A bit of notation first: ,/' - integral symbol; [a~b] - a to b, included; inf - infinite; Let's accept the challenge for: Sum[n=1~inf] 1/n⁶ Know: Sum[n=1~inf] 1/n² = pi²/6 Sum[n=1~inf] 1/n⁴ = pi⁴/90 as the result in this video Best trial is: f(x) = x³ since [f(x)]² = x⁶ a[0] = 1/2pi ,/'[-pi~pi] x³ dx = = 1/2pi (x⁴/4)[-pi~pi] = 0 a[n] = 1/pi ,/'[-pi~pi] x³ cos(nx) dx Use DI method: D: x³ to 6 to 0; I: cos(nx) to cos(nx)/n⁴ Simplify using: sin(n.pi) = 0 cos(n.pi) = (-1)ⁿ And we find: a[n] = 0
So b[n] is our hope. b[n] = 1/pi ,/'[-pi~pi] x³ sin(nx) dx Use DI method: D: x³ to 6 to 0; I: sin(nx) to sin(nx)/n⁴ Simplify using: sin(n.pi) = 0 cos(n.pi) = (-1)ⁿ And we find: b[n] = 12(-1)ⁿ/n³ - 2pi²(-1)ⁿ/n
Parseval, 1st (left) part: 1/pi ,/'[-pi~pi] x⁶ dx = = 1/pi (x⁷/7)[-pi~pi] = 2pi⁶/7 Parseval, 2nd (right) part, as a[0] = a[n] = 0 then: Sum[n=1~inf] b[n]² = Sum[n=1~inf] (12(-1)ⁿ/n³ - 2pi²(-1)ⁿ/n)² Simplify using: (-1)²ⁿ = 1 And we find: Sum[n=1~inf] 144/n⁶ - 48pi²/n⁴ + 4pi⁴/n² Now replace the Sums by known values: 1/n² per pi²/6 and 1/n⁴ per pi⁴/90 And make Parseval 2 parts altogether: -24pi⁶/45 + 2pi⁶/3 + 144 Sum[n=1~inf] 1/n⁶ = 2pi⁶/7 Solving the fractions we can find the result: Sum[n=1~inf] 1/n⁶ = pi⁶/945 Is that correct?
Pikachu I choose you..... Pikachu 'thunder bolt attack'..... 'The sum goes from 1 to infinity of 1/n⁴' gets a shock and died 'The sum goes from 1 to infinity of 1/n⁴' is solved
Let me get those people names for a second. To solve a series I need to use: - series by Fourier: ok, a historical French guy, good one by the way; - theorem by Parseval: another historical French guy, a bit more obscure, not a knight of Round Table, not Wagner's opera character; - integration method by Lu Chen the inverse of Chen Lu: both (fictional Chinese???) characters, I didn't find anything relevant on Math in Wikipedia with these names, in order to memorize some Calculus methods; - praying, may God quotient rule would not show up or probably another (fictional Chinese???) character called Quo Chen Lu should be memorized; - and now Pikachu, a Japanese manga character... Too much people for me, almost a Legion to defeat a series.
This was a question on a real analysis exam I had. I really love how this is derived and used. I got that one right. Four years later this has helped me, thank you so much.
Tears of joy. I learned Parseval's theorem, Fourier series and odd/even functions -- all from this single video. Thank you for existing, you incredibly adorable Pikachu.
I have never seen the fourier series written like that... Shouldn't it be the sum from t = negative infinity to t = infinity of a_n * e^int Where the a_n coefficient is 1/2pi integral of f(x) e^-inx dx
BPRP Pokemon's theorem: The number(n) of likes in this video is directly proportional of the number of Pikachu occurring in the future, where n is natural number and tends to positive infinity
Thank you, very very interesting, and the Pikachu outfit is really cool, we want more! I had one doubt about this proof: why does it work at all, considering that x^2 is not a periodic signal (hypothesis of the Fourier series and of the version of Parseval’s theorem you used)? Plotting the reconstructed signal with a0 and a dozen of an, it becomes clear that the function used in the proof is not x^2 over R: it is the repetition along R of the function x^2 defined over the closed interval -pi, pi, which is of course periodic. Thanks!
Hahah 😂 Oi pikachu, wise choice of doing it with Parseval theorem and y=x^2, the version without the theorem yet with the use of y=x^4 was a nightmare of a chan lu xD
AhmedHan Only when n is even. And, if you are in advanced mathematics and you are working with well-defined divergent summations, then there is also on for negative n. There is no formula for positive odd n, though, at least not yet.
Hey BPRP. This video got me so interested in if there is a way to calculate f(x) so the part of the sum (an)^2+(bn)^2 is a specific serie i wanna calculate?
Great video! But you forgot to say that the procedure is only true if we consider the function with period 2π, for any other period the coefficients would change a little
You can find the value of the zeta function at any positive even power using Fourier series, but there is no simple pattern to it. You can still find a general formula, though. In fact, if you’ve ever heard of the Bernoulli numbers before, you may be surprised to hear that they arise in this context.
If we want to approximate pi, we multiply both sides of the infinite sum by the denominator and take the nth root.so which sum is more accurate when approximating pi for the same number of terms The sum of n^6,n^4 or n^2?
If you want to use this method to find what a sum converges to in general, how do you pick the function? Do I just chose the square root of denominator like you did here? For example, if I wanted to find the infinite series of 1/n^6, would I choose x^3 for my function?
Does it matter if the problem specifies a different interval? Like in the video, integrals are from -pi to pi, but if the problem states x^2 is from 0 to pi, then should we integrate on that interval instead?
Mr. Cao, if f(x)=sqrt(x), what does this mean for the Riemann zeta function of 3? Does this mean that the Riemann zeta function of 3 cannot be written in closed form?
I've never seen Parseval's theorem before though. Is this something you prove on your course? I would also question how valid the proof is: given that Fourier series are 2pi-periodic, clearly the series can only converge on an interval of length 2pi. Is this sufficient to be able to use Parseval?
This infinite series converges to an irrational numerical quantity. For even powers of the series, the sum contains pi raised to the same even power. Is this the case with odd powers of the series? I salute the French Mathematicians.
I know it seems like a bit of a niche topic‚ but is there a way that you could teach some mathematical vocabulary in Chinese (or point me towards somewhere where I can learn it xD). 我从去年十一月起学习中文, 但是我还不会谈论数学。Which is a shame because it's one of my favorite topics. Edit: Though, granted, not one I get to talk about often, not being in school anymore xD
For the sum of reciprocals of sixth powers, let's just call it S let f(x)=x^3 a_0=0 because x^3 is odd a_n=0 because (x^3)(cos(nπ)) is odd Using DI method, we can find out that, b_n =cos(nπ)(-2π^2/n + 12/n^3) =(-1)^n × (-2π^2/n + 12/n^3) so using Parseval Theorem and substituting the sum of reciprocals of fourth powers (π^4/90) and the sum of reciprocals of squares (π^2/6), we can find out that, 2π^6/7 = 144S - 8π^6/15 + 2π^6/15 144S = 16π^6/105 S=π^6/945 so the sum of reciprocals of sixth powers is π^6/945.
This only works for sums of even powers of the denominator as square roots don't work for negative numbers UNLESS you want the sum of real numbers to give you a complex number
If it's popular do it again
We will see!!
Perhaps 99 more times?
A bit of notation first:
,/' - integral symbol;
[a~b] - a to b, included;
inf - infinite;
Let's accept the challenge for:
Sum[n=1~inf] 1/n⁶
Know:
Sum[n=1~inf] 1/n² = pi²/6
Sum[n=1~inf] 1/n⁴ = pi⁴/90
as the result in this video
Best trial is:
f(x) = x³ since [f(x)]² = x⁶
a[0] = 1/2pi ,/'[-pi~pi] x³ dx =
= 1/2pi (x⁴/4)[-pi~pi] = 0
a[n] = 1/pi ,/'[-pi~pi] x³ cos(nx) dx
Use DI method:
D: x³ to 6 to 0;
I: cos(nx) to cos(nx)/n⁴
Simplify using:
sin(n.pi) = 0
cos(n.pi) = (-1)ⁿ
And we find:
a[n] = 0
So b[n] is our hope.
b[n] = 1/pi ,/'[-pi~pi] x³ sin(nx) dx
Use DI method:
D: x³ to 6 to 0;
I: sin(nx) to sin(nx)/n⁴
Simplify using:
sin(n.pi) = 0
cos(n.pi) = (-1)ⁿ
And we find:
b[n] = 12(-1)ⁿ/n³ - 2pi²(-1)ⁿ/n
Parseval, 1st (left) part:
1/pi ,/'[-pi~pi] x⁶ dx =
= 1/pi (x⁷/7)[-pi~pi] = 2pi⁶/7
Parseval, 2nd (right) part, as a[0] = a[n] = 0 then:
Sum[n=1~inf] b[n]² =
Sum[n=1~inf] (12(-1)ⁿ/n³ - 2pi²(-1)ⁿ/n)²
Simplify using:
(-1)²ⁿ = 1
And we find:
Sum[n=1~inf] 144/n⁶ - 48pi²/n⁴ + 4pi⁴/n²
Now replace the Sums by known values:
1/n² per pi²/6
and
1/n⁴ per pi⁴/90
And make Parseval 2 parts altogether:
-24pi⁶/45 + 2pi⁶/3 + 144 Sum[n=1~inf] 1/n⁶ = 2pi⁶/7
Solving the fractions we can find the result:
Sum[n=1~inf] 1/n⁶ = pi⁶/945
Is that correct?
This is the best gaming channel in youtube hands down.
Pikachu I choose you.....
Pikachu 'thunder bolt attack'.....
'The sum goes from 1 to infinity of 1/n⁴' gets a shock and died
'The sum goes from 1 to infinity of 1/n⁴' is solved
I've seen this identity a lot but I've never seen such a nice, elegant proof for it. This video was great, keep it up!
Riley Wells thank you!!!
0:09
Pikachu used Stare:
*Its super effective*
RAJAS SURLIKAR hahahaha thank you!!!
@@blackpenredpen
^_^
Let me get those people names for a second. To solve a series I need to use:
- series by Fourier: ok, a historical French guy, good one by the way;
- theorem by Parseval: another historical French guy, a bit more obscure, not a knight of Round Table, not Wagner's opera character;
- integration method by Lu Chen the inverse of Chen Lu: both (fictional Chinese???) characters, I didn't find anything relevant on Math in Wikipedia with these names, in order to memorize some Calculus methods;
- praying, may God quotient rule would not show up or probably another (fictional Chinese???) character called Quo Chen Lu should be memorized;
- and now Pikachu, a Japanese manga character...
Too much people for me, almost a Legion to defeat a series.
The only acceptable way to teach math.
Eric Cintron hahahahaha thanks!!!!
This was a question on a real analysis exam I had. I really love how this is derived and used. I got that one right. Four years later this has helped me, thank you so much.
yeah, im pikachu:
Pretty
I good
K at
A calculus
C
H
U
X Ry yay!!!!!
Yeah I'm Pikachu
P why
I i
K hate
A trigonometry
C so
H hek
U Much
Lmao
Tears of joy.
I learned Parseval's theorem, Fourier series and odd/even functions -- all from this single video. Thank you for existing, you incredibly adorable Pikachu.
That's very nice . Continue uploading videos like this . It's awesome
you saved me, I needed this proof to write my monography in maths... I'm completely thankful :')
bprp: *wears costume*
audience: *surprised pikachu face*
Sensei hahahahahahaha!!!
Amazing video! Thanks for helping me prepare for my AP Calculus BC test dad
More videos like this please ! 😂😂😂 I LOVED the costume !!
ThisIsEduardo lol thanks!!!
But I thought Pikachu is a detective, not a maths professor.
WarpRulez hahahaha it can be anything!!
u should send the math professor pikachu idea to game freak
Idk why but u made me spill my water lmao
BTW nice video as always!
ianvideos314 lol ok!
😂😂😂 the first part. Great video. Saw u first time without spectacles
0:15 That's the second most lovely Pikachu I've ever seen
The No.1 is of course 0:08
How about a general formula for summation of I/(n^(2k)) as n goes from 1 to infinity. Now that would be a thunderbolt for Pikachu.
that needs Raichu
Or 1000 pichus
Hahahahahaha
@@blackpenredpen but seriously can we use Fourier series to find it out.
math.stackexchange.com/questions/1948206/sum-n-1-infty-frac1n6-frac-pi6945-by-fourier-series-of-x2
I have never seen the fourier series written like that...
Shouldn't it be the sum from t = negative infinity to t = infinity of a_n * e^int
Where the a_n coefficient is 1/2pi integral of f(x) e^-inx dx
A wild e^x appeared.
Pikachu used differentiate:
d/dx (e^x) = e^x.
It was not very effective...
pikachu used ∂/∂y,
∂/∂y(e^x)=0
it's super effective
This channel is just amazing.
Nacho thank you!!!
OMG i've never seen before the method you use at 2:58 to integrate by part, it's awesome !
You can check out my DI method video
Were you just in Dusseldorf?
There was Japan day, when you released this video. And you could see there many pikachus!
BPRP Pokemon's theorem: The number(n) of likes in this video is directly proportional of the number of Pikachu occurring in the future, where n is natural number and tends to positive infinity
Mak Vinci hahaha I hope so too
@@blackpenredpen Yeah sure you can Pika~ hahaha
Thank you, very very interesting, and the Pikachu outfit is really cool, we want more!
I had one doubt about this proof: why does it work at all, considering that x^2 is not a periodic signal (hypothesis of the Fourier series and of the version of Parseval’s theorem you used)? Plotting the reconstructed signal with a0 and a dozen of an, it becomes clear that the function used in the proof is not x^2 over R: it is the repetition along R of the function x^2 defined over the closed interval -pi, pi, which is of course periodic. Thanks!
I love this Pikachu fan who switches pen with a lightning
Wow...pikachu...😆😆
btw u r looking cute..😊
Saumy tiwari thanks!!!!
3:08 TECHNICAL DIFFICULTIES, PLEASE STAND BY. Error code: HTTP-404: Full Blue Pen Not Found.
Jared Kaiser lolll
Just what I needed to start my morning.
Detective Redpen solves the case again.
I really like your videos, could please make a video on the concept of locus, i'm pretty much confused about the topic.
THIS VIDEO WAS LIFE SAVERRRRR!!!! math methods made easy!
I wish so, so, so much that this man were my calc1 instructor.
Eddie Mercury awww I am tho, on YT
Hahah 😂 Oi pikachu, wise choice of doing it with Parseval theorem and y=x^2, the version without the theorem yet with the use of y=x^4 was a nightmare of a chan lu xD
Is there a general formula for sum of 1/x^n, for all x element of positive integers?
AhmedHan Only when n is even. And, if you are in advanced mathematics and you are working with well-defined divergent summations, then there is also on for negative n. There is no formula for positive odd n, though, at least not yet.
Hey BPRP. This video got me so interested in if there is a way to calculate f(x) so the part of the sum (an)^2+(bn)^2 is a specific serie i wanna calculate?
Alessandro Villanueva Cantillo I will prove it today.
Blackpikachuredpikachu!!!! :3
Dr Peyam hahahaha
Great video!
But you forgot to say that the procedure is only true if we consider the function with period 2π, for any other period the coefficients would change a little
Can you prove Parseval’s theorem
It's simple just square the Fourier expansion and simplify then just integrate from -π to π
Between the math and the costume I feel like I'm having a fever dream.
Fantastic ......
But I have got quastion
Do you think with solution or you found it when you searsh ?
Of course, a^2+b^2=(a+b)^2.
Profit.
Pikachu used Fourier series....
It was super effective!
hahaha thank you!!!
Is it possible to find a formula to work out the sum from 1 to infinity of 1/n^(2m)?
You can find the value of the zeta function at any positive even power using Fourier series, but there is no simple pattern to it. You can still find a general formula, though. In fact, if you’ve ever heard of the Bernoulli numbers before, you may be surprised to hear that they arise in this context.
I was thinking the same
Thankyou so so so much for helping out when I needed this explanation the most.... Just loved the way you explained it....😍🥳
And he continues to solve the world's problems...
OhnoItsFranc Hahahah yup
how can i deal with sigma 1/(2k-1)^4? k from 1 to infinity
Why the Pikachu costume???
If we want to approximate pi, we multiply both sides of the infinite sum by the denominator and take the nth root.so which sum is more accurate when approximating pi for the same number of terms
The sum of n^6,n^4 or n^2?
What about Onix next? Or maybe Snorlax
Nitro Zox maybe?!!!
Integral 1/(x^2-1)^2 dx please
integration of x^3/e^x-1..... Plz solve it
hey BPRP . what is the function for the exercise ? still x² ???
If you want to use this method to find what a sum converges to in general, how do you pick the function? Do I just chose the square root of denominator like you did here? For example, if I wanted to find the infinite series of 1/n^6, would I choose x^3 for my function?
how do i explain to my friends that a pikachu is teaching me calculus?
Take hmmm it’s kinda hard...
What if i want to choose n is a trigonometry function, will it work?
how can i do with (-1)^k/2k+1 and f:x? thanks
OH MY . . . Fourier and Parseval . . . seriously?
Have you EVER introduced them in your previous videos yet?
Peter Chan yes. You can see my description for links. I am almost done teaching my spring classes so I can do some other topics soon
Btw, I will prove the Parseval’s theorem soon
pika pika pikachu...it means i like your work.
Aman kashyap hahahah thanks!!!
When UA-cam pushes Pikachu content because of a movie but you have a math channel to run.
Which university do you teach at?
how to know which function f(x) I should take for a given sum? (for exemple sum (1/(2k -1)^2) as k -> infinity)?
Osea que con este método se puede hallar la sumatoria de cualquier serie ?
the more i look at the thumbnail, the more that pikachu creeps me out
Coud you explain another proof of this formula which involves decomposition of sin(x) into infinite product?
Борис Назаров
Max already did that for the sum of 1/n^2 two years ago. You can see that in my description
Can you please do a video on the cauchy condensation test for series?
Does it matter if the problem specifies a different interval? Like in the video, integrals are from -pi to pi, but if the problem states x^2 is from 0 to pi, then should we integrate on that interval instead?
I wanted a suprised pikakuchu meme reference xD.
Ps: nice video as always
*Dresses as Pikachu*
Holds a Pokeball 😂
love that vid keep it up
that "holds a pokeball" made me luahg!!! lolllll
Awesome bro. Very well explained 👏
I choose you, x^2!
I have a question!How can you use Fourier series for a non-periodic function??
I never thought that Pikachu was such a good maths teacher.
Love your video as always
JarjisH_yper Hahahah thank you!!!!
more fourier pls! Really enjoyed the video
(also more pikachu pls)
this makes me wonder. If you or Peyam got to play PMD, would either of you get Pikachu as the Pokemon you become?
Next: Zeta(pi)
U will need pikachu again ;D
Okay, but how will Bulbasaur solve this?
Can you prove it without using parseval's formula???
Mr. Cao, if f(x)=sqrt(x), what does this mean for the Riemann zeta function of 3? Does this mean that the Riemann zeta function of 3 cannot be written in closed form?
hey bro can you please integrate 4x/(1+x)(1+x^2)^2
My life is officially complete and everyone who watches this is immortal.
hahahaha, nice!
I've never seen Parseval's theorem before though. Is this something you prove on your course? I would also question how valid the proof is: given that Fourier series are 2pi-periodic, clearly the series can only converge on an interval of length 2pi. Is this sufficient to be able to use Parseval?
blackpenredpen: sin n pi
Me: senpai!
This infinite series converges to an irrational numerical quantity. For even powers of the series, the sum contains pi raised to the same even power. Is this the case with odd powers of the series? I salute the French Mathematicians.
For a0 its 1/2pi or 1/pi
I know it seems like a bit of a niche topic‚ but is there a way that you could teach some mathematical vocabulary in Chinese (or point me towards somewhere where I can learn it xD). 我从去年十一月起学习中文, 但是我还不会谈论数学。Which is a shame because it's one of my favorite topics.
Edit: Though, granted, not one I get to talk about often, not being in school anymore xD
I support the idea, I'm interested myself
Hey Bprp, have you done a video on the proof of Parsevel? I couldn't find one if you did. If not, you totally should
will newman dr. P might have already done that. I am not sure tho.
Yes. Try to make sum of 1/n^3 and 1/n^5 also...
So this is what teachers do after the school year is over
Volticat it’s not over yet......
Pikachu used *calculus skills*
Summation has been defeated
OH MY .... LORD! THE MATHS!!!
C. D. Chester hahaha thank you!!!!
Thank you!
Why does Pikachu sound so unusual? I expected a bunch of pika pikas
It's been too long since I watched these when I wasn't studying😂
so apparently blackpikaredpika can use the world to erase his board...
Well done pikachu,
For the sum of reciprocals of sixth powers, let's just call it S
let f(x)=x^3
a_0=0 because x^3 is odd
a_n=0 because (x^3)(cos(nπ)) is odd
Using DI method, we can find out that,
b_n
=cos(nπ)(-2π^2/n + 12/n^3)
=(-1)^n × (-2π^2/n + 12/n^3)
so using Parseval Theorem and substituting the sum of reciprocals of fourth powers (π^4/90) and the sum of reciprocals of squares (π^2/6), we can find out that,
2π^6/7 = 144S - 8π^6/15 + 2π^6/15
144S = 16π^6/105
S=π^6/945
so the sum of reciprocals of sixth powers is π^6/945.
This only works for sums of even powers of the denominator as square roots don't work for negative numbers UNLESS you want the sum of real numbers to give you a complex number