And i came up with a simpler solution. Ok, until 6/5
@mariomestre7490
Жыл бұрын
Genial
@elroeleykun2803
Жыл бұрын
Good
@yoav613
Жыл бұрын
Notice that 1
@mathcanbeeasy
Жыл бұрын
Unfortunately f(1) is 2, not 3.
@yoav613
Жыл бұрын
@@mathcanbeeasy 2+2/1-1=2 good to know
@mathcanbeeasy
Жыл бұрын
@@yoav613 auci, sorry, is to early for me. 😂😂😂 I don't know why i made 2+2/2-1. 😂 Congrats!🙂
@davidbrightly3658
Жыл бұрын
Best method so far!
@SuperYoonHo
Жыл бұрын
love this vid!!
@carstenmeyer7786
Жыл бұрын
2:05: Multiply everything by *a^2 + 1* to simplify the polynomial and get an expression for *a^6* dependent on lower powers of *a* , so you can get away with the approximation *a ∈ (1; 2)* : *a^7 + a = 2 * (a^2 + 1) => a^6 = 2 * (a + 1/a) - 1* Note the RHS is increasing for *a > 1* , so we use *a ∈ (1; 2)* to approximate *a^6* via RHS *a^6 ∈ (2 * 2 - 1; 2 * 2.5 - 1) = (3; 4) => ⌊ a^6 ⌋ = 3*
@Mephisto707
Жыл бұрын
Audio and video are frequently desynchronized in your videos, you should look into it. Awesome content by the way!
@lgooch
Жыл бұрын
Are you sure it isn’t you? Mine seems perfectly normal
@sirisaacalbertmravinszky2671
Жыл бұрын
On my device the video is way before the sound.
@federicoa3981
Жыл бұрын
Is a lot easier to notice that f(x)=-x^2+1+x-2=-x^2+x-1=0 mod(x^2-x+1) so f(x) is divisible by (x^2-x+1), do long division to find f(x)=(x^2-x+1)(x^3+x^2-x-2), and since the first factor is always positive, a has to solve a^3+a^2-a-2, and from that you can find by hand 3
@bait6652
Жыл бұрын
Some calculus and numerical bracketting nice
@lt97235
Жыл бұрын
Great Video ! Like all the others. And sir, I would like to ask you a question : How do you make your videos ? On which software do you write your solutions? And audio ? Thank you in advance
@patrickjefferson1438
Жыл бұрын
I do think he is using OneNote and just simply screen record
@justinnitoi3227
Жыл бұрын
I just found a range of possible a values and continued to shrink the range until 3.06< a^6 < 3.138.
@iainfulton3781
Жыл бұрын
Turn on postifications
@bait6652
Жыл бұрын
Is it safe /robust just to find a in (1,2)...then do the alg-manip to bound a^6 with the lower bound of a/(a^2+1) at a=2 thus bounding 3
@wonjonghyeon
Жыл бұрын
At least for me, it is not an easy solution...
@armacham
Жыл бұрын
A few things you can do. First use RRT to eliminate all rational roots, including all integers. You showed any solution must be positive using derivatives, but you can also do it with factoring. split into three cases: x < -1, x is between (-1, 0), and x > 0. (knowing that there are no integer solutions we don't need to consider any other cases) then you break the LHS into 3 parts and show that all 3 parts are negative, so they can't be equal to RHS which is zero in the first case, where a < -1, you know that a < 0 and -2 < 0. And clearly a^5 - a^3 < 0 in the second case, where a is between (-1, 0), you know that a^5 < 0 and -2 < 0. And clearly a - a^3 < 0 so you can eliminate all negative solutions. knowing that a must be a positive irrational number makes things easier going forward I also did the (a^2+1) thing to show that a^6 must be at least 3 to show that a^6 < 4 I did the following: a(a^4 - a^2 + 1) = 2 a^4 - a^2 + 1 = 2/a a^4 - 2a^2 + 1 = 2/a - a^2 (a^2-1)^2 = 2/a - a^2 this means 2/a - a^2 >= 0 2/a >= a^2 since we know a is positive we can multiply both sides by a without flipping the sign 2 >= a^3 4 >= a^6 thus a^6 must be between 3 and 4
@tianqilong8366
Жыл бұрын
how do you know "a" has to be irrational number? Is there any proof?
@Aman_iitbh
Жыл бұрын
@@tianqilong8366 if it has rationt root it will pass rrt test
@tianqilong8366
Жыл бұрын
@@Aman_iitbh i see, thank you👍
@armacham
Жыл бұрын
@@tianqilong8366 Yes. There is something called the "Rational Root Theorem" (RRT). When you have a polynomial with integer coefficients, you can use the RRT to get a complete list of all possible rational solutions. Then you can test everything on the list. When you are done testing them, you have found all rational solutions and you know that any remaining solutions MUST be irrational
@jamescollis7650
Жыл бұрын
My preferred solution, except you missed showing that a^6 doesn't equal 4 :)
@tianqilong8366
Жыл бұрын
Nice manipulation on the final part!
@user-fm1ri1se8p
Жыл бұрын
I found lower bound like in the video, and the upper bound this way: a³+2=a⁵+a⩾2a³ => a³⩽2 => a⁶⩽4, but equality holds only when a⁵=a, so a=0,±1 which are not roots. So a⁶
@holyshit922
Жыл бұрын
Polynomial from this equation can be factored into (a^2-a+1)(a^3+a^2-a-2) This polynomial has only one real root a^3=2+a-a^2 a^6 = (2+a-a^2)^2 a^6 = (a^2-(a+2))^2 a^6=(a^4-2a^2(a+2)+(a+2)^2) a^6 = a^4-2a^3-3a^2+4a+4 a^6 = a^3(a-2)-3a^2+4a+4 a^6 = (2+a-a^2)(a-2)-3a^2+4a+4 a^6 = -a^3+a^2+2a+2a^2-2a-4-3a^2+4a+4 a^6 = -a^3+4a a^6 = -(2+a-a^2)+4a a^6 = a^2+3a-2
@wieuchenst3499
Ай бұрын
widzę cię już prawie wszędzie, Jezu
@Utesfan100
Жыл бұрын
At the 4 minute mark you had bound 5/4 2(x^2+1)/x. But (x^2+1)/x = x + 1/x > x for all positive real numbers, including 3^(1/6). Hence f(x)
@himu1901
Жыл бұрын
This is crazy
@screamman2723
Жыл бұрын
where does that function come from
@gdtargetvn2418
Жыл бұрын
f(x) is the original equation, f'(x) is its derivative
@screamman2723
Жыл бұрын
@@gdtargetvn2418 ohh didn't know that but thanks
@petersievert6830
Жыл бұрын
If you invest a lot of time and sweat, you could come up with 241/200 < a by calculating the according values and accordingly 3 < (241/200)^6 < a^6 . But quite in the contrast I'd rather be interested if it is possible to show that f( 6th root of 3) < 0 and thus a > 6th root of 3 . It was rather easy to show f (6th root of 4) > 0 , so a solution like that would seem a reasonable approach to me.
@mojota6938
Жыл бұрын
The trick at 5:18 seems to come out of thin air. Can you give some insight into how you came up with it?
@vitalsbat2310
Жыл бұрын
I guess you could see that it is a part of the sum of cubes
@justanotherman1114
Жыл бұрын
Or geomtric sum with ratio -a^2.
@NicolasGuerraOficial
Жыл бұрын
Because is similar to: 1+a^3 = (a^2 - a + 1) (a + 1) From general formulas: a^3 + b^3 = (a^2 - a b + b^2) (a + b) a^3 - b^3 = (a^2 + a b + b^2) (a - b)
@tianqilong8366
Жыл бұрын
@@NicolasGuerraOficial That seems to be some nice intuition bro!
@mcwulf25
Жыл бұрын
I kicked myself when he did that. It's a well known result for factorising the sum of two cubes x^3+y^3 = (x+y)(x^2 - xy + y ^2). Appears a lot in number theory problems, usually as a factorisation. But here in reverse.
@predator1702
Жыл бұрын
Please explain what you are writing....!!!
@deejayaech4519
Жыл бұрын
Use newtons method to find the real root of the polynomial to a good degree of approximation and from there floor(a) is trivial
@user-hq7hi2sl2o
Жыл бұрын
asnwer=2 isit 🤣😅😅😅 mom nag hurt force🤣🤣😅study good bad
@philippenachtergal6077
Жыл бұрын
0:00 Hum. f(0) = -2 f(1) = -1 f(2) = 24 So there is a root between 1 and 2 which isn't precise enough. now, f'(x) = 5a^4 - 3a^2 + 1 , f'(1) = 3 So first iteration of newton's method tells me to try for a root around 1 - (-1)/3 = 1.3333 If we test at 1.3 we get f(1.3) > 0 If we test at 1.2 we get f(1.2) = -0.03968 1,2 ^ 6 = 2,985984 so argh we are so close to 3 that I can really say if floor(a^6) is 2 or 3 Not without doing calculs that would be quite annoying to do by hand Now, if I were to pretend that I would do this kind of calcul by hand, I would eventually figure out that f(1.201) < 0 and f(1.21)>0 so there is a root between those two. As 1.201^6 > 3 and 1.21^6 < 4 we have floor(a^6) = 3 Answer is 3 (at least one of the answers that is, there might be other roots) But yeah, I wouldn't do that by hand unless I was paid handsomely for it. So another approach is needed.
@philippenachtergal6077
Жыл бұрын
Now. I could actually build on what I found above. If I multiply everything by a, I get a^6 = a^4 - a^2 + 2a (this is of course not true for any a, it's only true for the one a we are hunting for) And now, evaluating the right hand side a at 1,2 and 1,3 reasonably gives that floor(a^6) = 3 as the right hand side evaluates between 3 and 4 for both 1,2 and 1,3. Only "reasonably" however as I think I would need to prove that the right hand side is an increasing function. Still not very satisfactory but more reasonably doable by hand than my first attempt.
@johnjiao4564
Жыл бұрын
I can get one solution a = e^(iπ/3). e^(iπ5/3) - e^(iπ3/3) + e^(iπ/3) - 2 = 0 and e^(iπ6/3) = 1
@johnjiao4564
Жыл бұрын
My answer is wrong. Should be simplified a^5 - a^3 + a - 2 to a^3 + a^2 - a - 2 = 0, get a^6 = 3.07
@tontonbeber4555
Жыл бұрын
"Suprisingly easy" ? lol !! I did it trying a=1.1,1.2,1.21 and finally 1.205 which is very close to the root ... and indeed answer is 3 ... a^6 probably close to 3.06
Пікірлер: 52