I look at these problems and my eyes glaze over. You make solving them seem easy, and always fascinating and informative too. Thank you so much.
@mathflipped
2 жыл бұрын
I love the problems where you have to be creative and find a way to crack them.
@parimtm
2 жыл бұрын
Rather solve some entrepreneurial real world problem.. more directly 🙃
@MathFromAlphaToOmega
2 жыл бұрын
If you really want a challenge, try summing 1/(n^6-1). For that, you might need to convert it into an integral. Interestingly, it also has 11/12 as part of the answer, but there's another term involving exponentials and pi.
@tonyhaddad1394
2 жыл бұрын
How can you change it into integral ????
@riadsouissi
2 жыл бұрын
Same question
@MathFromAlphaToOmega
2 жыл бұрын
@@tonyhaddad1394 My idea was to use partial fraction decomposition to write it as a sum of terms of the form 1/(n-(sixth root of unity)). Then using the fact that integral from 0 to 1 of t^(k-1)dt=1/k, we could rewrite each term as an integral. Then we can combine those using a geometric series. That trick works well for evaluating sums like 1-1/2+1/3-1/4+... and 1-1/3+1/5-1/7+... but now that I think about it, it probably won't work so well if the denominators are integers. Instead, you could use partial fractions to get it in the form A/(n-1)+B/(n+1)+(Cn+D)/(n^2+n+1)+(En+F)/(n^2-n+1). Then the terms with quadratics in the denominator can be handled using the fact that cot z=1/z+(sum from n=1 to infinity of 2z/(z^2-n^2*pi^2)) along with completing the square.
@0cgw
2 жыл бұрын
You can do this by taking f(z)= pi cot (pi z) /(z^6-1) and integrating around a large square in the complex plane. The sum of the residues is zero, and hence you get double series you want plus contributions from +1,-1 which are double poles, a contribution from zero and four contributions from the remaining sixth roots of unity (these give rise to terms like pi exp(i pi/3)cot(pi exp(i pi/3)). It's a bit of a mess to simplify in terms of hyperbolic trig functions, but should be okay.
@0cgw
2 жыл бұрын
In more detail: If n is an integer, not equal to 1 or -1, then Res( f(z), z=n)= 1/(n^6-1) We have Res( f(z), z=-1)=Res( f(z), z=1)= lim_{z to 1} (d/dz)[(z-1)²f(z)]= -5/12 And if w= exp(i pi/3), exp(-i pi/3), exp(2i pi/3) or exp(-2i pi/3) then Res( f(z),z=w)=lim_{z to w} (z-w)f(z)= (pi w/6) cot (pi w). Thus if S= sum_{n=2}^infty 1/(n^6-1) then 2S-1-2(5/12)+X=0 where X=pi/6[exp(i pi/3)cot( pi[1/2+isqrt(3)/2 )+exp(-i pi/3)cot( pi[1/2-isqrt(3)/2 ])+exp(2i pi/3)cot( pi[-1/2+isqrt(3)/2 ])+exp(2i pi/3)cot( pi[-1/2-isqrt(3)/2 ])] Now cot(pi[1/2+i sqrt(3)/2)=tan(-i sqrt(3)pi/2)=-i tanh(sqrt(3)pi/2) cot(pi[1/2-i sqrt(3)/2)=tan(i sqrt(3)pi/2)=i tanh(sqrt(3)pi/2) cot(pi[-1/2+i sqrt(3)/2)=i tanh(sqrt(3)pi/2) cot(pi[-1/2-i sqrt(3)/2)=-i tanh(sqrt(3)pi/2). Thus X=(4pi/6) sin(pi/3)tanh(sqrt(3)pi/2)=(pi/sqrt(3))tanh(sqrt(3)pi/2) and thus S=11/12-X/2=11/12-(pi/2sqrt(3))tanh(sqrt(3)pi/2)= 0.01759302638532157621375660449795520020
@hassanalihusseini1717
2 жыл бұрын
Thank you so much. In this video I was able to follow all the steps without feeling dumb. :-)
@mathhack8647
2 жыл бұрын
Amazing demonstration!
@kaiovieira230
2 жыл бұрын
Beautiful!
@goodplacetostop2973
2 жыл бұрын
21:26
@gavintillman1884
2 жыл бұрын
In your side calculations, your second set of partial fractions have quadratic denominators so surely your numerators should be linear. Cx+D and Ex+F rather than C and D?
@ezequielangelucci1263
2 жыл бұрын
the numerator was also lineal, i think that makes it function well
@jimschneider799
2 жыл бұрын
@0:24 - since you are able to construct a closed form expression for the partial sum of the series, and the limit of the closed form (as big-N approaches infinity) exists, is it necessary to first prove that the series converges? In other words, does the existence of the limit of the closed form, as big-N approaches infinity, necessarily mean that the series itself converges? Or is that circular logic?
@pierreneau5869
2 жыл бұрын
It was also possible to select the third alternative : (n3+1)=(n+1)*(n2-n+1) and (n3-1)=((n-1)*(n2+n+1) The product (n2+n+1)*(n2-n+1) = (n4+n2+1)
@nicolascamargo8339
Жыл бұрын
Es darle vueltas a lo mismo porque si son factorizaciones del mismo polinomio pues con manipulación algebraica se puede llegar de uno a otro con la segunda factorización pues ya le salio directo
@mikeschieffer2644
2 жыл бұрын
At 9:26 I thought when doing partial fraction decomposition with a non-factorable quadratic in the denominator that we needed to put a linear factor in the numerator rather than a constant. So shouldn't we use Cn + D and En + F instead of C and D for the (n^2 - n + 1) and (n^2 + n + 1) denominators?
@lexhariepisco2119
2 жыл бұрын
same question, I need enlightenment
@AlexandreRibeiroXRV7
2 жыл бұрын
Yeah, we should, but we'd get that the coefficients for the linear terms would be 0. Michael got off lucky this time lol.
@randomlife7935
2 жыл бұрын
Man, after all these years, I can still remember my instructor castigating some of my classmates for failing to do what you are asking during a quiz.
@ramzikawa734
2 жыл бұрын
I’m sure there must be some sort of formal justification, but I was hoping he’d explain that justification somewhere
@paulbooker
2 жыл бұрын
You can do this if you already know the answer.
@em_zon2643
2 жыл бұрын
It all looks like magic ... And the final result is surprising...
@mihaipuiu6231
2 жыл бұрын
I like this very interesting proof.... by Prof Michael P.
@dneary
2 жыл бұрын
My first approach was to factor n^6-1 into (n-w)(n-w^2)(n-w^3)(n-w^4)(n-w^5)(n-1) for w = e^{\pi i /3} and go straight for a linear partial fraction decomposition - it got messy, and I made some mistakes, so I dialed it back and went for n^6-1 = (n-1)(n+1)(n^2-n+1)(n^2+n+1) and went for partial fractions with A/(n-1)+B/(n+1)+(Cn+D)/(n^2-n+1)+(En+F)/(n^2+n+1) - whence you can find A=-B=D=-F=1/2, C=E=0 - which makes a nice telescoping sum that leaves 1/2(1 + 1/2 + 1/3) = 11/12
@dneary
2 жыл бұрын
I got that factorization via the intermediate factorization n^6-1 = (n^3-1)(n^3+1) (a difference of squares) and then further factoring with the well known sum and difference of cubes factorizations.
@nicolascamargo8339
Жыл бұрын
Si pero esas sumas parciales se le va mucho tiempo entre mas sean mas trabajo es por eso que es mejor hacerlo con polinomios de grado mayor para que sean menos fracciones.
@josemath6828
2 жыл бұрын
Buen trabajo.
@petterituovinem8412
2 жыл бұрын
I wish I knew when it was good place to stop. also I would like you to calculate an integral using residue theorem
@alainbarnier1995
2 жыл бұрын
Great and all so well explained ! Thanks a lot
@Bodyknock
2 жыл бұрын
3:09 This actually seems like just a roundabout way of noticing that n⁵ + n⁴ + n³ + n² + n + 1 = (n + 1)(n⁴ + n² + 1).
@anshumanagrawal346
2 жыл бұрын
True
@cernejr
2 жыл бұрын
Really nice, I like the part with using the finite sum, then taking it to the limit.
@stilqnpetrov1998
2 жыл бұрын
Can I ask is there a method for the factorization of n^x +- k?
@Danylux
2 жыл бұрын
The ones I know are If x is even an k is a perfect square that is substracted you can use the difference of squares, if x is divisible by 3 and k is a cube you can use cubes addition or substraction, but i don't know any others without including complex roots
@romajimamulo
2 жыл бұрын
For n^x -1, it always factors as (n-1)(1+n+...+n^(x-1)). If X is composite, you can do what he did here to get additional factorizations
@someuser257
2 жыл бұрын
Well, the (n^5+n^4+…+1) could be written as (n+1)(n^4+n^2+1), which would lead to the same result The same for the last factorization, where we could go further with the calculations
@CM63_France
2 жыл бұрын
Hi, 5:53 : to N, instead of to infinity, 6:11 : ok, correct, 9:11 : n⁴+n²+1=(n²-n+1)(n²+n+1) bravo! For fun: 17:47 : ok, nice (Côte d'Azur), 21:27 : nice photo composition with the splash screens.
@jo3d0om
2 жыл бұрын
satisfying
@philg4116
2 жыл бұрын
you will squarely face the horror of the sphere
@kh.h.3561
2 жыл бұрын
I always wait for you to say that's a good place to stop 😂
@xCorvus7x
2 жыл бұрын
have you seen the compilation made by A Good Place To Stop
@CanIHaveSomeMore
2 жыл бұрын
at 10:57, is it legal setting n = -1 and then n = 1, since the original condition requires n >= 2 ?
@martinepstein9826
2 жыл бұрын
It's a slight logical leap but it's fine. Two lines (y = 1 and y = A(n+1) + B(n-1)) must intersect at no points, one point, or every point. They can't intersect at all but two points.
@geniusgemini2924
2 жыл бұрын
That was a side calculation to decompose the fraction and thus, not bounded to that condition.
@Goku_is_my_idol
2 жыл бұрын
Nice one
@doctorb9264
2 жыл бұрын
check it out we get 11/12.
@manucitomx
2 жыл бұрын
I quite enjoy this type of problems, they are very eye opening. Thank you, professor!
@chillmathematician3303
2 жыл бұрын
ASMR mathematics
@mr.mckinnon5680
2 жыл бұрын
7. The answer is always 7.
@alexandrebatata4552
2 жыл бұрын
EITA PORRA .... KKKKKK
@sakethram538
2 жыл бұрын
this one was easy
@shalvagang951
2 жыл бұрын
please do start a course on combinatorial group theory i heard that this course is very interesting
@avalons2170
2 жыл бұрын
so so cool
@adityaekbote8498
2 жыл бұрын
Noice
@fouadamzil9490
2 жыл бұрын
Nice
@Dniektr
2 жыл бұрын
Here’s a quick way to solve simple (first degree polynomial) partial fractions decomposition: To get A, solve n sucht that the denominator equals 0, meaning n-1=0 => n=1, and replace the value of n into the full fraction, removing the denominator for A, meanning 1/(n+1)=1/2. For B, the denominator is n+1, so solving n+1=0 gives n=-1. Replacing it into the full fraction where the denominator part for B, n+1, is removed, you get 1/(n-1)=1/(-1-1)=-1/2. For second degree or more, it gets trickier and solving it the way it is in the videos seems to me the easiest one.
@tonyhaddad1394
2 жыл бұрын
Wowwwwwwwww , i solved it correctly but i spend discret 4 hour
@555AMC
2 жыл бұрын
The answer would have been way more interesting, if the question was : find S = -1 + sum((n^4+n^3+n^2-n+1)/(n^6-1)). Since the latter gives -1/12 = Zeta(-1).
@riadsouissi
2 жыл бұрын
Got a link with a proof of this ?
@sasharichter
2 жыл бұрын
this provides a constructive proof to the question whether a 2 minute problem can be solved in over in 20 minutes
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