Super elegant derivation of the formula for 1^2 + 2^2 + ... + n^2 using triangles. Enjoy! Thank you Sho Seto for the idea! Subscribe to my channel: / drpeyam
It looks much easier (and more fun) to show that the sum in each entry is unchanged whichever direction we travel in from it. Going in any direction you move towards or away from a vertex, meaning the entry in one triangle goes up, one goes down and one is unchanged
... and that quite clearly generalizes to the sum of k-th powers using k-simplexes. Veeeeery nice indeed. 3D-version for the sum of 3rd powers is now needed.
Thank you once again for sharing this very "Lit/Dank" Proof of your student. I feel enlightened when I see intuition of quadrature geometry combined with algebra - it is a miracle!
Could the "turning the set around" trick be some sort of pattern in those sums by any chance? e.g. The case of the sum of the integers is easily proven by "flipping" an ordered number set around (i.e. a 1d set). Now in this video youve shown that for the sum of i^2 you "flip" a triangular set. Would this extend to i^n with higher dimensions of sets ? I would really like to know if there's something already. Thx in advance!
You can also draw equivalence between value at square and N-distance from corresponding side and then you use the theorem that sum of distances from all 3 sides is always equal
That's a very nice pictorial derivation. So, the sum of integers it is half the sum of two rows. The sum of squared integers one third the sum of 3 triangles. Hence, the sum of cubed integers is one fourth of the sum of four tetrahedrons.
The sum from 1 to n over i =n(n+1)/2 was the first proof of my first math lecture at uni my prof used it as an introductory example. One of the homework questions on the first sheet was proof the sum over i^2 formula, at the time i simply proofed it by induction but this is way better! Awesome vid brings back some sweet memories gonna show this to all my friends :P
What is the value of this series when N->Infinity? Should it be the Zeta(-2)=0 ? I don't know why. This is probably a presentation that must be undeniably correct for all N < infinity.
Hey Dr. Peyam, here's another approach to this sort of problem that I learned from Stopple's 'A Primer of Analytic Number Theory' chapter 1. Let n[^]m denote the falling factorial: n[^]m = n!/(n-m)! = n*(n-1)*...*(n+1-m) The way these behave with regard to sums and differences is perfectly analogous to the way ordinary powers behave with regard to integrals and derivatives: (n+1)[^]m - n[^]m = m*n[^](m-1) a[^]m + (a+1)[^]m + ... + (b-1)[^]m = 1/(m+1) (b[^](m+1) - a[^](m+1)) So summing ordinary powers is easy if we first express them as linear combinations of falling factorials. The coefficients are called the Stirling numbers and we can generate them with a simple (but far from obvious) Pascal-like recursion.
Man la Jeez, I wish. I just tried it out, this method doesn’t work in general. You’d need to be able to use the symmetries of the tetrahedron, but what you’d be computing is not the sum of consecutive cubes.
Indeed, you can! It requires more algebra than the case for the power 2, however that's also nice. There are two steps you need to understand in order derive the formula for the power 3. To define the Tetrahedron, consider a Tetrahedron containing numbers in the following order: {{1}, {2,2,2}, {3,3,3,3,3,3} ... (N rows) , {S(1) N's)} } (equivalently, bowling pins stacked up in a Tetrahedron shape). Than from this definition, If we examine the rotation of the Tetrahedron on it's 4 vertices (just like the triangle is rotated 3 times) The sum of all Tetrahedrons 4S = (3N+1)(sum from n=1 to N of S(1) ). This was step 1. Step 2 is simply summing all the elements in the Tetrahedron, it's surprisingly equal to 0.5(sum from n=1 to N of (n^3 + n^2)) = 0.5(S(3) + S(2)), and therefore, S(3) = 2S -- S(2). and we already know S(2) and S. Using these two results from the two steps and some algebra we find that S(3) = (S(1))^2 or simplified: 1^3 + 2^3 + ... + N^3 = (1 + 2 + ... + N)^2 note: S(k) denotes the sum 1^k + 2^k + ... + N^k
@@erel2099 I'm not able to construct this tetrahedron correctly = in the way that it contains the right amount of "bowling pins" for each row etc... :/ have you put them only in the surfaces or inside its volume too?
When you see a n(n+1)/2 factor, you know that you have a factor of the multiple of the sum of the natural numbers from 1 to n. Methods I once learnt and have since forgotten: brilliant.org/wiki/sum-of-n-n2-or-n3/. Thanks to this video, I remember them once more.
Well, I "brute forced" the solution with lots and LOTS of summing, algebra and induction (to verify I was correct), but I also got (2n^3+3n^2+n)/6. It's a success for me, everyone has to start somewhere, right? I'm too dumb to make elegant proofs such as this one. Mine's really bulky.
Near at the third minute i got how to do next.Very interesting way.So,what about third power? I thought about 3d triangle (thetraedr).but it uses the summ of the squares...such as when doing this only with algebra,the same:( Even in your solution formula of the numbers (1+2+3...) was used... In any way,very nice methood,thank you a lot for making videos like this. So,sorry about my grammar mistakes and so on (i am from the Russia)
wow that was great. i've been trying to find a geometric clue of the result since... ever. now I wonder if a similar approach can be used for i^3 and higher. For i^3 I guess I'd need a triangular pyramid, to begin with. Then I'd probably need to consider all the copies that it produces by rotating it in a way that every angle starts with 1 once: 4 copies. But I need the pyramid to be build in a way that for every "row" you have i^3: for example in the second row you must have 4 twos, in the third row 9 threes and so on. I wonder if this is even possible. (Perhaps it doesn't have to be like that though. As long as each "row" amounts to 1, 8, 27, etc it should be fine.) And then, assuming you have found a way to do that, you must hope that the sum of the (i,j,k) terms in all 4 pyramids is invariant of (i,j,k). Well it doesn't sound simple...
@@drpeyam Because like, 1² + 2² + 3².... is 1/1^(-2) + 1/2^(-2) + 1/3^(-2) .... which is actually Zeta of -2, and therefor it is 0 (because -2 is a trivial zero of the zeta function) Correct me if I failed somewhere. Thanks!
"SUM" means "the sum from x = 1 to n, where n is a positive integer". This video is interesting but here's an easy algebraic method which allows us to recursively calculate SUM[x^k] (where k is a positive integer) from SUM[1] (= n), SUM[x] (= n(n+1)/2), SUM[x^2] (= n(n+1)(2n+1)/6), ..., SUM[x^(k-1)]. We'll illustrate by calculating SUM[x^3]. (y+1)^4 - y^4 = 4y^3 + 6y^2 + 4y + 1, by the Binomial Theorem. Thus, letting y = n, n-1,...,w-1,...,1 yields: (n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1; n^4 - (n-1)^4 = 4(n-1)^3 + 6(n-1)^2 + 4(n-1) + 1; ................................................................................................. w^4 - (w-1)^4 = 4(w-1)^3 + 6(w-1)^2 + 4(w-1) + 1; ................................................................................................. 2^4 - 1^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1; The sum of the left-hand column telescopes to (n+1)^4 - 1 and the sum of the right-hand column is 4SUM[x^3] + 6SUM[x^2] + 4SUM[x] + SUM[1]. Therefore, since we know SUM[1], SUM[x] and SUM[x^2] , SUM[x^3] = ((n+1)^4 - 1 - 6SUM[x^2] - 4 SUM[x] - SUM[1]) / 4 = (n(n+1))^2 / 4. I wish I could remember where I learned this method. I believe that there is no known formula for SUM[x^k] though it is easily seen (by this method) to be a polynomial in n of degree k +1 with rational coefficients, leading coefficient 1 / (k+1) and constant term 0. I'd appreciate any other references. P.S. I just noticed that blackpenredpen already has a video on this method.
Superb! Seems like you could visualize the sum of the 3 triangles as a sum of 3 planes [each a linear function of (x,y)] in (x,y,z)-space, whose gradients are 120º apart . . . And since the sum must again be a plane [a linear function of (x,y)], the 3 corners of the triangular region determine the entire interior. [3 non-collinear points in space determine a plane.] Then all you have to do to prove that the sum is constant, is to show that all 3 corners add to the same value. But this is obvious from the symmetry of the layout! [Each corner value is the sum of the same 3 numbers, just in a different order.] Fred
Geometric-related proofs with sums and series are among my favourites. Keep them coming Doctor πm!
Ditto. They are so ripe.
You are the happiest teacher I have ever seen.
Hope life keeps you and your loved one happy always :)
It looks much easier (and more fun) to show that the sum in each entry is unchanged whichever direction we travel in from it. Going in any direction you move towards or away from a vertex, meaning the entry in one triangle goes up, one goes down and one is unchanged
fantastic
That what I perceived. Nice Proof.
Gauss should be very proud of you. Thank you very much for the beautiful proof.
... and that quite clearly generalizes to the sum of k-th powers using k-simplexes. Veeeeery nice indeed. 3D-version for the sum of 3rd powers is now needed.
Incredibly simply proof! And I love your energy and enthusiasm!
Always fascinated by your alternative proofs Doc !
Goodness this is pretty! What a nice, little victory!
That was very cool! Thank you for the video.
you save my life !!
finally understand this formula and i dont need to recite it anymore
Thanks for your video!
it's Sho Time!!! Thanks for a really big Sho!
Wow! Super cool approach... Very much impressed! Thanks!
Nothing is sexier than clever mathematics😋
Beautiful deduction, this would be marvelous with the animations of 3Blue1Brown. Thanks Dr. Peyam.
You never fail to amaze us man !!!
Thank you once again for sharing this very "Lit/Dank" Proof of your student. I feel enlightened when I see intuition of quadrature geometry combined with algebra - it is a miracle!
simply amazing!
That's a very beautiful derivation. Surprising what you can do with a bunch of triangles.
beautiful! wonder what could've been the motivation behind this amazing proof
WOW ...I FELT LIKE MY NEURONS CONNECTING 🤩
Thats a really cool method thanks
Could the "turning the set around" trick be some sort of pattern in those sums by any chance? e.g. The case of the sum of the integers is easily proven by "flipping" an ordered number set around (i.e. a 1d set). Now in this video youve shown that for the sum of i^2 you "flip" a triangular set. Would this extend to i^n with higher dimensions of sets ? I would really like to know if there's something already. Thx in advance!
Not sure, but that’s a great idea
That was amazing
Awesome!!!
Looks good...
You can also draw equivalence between value at square and N-distance from corresponding side and then you use the theorem that sum of distances from all 3 sides is always equal
Bravo!
Then he can say in return "only if you rename your channel to Dr Peyam's Pi Jam."
Creative, I like it
Brilliant
Nice!
That's a very nice pictorial derivation. So, the sum of integers it is half the sum of two rows. The sum of squared integers one third the sum of 3 triangles. Hence, the sum of cubed integers is one fourth of the sum of four tetrahedrons.
The sum from 1 to n over i =n(n+1)/2 was the first proof of my first math lecture at uni my prof used it as an introductory example. One of the homework questions on the first sheet was proof the sum over i^2 formula, at the time i simply proofed it by induction but this is way better! Awesome vid brings back some sweet memories gonna show this to all my friends :P
What is the value of this series when N->Infinity? Should it be the Zeta(-2)=0 ?
I don't know why.
This is probably a presentation that must be undeniably correct for all N < infinity.
nice!
This is just fantastic. Is there one geometric proof for the sum of cubes?
Yep 👍 This will be part of another video
Too Much Positive Attitude
I like it
Wow!
Godlike
So what if N=infinity? Is it cov. or div. , and what's the answer?
Divergent, and it’s infinity
Hey Dr. Peyam, here's another approach to this sort of problem that I learned from Stopple's 'A Primer of Analytic Number Theory' chapter 1. Let n[^]m denote the falling factorial:
n[^]m = n!/(n-m)! = n*(n-1)*...*(n+1-m)
The way these behave with regard to sums and differences is perfectly analogous to the way ordinary powers behave with regard to integrals and derivatives:
(n+1)[^]m - n[^]m = m*n[^](m-1)
a[^]m + (a+1)[^]m + ... + (b-1)[^]m = 1/(m+1) (b[^](m+1) - a[^](m+1))
So summing ordinary powers is easy if we first express them as linear combinations of falling factorials. The coefficients are called the Stirling numbers and we can generate them with a simple (but far from obvious) Pascal-like recursion.
Cool
Thanks u
The last triangle has (1+2+3+....N) times (2N+1) so
(N (N+1)/2)(2N+1) then divide by 3
It seems this method can generalise to sum n^4 and then 5,6,7....
Man la Jeez, I wish. I just tried it out, this method doesn’t work in general. You’d need to be able to use the symmetries of the tetrahedron, but what you’d be computing is not the sum of consecutive cubes.
Yan Mong Chan Could you clarify what you mean?
Yan Mong Chan Thank you! That’s really cool.
Professor Shoo Seto?? I had that professor for complex analysis last year lol, what a coincidence you know him!
Cool!!! Yeah, it’s him :)
Hmmm... Can we do the same with a tetrahedron for n^3 ?
Not sure, but that’s a great question
Indeed, you can! It requires more algebra than the case for the power 2, however that's also nice. There are two steps you need to understand in order derive the formula for the power 3. To define the Tetrahedron, consider a Tetrahedron containing numbers in the following order: {{1}, {2,2,2}, {3,3,3,3,3,3} ... (N rows) , {S(1) N's)} } (equivalently, bowling pins stacked up in a Tetrahedron shape). Than from this definition, If we examine the rotation of the Tetrahedron on it's 4 vertices (just like the triangle is rotated 3 times) The sum of all Tetrahedrons 4S = (3N+1)(sum from n=1 to N of S(1) ). This was step 1. Step 2 is simply summing all the elements in the Tetrahedron, it's surprisingly equal to 0.5(sum from n=1 to N of (n^3 + n^2)) = 0.5(S(3) + S(2)), and therefore, S(3) = 2S -- S(2). and we already know S(2) and S.
Using these two results from the two steps and some algebra we find that S(3) = (S(1))^2 or simplified: 1^3 + 2^3 + ... + N^3 = (1 + 2 + ... + N)^2
note: S(k) denotes the sum 1^k + 2^k + ... + N^k
@@erel2099 I'm not able to construct this tetrahedron correctly = in the way that it contains the right amount of "bowling pins" for each row etc... :/
have you put them only in the surfaces or inside its volume too?
@@marce3893You fill the whole volume. However, picture is worth a 1000 words: prnt.sc/nr5kls .
You’re amazing!! Seems like you have lots of fun doing this videos and it’s so great to watch. Keep it up sir and have a great day
Sum of squares but it's with triangles
COOL
How about the cube sum of integer? （~.~)
AweSum
Awe-summ!
How to know only three triangles should be taken sir?
No reason
When you see a n(n+1)/2 factor, you know that you have a factor of the multiple of the sum of the natural numbers from 1 to n.
Methods I once learnt and have since forgotten: brilliant.org/wiki/sum-of-n-n2-or-n3/. Thanks to this video, I remember them once more.
Thanks D peyam السلام عليكم
Well, I "brute forced" the solution with lots and LOTS of summing, algebra and induction (to verify I was correct), but I also got (2n^3+3n^2+n)/6. It's a success for me, everyone has to start somewhere, right?
I'm too dumb to make elegant proofs such as this one. Mine's really bulky.
Near at the third minute i got how to do next.Very interesting way.So,what about third power?
I thought about 3d triangle (thetraedr).but it uses the summ of the squares...such as when doing this only with algebra,the same:(
Even in your solution formula of the numbers (1+2+3...) was used...
In any way,very nice methood,thank you a lot for making videos like this.
So,sorry about my grammar mistakes and so on (i am from the Russia)
Wow.
You can also divide any cube in 3 indentical pyramids.
Wow!!!!
What about the sum f N^3 then. A pyramid?
A tetrahedron?
You’ll see :)
wow that was great. i've been trying to find a geometric clue of the result since... ever.
now I wonder if a similar approach can be used for i^3 and higher.
For i^3 I guess I'd need a triangular pyramid, to begin with. Then I'd probably need to consider all the copies that it produces by rotating it in a way that every angle starts with 1 once: 4 copies.
But I need the pyramid to be build in a way that for every "row" you have i^3: for example in the second row you must have 4 twos, in the third row 9 threes and so on. I wonder if this is even possible. (Perhaps it doesn't have to be like that though. As long as each "row" amounts to 1, 8, 27, etc it should be fine.)
And then, assuming you have found a way to do that, you must hope that the sum of the (i,j,k) terms in all 4 pyramids is invariant of (i,j,k).
Well it doesn't sound simple...
Read my comment to "felineboy" bellow, It works for i^3!
Maybe this is also somehow a method to find numbers that are simultaneously square and triangular.
What's the sum of all natural numbers ? 🔥
Zyzzyzus How could be positif + positif + positif + .... be négatif ?
Zyzzyzus yes indeed 🔥
If you want a general formula for the sum of powers, I suggest you look into the bernoulli numbers.
en.wikipedia.org/wiki/Bernoulli_number
When n happens to tend to positive infinity, shouldn't it be 0 ?
No, why?
@@drpeyam Because like, 1² + 2² + 3².... is 1/1^(-2) + 1/2^(-2) + 1/3^(-2) .... which is actually Zeta of -2, and therefor it is 0 (because -2 is a trivial zero of the zeta function) Correct me if I failed somewhere. Thanks!
"SUM" means "the sum from x = 1 to n, where n is a positive integer".
This video is interesting but here's an easy algebraic method which allows us to recursively calculate SUM[x^k] (where k is a positive integer) from SUM[1] (= n), SUM[x] (= n(n+1)/2),
SUM[x^2] (= n(n+1)(2n+1)/6), ..., SUM[x^(k-1)]. We'll illustrate by calculating SUM[x^3].
(y+1)^4 - y^4 = 4y^3 + 6y^2 + 4y + 1, by the Binomial Theorem. Thus, letting y = n, n-1,...,w-1,...,1 yields:
(n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1;
n^4 - (n-1)^4 = 4(n-1)^3 + 6(n-1)^2 + 4(n-1) + 1;
.................................................................................................
w^4 - (w-1)^4 = 4(w-1)^3 + 6(w-1)^2 + 4(w-1) + 1;
.................................................................................................
2^4 - 1^4 = 4(1)^3 + 6(1)^2 + 4(1) + 1;
The sum of the left-hand column telescopes to (n+1)^4 - 1 and the sum of the right-hand column is
4SUM[x^3] + 6SUM[x^2] + 4SUM[x] + SUM[1]. Therefore, since we know SUM[1], SUM[x] and SUM[x^2]
,
SUM[x^3] = ((n+1)^4 - 1 - 6SUM[x^2] - 4 SUM[x] - SUM[1]) / 4 =
(n(n+1))^2 / 4.
I wish I could remember where I learned this method. I believe that there is no known formula for SUM[x^k] though it is easily seen (by this method) to be a polynomial in n of degree k +1 with rational coefficients, leading coefficient 1 / (k+1) and constant term 0. I'd appreciate any other references.
P.S. I just noticed that blackpenredpen already has a video on this method.
I imagine if you did this with tetrahedra you would get the sum of cubes?
👍
THIS IS GENIUS
I know the way called by Graham Knuth Patashnik in Concrete mathematics finite calculus
Fucking love this!!!!!!!!!!
The real answer is −n :)
Why though? I don't get it 😁
Superb!
Seems like you could visualize the sum of the 3 triangles as a sum of 3 planes [each a linear function of (x,y)] in (x,y,z)-space, whose gradients are 120º apart . . .
And since the sum must again be a plane [a linear function of (x,y)], the 3 corners of the triangular region determine the entire interior. [3 non-collinear points in space determine a plane.]
Then all you have to do to prove that the sum is constant, is to show that all 3 corners add to the same value.
But this is obvious from the symmetry of the layout! [Each corner value is the sum of the same 3 numbers, just in a different order.]
Fred
Sum of triangular numbers
Doctor πm hello. Thanks for loving my comments ☺️☺️☺️☺️very interesting video
n^3=sum (k=1 to n)
(k^3-(k-1)^3)
What a simple demonstration! Can you also simplify math to teach my dog how much shit i have to take every day?? Thank you in advance!
04:08 You really did not have to work hard from here:
Factor out 2n + 1, and the sum of this *unit triangle* is n(n + 1)/2. Then divide by 3.
First!
1^2+2^2+3^2+ ... +24^2 =70^2. Pythagoras 42^2+56^2=70^2 Illuminati triangles confirmed LoL
Усложняли...
It’s really confusing.. Can't you explain in simple terms and with the original formula
Is math a jock ?
Great proof sir
At the end of the vedio i was left with my. Mouth open😂
Wondering.........
JUST WOW IT SELF
but i² is -1
lol
not in this case.
lol
He is not using "i" as sqrt(-1)
sum from i=1 to infty of i^2 just means 1+4+9+16+25+... until infty
@@ab_lk ty, but it was a joke bro
@@ulqi oh lol xd sry
the way he is acting is distracting
Why do you speak like this