Dear Tejas, challenge ACCOMPLISHED! (Speed run, Uncut)
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- Опубликовано: 13 окт 2018
- We will solve a very interesting square root equation sqrt(5-x)=5-x^2 with infinitely nested square roots. This is a math speed run!
Solve this as a quadratic equation in terms of 5: 👉 • quadratic equation but...
This is my "equation of the year" in 2019.
To see others, please check out here 👉bit.ly/equationoftheyear
#equationoftheyear
Infinitely nested Michael Jordans, • Infinitely Nested Mich...
Tejas's video: • Video
You know shit got real when he pulls out the purple marker
account hahhahaha flyyyy
Just made your likes evil
Hahaha
😂😂😂😂
Does he ever break out the plaid marker?
*and here I just thought to power 2 both sides like an innocent child..*
¡Que vergüenza! (What a shame!)
Same
Why is it wrong? I haven't tried it yet
@@justinsantos5751 it's right, u can square both sides and then use polynomial division. Answers are the same
You technically could as there actually is a formula to solve fourth degree equations.
I don't recommend it however...
The teacher during the whole class: stories of his life, personal thoughts, etc
The teacher when I leave 4 mins to go to the bathroom:
best joke ever
B-but the stories are fun to hear!
😂😂😂😂😂
Just in case that anyone did decide to try to square the equation instead, 5 - x = (5 - x^2)^2, which would give you a quartic equation. Add to the equation the expression 4(x^2 - 5)y^2 + 4y^4 to obtain 4y^2·x^2 - x + 4y^4 - 10y^2 + 5 = (x^2 - 5 + 2y^2)^2 = 4y^2·x^2 - x + (4y^4 - 20y^2 + 5). Then divide by 4y^2 to get (x^2/2y - 5/2y + y)^2 = x^2 - x/2y^2 + (y^2 - 5 + 5/2y^2). We want x^2 - x/2y^2 + (y^2 - 5 + 5/2y^2) to be a perfect square, which implies we want 1/4y^4 = y^2 - 5 + 5/2y^2, or 1 = 4y^6 - 20y^4 + 10y^2, or 4(y^2)^3 - 20(y^2)^2 + 10(y^2) - 1 = 0. Then, we must solve this cubic for y^2. Thus, we really want to solve 4z^3 - 20z^2 + 10z - 1 = 0 and let z = y^2. To solve this, divide by 4 to get z^3 - 5z^2 + (5/2)z - 1/4 = 0. Let z = a + 5/3, so (a + 5/3)^3 - 5(a + 5/3)^2 + (5/2)(a + 5/3) - 1/4 = a^3 + 5a^2 + 25a/3 + 125/27 - 5a^2 + 25a/3 + 125/9 + 5a/2 + 25/6 - 1/4 = a^3 + 115a/6 + 1225/54 = 0. This is a depressed cubic equation. Now, notice that (u + v)^3 + 3uv(u + v) - (u^3 + v^3) = 0. Thus, if a = u + v, then 3uv = 115/6 and u^3 + v^3 = -1225/54. u = 115/9v, so u^3 + v^3 = (115/9)^3/v^3 + v^3 = -1225/54, which implies v^6 + 1225v^3/54 + (115/9)^3 = (v^3)^2 + (1225/54)(v^3) + (115/9)^3. Then v^3 = {-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2 or {-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2. Regardless of rhe choice, u^3 will be equal to its conjugate. This implies that a = cbrt[{-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2] + cbrt[{-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2], which implies z = -5/3 + cbrt[{-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2] + cbrt[{-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2]. Simplifying this would be useful. 54^2 = 9^2·6^2 = 3^4·3^2·2^2 = 2^2·3^6, while 9^3 = 3^6. To get the common denominator, the inner radicand would equal [(1225)^2 - (115·4)^2]/(54)^2 = [(1225)^2 - (460)^2]/54^2. (1225)^2 - (460)^2 = (1225 - 460)(1225 + 460) = (1 685)(765)/54^2 = 337·153·5^2/54^2 = 337·17·15^2/54^2 = 5 729·(15/54)^2. This leaves the outer cubic radicands -[1225 - 15·sqrt(5 729)]/108 and -[1225 + 15·sqrt(5 729)]/108 respectively.
As you can see, this is a treacherous path, and we are dealing with expressions far more complex than what was shown in the original problem and its solutions. Clearly, this is not the way to solve, and anyone would have given up halfway through.
Wow!!!
Man, one like for your commitment even though I didn't understood a single fucking line 🤣
I gained nothing from reading this.
I assumed it can be written as the product of two trimonials: (ax^2 + bx + c)(dex^2 + ex + c). I solved it with a system of equations, letting c=-4 so that every variable is an integer. This led to (x^2 - x + 4)(x^2 + 5x - 5)=0 and solving x from there was pretty straightfoward
i aint readin allat🔥🔥
Excellent solution ...nicely explained at 314 words per second! Cheers
tumak1 : )
Is this a pi joke lmao?
My math teacher wrote 314 faster than him though. I still love math despite facing a speed writer in math lessons in high school. He used right hand to write and the left hand to clean at the same time.
best comment with a good sarcasm
@@Rekko82 no way
The fabled purple pen was only mentioned in the legends. This video marks the speed of its glorious power to solve maths. On the hands of this man lies the power to overthrow deities.
Cool
Set playback speed to .75.
Yup it’s that of the legend
Teacher: you have 5 minutes before the exam begins
Me: *watch this video*
OMEGA LUL you made a mistake
OMEGA LUL hilarious
Misty Diablo I would've just looked at memes which would backfire cause I would be distracted while writing cause I'd keep internally laughing
Don't u have to write the exam
Call in a sub.... That is better...
This problem has a bit of notoriety in the math olympiad community for having a funny alternate solution:
sqrt(5 - x) = 5 - x^2
Square both sides
5 - x = 5^2 - 2x^2*5 + x^4
Rewrite this as a quadratic equation. But wait, how can we make a quadratic when there's an x^4 term? The key is to not write it as a quadratic in the variable x; write it as a quadratic in the variable 5:
5^2 - (2x^2 + 1)*5 + x^4 + x = 0
Use the quadratic formula to solve for the variable 5:
5 = (2x^2 + 1 +/- sqrt(4x^2 - 4x + 1)) / 2
5 = x^2 + x
or
5 = x^2 - x + 1
Finish by solving both quadratics. Remember to throw out the two extraneous solutions where 5 - x^2 is negative, due to our first step.
Thank you for widening our mathematical perspectives
Unbelievable! Thanks for this mind-blowing solution!
Who went straight here after watching the video inspired by this comment?
I always thought this video was about this solution, since it's pretty famous, so I never checked it out. I'm surprised to see it's not
nice man
black pen, red pen, blue pen, purple pen,...this is getting out of hand!!!
There is also a green pen, and am hoping the orange pen will make a return in a few days
there was no black pen in this video. maybe he's taking a holiday
@@leftysheppey well, there was Mr.blackpenredpen :-)
Sounds like Goku's hair color 😂
I might be too late but do you mean..
Getting out of pen
i’m beginning to feel like a rap god, rap god
Hahaha
All my people from front to the back knot back knot
@@jesusshrek1271 back nod*
@@avinavverma2315 soori for ma engris. Nyan nyan
@@jesusshrek1271 it ish awkay ma frind dunt wiry
'How do we do this? Let pull up the purple pen again.'
Thanks, now I just need to buy a purple pen for my exams.
Hahaha nice and thanks!
This man can now reedem his "top 10 rapper Eminem is afraid to diss" reward
"It's not good, it's bad it's dangerous, Infact, its a trap."
Mairis ̶ʙ̶ʀ̶ɪ̶ᴇ̶ᴅ̶ɪ̶s̶ Bērziņš yup yup
captain ackbar approves
Straight bars
Admiral X-bar
@@blackpenredpen how many languages do u know u also know maths and all the other science subjects maybe at Olympiad level...😶😶😶
When you look away for one second in class
lol underrated
Lmao
2^8th like!
(3×91)th like
This is the first time I've ever slowed down a video just to understand it.
: )
I checked twice if the video was on 2x speed
This is the reason why i subscribed. BPRP probably did it in his head in ten second but took so long just because he had to explain it to us.GREAT AS USUAL.
Jamie Handitye : ). Thank you.
I think he probably fell for the trap the first time he tried it.
Cuong Hoang no. I am a guy with experience.
: )
blackpenredpen the sheer top energy in the comment, wow, we stan
@@blackpenredpen true u r the man who created world record right!!
“Let’s just focus on this part right here”
*proceeds to circle the entire right side of the equation*
he didnt circ- rectangle the minus
when your name is Tejas and you see this on top of recommended
T3MPURR lol my name is Tejas as well
And you're like "bhai ye kaunsa scene kardia maine?"
You can also find the solution for the positive intersection of the right and the left side, by noticing that the left side is the inverse of the right side, therfore the intersesction of the right side and the left side must lie on the line y=x. so just put either the right side or the left side equal to x and solve.
How fast... Did you get MATH-ANPHETAMINE?
Crystal Math
10 seconds
we need more of this type of speedrun
OK!
Agreed
@@blackpenredpen Fun fact: x=1. The end.
It's not about the speed. It's about the accuracy that things happens. Bounded above increasing sequences comes into this. The topic is highly advanced. He explains this in another video.
Nested radicals is one thing but nested alternating radicals is another subject on itself. We need more content creators like this that raises interest in real analysis and higher advanced topics in mathematics.
Teachers like him are widely needed.
Never noticed you could solve algebraic equations recursively before. Neat.
Fixed point iteration and certain infinite continued fractions are similar. You do have to be careful you don't get a divergent limit when you do this though.
Changing the x = sqrt(5 - sqrt(5 - ....)) to simply x = sqrt(5-x) was such a clever move, I never would have thought of that! Who would of thought you could simplify it to a quadratic equation. Well done.
Would have*
actually once you've done a couple of infinite sums like these it is really easy to notice this. Nothing you would have easily thought of by yourself, but hey, that's what learning is for
I don’t get why, can someone explain?
@@brendanwoods4773 x= one half the square root of 5
@@stevefrei2588 suuuper old post I know - but what is the reason for this? Why can you do the substitution, is this just something you need to know?
Another way to solve this is to notice that two sides are inverse functions of eachother. The intersection of a function and an inverse function will lie on the line y = x. Therefore this can be solved simply by setting either side equal to y = x.
Excellent observation. Algebraically, this is f(x) = f^(-1)(x) => f(f(x)) = x. For a solution we can solve f(x) = x. This is what BPRP did. But there could in theory be more solutions, depending on f. E.g. if f is self inverse, such as 1/x, then any x in the domain is a solution.
So your argument is not water tight. If the graph of f meets y=x, this also solves f equals its inverse. But for f equals its inverse at x, we just require both (x, y) and (y, x) lie on the graph of f for some y, not that y=x.
@@MichaelRothwell1 could we use his argument if we do have more information regarding the solution?, in diaphontine equations for example?
This observation is extraordinary. But how does it find the second solution? I can't seem to find it.
Jifu Wen I couldn't either, since technically 5-x^2 is only the inverse function of root(5-x) for x > 0 and the other solution lies in the negative x.
hi, I have found another solution to this problem, let's say, which is based on geometric and symmetry considerations:
in fact we consider the curves in the Cartesian plane represented by the left and right sides of the equation:
y = 5-x ^ 2
y = sqrt(5-x)
the first is a parabola, with the vertex on the y axis, the second, is the same parabola, but rotated by 90° thus having the vertex on the x axis.
It is easy to verify (you can use a program that plots the curves in the Cartesian plane) that the 4 solutions of the equation are the 4 points of intersection of the two parabolas; these 4 solutions are placed on a heart-shaped figure symmetrical with respect to the straight line y = x; it is easy to verify that the first pair of solutions lies precisely on this last line.
therefore, to find the first pair of solutions, we can solve the following system of two equations:
y = 5-x ^ 2
y = x
substituting the second in the first, we obtain:
x ^ 2 + x -5 = 0
which admits the two solutions x = (-1 +/- sqrt (21)) / 2
to find the second pair of solutions, it is observed that they are symmetrical with respect to the line y = x and are found on the line y = A-x, where "A" is a constant to be determined. Therefore, the constant "A" must satisfy the two systems of equations simultaneously:
1) y = 5-x ^ 2
y = A-x
2) y = sqrt(5-x)
y = A-x
eliminating the "y" from the two systems of equations and rearranging, we obtain two equations of second degree in the unknown "x" and in the variable "A":
1) x^2-x+A-5=0
2) x^2+(1-2A)x +A^2-5=0
but, since the two equations must provide the same solutions, this happens, if and only if the single terms of the equations are identical and this occurs only when A = 1, therefore, the second pair of solutions is found simply by solving :
x^2-x-4=0
whose solutions are: x = (1 +/- sqrt (17)) / 2
As a speedrunner I love what you did.
...Will you speedrun a 4 part contour integral?
This is crazy to see you here, I am super interested in the 0xA community and saw your HMC video a while back, cool to see other people in that community in other places I visit as well.
playing it at 2x for even faster math. there is no limit to this man, he diverges.
i guess this is the reason why the 0.5 speed exists on youtube
sqrt(5-x) and 5-x^2 are inverse functions of one another. A function and its inverse will always meet one another on the line y=x, therefore setting either sqrt(5-x) =x or 5-x^2 = x will produce the same solutions as the original equation to be solved.
This is a very pretty argument and avoids the risk of divergent series. Top thinking!
This was understandable to me, in contrast to the original video. I would need a slow motion video. Is there the original version around, without speeding up?
That's not the case though? Solving 5-x^2 = x gets us the wrong answer.
It gets the first solution where x>0, but you're right that it doesn't find the second one where x
@@Liamdhall Haha, awesome you responded 3 years later 😀
First the Minecraft, and now the math itself.
These speedruns are getting wild.
You're a math rapper man
Hmm
That irrational equation is equivalent to the system formed by the following:
5-x >= 0
5-x^2 >= 0
5-x = (5-x^2)^2
The first two inequalities are solved by -sqrt(5)
Why x_1 and ×_4 aren't acceptable? >.
@@manuelrojas9547 because if you plug them in you get negative values under the radical. They are indeed solutions but they're on the imaginary plane
Thanks bro
Factorization wasn't really obvious. However this solution doesn't leave an open question for the alternating case. This is why I prefer it.
Well, once you applied your first trick to get x^2+x-5 = 0, you could also square the original equation and factor this polynomial out, getting (x^2+x-5)(x^2-x-4). You can then easily find all roots (and discard the irrelelevant ones)
(Neat trick btw)
Not only fast but also fairly well explained. Good job.
Justin - thanks!
Pro tip: Play at x0.75 speed
Pro tip: Play at x2 speed
Protip: MAKE SURE THE SYNTH AND THE VOCALS ARE IN THE SAME KEY.
Anyone?
Man...I want whatever kind of coffee you had before this video😂
Teacher : The test wont be soo hard it's only from what we studied in class
The Test:
What he's doing with ellipsis substitution is kind hand wavy tho. At no point does it actually extend out to infinity. At every step along the way, you have only ever done a FINITE number of substitutions...
The alternating/negative is even more sketch, because x somehow magically disappears from the right side entirely (via this "continued radical" identity, which is pretty neat).
I think what's going on here is that we are implicitly relying on the fact that as we do more and more substitutions, the influence of x goes to zero, because it comes under more and more radicals as we do more and more substitutions. Therefore, the right hand side can be replaced with lim as radical_count -> inf of F^radical_count(x) where F(x) = sqrt(5 - sqrt(5 - x)). So, one thing that we are missing is a proof that this limit even exists in the first place. If so, I think the rest is ok. Or maybe we can just proceed based on the assumption that the limit exists (which is what is implicitly going on in the video), and then double check at the end that the "solution" that we "dervied" actually works.
I think you can easily imagine a similar problem where "proceed based on the assumption that the limit exists" blows up in your face, and then, you'll be left wondering what went wrong. For example, let's change F to be F(x) = 2x. Then, F^n(x) = 2^n * x. Well, that does not converge except in the special case of x = 0. It only works for special values of F (such as F(x) = sqrt(5 - sqrt(5 - x))).
that was kinda cool math seems to be even more interesting than i thought
BPRP is a gateway drug
My idea is to let 5=a. Solve for a, then replace a with 5 and solve for x. Then throw out the extraneous x solutions.
That's so cool, I love seeing people rush things (well as long as they are still done correctly)
I never cease to be surprised by these recursive methods of solutions on your channel) it's magical!)
Please proof formula on 3:36 !!!!
I would love to see it!
There is a recursion, so that may help you
@@zzz942 Successive Square Roots with Alternating Sign - Bong Soriano
@niraj panakhaniya thanks!!!!
Expand sqrt (a-sqrt (a-x))=x. You will get x^4-2ax^2-x+a^2-a=0 which can be simplified to (x^-x-1)(x^2+x+1-a). Also note that x^2+x+1-a has two roots, and the positive one equals (sqrt (4a-3)-1)/2.
I CAN SWEAR BY JUST LOOKING AT YOU THAT YOU WERE DESPERATE TO BREATHE OUT THE SOLUTION......
NEVERTHELESS AWESOME SKILLS DUDE!!!
That is a genius thumbnail right there haha
I like how he makes a video later by squaring both sides, then sets up the quadratic formula as 5=... absolutely brilliant
0:01 Doraemon theme song.
1:07
The red marker: Please write more slowly !! I'm not enough to spend ink !!! Hellllp!
This is so interesting. GOOD JOB CONGRATULATIONS!!
Queria que tivesse um professor brasileiro de matemática que explicasse as coisas assim, sem enrolar e rápido
Queria que tuviese un profesor brasileño de matematicas que explicase cosas así, no entendí el resto
Esta bien la traducción???
I love his fast calculation abilities. Wait "super fast"
When BPRP starts saying 'is nothing but' you know he watches papa flammy's vids
Wild4lon hello, new subscriber
Thanks for letting me know!
These videos makes me fall in love math all over again ❤️
You need to do maths speedruns at the AGDQ. I'd donate money for that.
Squaring both sides would be easier. Everything you have to do that at first solving the equation wrt 5(wrt 5 the equation is quadratic) then wrt X
Nevermind, got it. I had never seen that before. Neat!
1:04 "And now I will just write this down again in red" Lmaaao 😂
"in fact, it's a trap" 😭😂😂😂 You the man, bprp
Two sides are inverse functions of each other => intersect at y=x
That was some kind of recursive witchcraft. That technique is new to me, thanks.
Nothing but respect for our favorite math legend!
You're amazing. I would like to see the alternating root series proof.
03:50 - Seeing this formula, I get three very important questions in my head:
1) Can you do the video with proof of this formula?
2) And what is the formula for the second alternating series of infinitely nested radicals?
I mean if the formula for the FIRST alternating series of infinitely nested radicals is:
*√{ a - √[ a + √( a - √[ a + √{ … } ] ) ] } = [√(4*a - 3) - 1] / 2*
then what is the formula for the SECOND (shown below) alternating series of infinitely nested radicals:
√{ a + √[ a - √( a + √[ a - √{ … } ] ) ] } = ???
I know it will be:
*√{ a + √[ a - √( a + √[ a - √{ … } ] ) ] } = [√(4*a - 3) + 1] / 2*
(Note *"+ 1"* - not "- 1")
but *how to prove it?*
3) And what are the general formulas:
*√{ a + b*√[ a - b*√( a + b*√[ a - b*√{ … } ] ) ] } = ???*
*√{ a - b*√[ a + b*√( a - b*√[ a + b*√{ … } ] ) ] } = ???*
and *how to prove them?*
I think he proved the formula in his infinitly nested Michael Jordan video
hey,are you still there? i think i prove it myself,please let me know if you would like to see the proving process
@@user-hw5gi3ci2v Sure!
Please.
@@ashwinraj2033 sorry, i have a lot work to do , so it was 5 days later till i saw your comment
@@ashwinraj2033 please click into my channel , it has the explaining video, i don't know why i couldn't paste the link
And I'm here doing my GCSEs. I wish I understood as much as these guys on RUclips. Are there any masters who want to give me some first hand tips? I like maths but I'm not too great at it. I wish I was so badly.
I'm an year late but bear with me,
The most important part is being able to simplify every step to make it easier to calculate. Factorization will help you a lot in every step. Its also good to practice mental math to speed up your calculations.
Beatiful. Thanks a lot!
coolest intergral prof love it
Him: How to solve without squaring both sides
Also him at 1:55
I don't comment a lot but this video was actually amazing
Γιωργος Γουργιωτης thanks!!!!!!
This is so beautiful wish you were also doing physics, I love your videos man
Dude u deserve more subs
First of all, i'm foreigner so i can't understand his English,but i realized that i can understand what he tryed to saying
Mathematics is "universal language"
My secret weapon:
×0.75
Con este man, subiré mi level de preparatoria.
Well well, a great approach for solving this
I thought the whole video was a rap.....
It wasn't?
Isn't it?
My first thought was like:
"Is he speaking english?"
I loved watching this. Very good
First time I see a video from you, so fast that I thought first that you were talking Chinese at some moments lmao very good channel btw
The term you have used to solve the first equation is called ' The formula of Sridhar Acharya'. He was an Indian and since I am an Indian too I liked your hasty process to work out this amazing equation...
Please make a video of the formula looks cool
yes please!
I don't know what video is more satisfying
Every other math tutorial I speed up the video, except his videos, I slow down the video
Oh and btw: degree 4 equations aren't terrible, you can always solve them with radicals :) although that wouldn't make it good for a speed-equation-solving
Professors hate him!
Learn how this man made thousands speedrunning math on RUclips just using this one small trick.
: )
so how can you prove the alternating term becomes that one?
Speed run... but you still explain it.. what a great teacher...
: )))))
Blue ped red pen purple pen YAY
jesusthroughmary yay!!!
Rip black pen
Speed of light!
I couldn't make out a single word you said but the math speaks for itself
Huh, as the video progressed your speech turned from incoherent speed talking to actual math and i love it.
Nice any% Speedrun, I hope to so your entry in the 100% run including proving the sqrt((4a-3)/2) formula.
Am i on some drugs or why is he talking so fast to me
Both.
Same.
I noticed the alternating nested square root formula looks similar to a quadratic formula solution. Interesting!
That was incredible 🔥
3:34 can't wait for the proof :>
"And of course, if we want to be cute--because we ARE cute" 😍 yall this man is a keeper we stan 💖
Great video! :D Formally speaking though, don't you first have to prove the existence of a definite finite solution for the infinitely nested radicals? I know this might've been too long for such a speedy video. Or is it somehow obvious? Thanks!