жыл бұрын
SyberMath 57 years 5months 16 days 5 hours 16 seconds at the time of replying , thats my guess, and your homework is to find the probability that it is correct
@Qermaq
3 жыл бұрын
@@SnapSnippetsHQ By the age of 85, most people in the world are dead. ;) When people ask my age sometimes I say "Of Enlightenment".
@SnapSnippetsHQ
3 жыл бұрын
@@Qermaq ok now I edited it
@Total_Syntheses
3 жыл бұрын
아 뭔가 익숙한 꼴이다 싶었는데 cos3x..
@barakathaider6333
2 жыл бұрын
👍
@MelomaniacEarth
3 жыл бұрын
I am damn sure u got sweated 😥after making this vdo😅....it has lot of substitutions 🤐
@SyberMath
3 жыл бұрын
I love substitution! 😁
@sabe607
3 жыл бұрын
For the very end, instead of realizing you’re gonna get the same values, just realize a cubic has only 3 roots
@SyberMath
3 жыл бұрын
That's right but it's also good to see the repetition, I think
@kavilwagh
3 жыл бұрын
I wish I did maths from 1st std
@user-lr8od4uz1n
3 жыл бұрын
와우
@SyberMath
3 жыл бұрын
Thank you! Glad you liked it!
@FarhadJahani
3 жыл бұрын
awful methods
@SyberMath
3 жыл бұрын
Why?
@GodbornNoven
2 жыл бұрын
x≈pi/3.3421
@sonalichakraborty6830
3 жыл бұрын
Nice video 👍👍
@SyberMath
3 жыл бұрын
Thank you 👍
@sonalichakraborty6830
3 жыл бұрын
@@SyberMath u're welcome 😊
@sonalichakraborty6830
3 жыл бұрын
Can I suggest problems?
@sonalichakraborty6830
3 жыл бұрын
I see that u r more interested about algebra stuff....do u mind doing some videos on combinatorics or number theory or elementary calculus or other maths contest topics?
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