The most FUN Series I EVER Evaluated! An Analytic Number Theory EXTRAVAGANZA!
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- Опубликовано: 11 июл 2019
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Outtakes: • Video
Polygamma: • An Introduction to the...
Int Digamma: • An Integral Representa...
Catalan: • Catalan's Constant and...
Geo Series: • The Geometric Progress...
pi^2/8: • This is SERIESLY NUTS!...
Leibniz Rule: • The Leibniz rule for i...
Today we are going to go extremely overboard! Let us use the Hurwitz Zeta funtction and the Trigamma function to solve this crazy series of 1/(4k+1)^2. This is just such a crazy beautiful and playful calculation extravaganza! At teh end we are going to end up with the trigamma function of 1/4 and Zeta(2,1/4) and catalan's constant! SO GOOD!
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2:40
This is it. This is the point when I can't recognise where the maths ends and the random scribbling begins.
Papa's Zeta functions be like ”了“
Chinese character for already
5:02
BREAKING NEWS: MAN MAKES MISTAKE WITH HAGOROMO CHALK. DESCRIBED AS "NOT A MATHEMATICIAN", CHALK CARTEL CONSIDERING OPTIONS
Varad Mahashabde the Hagomoro cooperation wanted to pay him loads of money so he would censor this, but he refused.
Just saw that.... man's fuccd
5 second rule, if you correct it in 5 secs it's not a mistake
Integrate ( 2arccot(x) - πe^{-x} ) / ( πx ) dx from 0 to infinity, answer should be very oily and macaroni
π²+8g is just 9g which is 100, just some engineering done right
Not even an engineer can agree with that, because g =/= G this is not about free falling euleroids. So it's approximately 17.
that sounds like a polygamous function
Oil macaroni constant
Love your videos man, you are killing it!
If I am not mistaken I think that in this particular summation since you have already talked about Catalan's constant you can completely avoid the whole analysis part by writing your sum as 1/2 * ( sum of reciprocals of all odds squared + the alternating series of all odds squared ) since this allows you to kill all the fractions containing odds that are equivalent to 3 mod 4.
Still awesome outtake on the subject, I always enjoy your analytical number theory videos
After evaluating the two integrals as infinite summation you get a result that's trivially equal to what you started with. You really didn't have to go through all of that
The Physicist Cuber Actually, there is a super easy derivation of this that can be done within 3 minutes in a video. Any infinite sum of the function 1/[(2^m)n + 1] can be written as a linear combination of the sums of the functions 1/[(2^(m - 1))n + 1] and (-1)^n/[(2^(m - 1))n + 1], for all natural m > 0. This is easy to prove. Then just express the two sums for their known values.
@@angelmendez-rivera351 that's exactly what I said :)
The Physicist Cuber I'm just clarifying what you were referring to.
I assume/hope this is going to be the “Chad” version he talks about at the beginning of the video
@@angelmendez-rivera351 I think this assumes absolute convergence, but that's true in this case so yeah
Well, no wonder I couldn't get it to pop out of sums & differences of index-manipulated forms of the full square harmonic series.
Failing that attempt, I was gonna guess π²/9. Close, but no cigar. Fine by me; I don't smoke.
Fred
Not bad my man, cool that you know all this stuff with these gamma functions and integrals, but...
If you look closely at what you did, its basically deriving that the sum of all the numbers 1/(4k +1)^2 is the same as
1/2(I - J), where I is the sum of 1/(2k +1)^2, and J is (-1) * the sum of (-1)^k/(2k + 1)^2.
These two sums contain basically the same terms, but in J every second term has a flipped sign.
So the terms of the form 1/(4k + 3)^2 cancel out, while only the ones of the form 1/(4k +1)^2 stay and are doubled, which gets cancelled by the 1/2.
i.e you get your original sum back.
So all of this work can be replaced by some simple algebraic manipulations :/
But hey, seems like you had a lot of fun, so thats nice :)
He needs to see this lol
Hilbert Black He doesn't. At the beginning of the video, he mentioned there are three methods to evaluate the series he evaluated. Ine of those methods he described as the bitchboi method. The method given by the OP is what he was referring to as the bitchboi method. In other words, Papa Flammy knows about this method. He just did not want to present it because 1. It is too boring. 2. The method from this video is far more educational and elegant as it involves non-trivial knowledge from previous videos of his, which he has carefully built from scratch. He even specified this in the video, too. Y'all are just making me repeat what he said in the video.
@@angelmendez-rivera351 Well okay maybe but I do not agree that the doing something overly complicated to solve a really easy problem can be considered "elegant" in any way. Complicated methods are for getting from some place to another, not for running in circles.
NAMEhzj You're not understanding something really basic that he said in his video, and you're literally making me repeat myself here. 1. It's not elegant because it's complicated. It's elegant because of the involvement of pre-built analytical number theory knowledge and due to all the non-trivial identities that the method exposes. 2. The method is not running in circles. Sounds to me like you are just too close-minded to even try to understand the method, hence you call it that.
Next time, please bother to pay attention to what he says in the intro of the video.
Love the "Čau" on the end
Very excellent indeed. But is the equation you derived not pretty trivial? If you add the series of 1/(2k+1)^2 to the corresponding alternating series of it, all terms in the form of 1/(4k+3)^2 will cancel out and you get double the series that you wanted.
Now plugging in the known results of both series and dividing by 2 yields the wanted result.
u 2πr This is one of the three methods he described in the intro of the video. He gave a reason for choosing the method he did, complicated as it may seem. People need to actually pay attention to what he says before commenting.
>Riot points
Now you're feeding both my addictions
Oh bruh, this is awesome
There was no need to do all that! You can just write the original series as sum of squares of reciprocals of odd numbers, minus that of the numbers congruent to 3 mod 4 and you get the same 2 series
Happy birthday of mine to you, Pappa Flammys express extravaganza
You could approach that much easier.
Through complex integration, you got the answer is equal to 1/2 * (Ⅰ + Ⅱ),
where Ⅰ = 1/1^2 + 1/3^2 + 1/5^2 + ... and
Ⅱ = 1/1^2 - 1/3^2 + 1/5^2 - ... .
It is so obvious that 1/2 * (Ⅰ + Ⅱ) is the same as the first one!
If your goal was to show those integration processes itself, it's a nice video!
Papa flammy, where do you get those math memes at the beginning of ur videos?
Can you evaluate the sum over all the natural numbers including 0 of 1/(8n + 1)^2, or alternatively, of (-1)^n/(4n + 1)^2 ?
Lol that early "thanks for watching" xD
Any body know the idea of evaluating an integral by solving a triangle? In Mikael Passare's paper on how to compute the sum 1/n^2 by solving triangles he transforms it into int -ln(1-e^-x)dx then evaluates it as e^-x+e^-y=1. Which makes sense as they are both evaluating the same function, but what is the theory of evaluating an integral as a triangle? It's a super cool proof and I even went through reference 4, but the book only had like 1 page on it, so it'll take a while to read the whole thing
Rito points? From the indie game Leek of Legons?
So, on a related note, I have some PDF files from Andrew Granville titled "A Pretentious Approach to Number Theory", so I decided to search for the videos with this title, and I found the 3 part talk by Andrew Granville, located here:
ruclips.net/video/RpuhtlFqFLw/видео.html
But I also found this talk here by Papa Flamey!
:))
17:42 German cartel leader threatens the opposition as negotiations go south in the 3rd European chalk dust war circa 2019, colorized.
This is gold
Have you proved the functional equation for zeta(s) in a video yet? I'd love to see that!
@@PapaFlammy69 It's very difficult, but would make a great video series ;)
Pretty sure if you do a similar method that you did for the sum k≥0 1/(2k+1)² you get a better solution
How is this shit monetized? How? Keep up the good work!
I would die without your videos
Ye
The summation converges to two terms.The second term is again an infinite series named G,the Catalon's constant of value 0.915965.
oil macaroni constant XD
I was wondering if there are multiple integer polygamma functions, there're also non-integer polygamma functions like an half-polygamma function or even something more crazy like a pi-polygamma function.
Ps:love you papa master of virgins.
Davide Pascu For each order n, there is a unique polygamma function. Also, the polygamma function is defined for complex orders, because the order is determined by the power of the derivative, and the derivative operator raised to a complex power is another well-defined operator called the complex fractional derivative.
Wow! Damit hast du so ziemlich den ganzen mathematischen Werkzeugkasten benutzt... und .. tadaaaaa... die "Crema Catalana"-Konstante! Soll noch mal einer sagen, Mathe wäre nicht lecker. :)
Holy macaroni man!
14:14 every math teacher
Can’t you compute the integral from 0 to 1 of ln(t)*t^(2k) dt by substituting u=-ln(x)? Way simpler imho
Yeah you're right
Flammable Maths Yeah -ln(t) I misstyped
Flammable Maths Yeah -ln(t) I mistyped
Ախպերս ես Էլ եմ սենց բանաձերով տունը զբաղվում, Հայսատանում ոչ մեկին չես զարմացնի մաթանալիզով:
well you sort of overcomplicated that. the two series at the end, the values of which you derived before, obviously sum to double your original series you wanted to evaluate. the odd terms are going to cancel, the even terms will give you that original summation(2/(4k+1k)^2). you could have skipped the integrals beyond noting that it is also the value of the triagamma function at 1/4ß
Jendrik Weise You obviously were not paying attention to the video. He knows this method is overly complicated. He chose it anyway for a specific reason, a reason which he gave literally in the introduction of the video. C'mon, people.
Angel Mendez-Rivera namely? is this what he means by "intuitive"? i still find that use of the word strange.
Jendrik Weise What are you getting at?
Angel Mendez-Rivera well the only other appraoch he mentions that could be this one is the "intuitive" approach. but i fail to see how that approach would be any less rigorous than this approach.
Jendrik Weise Is your English comprehension bad, or what? I already told you that he *gave a reason in the video.* Watch the goddamn video again, and you will notice that nowhere in his explanation did he mention rigor. In fact, he even said he will do a video on the method you are talking about. Did you conveniently ignore this? It looks like I'm going to have to explain his reasoning myself in the comments.
From 0:52 to 1:30 in the video, he explained there are three methods. One is the standard method he likes to use (his words, not mine). One is the intuitive, "bitchboi" method, on which he will make a video. And there is THIS method, which uses a bunch of cool analytical number theory identities and lemmas, and that the method is so amazing that we will dive right into it.
So it should be obvious why he did not choose his method or the intuitive method. It's right there. It has *nothing* to do with rigor. Don't @ me.
what I dont understand is
the series doesnt converge uniformly on lim1-, right?
it just converges uniformly if you close the interval. how can you proceed then?
Fractal It does not need to converge uniformly to 1-, because that is not part of the interval of integration. Whether the interval is open or closed is also irrelevant, since 1- is part of neither.
@@angelmendez-rivera351 setting the bounds of integration to 0 and 1 does not imply 1- is there?
am I getting the definition wrong?
Fractal I'm sorry, I was confused. For some reason, I flipped the 1- and 1+ around in my head, hence my argument.
The series only need converge uniformly for every 0 < x < 1. It need not converge at the bounds of integration. It is an improper integral.
@@angelmendez-rivera351 oh, ok. that explains it. thanks
Is this gonna be about the poly gamma function.
It is isn’t it
lel
Plugging the z is left as an exercise to the viewer
Someone a good book on number theory?
Good math
I think that you could have easily solved this if you just changed all of the k variables to xi
Can’t believe no one is talking about that cat in the wheelchair
riot points? daddy, you up for some duo queue on EUW ?
2:40 disgusting zeta tbh
69% votes. Reply with Nice.
Pinky swear kitty swear banana cherry strawberry swear.
Oh my god, Oh my no, Oh my wow!
You know what Papa Jens?? I am your son...
Huahahahahah
sum n from 0 to inf of 1/(3n)!
@Deniz Göksu please explain 🤔
@Deniz Göksu no, thats smaller than e
@Deniz Göksu e^3 = sum n from 0 to inf of (3^n / n! ), and not of 1/(3n)!
Varad Mahashabde I'm fairly certain he is joking. Look at how he wrote his letters.
Flammable Maths Can you do it?
I didn't know you played league?
@@PapaFlammy69 So you're one of those epic gamers I've heard about?
Hi could you help me
Bruh you could start with saying that Σ1/(4k+1)²=1/2(Σ1/(2k+1)²+Σ((-1)^k/(2k+1)²) 😂
Bro, hear me out. This has a very simple solution.
Sum k=0 to inf of (1/(4k+1)^2 ).
=Sum k=0 to inf of (1/(2(2k)+1)^2 )
Which is the same as:
Sum m=0 to inf of (1/(2m+1)^2 )
Where m iterates over even integers.
Where m=2k.
Which means:
Sum k=0 to inf of (1/(4k+1)^2 )=
Sum k=0 to inf of (1/2) * (1/(2k+1)^2 + (-1)^k/(2k+1)^2 ).
And after that everything is same as yours.
I mean no polygamma, no integration nothing required.
No hagaromo. I mean no homo.
breh!
Flammable Maths Is this some rich mathematician joke that I'm a poor engineer to understand.
@@8ball437 still breh, but numbers
subpole
¿Has entrado en la carrera?
sum(n=1..inf){1/n^2}=sum(n=1..inf){1/(4n)^2}+sum(n=1..inf){1/(4n-1)^2}+sum(n=1..inf){1/(4n-2)^2}+sum(n=1..inf){1/(4n-3)^2}.
sum(n=1..inf){1/(4n)^2}=(1/4^2)sum(n=1..inf){1/(n)^2}
sum(n=1..inf){1/(4n-2)^2}=(1/2^2)sum(n=1..inf){1/(2n-1)^2}
sum(n=1..inf){1/(n)^2}=sum(n=1..inf){1/(2n)^2}+sum(n=1..inf){1/(2n-1)^2}
so sum(n=1..inf){1/(2n-1)^2}=(1-1/2^2)sum(n=1..inf){1/(n)^2}
so sum(n=1..inf){1/(4n-1)^2}}+sum(n=1..inf){1/(4n-3)^2}=(1-1/2^2)sum(n=1..inf){1/(n)^2
"catalan thing" =sum(n=1..inf){1/(4n-3)^2-sum(n=1..inf){1/(4n-1)^2
so sum(n=1..inf){1/(4n-1)^2=(pi^2-8c)/16
and sum(n=1..inf){1/(4n-3)^2=(pi^2+8c)/16 ...
Roly polygamma
Bräh
You could just say it's (1/16)*Polygamma[1,1/4] :)
you can just say it is itself
Are you serious?
Sorry but you proofed in long way. This is have a very easy method
WTF are u married??? Im sad now :(
Tensor boi has a stepmom lol
@@PapaFlammy69 oh so we also have a mama
first
You could have split the sum instantly, without all the integration nonsense..