Blackpenredpen vs. Dr. Peyam on convergent series (uncut, unscripted)
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- Опубликовано: 11 дек 2018
- Blackpenredpen vs. Dr. Peyam on convergent series! Calculus teachers battle! Uncut, real version, not staged. Who wings? Enjoy this video? Then check out another one by Dr. P • Peyam’s reaction to bl...
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Yay!!!! 😄😄😄 We’re both maximal 😊
Dr. Peyam's Show yup!!!!!
dynamic duo
U r amazing
You're both monsters of math. I enjoy your videos!
Holder inequality, right?
TOP 10 MOST EPIC ANIME BATTLES OF ALL TIME
The world needs more people like these two nice gentlemen.
Thank you!
Dr Peyam is a flamboyant boy and I love him, bless his heart. Always in good mood.
I couldn’t agree more!
I love seeing interactions between you and Dr Peyam
Karyuu
Thanks!!!! Here's another one from last year.
I’ve never seen two guys happier about math. Almost too happy LOL. Great channel glad I found you thanks for the post
Dr. Peyam is such a nice guy. I wish I could develop a similar personality. Blackpenredpen you are still the best :)
be yourself lol
@@thedoublehelix5661 very true
Can you name a more iconic duo?
Dick and balls
Tyler Dewitt and PatrickJMT?
Bohr and yourself, Herr Einstein!
@@dreamybull1509 Who?
Heavy and Medic.
I took Calculus and real analysis 25 years ago so my memory of series analysis is spotty. I was able to solve the first one only in my head but that is amazing Peyam can solve all in one try on the whiteboard. Also great he didnt disturb the black and red pens. He wins the wings.
Yup, he won!!
Peyam for president
Peyam of the United States 😆
Dear Lord people would say Hitler should become president if he gave a speech on a RUclips channel
Like in real life
We definitely need both of you to do this more.
: ))))
i didn't realize math collabs are the greatest thing on earth. peyam and yourself are two math gods. thank you for sharing your time together.
Oh gawd I love Dr Peyam he's like my inner child
These two guys are my most favorite Math teaching RUclipsr...!!
Its Absolute fun to seeing both of u!!
PLEASE make this a regular thing. this made my day
This is one of my favourite videos on RUclips by far -- please do something like this again
You guys are making me fall in love with this beautiful subject even though I'm gonna finish my undergrad next year and have been studying this since the past two years 😊😃
I thoroughly enjoyed this just as much as you both probably did. I paused and was working those problems out along side yall and had fun too. Thanks for sharing.
Loved this! Need another episode of this. Next time you both try each other’s problems and see who can stump who.
I have a marvelous proof (and it actually fits on this comment):
(1/n - √a_n)^2 >= 0, therefore √a_n / n
Right!! Easier. But the triangle proof a bit more erudite and an advanced calc proof technique. Just put something very similar beginning with divergent harmonic greater than An and square root divided by n causes it to converge so it will squeeze anything smaller (which for some n is every convergent positive series)
the last series could be done by the following way as well :
(sqrt(a_n) - 1/n )^2 >=0
=> sqrt(a_n) / n
Epic math battles of history!
Dr.Payam is amazing he's so energetic and bright !! BTW you are also amazing :)
: ))))
Hope to see more videos like that!!! It's really cool!!!
You should continue such types of videos, it's funny, tricky, challenging and interesting.
You are the best channel on youtube honestly
"My name is Peyam! :D" bprp "cool" hahaha sickest reaction from Peyam so happy
Yay! A challenge! :D
This is so much fun to watch!
This was fun to watch. You should do some more unscripted videos.
Biggest battle of all time
I know this is old, but I love these videos where you guys just talk about math problems.
Using the dot product property to use the comparison test was just wonderful. Recently I did research into the idea that mathematics is rhetorical in nature and these videos are lovely examples of that.
The fact that Dr. Peyam came up with a five page solution that was valid and this one line solution is valid just goes to show that math is indeed rhetorical.
MORE ! Please it's so cool !
E7 TAYLZ
This one is from last year ruclips.net/video/XG3rOsATZZU/видео.html
This is great because in my next exam series is a big topic. Please more videos on series!!
Thank me later
tutorial.math.lamar.edu
www.pdfdrive.com/mathematical-methods-for-physics-and-engineering-e3061564.html
leangsimschoolboy.files.wordpress.com/2013/09/student_solutions_manual_for_mathematical_methods_for_physics_and_engineering.pdf
I wish I could be this Happy when I'm doing my Calculus, great energy guys!
This is one of my most favorite videos of blackpenredpen!!
Thanks!!!
7:03 me after doing literally any problem
I am grateful I love math, it allowed me to know both of you!
This is the sacred rite of passage that every mathematics enthusiast must go through before calling himself a true mathematician...
What a pleasure seeing these two gentlemen in one frame
This is the definition of happiness and love.
Wow,we lucky to see this video amazing thanks for uploading this video ❤️
I can NOT wait to see him in another video!
Please turn this into a series, like once a week prepare each other a couple of math questions (any math topic)
Unfortunately we don't live close by and we have busy schedules. I did help him to record a few more vids that day and they will be on his channel in a few weeks. : )
But if they turn it into a series, then we could just watch a few episodes and develop a formula to tell us exactly what a given future one would be, thus ruining those episodes. Furthermore, assuming the series converges, which you request to happen weekly, then we can determine the sum total of all videos they make together beforehand. Due to BPRP and Dr. Peyam both wanting to see us all enjoy the beauty of math together, they would never undertake a project which would spoil our enjoyment of them. Thus, we have the real reason it cannot happen. QED.
:D
OMG pls do more videos like this
the solution to the last question is a gem
For the last question, An =1/n^2 also works
[In a Maxwell Smart voice]: AHA!! I knew it! It's the old Cauchy-Schwarz trick!
Fred
ffggddss
I was thinking the comparison test : )
Could you do two direct comparisons for the last one? (Compare to sqrt(a(n)) and compare to a(n).) Anyways I always love these vids!!! So cheerful!
MUCH LOVE FROM GREECE
Awesome video guys!!
Such a good and interesting video!
Very impressing and funny! I would like to see some more fun math related to physics, you already did the Euler and how it relates to wave-functions. Perhaps some advanced mechanics: The Lagrange equation in a rotating frame of reference proving the Coriolis force....
The final is very cool and surprising. Thanks.
How about:
Since An converges consider Bn the harmonic which will be greater than An for some/all n > m. Root An is also < Root Bn. Root Bn over n (1/n^1.5) converges by denominator power over 1. So Root An /n squeezed between Root Bn/n and 0 so converges.
My solution to the last one: ∑a_n < inf, a_n > 0 implies a_n is smaller than an 1-series, i.e. a_n < 1/n √a < 1/√n. Therefore, √a/√n < (1/√n)^2 = 1/n, so it's also smaller than a 1-series, so it's convergent.
Funny and very instructive !!! What two brilliant minds together !!!
Both of you are genius! Thumbs up
This was really cool and took some rust off my series knowledge! I'm just really curious. How can we prove that Sum A(n) converges?
we can solve the last one in an easier way by saying that eventually an
You guys are Amazing!!!
In the last series, could u show its convergence aproximating An as a Reyman's series and when u do the product, u prove that the exponent of the n located in the denominator is bigger than one?
More of this please.
OMG, I haven't seen this video yet, how not? It's so wholesome
Cauchy-Schwarz is really helpfull in this case, but you could also solve it that way: a_n M (that has to be true because sum of a_n conv.) Because the squareroot is konstantly increasing, the sqrt(a_n)
In the second example, could we also have argued (by a comparison test) that since a_n > [sqrt(a_n))/n] for all n after a certain integer, then by virtue of the convergence of sum[a_n], the series with the inferior general term must also converge?
for the last question, an becoming smaller and smaller as N goes to infinity, so square root of an also becoming smaller, and it is also divided by N which expand into infinity
Reminds me of something I was working on for fun. I was trying to figure out different sets of polynomials that were orthogonal from -inf to inf relative to a weight function of 1/(1+x^2)^n
I figured out that you can make the polynomials orthogonal up to degree 2n-1 since you only have to deal with products and the product of the highest two degrees is always zero since it's odd and the product of the 2n-2 with itself is always 2 less than the highest degree in the denominator. As n approaches infinity, they start to turn into Hermite polynomials.
I don't know if that's useful for anything, but I had fun playing with it.
Fantastic minds !! Both of you.
This video is so so much interesting plz make more videos like this
PURE MATHEMATICS
This one is from last year ruclips.net/video/XG3rOsATZZU/видео.html
I'm probably wrong in my thinking, but could we use the comparison test for the last question to show convergence?
Sqrt(a_n)/n is less than or equal to a_n for all n>0.
=> the sum from n=1 to infinity of sqrt(a_n)/n (call this sum 1) is less than or equal to the sum from n=1 to infinity of a_n (call this sum 2).
We know sum 1 is positive because a_n is always positive and n is always positive, so sqrt(a_n)/n is always positive. This means that the series cannot go to -infinity. As well, sqrt(a_n)/n is a decreasing function as n increases, so we won't have to worry about sum 1 oscillating between two values without settling to some value. Since sum 1 is positive, less than or equal to sum 2 (which we claim is convergent), and it's not an oscillating series, then sum 1 is forced to converge.
The attitude of this guy is on point
I think the proof is simple .
Since aₙ >0 then ,
√aₙ /n < √aₙ < aₙ
=> √aₙ /n < aₙ
=> SUM ( √aₙ /n ) < SUM ( aₙ)
But since SUM (aₙ) is convergent . Therefore , SUM ( √aₙ /n ) is also convergent .
√(a_n) some m since the series converges
Will be blush every time he does math
Thank you: this was a lovely duel! You two have a beautiful character! Just one question: could not we answer the last question, just saying that: since
sqr[a(n)]/ n < a(n)/n < a(n) and
Sum of a(n) converges,
then Sum of sqr[a(n)]/ n converges?
Rather than using a scalar product? All the best. M
I was thinking the same, this solution is so easy that I think it's wrong hahah
OHH its wrong because a(n) can be less then 1. So √a(n) not necessarily is less then a(n).
Can you find a serie for which telling if it diverges or converges is impossible ?
Brilliant! Thank you!
They are both such jolly fellows. I would like to have them as teachers but the possibility of that is... well, I will let you work it out.
Well i really wanna understand this . could you suggest some course/book which is pre requisite for this.?
Can Cauchy Schwartz in infinite dimensional vector spaces be utilized for household purposes?
Love this kinda videos.
On the last one , what's wrong with the comparison test:
Let sqrt(a_n)/n ≡ b_n
If a_n > 0, sqrt(a_n) converges.
For the last question:
An > sqrt(An) / n Since n >= 1
Therefore converges by comparison test...?
This is the most beautiful vedio that I have ever seen on this channel 😍
Nothing that gets me more giddy than alternating series.
isnt sqrt(an)/n =2 ?
Another way to do the last exercise :
As sum 1/n^2 and sum an are convergent and as sqrt(a*b)≤(a+b)/2 (classic inequality) then we take a= an and b= 1/n² and so sqrt(an*1/n^2)=sqrt(an)/n is convergent.
Can someone please explain (or direct me to where) I could find what a two Series is?? I’ve been scouring the internet trying to understand this and I’m a little lost 😂
For the third question, I wonder:
sqrt(a_n)/n = a_n/n^2 = a_n * 1/n^2
Would that not be enough to show convergence already?
el mejor crossover que he visto... son geniales .....
Requesting an uncensored version.
why does the change in radius for an infinitesimal increase in x matter for the surface area calculation, but not for the volume calculation? Ie why can we sum up the volume of infinitely thin cylinders for the volume calculation, but not sum up the surface area of these infinitely thin cylinders for the surface area calculation?
For the case in which the question was whether anbn could diverge if an and bn converges, after a while, I thought about (-1)^n 1/sqrt(n). When Peyam realised about it I chuckled
He forgot "Thats it!"
The best anime crossover
I have no idea what's going on but it's so endearing
I think i am falling in love with mathematics 😀❤️ thank you 🔥
For first question: take an=0, bn=1 for odd n; an=1, bn=0 for even n.
There is no limitation on an and bn? Why not simply a=(0,2,0,2,0..) and b=(2,0,2,0..) or any other with same pattern
an = 1+(-1)^n
bn = 1 + (-1)^(n-1)
The limitation is the fact that the series conv or div. That restricts whitch kind of sequence is an and bn
hey dr peyam can you do a video about the solution that took you 5 pages, i mean just explaining how did you get to the answer
thanks
I can feel the happiness throughout the screen🎉