@0:43 So that means it's the square of the sum of n consecutive numbers! Its also the triangular sum of odd numbers! 1 = 1^3 3 + 5 = 2^3 7+9+11= 3^3 etc. So we can write it as an iterated sum!! Edit: I now see BPRP has put up a video about iterated sum, it's like you guys plan this =D haha
You could have derived that area of the big square is actually (N*(N+1))^2 too... By just noticing that single outer side has N+1 squares of side length N This would make proof even more awesome
A quick search of the OEIS (online encyclopedia of integer sequences) and google shows no results. I assume this is because of how wildly divergent that series is, but I'd love to see someone try and solve it!
Hmm... that brings back memories! 40 years ago I worked as a student in an exchange program in Germany and attended a party where I knew of Carl Friedrich Gauss that calculated the sum of the simple. The squares I also knew, but the cubes I did not. And then a passed a note that formulated the cube problem - with a questionmark. I'm not well trained in math. It passed around till it got to a mathematician. Immediately he said: I don't know. But then: 1) Within a minute the boys were dispatched to get pen, paper and calculator. That taught me how organised Germans are - teamwork is their second nature. 2) All the girls threw their hands in the air and moaned - they KNEW that the boys would have no interest in them - before the problem was solved! 3) My contribution was that I remarked it was the simple squared. 4) But I never found the general formula.
It's not sooo difficult. He knows the sum of first powers, second powers, third powers and so on. This relation between summing up cubes and summing up natural numbers and square them is not secret result. Because of the square on the right hand side it is natural to consider the area of squares. Before somebody misunderstood me... I don't want to say it is "easy" in conventional sense. But his level allows him to think about in a mathematical "natural" way. That's makes it quite "easy". That's all.
@@easymathematik Understandable but yeah, there was a slight misunderstanding. My question was focused on why someone who makes videos about linear algebra and integration is able to spot a relation like that without writing down and attempting a full proof
@@easymathematik Yeah, he could've taken time to prove it. I just think that it's unlikely that he took time out of his day to actually prove it which makes me think that he actually came up with this in his head, hence my amazement
Wow, weirdly this video literally launched itself on my computer, I wasn't even using youtube at the time it suddenly started playing unbidden. But I love his enthusiasm for maths and will look out for more of his videos for sure.
Actually proving this algebraically is quite nice too. Expanding the square of sum you get sum of squares from 1 to N plus sum[2 to N](i^2(i-1)), leaving 1 out, the rest simplifies to sum[2 to N] i^3, but 1=1^3 so you get the result. That said, you still win because of the nice picture ~
When I learnt about this identity, I checked on the proofwiki "Sum of Sequence of Cubes " and it turned out there were many different proofs for it. You chose the visual demonstration but my favorite is the one by recursion.
A few weeks ago, tried to prove this without using induction. Didn't think of this cool geometric proof. Did find a nice double summation. Sum_{i=1,...,n-1}Sum_{j=1,...,i} j^2 = n×(Sum of squares) - Sum of cubes. So double sum gives n×n(n+1)(2n+1)/6 - [n(n+1)/2]^2 = n^2 (n^2 - 1)/12
Hello from France (thank you Dr Peyan, I love what you do, I am a highschool student, and when I can understand what you are doing I am very impressed)
Muchas gracias. En estadística cuando hacemos una transformación a rangos, esta suma se vuelve fundamental. Luego ser posible que mis alumnos puedan ver este vídeo es muy importante y motivante a la vez. Muchísimas gracias.
Thank you, this is so fascinating and beautiful! I wanted to share my summary that helped me wrap my head around what your described: 4 1x1 squares arranged in a 2x2 square 8 2x2 squares surrounding the previous inner 2x2 square 12 3x3 squares surrounding the previous inner 6x6 square 16 4x4 squares surrounding the previous inner 12x12 square 20 5x5 squares surrounding the previous inner 20x20 square ... 4n nxn squares surrounding the previous inner 2Tnx2Tn square where Tn = n(n+1)/2 = n-th triangular number Each successive layer consists of: + 4 nxn corners + n-1 nxn squares extending of each of the 4 sides = (4 + 4*(n-1)) nxn squares = 4n nxn squares Note: each side extension of each layer "fits" because it is subdividing (n-1)x(n-1) squares into n-1 pieces
sum[1 to n](k^4) = sum[0 to n-1]((k+1)^4) = sum[0 to n-1](k^4 + 4k^3 + 6k^2 + 4k + 1) = five sums. The last three sums can be determined (as each is dependent on the next) to be n(n-1)(2n-1) (6 * sum of k^2), 2n(n-1) (4 * sum of k), and n (sum of 1), also considering the index which changes things from the normal setup. so we have (sum of fourths from 1 to n) = (sum of fourths from 1 to n-1) + 4(sum of cubes from 1 to n-1) + n(n-1)(2n-1) + 2n(n-1) + n; therefore (subtracting the ^4 sums from 1 to n-1) we get n^4 = 4(sum of cubes from 1 to n-1) + n(n-1)(2n+1) + n = 4(sum of cubes from 0 to n-1) + (2n-1)n^2, subtracting, dividing, and factoring we get (n(n-1)/2)^2 = sum of cubes from 0 to n-1, and finally with a shift of index (just adding n^3 to the end) we get that the sum of cubes is (n(n+1)/2)^2
Great video! Σn can be also be derived using the area of rectangle and Σn² using the volume of cubes. I thought Σn³ would require 4th dimension but you amazingly did it with the area of squares!
Dr. Peyam I must admit that you noticed something ... Mathematics is even more beautiful when you can find a representative graphical explanation. After all, this is how the Greeks and physicists got used to thinking a long time ago.
Neat demo, thanks! The only thing I miss is at 2:30. You assume it is 4N squares and do not demonstrate it. There is a mathematical explanation showing that adding exactly 4N squares of size N to the (N-1)th figure (a square of size N(N-1)) makes it the Nth figure (a square of size N(N+1)) edit: after redrawing your construction myself, I got the point: your construction works because the edge size of the nth square is (n+1), which stems from the well-known formula (1+...+n) = n(n+1)/2 you (neatly!! I appreciated that!) demonstrated at the end. Some comments complained that this last demo ((1+...+n) = n(n+1)/2) was useless, because you had under your eyes a square whose size is n(n+1) under your eyes. The flow is the other way around: you have it under your eyes and were able to build it precisely because 2(1+...+n) = n(n+1); the fact is only that this demonstration comes too late in your video. Thanks a lot again for the nice video :)
Very nice. Yesterday I had just done a similar visual proof with actual blocks (I have a collection of original Wooden Deines Blocks). Just imagine that your squares all have a height of 1 then you can make cubes out of them and your square are the slices of height one for each cube of sides 1,2 ,3 3 etc. I am wodering if there are visual proof for all powers beyond 3?
Hi, I will next year be a college student in physics. And I was asking myself the question : would this proof be considered as formal and so, would you give all the points to a student making this kind of proof? Thank you for recording this video because I lately asked myself the question of why it was the case ^^
No it does not. Why? This values of zeta have nothing to do with the sum of cubes. zeta(-3) = 1/120. That's okay so far. But zeta(-3) != 1^3 + 2^3 + 3^3 ... Because for -3 zeta is not defind. zeta(-3) is defind via analytic extension. It has nothing to do with summing up cubes. Clear? :)
your proof makes me think... is there an analogue in 3d? is it truly possible to guarantee that always there will be an integer number of cubes of side n+1 that can envelop the previous iteration composed of n? 8 * 1^3, + (27-1) * 2^3, + (4^3 - 2^3) * 3^3, it appears so. the sum from 0 to N of the terms: ((k+1)^3 - (k-1)^3) * k^3 = 8 * the cube of the sum of integers. In 2d, there seems to be this guarantee that ((k+1)^2 - (k-1)^2) = 4k, which is very simple, but let's see for cubes: k^3 + 3k^2 + 3k + 1 -k^3 + 3k^2 -3k +1 = 6k^2 + 2, less easy. So, we have a two-termed sum: sum from 0 to N of (6k^5 + 2k^3) = (n(n+1))^3, and you just calculated the sum of cubes to be (n(n+1)/2)^2, so we can subtract twice that value from both sides to very easily (and i think very beautifully) calculate the sum of fifths! this is absolutely amazing! maybe you can do a video about some of these? or maybe there's a bigger pattern here? 6 * the sum of k^5 = (n(n+1) - 1/2) * (n(n+1))^2, so the final result is something to the order of (2n^2+2n-1)(n(n+1))^2 /12, which is indeed correct.
i wonder... yes. this seems to be generally true as well: you only need an N dimensional space to calculate the sum of k^(2N-1), as the volumes of the partial cubes are k^N, but their quantity is proportional to the *surface* of said cubes, which is k^(N-1), giving the k^(2N-1) term. So in 1d, you could sum the integers in order to calculate... the sum of integers. :P ((k+1)-(k-1)) * k = 2k, nothing special XD, but still the pattern exists.1->1, 2->3, 3->5, 4->7
My one complaint is that at the end you show a standard proof of the triangular sum formula, but there's already a proof of it staring us in the face on the board! The side length of the square you drew is n+1 segments of length n.
Mathmagician that does not evaluate to anything. Infinite titrations do not work for all inputs, there is an interval of real numbers such that only those numbers can be inputs of an infinite titration. -4 is not in that interval.
Mathmagician I just tried your problem. Using your corrected problem and properties of the Lambert W function, the answer should be -1/2. I checked convergence over the real numbers and it does converge to -1/2 (or -0.5, whichever way you prefer to write it.)
Clever but over complicated and embellished to my way of thinking and the reasoning gets muddled. Far, far simpler is to take the 'difference of squares' 4^2 + 3^2 = 7. Multiply thro' by 7^2 to give 28^2 - 21^2 = 7^3. 28 & 21 are of course triangular numbers. 'SIMPLES'!
![แฉ [2/4] 17 ก.ค. 67 “ลิลลี่ เหงียน” เปิดใจหลัง “เอิร์ก เลเดอเรอร์” ตอกกลับเรื่องโกงเงิน 50 ล้าน](http://i.ytimg.com/vi/s8KapfEO5NI/mqdefault.jpg)
Oh my goodness... You solved a question that has puzzled me for long time ... I mean... ever since my high school days. Thank you very very much.!!!
Wow this is a really clever way to get to that equality. I had no idea there could be such intuition behind this problem.
@0:43 So that means it's the square of the sum of n consecutive numbers!
Its also the triangular sum of odd numbers!
1 = 1^3
3 + 5 = 2^3
7+9+11= 3^3
etc.
So we can write it as an iterated sum!!
Edit: I now see BPRP has put up a video about iterated sum, it's like you guys plan this =D haha
You could have derived that area of the big square is actually (N*(N+1))^2 too... By just noticing that single outer side has N+1 squares of side length N
This would make proof even more awesome
Totally agree... Now he has to update his video! :)
This is an even cooler observation in my opinion
I also spotted it :)
Why is this channel not famous enough?
4:10 "and 4 times N cubed... and gamecube"
A man with culture
I got it as soon as you drew the squares. Thanks Dr Peyam.👍🏾👍🏾👍🏾
Is there a formula for 1+(2^2)+(3^3)+(4^4)...+n^n?
A quick search of the OEIS (online encyclopedia of integer sequences) and google shows no results. I assume this is because of how wildly divergent that series is, but I'd love to see someone try and solve it!
@@HasXXXInCrocs Indirectly I was also benefitted by your commentary. Thank you.
@@MegaUpstairs That's n^n, not the sums
there is, however, a formula for the sum of k^p
No.
Hmm... that brings back memories!
40 years ago I worked as a student in an exchange program in Germany and attended a party where I knew of Carl Friedrich Gauss that calculated the sum of the simple. The squares I also knew, but the cubes I did not. And then a passed a note that formulated the cube problem - with a questionmark. I'm not well trained in math.
It passed around till it got to a mathematician. Immediately he said: I don't know. But then:
1) Within a minute the boys were dispatched to get pen, paper and calculator. That taught me how organised Germans are - teamwork is their second nature.
2) All the girls threw their hands in the air and moaned - they KNEW that the boys would have no interest in them - before the problem was solved!
3) My contribution was that I remarked it was the simple squared.
4) But I never found the general formula.
That proof was really cool! Added to my favourite videos list.
This explanation is far better than mashing several Play-Doh cubes together to try to figure out what's going on.
Great video. The algebra trick for the sum of all integers is so neat.
How do you even spot this kind of stuff? Are you secretly a wizard?
That requires some deep-thinking and good observation.
It's not sooo difficult.
He knows the sum of first powers, second powers, third powers and so on.
This relation between summing up cubes and summing up natural numbers and square them is not secret result.
Because of the square on the right hand side it is natural to consider the area of squares.
Before somebody misunderstood me... I don't want to say it is "easy" in conventional sense. But his level allows him to think about in a mathematical "natural" way. That's makes it quite "easy". That's all.
@@easymathematik Understandable but yeah, there was a slight misunderstanding. My question was focused on why someone who makes videos about linear algebra and integration is able to spot a relation like that without writing down and attempting a full proof
@@benjaminbrady2385 I can't follow you, pardon.
He shows a (one out of different) possibility to proof it.
Where is your problem?
@@easymathematik Yeah, he could've taken time to prove it. I just think that it's unlikely that he took time out of his day to actually prove it which makes me think that he actually came up with this in his head, hence my amazement
This is the most elegant proof of this identity I have seen yet!
Well take a look at the comment I have just posted.
I always wondered the same thing thank you for clearing this up :)
25.000th suscriber, really like your videos and proofs
Wow, weirdly this video literally launched itself on my computer, I wasn't even using youtube at the time it suddenly started playing unbidden. But I love his enthusiasm for maths and will look out for more of his videos for sure.
It’s a sign! 😄
Like your enthusiasm as well. Thanks for uploading your math Dr. Peyam. 👌😉
Actually proving this algebraically is quite nice too. Expanding the square of sum you get sum of squares from 1 to N plus sum[2 to N](i^2(i-1)), leaving 1 out, the rest simplifies to sum[2 to N] i^3, but 1=1^3 so you get the result. That said, you still win because of the nice picture ~
Bravo Dr. Peyam
When I learnt about this identity, I checked on the proofwiki "Sum of Sequence of Cubes
" and it turned out there were many different proofs for it.
You chose the visual demonstration but my favorite is the one by recursion.
Recursive proof FTW
He did mention the inductive proof.
Watching Your Videos make me Happy☺
A few weeks ago, tried to prove this without using induction. Didn't think of this cool geometric proof. Did find a nice double summation.
Sum_{i=1,...,n-1}Sum_{j=1,...,i} j^2 = n×(Sum of squares) - Sum of cubes.
So double sum gives n×n(n+1)(2n+1)/6 - [n(n+1)/2]^2 = n^2 (n^2 - 1)/12
Hello from France (thank you Dr Peyan, I love what you do, I am a highschool student, and when I can understand what you are doing I am very impressed)
Salut!!! 😄
Muchas gracias. En estadística cuando hacemos una transformación a rangos, esta suma se vuelve fundamental. Luego ser posible que mis alumnos puedan ver este vídeo es muy importante y motivante a la vez. Muchísimas gracias.
Neat solving methods like this is one thing I love about math
Perfectly clear ! Thank you Dr 𝜫M
So cool!
U are solved a thing that I cound't for a long time Dr. peyham yeah:)
Thank you, this is so fascinating and beautiful!
I wanted to share my summary that helped me wrap my head around what your described:
4 1x1 squares arranged in a 2x2 square
8 2x2 squares surrounding the previous inner 2x2 square
12 3x3 squares surrounding the previous inner 6x6 square
16 4x4 squares surrounding the previous inner 12x12 square
20 5x5 squares surrounding the previous inner 20x20 square
...
4n nxn squares surrounding the previous inner 2Tnx2Tn square where Tn = n(n+1)/2 = n-th triangular number
Each successive layer consists of:
+ 4 nxn corners
+ n-1 nxn squares extending of each of the 4 sides
= (4 + 4*(n-1)) nxn squares = 4n nxn squares
Note: each side extension of each layer "fits" because it is subdividing (n-1)x(n-1) squares into n-1 pieces
sum[1 to n](k^4) = sum[0 to n-1]((k+1)^4) = sum[0 to n-1](k^4 + 4k^3 + 6k^2 + 4k + 1) = five sums. The last three sums can be determined (as each is dependent on the next) to be n(n-1)(2n-1) (6 * sum of k^2), 2n(n-1) (4 * sum of k), and n (sum of 1), also considering the index which changes things from the normal setup. so we have (sum of fourths from 1 to n) = (sum of fourths from 1 to n-1) + 4(sum of cubes from 1 to n-1) + n(n-1)(2n-1) + 2n(n-1) + n; therefore (subtracting the ^4 sums from 1 to n-1) we get n^4 = 4(sum of cubes from 1 to n-1) + n(n-1)(2n+1) + n = 4(sum of cubes from 0 to n-1) + (2n-1)n^2, subtracting, dividing, and factoring we get (n(n-1)/2)^2 = sum of cubes from 0 to n-1, and finally with a shift of index (just adding n^3 to the end) we get that the sum of cubes is (n(n+1)/2)^2
Ok now this is epic
Great video! Σn can be also be derived using the area of rectangle and Σn² using the volume of cubes. I thought Σn³ would require 4th dimension but you amazingly did it with the area of squares!
very cool :) thanks for sharing
Excellent!
Dr. Peyam
I must admit that you noticed something ... Mathematics is even more beautiful when you can find a representative graphical explanation.
After all, this is how the Greeks and physicists got used to thinking a long time ago.
Last step also follows from the length of the side expressed in just the largest of the squares, which gives N(N+1).
0 dislikes 100% deserved so elegant so pleasurable
SOMEBODY DARED DISLIKE
Beautiful!
super awesome derivation!
Nice geometry proof!
Neat demo, thanks!
The only thing I miss is at 2:30. You assume it is 4N squares and do not demonstrate it. There is a mathematical explanation showing that adding exactly 4N squares of size N to the (N-1)th figure (a square of size N(N-1)) makes it the Nth figure (a square of size N(N+1))
edit:
after redrawing your construction myself, I got the point: your construction works because the edge size of the nth square is (n+1), which stems from the well-known formula (1+...+n) = n(n+1)/2 you (neatly!! I appreciated that!) demonstrated at the end.
Some comments complained that this last demo ((1+...+n) = n(n+1)/2) was useless, because you had under your eyes a square whose size is n(n+1) under your eyes. The flow is the other way around: you have it under your eyes and were able to build it precisely because 2(1+...+n) = n(n+1); the fact is only that this demonstration comes too late in your video.
Thanks a lot again for the nice video :)
That's really cool sir :)
Very nice. Yesterday I had just done a similar visual proof with actual blocks (I have a collection of original Wooden Deines Blocks). Just imagine that your squares all have a height of 1 then you can make cubes out of them and your square are the slices of height one for each cube of sides 1,2 ,3 3 etc. I am wodering if there are visual proof for all powers beyond 3?
Hey mate, this is dope! You're on @3blue1brown 's path... Please don't stop! :)
Thank you. I've always found the induction method to be vaguely unsatisfying .
Excellent geometrical approach, receive a big hug in my behalf dear friend 🍻 Cheers. ( I don't know if that is well expressed)
I was wondering about whether or not this was a coincidence or not since Calc II days, thanks Dr. πm
That was funny
that's rather nice
Thank you sir!
I was just about to try that! XD
Hi, I will next year be a college student in physics. And I was asking myself the question : would this proof be considered as formal and so, would you give all the points to a student making this kind of proof?
Thank you for recording this video because I lately asked myself the question of why it was the case ^^
Thank you very very much sir . You are awesome .
That's a classic
Hello Dr Peyam's. Great idea. For me I've used a recursive serie for finding the result plus more work. Peyam's^3 !
8:00 That's a cool proof!
Very elegant! Now please the closed form for Sum (1/i^3)
Open problem!
@@drpeyam I was kidding :D
So geometry is not so bad😉
geometry is cooler when you understand it
Cool! I thought it's just a coincidence!
'series'-ously :)
Plugging -3 into the Reimann Zeta function gives 1/120, which is different from (-1/12)^2. Does that contradict this at all?
No it does not.
Why?
This values of zeta have nothing to do with the sum of cubes.
zeta(-3) = 1/120. That's okay so far.
But zeta(-3) != 1^3 + 2^3 + 3^3 ...
Because for -3 zeta is not defind.
zeta(-3) is defind via analytic extension. It has nothing to do with summing up cubes.
Clear? :)
Let me introduce you to the Faulhaber's formula, the mighty beast that does the same trick but not only for k=3 but for every positive integer k.
Thank you for this comment, I've just read the Wikipedia page about it, very interesting. You should make a video about it! :)
your proof makes me think... is there an analogue in 3d? is it truly possible to guarantee that always there will be an integer number of cubes of side n+1 that can envelop the previous iteration composed of n? 8 * 1^3, + (27-1) * 2^3, + (4^3 - 2^3) * 3^3, it appears so. the sum from 0 to N of the terms: ((k+1)^3 - (k-1)^3) * k^3 = 8 * the cube of the sum of integers. In 2d, there seems to be this guarantee that ((k+1)^2 - (k-1)^2) = 4k, which is very simple, but let's see for cubes: k^3 + 3k^2 + 3k + 1 -k^3 + 3k^2 -3k +1 = 6k^2 + 2, less easy. So, we have a two-termed sum: sum from 0 to N of (6k^5 + 2k^3) = (n(n+1))^3, and you just calculated the sum of cubes to be (n(n+1)/2)^2, so we can subtract twice that value from both sides to very easily (and i think very beautifully) calculate the sum of fifths! this is absolutely amazing! maybe you can do a video about some of these? or maybe there's a bigger pattern here? 6 * the sum of k^5 = (n(n+1) - 1/2) * (n(n+1))^2, so the final result is something to the order of (2n^2+2n-1)(n(n+1))^2 /12, which is indeed correct.
i wonder... yes. this seems to be generally true as well: you only need an N dimensional space to calculate the sum of k^(2N-1), as the volumes of the partial cubes are k^N, but their quantity is proportional to the *surface* of said cubes, which is k^(N-1), giving the k^(2N-1) term. So in 1d, you could sum the integers in order to calculate... the sum of integers. :P ((k+1)-(k-1)) * k = 2k, nothing special XD, but still the pattern exists.1->1, 2->3, 3->5, 4->7
My one complaint is that at the end you show a standard proof of the triangular sum formula, but there's already a proof of it staring us in the face on the board! The side length of the square you drew is n+1 segments of length n.
Yep, I spotted that too.
The fact you point out is also the missing step in Dr. P's argument.
Yes, it would have been so neat to have the sum of integers, the sum of cubes and the link between them in just one graph!
And there i was, expecting a geometric proof using hypercubes :(
Can we take it as rigours proof ?
Yeah
Woow
Sir i am having problem in finding the roots of x^3+48x^2+49x+1
Please help
en.m.wikipedia.org/wiki/Cubic_function
Sir but according to rational root theorem the rational root satisfying the equation is either +1 or -1 only but both does not satisfies the equation
There are no sign changes therefor only imaginary roots. See Deacates sign rule.
เจ๋ง
Now do sum of n^4 in 2d.
Can you then do this?:
sum(i=1, n; i^5) =
(sum(i=1, n; i^3)^2 =
(sum(i=1, n; i)^4
Sadly no, it’s not that easy
No dislikes
sum of cubes but now with squares
Sir pls solve value of
-4^-4^-4^-4........ infinity =????
Mathmagician that does not evaluate to anything. Infinite titrations do not work for all inputs, there is an interval of real numbers such that only those numbers can be inputs of an infinite titration. -4 is not in that interval.
Mathmagician if ur wondering, that interval for the domain is e^(-e)
My question is (-1)*4^(-1)*4^(-1)*4^........ infinity
i have solution i prove this.......
(-4)^(-4)^(-4)^....... infinity this question have no solution but
(-1)*4^(-1)*4^(-1)*4^........ have solution just try one time
Mathmagician I just tried your problem. Using your corrected problem and properties of the Lambert W function, the answer should be -1/2. I checked convergence over the real numbers and it does converge to -1/2 (or -0.5, whichever way you prefer to write it.)
blackpenredpengreenpen hahah
Has someone been steeling bprp's black markers 😂
Hahaha
Think Twice wants to know your location
Berk... er, I mean, nobody give his location to anybody else he is precious!!!
Clever but over complicated and embellished to my way of thinking and the reasoning gets muddled. Far, far simpler is to take the 'difference of squares' 4^2 + 3^2 = 7. Multiply thro' by 7^2 to give 28^2 - 21^2 = 7^3. 28 & 21 are of course triangular numbers. 'SIMPLES'!
Excellent!