a complex approach to a classic sum

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[@hyperboloidofonesheet1036]

The complex logarithm is a multi-valued function. How do you know the principal value is the correct value to use at the end?

[@ayushrudra8600]

because arctan gives the principal value it’s defined to give a value between -pi/2 and pi/2

[@hyperboloidofonesheet1036]

@@ayushrudra8600 But those values are being summed; there's no reason to believe the sum of those values necessarily lies within that range.

[@ayushrudra8600]

@@hyperboloidofonesheet1036 oh good point. maybe you can argue that the sum is less than some geometric series or smthing to find a maximum value? but i’m not entirely sure

[@xinpingdonohoe3978]

Just like ln is multivalued, arctan is also multivalued, and it's a logarithm based function. At the start, we fixed the branch where ln(1)=0 by assumption. This branch equates to arctan(1)=π/4, and arctan(1) is the expression we got at the end, just written in logarithmic form.

[@andyneeman4378]

Just by definition, we start by stating that we take the principal branch of the logarithm, which gives us the identity θ=1/i*log(z/|z|). From here we just use the same branch for the entire question, as simply manipulate the same operator. It is not that we start by assuming the principal branch, and then at the end picking again the principal branch. Rather it is we simply use only the principal branch operator for the entire question.

[@MichaelMaths_]

I recall trying this sort of approach the first time I saw this problem. Though partway through, I remembered the arctan subtraction formula and switched to that instead to finish it up over the reals. Nice to see the resolution using complex numbers 🔥

[@varun3282]

Beautiful. I really adore when complex numbers are applied to solve something.

[@jonathantorres913]

Love your work man!

[@alipourzand6499]

As would say Dr. Peyam:"pourquoi faire simple quand on peut faire compliqué?". ☺ But there are great pedagogic advantages when using this kind of alternative approachs.

[@tomholroyd7519]

Happy Thanksgiving, Michael! ❤

[@goodplacetostop2973]

13:00

[@dneary]

4:00 May I suggest (before doing this simplification_ writing 2n^2 + 1 as 2n^2 - i^2 and showing that you're basically doing a difference of squares factorization

[@skylardeslypere9909]

Two things. 1, he did explain it in words what was going on, but I agree it was kinda confusing. 2, Neither Michael nor the editor ever look at the comments or suggestions so I fear that this comment will never be seen by Michael

[@zohidtugyonov9494]

another way: arctan(2n+1)-arctan(2n-1)

[@BenfanichAbderrahmane]

🤔

[@Bruno-j6x]

how

[@MayureshChaudhary]

for anyone interested how,
=arctan(2/4n²)
= arctan(2/1+4n²-1)
= arctan(2/1+(2n-1)(2n+1)) now write 2 as difference of 2n+1 and 2n-1 and recall the formula that arctan(x-y/1+xy) = arctan(x)-arctan(y) , so our sum becomes a telescopic 🔭 series

[@gp-ht7ug]

Love the use of complex numbers

[@rainerzufall42]

Very nice approach!

[@martincohen8991]

Using Wolfram Alpha and computing the first few partial sums leads to the conjecture that the sum to m terms is (1/2) arctan(2m(m+1)/(2m+1)). Using the arctan addition formula and Wolfy again, shows this is true by induction. This would have been a pain to do by hand, since the final formula leads to a fraction with 4 m^4 - 8 m^3 + 8 m^2 - 4 m + 1 in both numerator and denominator which cancel out leading to the result.

[@martincohen8991]

Further investigation shows that (1/2) arctan(2m(m+1)/(2m+1))=arctan(1-1/(m+1)) which makes the induction much easier.

[@letis2madeo995]

At 3:23 should be sqrt(4n^4+1)

[@davidemasi__]

4:20

[@rainerzufall42]

He changed that later. Noticed that as well...

[@Alan-zf2tt]

It is beautiful math method

[@artichaug1719]

so cool

[@alexchan4226]

0

[@damyankorena]

11:55 technical mistake. Considering a product up to n notice that parts from the numerator and denominator dont get canceled out so a limit is needed.

[@hydropage2855]

8:27

[@martincohen8991]

What is left is the last fraction which goes to 1.

[@monzurrahman8307]

He says it verbally at 11:24

[@davidemasi__]

Very elegant solution, great job as usual 👏🏼

[@zh84]

It delights me that, if you use complex numbers, 4n² + 1 is a difference of squares: (4n - i)(4n + i)

[@rainerzufall42]

More likely (2n - i) (2n + i)...