@@DrBarkerI think there is every more geometry here with stacking cubes inside cubes and filling the gaps with squares. A cube has the volume of the next cube down + 3 square faces + 1 unit square. A completely geometric proof without algebra would be possible here.
What a neat proof! Inserting the sum of the first n integers was such an unexpected but great trick over there too; I completely though you'd abuse that "telescopation" in a different way, or you'd exploit reversing the summation order (to go column by column instead of row by row), but this was a very nice one!
By the Fubini's theorem when both limits in the first final summation are independent the double sum is the product of the simple sums, which gives sum of i from 1 to n, the whole summation squared
Your videos inspire me. I guess my level of maths is a lot lower than your viewers. In the past few months I have started running puzzle evenings in a local bar. A vastly simplified version of this could make it into the next evening! Thanks.
@@Sasseater I did the first one a few weeks ago. It’s rally small bar who needed to attract business on Thursday evening’s. A few maths questions and word questions. The people loved it and the next one is selling out fast. An example of a question. An 11 digit number. The first digit is perfect square, the first two digits are a perfect square, the first 6 digits are a perfect square and all 11 digits are a perfect square? No one got it right! And calculators were allowed!
More precisely : when you multiply both side of the equation with n^2/4, you get n^3 = n^2*(n+1)^2/4- (n-1)^2*n^2/4 = (n(n+1)/2)^2)-((n-1)n/2)^2 (you can recognize the sum of terms between 1 and n or n-1) Given an integer k, let's call S(k) = ((k(k+1))/2)^2 To sum up, n^3 = S(n) - S(n-1). We easily see a telescopic sum : n^3 + ... + 2^3 + 1^3 =S(n) - S(n-1) + S(n-1) -... +S(0) =S(n) - S(0) =S(n) =(1+2+...+n)^2
There is a proof that goes like this: If a1 < a2 < a3 < ... < an, where ai is the natural number, and (a1 + a2 +...+ an)^2 = a1^3 + ... + an^3, then ai = i.
Michael Penn sometimes dons a shirt with this particular integer identity. A suitable definition for a nerd among the nerds. Nice to see non induction way of showing this.
I first stumbled on this equivalence one night while trying to go to sleep... Not only did it cost me that night's rest, I wasted most of the next couple days trying to figure out why it was true. It seems like the sort of thing you should be able to do with basic grade school algebra, but I quickly got frustrated and wound up looking up some proofs online. Most of these - at least, the ones I understood - were of the visual/geometric variety. Which was fine, but a bit unsatisfying... Like they did not so much show the "why" of it as just restate it in building blocks instead of numbers. So, this proof was exactly the sort of thing I was looking for! Tbh I still find the equivalence itself fairly incredible (in several senses of the word), but it's nice seeing it proved with calculations I can actually follow.
My proof is that (1+2+3+4………….+n)=n(n+1)/2 (1+2+…..n)^2=(n(n+1)/2)^2=(n^4+2n^3+n^2)/4 (1+2+3+4…..+n-1)=(n-1)(n+1-1)/2=(n-1)n/2 (1+2+3……..+n-1)^2=(n^4-2n^3+n^2)/4 (Sum of 1 to n)^2-(sum of 1 to n-1)^2=n^3 (Sum of 1 to n-1)^2- (sum of 1 to n-2)^2=(n-1)^3 (Sum of 1 to n)^2=n^3
Hi. This is a fantastic video, but I'd just like to point out that the sum from 1 to 0 of anything is absolutely well-defined, and is equal to zero. (It is the sum over all integers i satisfying 1 ≤ i ≤ 0, that is it is the sum over the elements of the empty set (so an empty sum), which is by convention defined as the neutral element for addition, i.e. 0)
Imagine a world in which, instead of firing missiles at each other, countries simply sent Olympiad Math Challenges at each other. Or angry Dance Off challenges. Take your pick.
You should have timestamped that to 3:45. And no, summing terms for 1 to 0 is not well defined. One could easily argue that it should by the negative of the sum for 0 to 1, just like you flipped integrals that go backwards.
This is a delightful proof! I think it could be nice to run it backwards. Start with the the (sum of the integers from 1 to n)² and use your diagram to split this into two equal triangular double sums and the sum of the squares of the integers from 1 to n, then convert the double triangular sum to the sum of k³-k² for k from 2 to n, by again running your argument backwards, and we are done. Here are the details. We want to prove that ∑(k=1 to n)k³=[∑(k=1 to n)k]² For this proof we'll manipulate the RHS until it becomes the LHS. [∑(k=1 to n)k]² =[∑(k=1 to n)k][∑(i=1 to n)i] =∑(k=1 to n)∑(i=1 to n)ki =∑(k=1 to n)[∑(1≤i
i like how some geometry still snuck in there with the multiplication table
Yes! While it's technically a purely algebraic proof, that particular step can be seen in quite a nice, geometric way.
@@DrBarkerI think there is every more geometry here with stacking cubes inside cubes and filling the gaps with squares. A cube has the volume of the next cube down + 3 square faces + 1 unit square.
A completely geometric proof without algebra would be possible here.
At one time algebraic questions were geometrically solved before mathematical notations for operators were developed
More like matrix than geometry.
This is somewhat more satisfying than the induction proof.
What a neat proof! Inserting the sum of the first n integers was such an unexpected but great trick over there too; I completely though you'd abuse that "telescopation" in a different way, or you'd exploit reversing the summation order (to go column by column instead of row by row), but this was a very nice one!
By the Fubini's theorem when both limits in the first final summation are independent the double sum is the product of the simple sums, which gives sum of i from 1 to n, the whole summation squared
Your videos inspire me. I guess my level of maths is a lot lower than your viewers. In the past few months I have started running puzzle evenings in a local bar. A vastly simplified version of this could make it into the next evening! Thanks.
This is great to hear! Maybe an example just using a specific value of n could be made to work. Do let me know if it makes it into the next evening!
Puzzle night at the bar sounds like a blast! I may be a math major but after a few drinks a simplified version is likely right up my alley! 🍺
@@Sasseater I did the first one a few weeks ago. It’s rally small bar who needed to attract business on Thursday evening’s. A few maths questions and word questions. The people loved it and the next one is selling out fast. An example of a question. An 11 digit number. The first digit is perfect square, the first two digits are a perfect square, the first 6 digits are a perfect square and all 11 digits are a perfect square? No one got it right! And calculators were allowed!
@@sr6424 16,646,418,441. What do I win?
nice one! I’ve also seen ways to prove this using finite differences
That's a nice one. Thx a lot.
Awesome 👍
Loved it!
A beautiful proof. :)
Thank you for the video
Muy bella prueba.
beatiful proof
That serious look at the end makes me giggle.
不能直接积分吗？
By using the closed form for Σk arent you silently using induction in the proof?
¡Qué interesante!
But can you prove the inner sum without induction?
good one
(n + 1)² - (n - 1)² = 4n. Now multiply by n²/4
More precisely :
when you multiply both side of the equation with n^2/4, you get
n^3 = n^2*(n+1)^2/4- (n-1)^2*n^2/4
= (n(n+1)/2)^2)-((n-1)n/2)^2
(you can recognize the sum of terms between 1 and n or n-1)
Given an integer k, let's call S(k) = ((k(k+1))/2)^2
To sum up, n^3 = S(n) - S(n-1).
We easily see a telescopic sum :
n^3 + ... + 2^3 + 1^3
=S(n) - S(n-1) + S(n-1) -... +S(0)
=S(n) - S(0)
=S(n)
=(1+2+...+n)^2
Nice
i wanna die after seeing how you do your sum sign 😂😂
How about the long second between 2:00 and 2:01 with the failed minus sign?
this question apered in a similar way in a braziliam olympiad, all tho the exponents were indetermined, and u had determin them
Neat.
There is a proof that goes like this: If a1 < a2 < a3 < ... < an, where ai is the natural number, and (a1 + a2 +...+ an)^2 = a1^3 + ... + an^3, then ai = i.
Michael Penn sometimes dons a shirt with this particular integer identity.
A suitable definition for a nerd among the nerds.
Nice to see non induction way of showing this.
I first stumbled on this equivalence one night while trying to go to sleep... Not only did it cost me that night's rest, I wasted most of the next couple days trying to figure out why it was true. It seems like the sort of thing you should be able to do with basic grade school algebra, but I quickly got frustrated and wound up looking up some proofs online. Most of these - at least, the ones I understood - were of the visual/geometric variety. Which was fine, but a bit unsatisfying... Like they did not so much show the "why" of it as just restate it in building blocks instead of numbers.
So, this proof was exactly the sort of thing I was looking for! Tbh I still find the equivalence itself fairly incredible (in several senses of the word), but it's nice seeing it proved with calculations I can actually follow.
I'd say visual and geometric proofs show "why" just as much as algebraic ones.
My proof is that (1+2+3+4………….+n)=n(n+1)/2
(1+2+…..n)^2=(n(n+1)/2)^2=(n^4+2n^3+n^2)/4
(1+2+3+4…..+n-1)=(n-1)(n+1-1)/2=(n-1)n/2
(1+2+3……..+n-1)^2=(n^4-2n^3+n^2)/4
(Sum of 1 to n)^2-(sum of 1 to n-1)^2=n^3
(Sum of 1 to n-1)^2- (sum of 1 to n-2)^2=(n-1)^3
(Sum of 1 to n)^2=n^3
Hi. This is a fantastic video, but I'd just like to point out that the sum from 1 to 0 of anything is absolutely well-defined, and is equal to zero. (It is the sum over all integers i satisfying 1 ≤ i ≤ 0, that is it is the sum over the elements of the empty set (so an empty sum), which is by convention defined as the neutral element for addition, i.e. 0)
Imagine a world in which, instead of firing missiles at each other, countries simply sent Olympiad Math Challenges at each other. Or angry Dance Off challenges. Take your pick.
Except the loser of the challenge still has missiles…
4:00 you say the sum is not defined but it is and is equal to 0
You should have timestamped that to 3:45. And no, summing terms for 1 to 0 is not well defined. One could easily argue that it should by the negative of the sum for 0 to 1, just like you flipped integrals that go backwards.
@@jursamaj its a convention, the sum from 0 to -1 is indeed 0 you can ask anybody
at ~ 7:00 you omitted the induction 😿
n(n+1)/2 = 1+2+3+...+n
1) It's a simpler induction
2) It can be proven geometrically
Those upside down sigmas are disturbing. 😁
Answers to Questions that no one asked
Mad?
Don’t listen to the person who has the answers, listen to the person who has the questions. - Albert Einstein
This is such a nice proof but I can't get over how janky you are drawing the sigmas lmao
This is a delightful proof! I think it could be nice to run it backwards. Start with the the (sum of the integers from 1 to n)² and use your diagram to split this into two equal triangular double sums and the sum of the squares of the integers from 1 to n, then convert the double triangular sum to the sum of k³-k² for k from 2 to n, by again running your argument backwards, and we are done.
Here are the details.
We want to prove that
∑(k=1 to n)k³=[∑(k=1 to n)k]²
For this proof we'll manipulate the RHS until it becomes the LHS.
[∑(k=1 to n)k]²
=[∑(k=1 to n)k][∑(i=1 to n)i]
=∑(k=1 to n)∑(i=1 to n)ki
=∑(k=1 to n)[∑(1≤i
P.S. I have edited my comment to include the "reverse" solution in detail.
John Berry's *perfect cube...*
𝑤³ + 𝑥³ + 𝑦³ = 𝑧³; 3³ + 4³ + 5³ = 6³
27 + 64 + 125 = 216
216 = 216 ✅ ; SO-... if we set-up the 𝑛=5 case minus the 𝑛=2 case...
(1 + 2 + 3 + 4 + 5)² - (1 + 2)² = 225 - 9 = 216
216 = 216 ✅ ; AND... if we set-up the 𝑛=6 case minus the 𝑛=5 case...
(1 + 2 + 3 + 4 + 5 + 6)² - (1 + 2 + 3 + 4 + 5)² = 441 - 225 = 216
216 = 216 ✅
Nice