the craziest trig simplification I have ever seen!!

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Опубликовано:

20 дек 2021

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Комментарии : 106

@goatmeal5241 2 года назад

5:19 shouldn't that be sqrt(4) = 2 left in the denominator on the RHS?

@Happy_Abe 2 года назад

Noticed that too

@Pengochan 2 года назад

also the 14 (which would be 28, replacing 4 with 2) would appear in the denominator. Done correctly R=28/27=56/54 => -27c=-27

@fizixx 2 года назад

Yes

@kiryls1207 2 года назад

oh yeah

@benbooth2783 2 года назад

Me a physics graduate: I know trigonometry, this will be fine. Michael: The roots of a cubic polynomial can be written as a trigonometric function. Me: What sorcery is this?

@dlevi67 2 года назад

They have curved arcs. Curved arcs.

@anshumanagrawal346 2 года назад

Don't have to do all that, let the argument in cosine be theta then use the identity of cos 3theta

@glumbortango7182 2 года назад

Look up Chebyshev polynomials if you want to know how to do that

@leif1075 2 года назад

WHO IN THE WORLD wouldve discovered thst fact that a root of a cubic wpuld invovle a cosine function..I don't see why anyone would..though I imagine you could fit any trig function to fit a root solution of s cubic curve as long as you set it to equal the right values maybe?

@shalvagang951 2 года назад

@@leif1075 maybe mathematicians can because they do that things that are impossible to think for other people like Langland program a connection between number theory and harmonic analysis

@roger12321 2 года назад

5:36 I think the 14 should be dividing

@jimschneider799 2 года назад

There were a series of calculation errors starting at @5:11, and culminating in rather interesting equation 63*a*b - 189*c - 2*a^3 = 1 @6:19 (which, using the previously calculated values of a = -1 and b = -9, gives c a value of 568/189 = 3 + 1/189). Fortunately, it appears that Prof. Penn calculated the true value of c off-screen, so no lasting harm was done to the solution.

@romajimamulo 2 года назад

You need a follow-up where you prove that fact because that seems absolutely insane

@yoav613 2 года назад

If you want to prove the fact it is not hard: we have cos(teta/3)=(3x+a)/(2sqrt(a^2-3b)). Now just use the identity: cos(teta)=4(cos(teta/3))^3-3cos(teta/3). We have cos(teta) and cos(teta/3) in terms of a,b,c so plug it to the equation of the identity and after algebra you will get: x^3+ax^2+bx+c=0.

@jonathanarrington1595 2 года назад

At 5:20, when divided by 7^(.5), you be left with 2 not 4

@jesusalej1 2 года назад

Jajaja, I dont know how you did. R=28/27, not 1/7*54. Then you multiplicate the R expression by 7, and forgot 2... but, after all those errors, you got c=1... you are very amazing. I love you for that.

@leonhardeuler9105 2 года назад

God damn holy shit

@saxbend 2 года назад

Also took a factor of four outside a square root and left it as 4.

@jesusalej1 2 года назад

@@saxbend that is way he got 14 instead 28. Then, he put it above multiplying 27 instead below, dividing 27. Then he did not multiplicate 2 times 7. He is a crack, aint no doubt about it.

@joshmyer9 2 года назад

5:23 Is the c value derived here based on this math, or did the 4-2switcheroo here get fixed before getting c=1?

@frabol02 2 года назад

the equation written at 6:17 doesn't give c=1 as the answer, because he made some mistakes previously. However, if you did the semplifications correctly, that would be the result

@jimbolambo103 2 года назад

The correct value of R is 28/27 which gives c=1. Whatever craziness went on for the minute before was ignored in writing down the value for c.

@graikoyt 2 года назад

Hi, there are clearly mistakes in the video on how you got the value for c. The conclusion that c=1 is correct, but there are some mistakes. It starts with sqrt(7) and sqrt(28) cancelling to 4 instead of sqrt(4)=2. Also, when you simplify the 4/56 (the 4 is already incorrect), you simplify it to 14 instead of 1/14. This all leads to your final equation of 63ab - 189c - 2a^3 = 1 actually giving an incorrect value of c being approximately 3. I suggest editing the video and putting a note that the process is incorrect, but if you do it correctly, you'll get c=1. Or, you could replace that part of the video with a corrected part. Just suggestions, but I think it would benefit viewers to know that there are mistakes, rather than wonder why they don't understand math because they have no idea how you did those incorrect steps (I think most viewers assume everything is correct).

@noeticresearch 2 года назад

You solved this problem very elegantly and simplistic! This was a very nice problem. I heard of de Moivre's numbers before, but I don't think I ever heard of the nth primitive root of unity. Nice to know it's the same thing. Thanks Dr. Penn. Excellent math problems as always. 😊

@goodplacetostop2973 2 года назад

12:34 Alright, 5K subs to reach 200K. You know what to do 😉

@goatmeal5241 2 года назад

round up?

@goodplacetostop2973 2 года назад

@@goatmeal5241 Yep

@kobethebeefinmathworld953 2 года назад

4:40 that's where the calculations start going wrong

@yoav613 2 года назад

@teeweezeven 2 года назад

Quite a bit of errors around 5:19. I hope the original creator of this problem didn't make those.

@alnitaka 2 года назад

I entered his crazy number into Wolfram Alpha and I got y^3-y^2-9y+1=0, like he did. So something happened besides writing 4 instead of 2. By the way, the discriminant of this equation is 3136, which is a square, so the Galois group of this equation is Z_3, so that if x_1 is a root, then x_2 will be a quadratic function with rational coefficents of x_1; i.e., including a root of the equation will automatically pick up the other two roots.

@tomatrix7525 2 года назад

You always come out with some aweesome problems

@nuranichandra2177 2 года назад

Where is the remotest intuition for the substitution x=y+1?

@CaradhrasAiguo49 Год назад

10:41 It's easy to see as x ends up being equal to 4 * cos(Arg(z)) + 1, where z = exp(2*pi / 7), exp(4*pi / 7) or exp(6*pi / 7), but the 4 pi / 7 and 6 pi / 7 arguments will result in a negative Cosine, whereas the original expression is, with one simple inequality below, > 3 theta = arccos(1 / sqrt(28)) --> theta is in the first quadrant, implying 0 < theta / 3 < pi / 6 as well as 1 > cos(theta) > sqrt(3) / 2, and the "irrational part" of the original expression is strictly greater than 4 sqrt(7) / 3 * sqrt(3) / 2 or 2 * sqrt(21) / 3 > 2 / 3 * 4 = 8 / 3, meaning the expression as a whole is > 8 / 3 + 1/3 = 3

@minamagdy4126 2 года назад

There's a sign error around 4:35, when you multiplied -Q by sqrt (-Q). That yields -sqrt (-Q^3), the negative of the desired result.

@brockobama257 2 года назад

I also can’t get over this, but maybe I’m trippin. Please read. This calculation *does* support him: sqrt(-Q) * -Q sqrt(-Q) * sqrt((-Q)^2) sqrt(-Q * (-Q)^2) sqrt(-Q * Q^2) sqrt(-Q^3) But this calculation *doesn’t* support him: sqrt(-Q) * -Q -sqrt(-Q) * Q -sqrt(-Q) * sqrt(Q^2) -sqrt(-Q * Q^2) -sqrt(-Q^3) Both calculations seem valid but together they infer: sqrt(-Q^3) = -sqrt(-Q^3) Does the confusion come from: sqrt(Q^2) = +-Q Because that feels like the most likely reason.

@maxsch.6555 2 года назад

@@brockobama257 The first one is correct, because -Q is positive. It's because √(x)^2 = |x| for x ∈ ℝ. So in this case -Q = |-Q| = √(-Q)^2, but Q ≠ √(Q)^2 = |Q| = -Q

@Nothingtonnobodson 2 года назад

That feels so absurd why would someone just happen to know what that particular form is. unless it is from a different discipline that I don't know about :?)

@TJStellmach 2 года назад

As gnarly as it is, that trigonometric substitution is one of the standard closed-form solutions for cubic equations. Not as clean as Cardano's formula (which isn't all _that_ clean in the first place, of course), but it works well in exactly the cases where Cardano's formula doesn't (where by "works well," I mean "finds a clear real-number solution").

@Nothingtonnobodson 2 года назад

@@TJStellmach woah thanks!

@Nothingtonnobodson 2 года назад

@@TJStellmach can you please source a place where it is proven

@TJStellmach 2 года назад

@@Nothingtonnobodson I don't have a good reference on the proof, but I can tell you that the method was developed by François Viète based on a geometric construction and trisection of an angle. Hopefully that will help you find it.

@Nothingtonnobodson 2 года назад

@@TJStellmach I think I can actually go ahead with this hint don't know if this is how he did it but I can prove it with your hints thanks :)

@johnnath4137 2 года назад

A more direct route to the solution: Set u =(1/3)arccos(1/√28), v = cosu. Now 1/√28 = cos3u = 4cos³u - 3cosu = 4v³ - 3v → 4v³ - 3v - 1/√28 = 0, a depressed cubic for which there is a not too complicated standard solution formula. Once we know v, we can easily deduce u and hence 4√7/3cosu + 1/3, which is what we are after.

@curtmcd 2 года назад

Reminds me of that math cartoon where's the main step is "and then a miracle happens". Except about 3 miracles happened.

@icew0lf98 2 года назад

4:15 shoudln't you set R/sqrt(-Q^3) to be equal to 1/3 * arccos(1/sqrt(28))?

@MDMajor 2 года назад

He kinda glossed over what he did there. By the fact provided, θ=arccos(1/sqrt(28)). So when we take cosθ=R/sqrt(-Q^3) the cos and arccos cancel out, leaving just the 1/sqrt(28).

@MrRyanroberson1 2 года назад

5:40 woa woa wait wait 4/56 = 1/14, nit 14

@Happy_Abe 2 года назад

It’s 4/56=1/14 not 14

@The_Shrike Год назад

I’ve got another method, but give me a second let me get my calculator

@roberttelarket4934 2 года назад

Crazy indeed but ingenious and wonderful! Should be an IMO question or was it already?

@orenfivel6247 2 года назад

4*sqrt(7)/3*cos(1/3*arccos(1/sqrt(28))) + 1/3-(4*cos(2*pi/7)+1)=0 as per woafarmalpha, i assume that some of the errors on the board [e.g., sqr(28)/sqrt(7) cancel to 4 instead of sqrt(4)=2] are not accumulated to the cacluation proccess, and some other offine calculation were shown. besides that, gr* video. it would be nice video of prooving the "fact" of cubic's root, and to prove a negeral solution of cubic Eq.

@jamescoughlin8856 2 года назад

I'm confused, what happened to the factor of 1/3 inside the cosine function?

@vgstep 2 года назад

Yeah its micheal

@davidvelasco4423 2 года назад

It's 4 \cos({2\pi} \over {7}) + 1.

@nikolatesla6662 2 года назад

This is crazy

@michaelempeigne3519 2 года назад

sqrt 28 = 2 * sqrt 7

@DeanCalhoun 2 года назад

how do you know they are the same root?

@leonhardeuler9105 2 года назад

god damn holy shit

@jesusalej1 2 года назад

Genius, no place to discuss.

@juliang8676 2 года назад

The click bait! Love the thumbnail

@bilalabbad7954 Год назад

5:20 something went wrong I think 2 is the correct value not 4

@aakksshhaayy 2 года назад

It's easy to get the result using trig identities alone..

@hbowman108 2 года назад

Riemann extended the trigonometric cubic solution to quintics using doubly periodic functions of a complex variable (elliptic functions). One questions if this is "solved".

@MrRyanroberson1 2 года назад

It solves a subset of quintics, as is generally applicable

@ezrasteinberg2016 2 года назад

OK, great! :-)

@mcwulf25 2 года назад

What happened to the 1/3 inside the cos? Surely it means 3*theta = arccos(...)

@crystalline6755 2 года назад

In the root's equation there's theta/3 which means theta=arccos(1/√28)

@mcwulf25 2 года назад

@@crystalline6755 I thought I checked that. Will take another look ...

@mcwulf25 2 года назад

@@crystalline6755 I thought I checked that. Will take another look ...

@liorsilberman6757 2 года назад

Seems over complicated. Let's just set x=\cos(\frac13\arccos\frac1{\sqrt{28}}). Then by the triple angle formula 4x^3-3x=\cos(\arccos(\frac1{\sqrt{28}})=\frac1{2\sqrt7}. Multiplying by 2\sqrt{7^3} and setting y=2\sqrt{7}x we get y^3+21y=7 or y^3+21y-7=0. This polynomial is Eisenstein at 7 so irreducible and defines a cubic irrationality. It can be solved with the cubic formula. The number we want is z=(2y+1)/3.

@tomatrix7525 2 года назад

Is the c=1 calculation correct despite his errors?

@renerpho 2 года назад

c=1 is the correct result IF the calculations are done correctly. The calculations in the video contain a multitude of errors.

@padraiggluck2980 2 года назад

27*R/16 = 1

@crystalline6755 2 года назад

there are some mistakes done while calculating the c value. I got that the correct value should be -5

@Nothingtonnobodson 2 года назад

PLEASE PROVE IT FOR GOD'S SAKE!! THAT FORM IS DRIVING ME NUTS

@someuser257 2 года назад

It’s actually pretty impressive, coming from the fact about a right triangle connected with a third-degree polynomial. Search RU-vid for that

@alexey.c 2 года назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aAYTk-VlKQI.html

@Nothingtonnobodson 2 года назад

@@alexey.c thank you so much

@yoav613 2 года назад

@mmmmmark9751 2 года назад

More arithmetic errors.....and a big jump for cos theta = 1/root28.......not good enough, try writing it down or rehearsing I get the impression that this is showing off, not instruction....you know, like the back-flips

@graikoyt 2 года назад

There are errors in the process, but if done correctly, you'll end up with the same values. As for cos(theta) = 1/sqrt(28), I don't think it's that big of a jump. From the form, we see that cos(theta/3) = cos((1/3) arccos(1/sqrt(28))) So, theta/3 = (1/3)arccos(1/sqrt(28)), which gives theta = arccos(1/sqrt(28)). Thus, cos(theta) = cos(arccos(1/sqrt(28))), and by inverse property, cos(arccos(1/sqrt(28))) = 1/sqrt(28)

@d4slaimless 2 года назад

Kind of funny how this guy makes such amateur mistakes after showing so much tough math in other videos. Had to pause the video few times to check wtf is happening.

@maximilianludwig4301 2 года назад

Typical maths lecture. Nothing to be understood

@GregShyBoy 2 года назад

That's a worst place to stop! Do you mind proving that crazy ass "fact" about one of the roots of cubic equation? No? Ok, we'll trust you here, you're guru after all. But how about the fact that equation x^3 - x^2 - 9x + 1 = 0 has three different real solutions? (Don't believe me, build a plot of that function by yourself). How do you prove that the solution you've found is THE solution? Right, you don't...