Killer Problem With A Golden Answer
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- Опубліковано 3 лис 2024
- Can you solve this "coffin" problem? Thanks to Rahul for the suggestion! Special thanks this month to: Kyle, Mike Robertson, Michael Anvari. Thanks to all supporters on Patreon! / mindyourdecisions
Adapted from problem 4 of Jewish Problems by Tanya Khovanova, Alexey Radul
arxiv.org/abs/...
Coffin Problem ft. blackpenredpen: Cube and Cube Root (Mu Prime Math)
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When you mention 'Golden' in title. I always feel The Answer is gonna be Golden Ratio.
Edited: Indeed it was!
Yes
that is indeed the answer!
Me too
I predicted the same outcome (except -1)
And yes....the ans was the golden ratio as i expected it to be!!!!
Holy moly smartass kumble! thanks a lot whenever I hear "golden", I will think of the golden ratio!
Hi,can you watch the this trigonometri video ua-cam.com/video/dfTM2YKxg1s/v-deo.html🤛
Me : abc formula
Normal People: Quadratic Formula
Presh: BRAHMAGUPTA
It's actually Shree Dharacharya Formula and not Brahmagupta
Yeah it is actually shri Dharachrya formmula
He wants to recognise contemporary mathematician
inferiority complex : he wants to show that indians are good in math (to compare with Euler, Gauss, Hamilton, Bernouli, Laplace, Lagrange, Fourier, Euclide, Pascal, Pythagore, Descartes etc...
Hi,can you watch the this trigonometri video ua-cam.com/video/dfTM2YKxg1s/v-deo.html🤛
Instead of just trying "convenient values", it's quicker to use the rational root theorem. We only need to test +1 and -1.
What is rational root theorem?
@@madhukushwaha4578 is the channel's name "your math and physics guy"?
@@dhvanilgheewala8036 In a polynomial equation if you find the factors of the leading coefficient and the constant and then divide the factors of the constant by the factors of the leading coefficient (plus or minus) then you will find all possible rational roots. Other roots may exist but if a root is rational it can be found this way.
Example.
(3x^3 + 4x^2 - x + 6)
Lead coefficient = 3
Factors of 3 = (1,3)
Constant = 6
Factors of 6 = (1,2,3,6)
I randomly chose this polynomial so I do not know if it has any rational roots however if they exist they will be
(Plus or minus)
1/1=(+1,-1)
2/1=(+2,-2)
3/1=(+3,-3)
6/1=(+6,-6)
&
1/3=(+1/3,-1/3)
2/3=(+2/3,-2/3)
3/3=(+1,-1)
6/3=(+3,-3)
Some obviously came up more than once but in that list of plus and minus are all possible rational roots. If you plug them in you will either find a zero or it could be all the roots are irrational.
So in some cases it pays to use this method if there are only a few factors but with larger numbers come longer lists. In the case of this video the only possibly options were (+1,-1) which is why checking zero was a waste... it was impossible to begin with according to the rational root theorem.
@@MegaMisch tq
And also integral root theorem
I can't believe there was a time when i could solve all this easily
bcz then u were practicing for olympiads
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@@madhukushwaha4578 rehne do
Do you think you not solve it now then you are wrong, people think that mathematics is hard. But mathematicians think mathematics is enjoyable and easy.
Be a Mathematician.
I solved it on thumbnail you can also do it.
Manish, no there wasn't. Your ego is fooling your memory.
I was understanding everything beautifully and then 'f' happened😭
ua-cam.com/video/OmSIcFQ3el4/v-deo.html
F
No u probably understand it's just that f stands for function
In fact, you only prove that if the equation have solutions they have to be on the y=x line, but you should check the answer you get in order to be rigorous
Yeah you're right.
@@djfranz1 all values are verifying brother ,, do it urslf
I learned some equation problems that involve a function and its inverse, so that experience helps me so much in this problem. Similar to what Presh presents in this video, we obtain a function that has inverse. We then want to solve a function that equals to its inverse. Since the graph of an inverse is the reflection of its original function with respect to line y=x, we can simply solve an equation of that function equals to x. We technically work on different equation but they have same solutions.
LHS : Cube me I am ready... Fast fast what are u thinking...
RHS : In coma!
😂
It took me some time to understand😂😂
@@madhukushwaha4578 Thank u.. u helped a 'math for fun' guy... U should visit 'blackpenredpen' now
@@madhukushwaha4578 hey why are you promoting other channels here.
Hey I didn't understand the joke.. Can anyone explain it?
Eyes:I have learned it.
Brain:No you haven't.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
That's not Brahmgupta's Equation
That is Shreedhar Acharya relation
It’s the quadratic formula, and it works. It doesn’t matter who made it.
I mean sure, but proper attribution is still important.
@@DepFromDiscord Ever heard of credit or copyright
@@DepFromDiscord it matters who made it, they're great people who served humanity with their effort and ingenuity, it's our responsibility to remember and appreciate them
Bramhagupta was the teacher of Sri dhar acharya. Even Sri dhar acharya admitted that his teacher was the founder of the formula. But as we have no proof about it and as it was first seen under the name of sridhar acharya in a sanskrit book. We call it sridhar acharya formula. But actually Bramhagupta was the founder of the formula.
Divide by 2 on both sides, LHS and RHS are inverse functions of each other. Can then directly substitute (x^3-1)/2 = x and then see that -1 is a root of the cubic and factorize and get the required answer!
ua-cam.com/video/OmSIcFQ3el4/v-deo.html
@@amalwijenayaka410 (d) 1
@@subhadeeproy ua-cam.com/video/t3I3rrS0L3A/v-deo.html
@@amalwijenayaka410 star:25 triangle:20 heart:2 circle:7
I am a huge fan of you. My birthday is on 4th December! I am very excited about it as after 4 long years all of our family is finally going to be together in one day, after soo many family wars and problems. I just can't wait!!!
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
Used to do this much in test of algebra chapter for 5 marks. Teacher never gave full marks.
In Maths, u r supposed to get full marks if write the correct answer. Not the method u use
@@whatdidyousay1235 step marking hoti thi. Aur agar teacher ka method use ni kiya to its suppose to get less marks than the student who just memorise teacher's solution instead of solving.
@@binga4026 probably because you didn’t do the problem right
@@SauceGodGaming umm that wasn't the case. They were like 'you didn't used my method so you dontget full marks' kind of attitude. Its ok... i got to explore more methods for specific problems.
There are basically 2 types of tests:
- finding by yourself a way to solve a problem
- prove that you have mastered the method previously learned in solving the problem
Your test was the second type.
It is quite easy to see that (x^3-1)/2 is the inverse function of 3√(2x+1)[ note that it is the cube root] . Now we know that f(x) and it's inverse if meets they meet at the line y=x, so we can just write (x^3-1)/2=x and thus we have obtained x^3-2x-1=0 which we can easily solve out .
The novel solution, especially the derivation of symmetric equations, was very interesting. Thank you very much.
From the symmetry of the formula, how about the following calculation method?
1/2 (x ^ 3-1) = y… ①
1/2 (y ^ 3-1) = x… ②
Calculating ①－②, 1/2 (x ^ 3-y ^ 3) + (x-y) = 0
(x-y) (x ^ 2 + xy + y ^ 2) + 2 (x-y) = 0
(x-y) (x ^ 2 + xy + y ^ 2 + 2) = 0
x-y = 0 or x ^ 2 + xy + y ^ 2 + 2 = 0, but from x ^ 2 + xy + y ^ 2=(x+1/2y)^2+3/4y^2≧ 0,
x ^ 2 + xy + y ^ 2 + 2> 0. .. Therefore, x-y = 0, that is, x = y. The rest is the same as the explanation.
Fabulous! I can't understand the method shown in the video, and I saw your comment and understood really well. You are really a genius!
By the way,
Can you please explain how you deduced that at last, x²+xy+y²+2 > 0. I can't understand that last part, so if you can explain it more, it would be really helpful for me😊.
@@lexus_bkl Because as he shows in his second to last sentence [x ^ 2 + xy + y ^ 2= (x+1/2y)^2 + 3/4y^2] x²+xy+y²+2 is the sum of 2 squares.
@miguel angel Gregorio I think that the complete number of solution is 9, but only 3 of them are real (which is required here).
A different way to solve it: notice that if x is a solution then ∛(2x+1) is also a solution. You can spot this if you rearrange the equation in a way that leaves only x on the right hand side. The equation cna be reduced to a polynomial of degree 9 so it can't have more than 9 solutions. ∛(2x+1) is a strictly increasing function so if x was either smaller or bigger than ∛(2x+1) we would have an infinite number of solutions. Therefore ∛(2x+1)=x. From here, we proceed the same way as seen in the video. Also note that you don't have to formally divide the third degree polynomial by (x+1), you can conveniently just rearrange it: x³-2x+1=(x³+1)-2x-2=(x+1)(x²-x+1)-2(x+1)=(x+1)(x²-x-1).
This is a famous coffin problem or a next of kin. You never forget the inverse function trick, it is so amazing.
Instead of using contradiction we could have divided two on both sides and show that the left-hand side is the inverse off of the right hand side. By definition, the left-hand side is a reflection of the right hand side across the line Y equals X. Hence we can just replace the left inside with X.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
x^3 -2x -1 = 0
How can we solve this equation?
Easy! We can simply substitute the coefficients into the cubic formula.
*proceeds to prove the cubic formula*
@0:39 if I was solving this without watching the video I could see myself cubing both sides and ending up with a 9th degree polynomial
But once you see that degree, you terminate this method amd search for another one
9th or 6th ?
Please bring the same type of interesting questions (equations)
How many guessed that the answer is golden ratio
ua-cam.com/video/OmSIcFQ3el4/v-deo.html
There was one other root. It was a cubic problem. the Golden Ratio only takes care of two of the solutions.
If we choose to guess the roots using popular numbers, golden ratio is one of the most popular numbers, so we could guess it directly :) And since we solve by guessing, we can try to guess immediately :) But the trick to simplify the equation was brilliant. By the way, there's a cubic formula that we can use to solve it (derived by Girolamo Cardano in XVI century).
By the way, the title of this video is a hint which number we should guess.
There is a similar question in the book "problems in calculus of 1 variable" by "I A Maron". Awesome book 👍👍👍.
And as always, great presentation by you ❤️❤️❤️
I found a jee aspirant 😂
@@prachi4110 , not aspirant but teacher 😀
@@prachi4110 yes, guessed same
Try this channel... Always amazing contents... very impressive math channel... ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Hi,can you watch the this trigonometri video ua-cam.com/video/dfTM2YKxg1s/v-deo.html🤛
5:16 It's Shridharacharya's Formula not Brahmagupta's.
Sridhara was born two centuries after Brahmagupta died, and the quadratic formula was clearly written out in Brahmasphutasiddhānta.
@@cortexauth4094 The Babylonian equations you're talking about were specific formulas in the special case where a=1 and c
@@cortexauth4094 what problem do you have if someone gets the credit for their work?
Hi,can you watch the this trigonometri video ua-cam.com/video/dfTM2YKxg1s/v-deo.html🤛
It was proven by both of them! just fact is brahmagupta did it earlier
You can go for a short cut method for factorization
*Presh : NO, HOLD MY LONG DIVISION*
Do you mean synthetic division?
The long division was the shortcut. Otherwise you would have to do a cubic formula which are really annoying.
Who the hell would.ever solve it this way..is there any other way..i tried difference of cubes but doesnt work..then this problem is really impossible unless you happen to try graphing and im not sure that's right either because the two graphs are not the same...
@@crobodile He means to say synthetic division
@@leif1075 I'm sure that the person who commented is implying to say that synthetic division is kind of a shortcut for long division
I used to be able to follow this logic. At this time, it went straight over my head
5:17 that is not brahmaguptas ratio, it is sreedharacharya's ratio
True 💯
-1 for Presh, +1 for you
True😂
Indian mathematician beef 😂
Yeah...
since x^3=2*y+1 and y^3=2*x+1, subtract and y^3-x^3=2*(x-y) or (x-y)*(x^2+x*y+y^2+2)=0, but the quantity x^2+x*y+y^2+2 is always >0 so x=y
Exactly!
Just hint for completeness of proof :)
x^2 + x*y + y^2 = (x^2 + y^2 + (x+y)^2)/2 >= 0
In the end instead of doing polynomial division it is much easier to do some simplifications:
Rewrite x^3 -2x -1
As x^3 -x -x -1
Factor x in the first two terms and - on the second ones
x (x^2 -1) - (x+1)
Difference of squares
x(x+1)(x-1) - (x+1)
Factor (x+1) in both terms
(x+1)( x(x-1) - 1)
And then solve the quadratic inside
Actually, I encourage crediting those who derived the formulae first (as far as we know).
Read the description.
You don't need to use the formula at all; you can solve it by "completing the square" (who first used that so we can name it?):
x^2 - x- 1 = 0
》(x - 1/2)^2 - (1/2)^2 - 1 = 0
》(x - 1/2)^2 - 1/4 - 4/4 =0
》(x - 1/2)^2 - 5/4 = 0
》(x - 1/2)^2 = 5/4
》x - 1/2 = +/- sqrt(5)/sqrt(4)
》x = 1/2 +/- sqrt(5)/2
》x = (1 + sqrt(5))/2 or x = (1 - sqrt(5))/2
I'm guessing the answer will be the golden ratio.
Edit: Ay I was right
Lol I thought the same
AN answer is the Golden Ratio. There are two other answers.
I thought that ans. Will be golden ratio
Even before this video is uploaded, hahh now wat!!
Wait, there was a third solution, wasn't there. X=-1
Without his hint I would have only found the obvious x=-1 solution .
I elevated both terms to the 3-power, got a 9 degree polynomium, dived by (x+1), then by (x**2-x-1), and showed that the 6 degree polynomium was always >0.
i tried going this direction as well and got stuck - how did you know to divide by (x^2-x-1)?
For this problem, I prefer manipulating the equations. For example, to prove y=x, simply subtracting x^3-2y -1=0 from y^3-2x-1=0, we have y^3-x^3+2(y-x)=0. An easy factorization gives (y-x)(y^2+xy+x^2+2)=0, and hence, y-x=0, because y^2+xy+x^2+2 > or = 2.
Very good problem!
But it is an other way to fact this:
x^3-2x-1=x^3-x-x-1=x(x^2-1)-(x+1)=x(x-1)(x+1)-(x+1)=(x+1)(x^2-x-1)
In Brazil - and I don't know why - the quadratic formula is credited to Bhaskara (fórmula de Bháskara).
Its actually sridharacharyas formula😂😂😂
Hello Sir.... the formula used in this problem is actually "Shridhar Acharya's formula"....... Anyway your videos are very helpful..... thankyou so much
I watched this in break time of my online classes and that to before maths period 😂😂😂😂
Same, and I have a test about graphics
Try this channel... Always amazing contents... very impressive math channel... ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
then you become math killer ISI of the class ..lol
Bhahmagupta found *area of cyclic quadrilateral* - en.wikipedia.org/wiki/Brahmagupta%27s_formula | *Sridhara Acharya* was the first who found quadratic formula - en.wikipedia.org/wiki/Sridhara
I tried sustitute the solutions with 'golden ratio' and is shy of the real solution,but x=-1 is correct.All the best and keep the good job .A top level;-)
Whenever someone says..incredible answer,shocking answer
My brain :answer is π.
Try setting the video Speed to .5 then the video becomes much more understandable.
You must reject -1 and (1-\sqrt(5))/2 because x must be greater or equal from -1/2.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
Not true. Unlike square roots, cube roots of negative real numbers are again real numbers.
I had no chance. This was a crazy problem.
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun ...
I will ask for simple method to my maths teacher during online class
Good morning sir. I am Rahul of class 9 from India and just started watching your videos as I am fond of maths. My uncle told me about this and said that, "as you love maths, this a biggest gift for you on your birthday!". I was very happy after seeing this!
"May our nationality, religion, or any thing would be different, but in maths, we all belong to one family, one religion, one nation, and we are same!"
~Rahul Singh
This is the beauty of math with thinking out of the box
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@@madhukushwaha4578 thanks for your recommendation
There is some mistake in here, because for instance, -1 couldn't be under the cube root as it will give (-1)⅓. It is the same for (1 - (5)½) /2 !
Unlike the square root, the cube root is defined for all real numbers, even the negative ones. As for the other contention, the minus sign is outside of the square root. It's not taking the root of -5, but rather taking the root of 5 and subtracting it from 1.
I do not like so much mathematic but all videos in this channel are truly interesting
Using the fact that 2·ϕ + 1 = ϕ³ , the equation in this problem transforms into 2·ϕ = 2·ϕ
Furthermore, using the fact that -1/ϕ = 1 - ϕ as well as (-1/ϕ)³ = 3 - 2·ϕ , the equation in this problem transforms into -2/ϕ = -2/ϕ
We therefore conclude, that both x = ϕ and x = -1/ϕ are solutions to this equation. QED :-)
This is very primitive try this
ua-cam.com/video/igdy05LZj90/v-deo.html
It’s been along time since Highschool but I would never be able to figure this out. Wow
Hii, If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
This is of the form f(x)=f^-1(x) equating a function with its inverse. So solve f(x)=x.
Then you have to find the other solutions or show there aren't sny more. Here he did the latter.
Yes, you don't need to assume f is increasing (though in this case it is). If f(x_1) = f(x_2) implies x_1=x_2, that's equivalent to saying that f has an inverse. So a simpler proof that x=f(x) is that if f(x) = f(f(x)) , then x=f(x). QED
@@drewmcdermott6798 Your argument is nice but unfortunately doesn't apply here as we have f(f(x))=x and not f(f(x))=f(x). You might not need f to be increasing but you do need some condition. Consider, for example, the case f(x)=-x which satisfies f(f(x))=x for all x but f(x)=x only for x=0.
Hi,can you watch the this trigonometri video ua-cam.com/video/dfTM2YKxg1s/v-deo.html🤛
Really enjoyed solving this one!
Try this channel... Always amazing contents... very impressive math channel... ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
@@mathematicsmath6724 thanks for the recommendation!
after tinkering with notation for a bit I reached the same conclusion, i.e t 2x + 1 = x^3, then adding x^2 to both sides gives x^2 + 2x +1 = (x+1)^2 = x^3 + x^2 = x^2(x + 1) and the solution just follows.
4:00 we can also use integral root theorem
What's that❓
another way:
if you assume (2x+1)^(1/3)=X
then you would end up to the same equation as before
so you'd conclude that X=x
and therefor to x^3-2x-1=0
which gives yo three x as follow:
x=-1
x=(1-sqrt(5))/2
x=(1+sqrt(5))/2
Hello, I think the final step at 5.16 after substitution should give -1 in the equation for the golden ratio, not 1. Cause -b ( in that formula) gives - (-(-x)) ...you excluded the negative sign before x.
f(y) and f(x) are inverse functions, so f(x) is just f(y) reflected across y=x and so they're equal on the actual line y = x, just an easier way to prove that y=x (i'm not gonna make it like i solved the problem on my own but i felt kinda proud of this logic)
1, φ, and -φ
Cool problem. I think there is something about how y is a function of x the same way that y is a function of x that should lead us to the golden mean.
y=x is immediate since the function f(x)=(x^3-1)/2 is one to one. You need not show that y less than x and x less than y both result in a contradiction.
I understood a fraction of that.
The fraction is x/1 where x is equal to the number of prime numbers between 547 and 557.
cubed ,minus1 and then divided by 2 ; Multipled by 2 , plus 1 then cube rooted . They are completely opposite function ! Good observation !
I did a different method to the one shown in the video, not sure if it is valid but I got the answers though. I turned the equation where 2(2x+1)⅓ = x³ - 1 into x³ = 1 + 2(2x+1)⅓ and then I cube rooted both sides to get x = (1+2(2x+1)⅓)⅓. Then I turned it into x = (1+2(1+2(x³)⅓)⅓)⅓ and substituted x³ = 1+2(2x+1)⅓ , and repeated this process over and over, getting this infinite radical of x = (1+2(1+2(1+2(1+2(...)⅓)⅓)⅓)⅓)⅓
And substituted x into it once again x = (1+2x)⅓ and cubed both sides to get the cubic equation
If you want more harder questions then I will highly recommend you this channel's latest videos #mathsandphysicsfun
@@madhukushwaha4578 looks interesting, thankss
Appreciate your method, but I found it very tough 😬
What a beautiful problem and solution!
Is just easier to put the -1 in the other side and I saw it was the same which meant you could say cubirroot(2x+1)= x and then carry on normally
Well it was easy enough to notice that the equation implied that a function is equal to its inverse and hence plotting the graph clearly showed that all real solutions must lie on the line y=x.
Try this channel... Always amazing contents... very impressive math channel... ua-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Another way to solve this is to factor the polynomial (x³-1)³ - 8(2x+1) = x⁹ - 3 x⁶ + 3 x³ - 16 x - 9 = (x + 1) (x⁸ - x⁷ + x⁶ - 4 x⁵ + 4 x⁴ - 4 x³ + 7 x² - 7 x - 9) = (x + 1) (x² - x - 1) (x⁶ + 2 x⁴ - 2 x³ + 4 x² - 2 x + 9), and the last factor is = x⁶ + x⁴ + (x²-x)² + 2 x² + (x-1)² + 8 ≥ 8 > 0, so it does not contribute any zeroes.
Well, there exists a much more elegant solution:
2 * (2x+1) ^ (1/3) = x^3 - 1 = x^3 -1 + 2x - 2x
2 * (2x+1) ^ (1/3) = x^3 + 2x - (2x+1)
denote y := (2x+1) ^ (1/3) , then:
2*y = x^3 + 2x - y^3 ==> x^3 - y^3 + 2x-2y = 0
(x-y)*(x^2 +xy + y^2 + 2) = 0
and then, x = y or x^2 +xy + y^2 + 2 = (x+y/2)^2 + 3y^2/4 + 2 = 0 - no solution
Hence, x=y is the only answer, x = (2x+1) ^ (1/3). and we know how to solve from this point
해설(아님말고):저걸 y로 하면 연립방정식의 해가 구하는 답이 되는데, 봤더니 역함수네? 증가함수여서 역함수와의 교점은 y=x위에 있을 테니 방정식 f(x)=x 를 조립제법으로 인수분해 후 이차식은 근의 공식으로 풀기.
결국 방정식의 해를 찾는 것을, 적당한 식으로 나눠(y) 두 함수의 교점을 찾는걸로 바꿔 생각하는게 핵심.
3:45: instead of trying to guess one of the solutions, just do this: x^3-2x-1=0 -> x^3-x-x-1=0 -> (x^3-x)-(x+1)=0 -> x(x^2-1) - (x+1)=0 -> x(x-1)(x+1) - (x+1)=0 [I used the formula x^2-1 = (x-1)(x+1)]
then we can take (x+1) as factor (x+1)(x(x-1) - 1)=0 -> (x+1)(x^2-x-1)=0
i solve this problem by wolfram mathematica by Solve Function and the answer was x = 1/2 (1 + Sqrt[5]), x = -0.965776 - 1.18647 i, x = -0.965776 + 1.18647 i
The easy way!
Since inverse of f(x) is same as f(x),
i.e f`f(x)=f(x)=y
Therefore x=y
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This was probably the easies problem posted on this channel
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The formula of vanishing method can be used for more sufficiency
The proof that x=y is only valid because we are able to assert that the original equation has real solutions.
Sir x=-1 cannot be in the solution because it doesn't come into the domain of (2x-1)^1/2 where x should be greater than x=(-1/2)
Therefore there should only be 2 solutions of the equation
I never noticed on this since childhood while taking factors
At 4:22 you can see x=-1 or x+1=0, hence here we are dividing x³-2x-1 by 0 here
Doesn't that mean we are making it undefined?🤔🤔
If anyone have a resonable point plz explain to me.
very incredible!! I'm from Korea. umm.. I started to be interested in this videos.. I like these problems and solutions..
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어 님 많이봤는데....ㅋㅋ
Pressure locker again 🤣
Dude love your videos
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my brother pressure locker always makes my day
It's actually Shree Dharacharya formule . He is an Indian mathematician.
This was actually amazing. Thanks for sharing
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The first 1:10 of the video looks like a super specific special case. After that it gets easy. Does the first 1:10 generalize if you say hack off the 2 or disrupt the symmetry? If not it looks like a problem built backwards from a solution, a problem whose general form would normally be solved numerically.
Very interesting exercise and proof.Thank you.
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@@madhukushwaha4578 Thank you for your info .I will take it into account.
I have just another way. Let's see. X3-1=2×cube root of 2x+1. Now x= cube root of[2.cube root of (2x+1) + 1] now expand the cube root of 2x+1 with respect to the found value of x. You will find that it is equal to x(actually it continues to grow and another cube root of 2x+1 will appear) . Now the equation becomes x=cube root of (2x+1). Cubing both sides we have the equation x3-2x-1=0. And we get the solution.
So much cleaner than cubing both sides and brute forcing a degree 9 polynomial.
There’s something the equations present in these video, whose solutions are phi, that portends a mathematical revelation or doorway.
5:17 as far as I know, it is Sridharacharya's formula and not that of Brahmagupta's.
Can you do a video on why polynomial long division works?
and why the equation of a perpendicular line is the negative reciprical of the equation
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-1 is answer by putting in equation
yeah exactly
Superb explanation
That's indeed a lovely one! And clearly not so trivial to solve, at least to me
1=y^3-2x=x^3-2y so y^3+2y=x^3+2x so x=y.
One can show x = y by assuming the contrary which leads to a a complex value for x which is clearly wrong.
Huh boy, now it’s Brahmagupta’s quadratic formula? I thought we weren’t naming things after people anymore.
Imagine ' The formula of Right-angled triangles'
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Quadratic Formula, Sine Rule, Cosine Rule, and Pythagoras' Theorem are for normal people (I know, the last one is "named" by tradition). The implication has to be that Presh is *not* normal.
@@piman9280 lmao
As someone else already pointed out, you should always check your solutions to be rigorous. In this case, both x = -1 and x = (1-sqrt(5))/2 lead to a negative number under the cuberoot, which, if we are mathematically strict, isn't well defined since it would break exponentiation laws (Example: -2 = (-8)^(1/3) = (-8)^(2/6) = ((-8)^2)^(1/6) = 64^(1/6) = +2). So from a strict mathematical standpoint, the equation itself only has one solution, x = (1+sqrt(5))/2
"Strict mathematical standpoint" is a nonsense phrase, and claiming that a cubic function with three real roots only has one solution is very far from a reasonable stance, because it artificially hides real information about the function. The example you give requires the assumption that the sixth root of a number is to be interpreted as having a single solution, which is not rigorous, and the real source of the problem. Limiting yourself to the reals, 64^(1/6) has two solutions, and pretending that one of them doesn't exist is a good way to lose about half of the information contained in a polynomial.
@@michaelleue7594 No, the real problem here is that cuberoots of negative numbers can't be well-defined. Of course, (-2)³ = -8, but the inverse got the problem I already pointed out: Assuming cbrt(-8) does exist, one can prove it has to be +2. Your point I'd assume 64^(1/6) only has one solution isn't true exactly. You have to be very careful here. Example: sqrt(4) IS 2, not -2. Sure, -2 fulfills (-2)²=4, but the squareroot is DEFINED as sqrt(x²) = abs(x). The EQUATION x² = 4 has two solution, the function sqrt(x) (which is defined from R+ to R+) evaluated at x = 4 only has one value.
I tried solving by isolating the radical, cubing, and getting a polynomial with x^9 and then tried solving with synthetic division. Figured it wouldn't end up working by hey I got x=-1. 1/3 lol.
I remember having to solve questions like this in high school. Of course I forget all of it after not using math for over a decade.
Excellent 👍
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Gandooo
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Once you start to scratch your head, you are lost. 😂
Super killer question ..... That's ends with golden ratio constant , that's something golden in it 😃 ...
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