Because of this video i found an interesting formula, could you take a look? Summation from n=1 to infinity of ( n / a^n ) equals to a / (a-1)^2 For any a s.t. |a|>1
Hi blackpenredpen I have a doubt... i^4 = 1 right => Now we raise both sides to 1/4 th power => (i^4)^1/4 = (1)^1/4 => i^1 = 4th root of 1 => i^1 = 1 => i = 1 => sqrt (-1) = i = 1 How is this possible?
@@rodrigolopez3874 i found this formula watching another summition video about 0.01+0.002+0.0003..... its nice to see that his vids make us create some cool formulas
[from my latest post] Hi all, The doctor is confident to say that Lars is a stage 3 cancer survivor!!! This is such a great news and we are beyond happy!!! Here's the video message from Lars to everyone: th-cam.com/video/LB1xWStHFPc/w-d-xo.html My video message to him: th-cam.com/video/LX-BE_o5y_U/w-d-xo.html Lars, Peyam and I are planning an livestream interview. So please leave a question that you'd like to know about Lars. Also if you have a good way to do a livestream interview, please let me know as well. Thank you! Backstory Lars is a long-term viewer of mine and@Dr Peyamfrom Belgium. About 4 months ago, he left a comment on my video saying that he was battling with cancer and was also watching my videos in the hospital. I felt very touched since I never thought that my videos could provide spiritual support to help him feel better. I had an idea of making a video dedicated to him to support him to my fullest. He told me that he really liked my "easiest integral on youtube" video but couldn't solve the integral of (tan(x))^(2/3). I tried it and got a solution. However during that time, I was preparing my calc 2 students and many other viewers to get better with integrals. I had done a video on the techniques/strategies th-cam.com/video/M5MaGUO0JDs/w-d-xo.html but many of you guys wanted more. Then while I was preparing for another similar video, a thought came to my mind: "** it, why don't I just solve 100 of them!". I postponed the "Lar's Integral" but my 100 integrals video was born. I really want to thank everyone who had left a supporting comment that help to put a smile on Lar's face during the tough time. *LARS IS A STAGE 3 CANCER SURVIVOR NOW AND I AM BEYOND HAPPY* Thank you all, and I wish everyone have a great great life! 100/(1-x)
Thank you very much blackpenredpen, you are the first one giving me a true intuition on why the inversion of order of summation is done like that. It's been bugging for a while and I feel freed now! Thank you for this awesome video and all of your other content!
I am very glad to help!!! This is actually really cool bc it's similar to change the order of a double integral. You can check out this longer version: th-cam.com/video/u1BtAhtCRcw/w-d-xo.html
Or you can draw a table m 1 2 3 4 ... n 1 x 2 x x 3 x x x 4 x x x x . . . changing horizontal summation to vertical summation will get you the upper and lower limits of summation.
I knew it wasn't an integral sign but I still got excited, because I've been banging my head against this for too long! I tried this among other ways, which (I now know) were wrong. Now I know which is right, and why - hence the excitement! Excellent!
Great video! I almost thought that there _was_ an integral at 5:58 and I got suprised... Congratulations on 300,000 subscribers! I'm glad to be one of them.That was a very interesting concept, and I can't wait for the next video!
It might be easier to do this way: First take 1/2^m out side and inside is geometric summation so (1/2*(1-(1/2)^m))/(1-1/2)=(1-(1/2)^m) then you are left with two geometric series and you get 1-sum(1/4^m) and you get 1-1/3=2/3 so YAY
Without changing the order of the double summation, first take the 1/2^m out of the inner summation since the index variable that changes is n, then use the fact that the inner summation is a finite geometric series with common ratio r = 1/2 to express it in terms of m (there's a formula for this), and finally evaluate the infinite geometric series.
I did this... really poorly, but was pleased when I also got the correct answer! I expanded out a handful of terms from the series and looked for a pattern. I found that each exponent of 2 was repeated floor(exponent/2) times. So 2^-2 and 2^-3 showed up once, 2^-4 and 2^-5 showed up twice, 2^-6 and 2^-7 showed up thrice, etc. This let me change from a double sum into a single sum: Sum of i=2 -> Infinity of 2^-i * floor(i/2), but that's not something I could work with nicely. So I expanded out sample terms from that and started simplifying, getting an easy 2^-1+2^-3+2^-5+2^-8, oh, so close. Expanding more terms got me the 2^-7 and a 2^-10 with a long tail of numbers ready to collapse it. Good enough for me to suspect my pattern was close, so I wrote down Sum of j=0 -> Infinity of 2^(2j+1) and then solved that infinite geometric series.
Thank you friend! Helped me a lot with a formula with probability generating functions where I couldn't understand why indices change when you change the order of summation
When you sum the n first, the sum over m becomes SUM 1->inf[(1/(2^m)) - (1/(2^(2*m)))] = SUM 1->inf[(1/(2^m)) - (1/(4^m))]. Both pieces converge, so you can sum them independently to be (1/(2-1)) - (1/(4-1)) = 1 - 1/3 = 2/3.
Oh my good. You helped me a lot. I was totaly confused how I can exchange the order of a double summation. Zero information in the internet in my mother tongue: german.
General Formula: ∑{α ≤ m ≤ ∞} ∑{β ≤ n ≤ m} c ᵐᐩⁿ = c ᵅᐩᵝ / (1 - c)² (1 + c) Iff for (α, β, c) = (1, 1, 1/2) then it turns out as special case the ▶-Formula: 2/3 ≈ 0.666 ...
I changed the order of summation and took out the factor of 1/2^n; then I used the identity Sum[1..inf](1/2^m) = 1 and Sum[n..inf](1/2^m) = Sum(1..inf)(1/2^m) - Sum[1..n-1](1/2^m) to get 1 - Sum[n=1..inf](Sum[m=1..n-1](1/2^(m+n)) = 1 - Sum[n=1..inf](Sum[m=1..n](1/2^(m+n)) - 1/2^(2n)). Then if S1 is the original sum, S1 = 1 - S1 + Sum[n=1..inf](1/2^(2n)). Let S2 = Sum[n=1..inf](1/2^(2n)), then 2S2 = 2Sum[n=1..inf](1/2^(2n)) = Sum[n=1..inf](1/2^(2n-1)). Then 3S2 = Sum[n=1..inf](1/2^(2n)+1/2^(2n-1)) = 1 so S2 = 1/3 and 2S1 = 1 + 1/3 and S1 = 2/3.
0:00-0:04 Let me tell you guys that the answer to this right here is... You do it. : ). #bprpYAY Speaking of sums, remember the video where you used the Fourier Series and Parseval's Theorem to calculate sum(1/(n^4)) for n=1->infinity and got the sum of pi^4/90? what if f(x)=sqrt(x)? Can this prove that there is no closed form for Apery's constant?
Everything he did here can be done by drawing a triangle. For any other shape, it's basically the same method... you just need to find the reverse function.
The first term in the summation is 1/2^m, and then each successive term is the previous term times 1/2. When that happens you have a geometric series (google that for more) which has an infinite convergent sum of a_1 / (1-r), as long as r is smaller than 1. Since r=1/2, it does converge so bprp can replace the summation with the fraction he does.
Actually, the original summation order is also pretty easy, after noticing that ∑[n=1,m] 1/2ⁿ is always short of 1 by exactly the last included term, so it's 1 - 1/2ᵐ : ∑[m=1,∞] ∑[n=1,m] 1/2ᵐ⁺ⁿ = ∑[m=1,∞] (1/2ᵐ ∑[n=1,m] 1/2ⁿ) = ∑[m=1,∞] 1/2ᵐ [1 - 1/2ᵐ] = ∑[m=1,∞] (1/2ᵐ - 1/4ᵐ) = 1 - ⅓ = ⅔ Also, another way to find the limits for the re-ordered sum, is to draw m,n axes; plot the grid points called out by the original summation; then notice how the indices have to be bounded when summed in the new order, so that the same grid points are covered. Fred
Hey dude. i was navigating through internet and i depared myself with this problem: int_0^1 e^x(1+e^x/(cosx)^2) dx. After several tries, i got nowhere, so can you solve it in a video?
Can i get help how to solve this question: Find Xs such that when you divide X by 2 the remainder is 1 and when divided by 3 the rem. is 2 and when divided by 4 the rem.is 4 and when divided by 5 the rem. is 4 and when divided by 6 the rem. is 5 And it a pure number theory/congruence question (modulos aren't relatively prime) And if #BPRP can explaim this topic in a video it would be so cool Thanks guys❤️
The lowest answer is 59, but the general solution is 60*n - 1 for any integer n greater than 0. This and any problem like this can be resolved by the chinese remainder theorem, take a look here: en.wikipedia.org/wiki/Chinese_remainder_theorem
Hey, can you solve this integral: (x^(-1/2) *ln(x))/(x-1) dx from 0 to infinity. But solve the integral using residues. That is a really interesting thing.
I've thought about a simple problem this week and I couldn't solve it. The problem is the following: "can we find a general expression for the coeficients of the polynomial P(x)=(x-1)(x-1/2)(x-1/4)...?". Please, I'm really getting crazy
Do you mean the switch of the summation? That works generally. If you mean pulling one term to the front, that works as long as the term you're moving does not contain the inner index. So it would also work with (1/2^m)*(1/3^n)
It looks like Cauchy product of sum(1/4^(m)*1/2^(n-m),m=1..infinity,n=1..infinity) (sum(1/4^(m),m=1..infinity)*sum(1/2^n,n=1..infinity) Whats wrong with this approach ?
Longer version: th-cam.com/video/u1BtAhtCRcw/w-d-xo.html
Because of this video i found an interesting formula, could you take a look?
Summation from n=1 to infinity of ( n / a^n ) equals to a / (a-1)^2
For any a s.t. |a|>1
Hi blackpenredpen I have a doubt...
i^4 = 1 right
=> Now we raise both sides to 1/4 th power
=> (i^4)^1/4 = (1)^1/4
=> i^1 = 4th root of 1
=> i^1 = 1
=> i = 1
=> sqrt (-1) = i = 1
How is this possible?
@@supremeassassin3478 When u do the 4th root of 1 you get 4 results {1, -1, i, -i}, you are equaling 2 diferentes roots
@@rodrigolopez3874 oh... Thx so much
Ur a God for me
My brain has been exploding because of this question for weeks
Thx
@@rodrigolopez3874 i found this formula watching another summition video about 0.01+0.002+0.0003.....
its nice to see that his vids make us create some cool formulas
"This is not an integral sign so don't get too excited." 😄
I literally laughed out loud after he said that. :P
Right! But integrals function the same way because it has summation in it too.
😂😂
I WAS SO EXCITED. XD /j
[from my latest post] Hi all,
The doctor is confident to say that Lars is a stage 3 cancer survivor!!! This is such a great news and we are beyond happy!!! Here's the video message from Lars to everyone: th-cam.com/video/LB1xWStHFPc/w-d-xo.html
My video message to him: th-cam.com/video/LX-BE_o5y_U/w-d-xo.html
Lars, Peyam and I are planning an livestream interview. So please leave a question that you'd like to know about Lars. Also if you have a good way to do a livestream interview, please let me know as well. Thank you!
Backstory
Lars is a long-term viewer of mine and@Dr Peyamfrom Belgium. About 4 months ago, he left a comment on my video saying that he was battling with cancer and was also watching my videos in the hospital. I felt very touched since I never thought that my videos could provide spiritual support to help him feel better.
I had an idea of making a video dedicated to him to support him to my fullest. He told me that he really liked my "easiest integral on youtube" video but couldn't solve the integral of (tan(x))^(2/3). I tried it and got a solution. However during that time, I was preparing my calc 2 students and many other viewers to get better with integrals. I had done a video on the techniques/strategies th-cam.com/video/M5MaGUO0JDs/w-d-xo.html but many of you guys wanted more. Then while I was preparing for another similar video, a thought came to my mind: "** it, why don't I just solve 100 of them!".
I postponed the "Lar's Integral" but my 100 integrals video was born. I really want to thank everyone who had left a supporting comment that help to put a smile on Lar's face during the tough time.
*LARS IS A STAGE 3 CANCER SURVIVOR NOW AND I AM BEYOND HAPPY*
Thank you all, and I wish everyone have a great great life!
100/(1-x)
We love u Lars
This concept has been crushing me. Great vid. Seeing the inequalities made it clear.
Fubiiini!!! 😄
Thank you very much blackpenredpen, you are the first one giving me a true intuition on why the inversion of order of summation is done like that. It's been bugging for a while and I feel freed now! Thank you for this awesome video and all of your other content!
I am very glad to help!!! This is actually really cool bc it's similar to change the order of a double integral. You can check out this longer version: th-cam.com/video/u1BtAhtCRcw/w-d-xo.html
@@blackpenredpen Multivariables Calc exam coming soon and here I am doing my revisions with help of your videos^^, I'm gonna check it out, thank you!
Or you can draw a table
m 1 2 3 4 ...
n
1 x
2 x x
3 x x x
4 x x x x
.
.
.
changing horizontal summation to vertical summation will get you the upper and lower limits of summation.
I like that visualization
I didn't understand that
Please explain that to mw
"Let me tell you that the answer for this iiiiis" :
:)
I knew it wasn't an integral sign but I still got excited, because I've been banging my head against this for too long! I tried this among other ways, which (I now know) were wrong.
Now I know which is right, and why - hence the excitement! Excellent!
Great video! I almost thought that there _was_ an integral at 5:58 and I got suprised... Congratulations on 300,000 subscribers! I'm glad to be one of them.That was a very interesting concept, and I can't wait for the next video!
Thank you!!!!!
It might be easier to do this way: First take 1/2^m out side and inside is geometric summation so (1/2*(1-(1/2)^m))/(1-1/2)=(1-(1/2)^m) then you are left with two geometric series and you get 1-sum(1/4^m) and you get 1-1/3=2/3 so YAY
Finally an easier explanation.Thank you
Without changing the order of the double summation, first take the 1/2^m out of the inner summation since the index variable that changes is n, then use the fact that the inner summation is a finite geometric series with common ratio r = 1/2 to express it in terms of m (there's a formula for this), and finally evaluate the infinite geometric series.
Congrats with 300k!! I love double summation! Nice video!:)
❤️❤️ math for ....
Hey blackpenredpen ,could you do a video on divergent series?
I really seem to be having a trouble in grasping the concept.
I did this... really poorly, but was pleased when I also got the correct answer! I expanded out a handful of terms from the series and looked for a pattern. I found that each exponent of 2 was repeated floor(exponent/2) times. So 2^-2 and 2^-3 showed up once, 2^-4 and 2^-5 showed up twice, 2^-6 and 2^-7 showed up thrice, etc. This let me change from a double sum into a single sum: Sum of i=2 -> Infinity of 2^-i * floor(i/2), but that's not something I could work with nicely. So I expanded out sample terms from that and started simplifying, getting an easy 2^-1+2^-3+2^-5+2^-8, oh, so close. Expanding more terms got me the 2^-7 and a 2^-10 with a long tail of numbers ready to collapse it. Good enough for me to suspect my pattern was close, so I wrote down Sum of j=0 -> Infinity of 2^(2j+1) and then solved that infinite geometric series.
Thank you friend!
Helped me a lot with a formula with probability generating functions where I couldn't understand why indices change when you change the order of summation
Great video. This explanation was the best !!!! 😃
like switching order of double integral also can draw graph
Spherical microphone is awesome
Let's see the sum of cubes as an iterated sum!!
Hahaha, did you just come from Dr. P?
@@blackpenredpen I could not resist!!
Hahahaha, I was like “Hey, this sounds familiar!” 😂
@@drpeyam
And here's the man!!!
When you sum the n first, the sum over m becomes SUM 1->inf[(1/(2^m)) - (1/(2^(2*m)))] = SUM 1->inf[(1/(2^m)) - (1/(4^m))]. Both pieces converge, so you can sum them independently to be (1/(2-1)) - (1/(4-1)) = 1 - 1/3 = 2/3.
Thanks for double summation
Love the magic sound effect 😁
Good video on double summation 👍
Hahahah, thank you Vinci!
wonderful explanation, congrs
Another great BPRP video. Thanks
Oh my good. You helped me a lot. I was totaly confused how I can exchange the order of a double summation. Zero information in the internet in my mother tongue: german.
5:58 this is not an integral sign, so don't get too excited.
*sad noises*
Thanks
General Formula: ∑{α ≤ m ≤ ∞} ∑{β ≤ n ≤ m} c ᵐᐩⁿ = c ᵅᐩᵝ / (1 - c)² (1 + c)
Iff for (α, β, c) = (1, 1, 1/2) then it turns out as special case the ▶-Formula: 2/3 ≈ 0.666 ...
5:58 oh ding a ling I was really excited
i did it without changing the order and i got 2/3 ,it fells great to get it right
poor example in my opinion as this is symmetrical if you swap m and n... however enjoyable and well explained
Can you please do a video on the Riemann rearrangement theorem? I would like to understand more about it. 😃
2:02, can someone please explain this step again? Why des m=n there?
3:23 what if m was starting at 2? what would we do?
I changed the order of summation and took out the factor of 1/2^n; then I used the identity Sum[1..inf](1/2^m) = 1 and Sum[n..inf](1/2^m) = Sum(1..inf)(1/2^m) - Sum[1..n-1](1/2^m) to get 1 - Sum[n=1..inf](Sum[m=1..n-1](1/2^(m+n)) = 1 - Sum[n=1..inf](Sum[m=1..n](1/2^(m+n)) - 1/2^(2n)). Then if S1 is the original sum, S1 = 1 - S1 + Sum[n=1..inf](1/2^(2n)). Let S2 = Sum[n=1..inf](1/2^(2n)), then 2S2 = 2Sum[n=1..inf](1/2^(2n)) = Sum[n=1..inf](1/2^(2n-1)). Then 3S2 = Sum[n=1..inf](1/2^(2n)+1/2^(2n-1)) = 1 so S2 = 1/3 and 2S1 = 1 + 1/3 and S1 = 2/3.
hi thank's, you done a good job. can you make a video about hypergeometric functions ?
Was very hyped for big integral. Time to invent new maths then
When Guido and our best friend shake hands.
But what about a triple summation?🤔
Richik Haldar It generalizes fairly nicely, although it becomes easier to express the inequality as a system of inequalities instead.
That's it!
Yup!
Excellent！
0:00-0:04 Let me tell you guys that the answer to this right here is...
You do it. : ).
#bprpYAY
Speaking of sums, remember the video where you used the Fourier Series and Parseval's Theorem to calculate sum(1/(n^4)) for n=1->infinity and got the sum of pi^4/90? what if f(x)=sqrt(x)? Can this prove that there is no closed form for Apery's constant?
Nobody’s talking about that voice crack at 0:50 lol
Doraemon intro music. Nice touch.
You SHOULD EXPLAIN OTHER CASES, WHEN THE INDICES START FROM 0, 2, 3, ETC TO n-1, n-2, ETC.
I did without changing values, just using Sum (k from 1 to n) (q^k) = (q-q^(k+1)) /(1-q)
very awesome but can you solve poincare problem?
Amazing!
I love your videos!!!
Really cool and interesting video. Series are my favorite math concepts!
Find to formula y=r Cartesian and polar coordinats
Amazing
@blackpenredpen
It seems like this problem has a sort of symmetry with m and n. What if it didn't, specifically what if it was originally 1/2^(m-n)?
And yet, I didn't understand how we changed the summation order.
Imagine if you write my mathematics exam paper.....
i'll get 200 out of 100....
I owe you
Hi sir. How to convert product of double summation into single summation for general limits. Give me an idea.
wow, i just read an article about fubini principle. What a coincidence 🤔
What if the limits of summation were different for the double summation? How would we switch the order then?
Everything he did here can be done by drawing a triangle.
For any other shape, it's basically the same method... you just need to find the reverse function.
Can someone explain the step that he made at 5:40? When summing the black series?
The first term in the summation is 1/2^m, and then each successive term is the previous term times 1/2. When that happens you have a geometric series (google that for more) which has an infinite convergent sum of a_1 / (1-r), as long as r is smaller than 1. Since r=1/2, it does converge so bprp can replace the summation with the fraction he does.
Actually, the original summation order is also pretty easy, after noticing that
∑[n=1,m] 1/2ⁿ is always short of 1 by exactly the last included term, so it's
1 - 1/2ᵐ :
∑[m=1,∞] ∑[n=1,m] 1/2ᵐ⁺ⁿ
= ∑[m=1,∞] (1/2ᵐ ∑[n=1,m] 1/2ⁿ)
= ∑[m=1,∞] 1/2ᵐ [1 - 1/2ᵐ]
= ∑[m=1,∞] (1/2ᵐ - 1/4ᵐ)
= 1 - ⅓
= ⅔
Also, another way to find the limits for the re-ordered sum, is to draw m,n axes; plot the grid points called out by the original summation; then notice how the indices have to be bounded when summed in the new order, so that the same grid points are covered.
Fred
Evaluate: lim x tending to 1+ 2^(-2)^(1/1-x).
Hey dude. i was navigating through internet and i depared myself with this problem: int_0^1 e^x(1+e^x/(cosx)^2) dx. After several tries, i got nowhere, so can you solve it in a video?
Hey please do more double summation videos, please...
Black pen redpen yay:)!!
I wonder if this allows for weird nifty refactoring of nested loops in coding?
can u use a double summation as a series of comparison in limit or direct comparison test?
*double series
also with a single series
triple summation to the power of a double summation or else i will crunch you
Can i get help how to solve this question:
Find Xs such that when you divide X by 2 the remainder is 1 and when divided by 3 the rem. is 2 and when divided by 4 the rem.is 4 and when divided by 5 the rem. is 4 and when divided by 6 the rem. is 5
And it a pure number theory/congruence question (modulos aren't relatively prime)
And if #BPRP can explaim this topic in a video it would be so cool
Thanks guys❤️
The lowest answer is 59, but the general solution is 60*n - 1 for any integer n greater than 0.
This and any problem like this can be resolved by the chinese remainder theorem, take a look here:
en.wikipedia.org/wiki/Chinese_remainder_theorem
I was trying this myself and i get 3/2 times sum(m=1 to infinity) of m/4^m... How can i resolve this?
Very interesting
Could you give more détails make clear the last two lines? Get me update at This please!
When I don't pay too much attention you speak Chinese
Explain
Hey, can you solve this integral: (x^(-1/2) *ln(x))/(x-1) dx from 0 to infinity. But solve the integral using residues. That is a really interesting thing.
I'm trying your summation-swap method for the sequence (-1)^(2*m+2*n -1). But something is not working!!
That just a bunch of negative ones so it diverges cant use techniques on divergent series
I can't understand the fifth line... 😢
Can you make a video about it? Please?
Search for the proof, that the geometric Series converges in the internet. You should find all the calculation steps to the 5th line.
@@ambigramm thank you, man! I figured it out! And it's very simple... -_- (shame)
You look like my math teacher.
I've thought about a simple problem this week and I couldn't solve it. The problem is the following: "can we find a general expression for the coeficients of the polynomial P(x)=(x-1)(x-1/2)(x-1/4)...?". Please, I'm really getting crazy
M O R E
F U B I N I
Brooooo, one question
Do you study these techniques or you play with these numbers??
Both?
The voice crack at 0:51 though...
BOOO
@@amaankazi6406 ?
5:56 "This is not an integral sign so please don't get too excited." xD
Now you've said that, you need to make a summation video involving integrals!
David yea, see description
@@blackpenredpen :o Yes!!
Please make gmat quant video also
Make a video on how to do integration by parts in definite integral
Why did you interchange the order of summation?
@@amaankazi6406 To me, it seems easier to solve it without exchanging the sums.
Brilliant has youtube learning by the Balls!
Sometimes, the double summation is more complicated then the double integral. Isn't it?
Doraemon theme song
You should change your name to blackpenredpenbluepen
where is ur question from？ i mean how do u get this question？
A lot of these exercises are the ones that I have done before as a student. This is just an extension of my calc 2 series question.
@@blackpenredpen i like ur video so much，because when i get to the usa，every math question is easy，i can get some new logical thinking from u ，thanks！
Shouldn't you prove there's a bijection between the first indices set and second (to justify your swapping of summations)?
0 dislikes, it says how wonderful you are
Pretty
This only works because the 2's are the same right? I.e. it wouldn't work if it was (1/2^n)*(1/3^n)
Do you mean the switch of the summation? That works generally. If you mean pulling one term to the front, that works as long as the term you're moving does not contain the inner index. So it would also work with (1/2^m)*(1/3^n)
It looks like Cauchy product of
sum(1/4^(m)*1/2^(n-m),m=1..infinity,n=1..infinity)
(sum(1/4^(m),m=1..infinity)*sum(1/2^n,n=1..infinity)
Whats wrong with this approach ?
0:51 I didn't hear that :)
riots
Hahahha thanks
Sorry; Can you give us some exercises for this ?
Are you a programmer blackpenredpen?