Evaluating an Infinite Series

[@mrigayu]

Of course the problem is a really nice one with the generous cancellation and the consideration of the limit as k tends to inf. But I think the marvel of this video lies in your consideration of the identity and for which values of x and y the identity is true, which requires a very thorough understanding of the tangent and arctangent functions. Brilliant!

[@DrBarker]

Thank you! Yes, I thought it would have been a shame to just use the identity without going into the details of when/why we can use it.

[@davidbrightly3658]

Argument also goes through marginally more naturally with Xn=1/n√3 and Yn=1/(n+1)√3. These are the reciprocals of the video values transposed, the angles then being the π/2 complements of the video angles. The telescoping sum then works on the upward slanting diagonals leaving the initial term arctan(1/√3) = π/6 and the final term -arctan(Yn)→ 0 as n→∞.

[@johnporter7915]

Great video :D could you do some crazy equations solving for x like a septic equation?

[@DrBarker]

Thank you! I made a video a while ago on solving a quintic, but it would be interesting to try even higher degree equations. A septic would need to be set up so we can use some tricks to simplify the problem, since there isn't really a general method that works for every septic.

[@islandfireballkill]

Cute infinite sum but the solution definitely felt like tricks just falling from the sky with that difference formula. 😅

[@italyball2166]

I mean, both the properties of telescopic series and that arctan formula are usually learnt in highschool and uni

[@DrBarker]

This is a fair point - definitely most of the time spent working on this problem would be working out what method to apply. Then it all just falls into place quite quickly, once we realise we need to use the arctan identity and telescoping series property.

[@aik21899]

At 15:11 : You could directly cancel the second half of the equation via tan(arctan(z))=z without having to prove arctan(x-y/(1+xy)) in (- π/2, π/2)

[@nathanisbored]

16:38 fails for (x,y) = (-1,1) as the denominator would equal 0. this corresponds to a LHS of -pi/2. its fine in our special case since x would be positive

[@soyoltoi]

More generally, onto functions like tan always have a right inverse but they don't generally have a left inverse. However, if you restrict the domain of tan to make it one-to-one, then there is a left inverse, and it is equal to the right inverse.

[@0cgw]

To compute the sum, you could use that exp (2i S_k) = prod_{n=1}^k [(n+i/√3)(n+1-i/√3)]/[(n-i/√3)(n+1+i/√3)] = [ (k+1-i/√3)/(k+1+i/√3) ] (1+i/√3)/(1-i/√3) where S_k is the sum to k terms (as the product telescopes) which tends to (1+i/√3)/(1-i/√3) = exp (2i tan^{-1} 1/√3 ) as k tends to infinity. Therefore, the original sum is tan^{-1} 1/√3 = π/6.

[@0cgw]

Of course, you have to be slightly mindful of the range of the argument in the final step (though thinking about where we are in the complex plane helps (Re exp (2iS_k)>0).

[@anestismoutafidis4575]

=> Σ (inf/n=1) arctan (1,732/7) 13,874; Σ(inf/n=1)+n+1 3^1/2/n+1+1...× ♾️ Limit value: Σ arctan 3^1/2 =3^(1/2)/3n^2+3n+1 with (n+1) +1)×..♾️ =13,99'

[@JohannNeumann-ic3gy]

Your videos help me a lot. Do you ever plan on doing calc 1 or linear algebra courses?

[@DrBarker]

There are already some very good free courses out there, e.g. bprp has one for calculus 1, and MathDoctorBob for linear algebra. I don't feel like I'd be adding anything particularly new by making my own courses, so I'm more focussed on continuing with random videos like this on different topics.

[@only10mileshigh]

I think you dropped out the root 3 in the first term of the original sum. Shouldn't it be (sqrt3*pi)/2?

[@justzs4866]

I think that he just wrote it funny. That root 3 should be inside the arctan I believe, so then it would be arctan(sqrt(3)(k+1)) which when k goes to infinity would be that pi/2 shown in the video.