The Answer is Surprisingly Easy! | India National Mathematics Olympiad 2003

[@mathcanbeeasy]

And i came up with a simpler solution. Ok, until 6/5

[@mariomestre7490]

Genial

[@elroeleykun2803]

Good

[@yoav613]

Notice that 1

[@mathcanbeeasy]

Unfortunately f(1) is 2, not 3.

[@yoav613]

@@mathcanbeeasy 2+2/1-1=2 good to know

[@mathcanbeeasy]

@@yoav613 auci, sorry, is to early for me. 😂😂😂 I don't know why i made 2+2/2-1. 😂 Congrats!🙂

[@davidbrightly3658]

Best method so far!

[@carstenmeyer7786]

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*

[@federicoa3981]

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

[@Mephisto707]

Audio and video are frequently desynchronized in your videos, you should look into it. Awesome content by the way!

[@lgooch]

Are you sure it isn’t you? Mine seems perfectly normal

[@sirisaacalbertmravinszky2671]

On my device the video is way before the sound.

[@justinnitoi3227]

I just found a range of possible a values and continued to shrink the range until 3.06< a^6 < 3.138.

[@armacham]

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]

how do you know "a" has to be irrational number? Is there any proof?

[@Aman_iitbh]

@@tianqilong8366 if it has rationt root it will pass rrt test

[@tianqilong8366]

@@Aman_iitbh i see, thank you👍

[@armacham]

@@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]

My preferred solution, except you missed showing that a^6 doesn't equal 4 :)

[@holyshit922]

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]

widzę cię już prawie wszędzie, Jezu

[@SuperYoonHo]

love this vid!!

[@Эдвард-ч3э]

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

[@Utesfan100]

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)

[@wonjonghyeon]

At least for me, it is not an easy solution...

[@iainfulton3781]

Turn on postifications

[@bait6652]

Some calculus and numerical bracketting nice

[@bait6652]

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

[@lt97235]

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]

I do think he is using OneNote and just simply screen record

[@tianqilong8366]

Nice manipulation on the final part!

[@philippenachtergal6077]

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]

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.

[@petersievert6830]

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.

[@와우-m1y]

asnwer=2 isit 🤣😅😅😅 mom nag hurt force🤣🤣😅study good bad

[@himu1901]

This is crazy

[@Ritesh_3.14]

Ive done using calc , i ruined the beauty of the q😢

[@mojota6938]

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]

I guess you could see that it is a part of the sum of cubes

[@justanotherman1114]

Or geomtric sum with ratio -a^2.

[@NicolasGuerraOficial]

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]

@@NicolasGuerraOficial That seems to be some nice intuition bro!

[@mcwulf25]

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.

[@screamman2723]

where does that function come from

[@gdtargetvn2418]

f(x) is the original equation, f'(x) is its derivative

[@screamman2723]

@@gdtargetvn2418 ohh didn't know that but thanks

[@JinhaoPan-np7zy]

写过

[@deejayaech4519]

Use newtons method to find the real root of the polynomial to a good degree of approximation and from there floor(a) is trivial

[@predator1702]

Please explain what you are writing....!!!

[@johnjiao4564]

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]

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]

"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

[@billiondream927]

Let's follow mathcreator #mronkoshorts & #mathmarvelasmr