i was trying to find since mid 2024 that have someone did this observation, and i was surprised to find how can peoples not see this that the year 2025 can be written in this form (20+25)^2 . Because there is no 4 digit year or may be 3 too that can be written in this format till this day. (whole square of sum of first and last 2 digits of an year is exactly that year. I feel mathematicians gonna remember this year when they do realize. Such an amazing year.
The fact that 2025 is a square also means it has unusually nice factorisation, 2025=3⁴·5². I remember 2023=7·17², and 2016=2¹¹-2⁵=2⁵·3²·7 as other years with surprisingly small primes. Factorising the year is tradition when preparing for the IMO, since it is common to be used in a problem.
The first perfect square year since 1936 (44^2), and we won’t get another one until 2116 (46^2). What a great year. 😁 (Also, anyone born in 1980 exactly will turn 45 in the year 45^2)
This video is a true gem! ✨ Connecting the New Year of 2025 to the sum of the first nine positive cubes and integers is fascinating. The visual demonstrations and proofs are brilliantly done, making complex concepts easy to grasp. Thank you for making math so engaging and understandable! Can't wait for more amazing content. 📚👏
Fun Fact: 2025 is a perfect square of 45 and it is the only year with perfect square since the year 1936 which was a perfect square of 44. Edit: The next perfect square year will be that of 46, which is the year 2116. I will be most likely dead by that point of time. 😅
@@MathVisualProofsBUT arent you lewving oitnthe buggest issie HOW wluld AMYONE arrive at the formula you prove ny induction ar around 3:00? How epuld somepne first deduce that the sum ld.cubes equals thenswuare pfnthe sum of the integers Why did ypu leave out the most inprotsnt step all due respect? Thanks for sharing
@@leif1075 I guess the main way would be just to study sum of cubes and realize they are all squares. Once you square root you realize they are squares of triangular numbers. The visual shows this happening so that was the motivation for this video.
2025 is going to be complex ,odd and unhappy But ,wholly Real ,natural and composite of Abundant positiveness . believe in your self but don't be a Narcissist squarely.
Also, 2025 is among the first few elements in this self-referencing iterated sequence, via Domotro from Combo Class. Where T(n) is a triangular number, and with n > 1: T(2) = 3 3^2 = 9 T(9) = 45 45^2 = 2,025 T(2,025) = 2,051,325 2,051,325^2 = 4,207,934,255,625 T(4,207,934,255,625) = 8,853,355,349,833,265,389,198,125 Tn^2 = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625 k = sum of first n cubes n = 2, k = 9 n = 9, k = 2,025 n = 2,025, k = 4,207,934,255,625 k = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
2:00 I find doing induction without intermediate step to be more beautiful: I'll use Sum[m,n](a_k) to denote sum of a_k from k=m to n. It is obvious that Sum[1,1](k^3)=(Sum[1,1](k))^2 Now assume (Sum[1,n-1](k))^2=Sum[1,n-1](k^3)---(1) Consider (Sum[1,n](k))^2=(Sum[1,n-1](k)+n)^2 =(Sum[1,n-1](k))^2+2n*Sum[1,n-1](k)+n^2 (By(1)): =Sum[1,n-1](k^3)+2n*n(n-1)/2+n^2 =Sum[1,n-1](k^3)+n^3-n^2+n^2 =Sum[1,n-1](k^3)+n^3 =Sum[1,n](k^3)
We have the formula for sigma i^3. We have the formula for sigma i. So, aren't these enough to show the result? I mean, we don't need induction to prove this right? We need induction to prove the formulas of sigma i and sigma i^3?
Liverpool lost only ONE game in the whole 2024 !! Leading the Premier League w/8 pts and leading Champions invictus !! ⚽️🔥 PS: and Salah has broken ... I don't know how many records this year alone ... and it's only half season!!! 😳 🥳🥳🥳
It's still 2024 and I haven't heard anyone but myself point out that 2024 is a betrothed number, affianced to 2295. The lesser others are 1925 and 1050, 1575 and 1648, 195 and 140, and 75 and 48. So 2024 is the 9th betrothed number.
Too early for this, otherwise a very interesting property of 2025. Also, why does it say "2025!" at the end of the video? Is that a factorial? (I'm joking.)
Great analysis, thank you! Could you help me with something unrelated: I have a SafePal wallet with USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How should I go about transferring them to Binance?
As usual, you are not proving anything. You take something that works, plot it and say "look, it works". As a consequence, you induce people who do not know maths into thinking they have a proof, when the only thing they have is the observation of something true, which had no way to be "not working" because, guess what, you were not trying to prove false things. So far, the only videos that could be claimed as "visual proofs" in this channel are maybe infinite sum ones, under the condition that all steps were simple enough to be self obvious, and not just the observation of something that could not be not working.
This is specially for all of you: In the new year eve, i thought , as a math enthusiast, why don't I give my math friends a math problem and wish them new year through that?And then , working for about an hour , I made a crazy problem myself. I sent it to some of my friends and some solved it too and were really impressed. This is a finalized version for you. Try it out yourselves. Let f(x) = 45.e^(-x/4).[cos (2x)/{(4/65).(e^(-π/8) - e^(π/8)}] Now find the integral of this function over the interval [-π/2 , π/2] , square the answer and see what you get!
Fun fact: 2025 = (20+25)^2
Noice
@AnshulPrajapati-n9lnoise? Not noice!! 😂
i was trying to find since mid 2024 that have someone did this observation, and i was surprised to find how can peoples not see this that the year 2025 can be written in this form (20+25)^2 . Because there is no 4 digit year or may be 3 too that can be written in this format till this day. (whole square of sum of first and last 2 digits of an year is exactly that year. I feel mathematicians gonna remember this year when they do realize. Such an amazing year.
The fact that 2025 is a square also means it has unusually nice factorisation, 2025=3⁴·5².
I remember 2023=7·17², and 2016=2¹¹-2⁵=2⁵·3²·7 as other years with surprisingly small primes.
Factorising the year is tradition when preparing for the IMO, since it is common to be used in a problem.
2016 was an amazing number
The first perfect square year since 1936 (44^2), and we won’t get another one until 2116 (46^2). What a great year. 😁
(Also, anyone born in 1980 exactly will turn 45 in the year 45^2)
This video is a true gem! ✨ Connecting the New Year of 2025 to the sum of the first nine positive cubes and integers is fascinating. The visual demonstrations and proofs are brilliantly done, making complex concepts easy to grasp. Thank you for making math so engaging and understandable! Can't wait for more amazing content. 📚👏
Thanks! Glad you liked it.
what a numerical year this is!!!!!
Fun Fact: 2025 is a perfect square of 45 and it is the only year with perfect square since the year 1936 which was a perfect square of 44.
Edit: The next perfect square year will be that of 46, which is the year 2116.
I will be most likely dead by that point of time. 😅
Another fun fact along the same line: people who are born in 1980 exactly will turn 45 in the year 2025 (45^2). Neat right?
@Ninja20704 The leap year blessed 2025 with crazy coincidences 😅
Happy New Year to all watching this video and the creator of the wonderful channel 😊
Thanks!
7:20 i thought it was factorial of 2025
I hope 45^2 goes well. 🙏
👍
@@MathVisualProofsBUT arent you lewving oitnthe buggest issie HOW wluld AMYONE arrive at the formula you prove ny induction ar around 3:00? How epuld somepne first deduce that the sum ld.cubes equals thenswuare pfnthe sum of the integers
Why did ypu leave out the most inprotsnt step all due respect? Thanks for sharing
@@leif1075 I guess the main way would be just to study sum of cubes and realize they are all squares. Once you square root you realize they are squares of triangular numbers. The visual shows this happening so that was the motivation for this video.
2025 is going to be complex ,odd and unhappy But ,wholly Real ,natural and composite of Abundant positiveness . believe in your self but don't be a Narcissist squarely.
...Wow. That makes 2025 even more special for me. I noticed that it's 45², but not that 45=1+2+3+4+5+6+7+8+9.
Encloses it:
(1+2+3+4+5+6+7+8+9)²
Also, 2025 is among the first few elements in this self-referencing iterated sequence, via Domotro from Combo Class.
Where T(n) is a triangular number,
and with n > 1:
T(2) = 3
3^2 = 9
T(9) = 45
45^2 = 2,025
T(2,025) = 2,051,325
2,051,325^2 = 4,207,934,255,625
T(4,207,934,255,625) = 8,853,355,349,833,265,389,198,125
Tn^2 = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
k = sum of first n cubes
n = 2, k = 9
n = 9, k = 2,025
n = 2,025, k = 4,207,934,255,625
k = 78,381,900,950,421,300,982,881,904,787,876,752,731,430,503,515,625
This math made me happy, cause i like math, thank you! :]
😀
Happy new year: Math Edition 🗣🔥🔥🔥🎉🎉🎉🎉🎉🎉
😀
Bro this doesn’t work in the new math update! 💀
i want to see you speedrun phytagorean theorem proofs, but some of them may be not seen as visual in a some sense
I have a video with 10 visual proofs in under 8 mins on my channel. 😀
2:00 I find doing induction without intermediate step to be more beautiful:
I'll use Sum[m,n](a_k) to denote sum of a_k from k=m to n.
It is obvious that Sum[1,1](k^3)=(Sum[1,1](k))^2
Now assume (Sum[1,n-1](k))^2=Sum[1,n-1](k^3)---(1)
Consider (Sum[1,n](k))^2=(Sum[1,n-1](k)+n)^2
=(Sum[1,n-1](k))^2+2n*Sum[1,n-1](k)+n^2
(By(1)): =Sum[1,n-1](k^3)+2n*n(n-1)/2+n^2
=Sum[1,n-1](k^3)+n^3-n^2+n^2
=Sum[1,n-1](k^3)+n^3
=Sum[1,n](k^3)
Yeah that’s nice but it still needs the sum of integers formula so that’s why I split them here 😀
the sum of cubes of natural numbers from 1 to n is equal to the square of the sum of natural numbers from 1 to n, where n is any number
The sum of cubes being equal to the squared sum of integers is kinda reminiscent of Stokes Theorem in a weird way.
2025 is also a Harshad number!
Actually 2022,2023,2024 and 2025 are harshad numbers
Do you know the longest consecutive streak is 20? Will never find 21 in a row :)
@@MathVisualProofs wow, that's so cool!
2025 , when mathematics starts to take over this world ☺️
happy new year
You too!!
I turn 45 next year
So according to Matt Parker, you were born in 1980.
Congrats, you turned your age in the year of your age squared
You posted this video too early it’s still December 26th
I will try to live to 46^2 year to remember this video.
I will only watch this clip upto 20:25
😀
We have the formula for sigma i^3. We have the formula for sigma i. So, aren't these enough to show the result?
I mean, we don't need induction to prove this right?
We need induction to prove the formulas of sigma i and sigma i^3?
Anything special about 2024?
Tetrahedral number! (Go check my short from last near around new years).
@@bagelnine9 2024 = 45² - 1² = (45-1)(45+1) = 44 × 46
Liverpool lost only ONE game in the whole 2024 !! Leading the Premier League w/8 pts and leading Champions invictus !! ⚽️🔥
PS: and Salah has broken ... I don't know how many records this year alone ... and it's only half season!!! 😳 🥳🥳🥳
If 45^2 is only (2024)^(1/2) as good for Liverpool, it'll be awesome !!!
2025 is also 12×3+(4+5)×(6+7)×(8+9).
It's still 2024 and I haven't heard anyone but myself point out that 2024 is a betrothed number, affianced to 2295. The lesser others are 1925 and 1050, 1575 and 1648, 195 and 140, and 75 and 48. So 2024 is the 9th betrothed number.
ein Beitrag des Montages, 30. Dezember 2024
What are "betrothed numbers"?
to y´all: Merry Christmas and a happy new year "45²".
Too early for this, otherwise a very interesting property of 2025. Also, why does it say "2025!" at the end of the video? Is that a factorial? (I'm joking.)
This video has 314 views as of posting this comment (pi)
2025 is a mathematical year
ua-cam.com/users/shortsVrgbYYMQlc4?si=IqVd0CCH9_E18552
ein Beitrag des Montages, 30. Dezember "45²-1"
I have found out a correllation between π [= "Pi"] and the number 2025.
2(π³)² + 2π² + 3π³ + 3π ~ 2,025.222 ~ MMXXV
π ~ 3.142 ~ III
π³ ~ 31.006 ~ XXXI
π^π ~ 36.462 ~ XXXVI
(π³)² ~ 961.389 ~ CMLXI
2(π³)² + 2π² ~ 1,922.778 ~ MCMXXIII
2(π³)² + 2π² + π³ ~ 1,953.78 ~ MCMLIX
2(π³)² + 2π² + 2π³ ~ 1,984.791 ~ MCMLXXXV
2(π³)² + 2π² + 3π³ ~ 2,015.997 ~ MMXVI
cool
👍
👏🏽👏🏽👏🏽
😀
@@MathVisualProofs You work is fantastic and necessary for all students!! Thank you very much!!
@ thanks!
3:58: Why do you think the story is a myth? I think it's well corroborated, especially since Gauß himself was fond of telling it.
Bryan Hayes did a detailed story about it and how it’s not so crystal clear what the story was.
So it is likely rooted in truth but the details are hard to know for sure.
Great analysis, thank you! Could you help me with something unrelated: I have a SafePal wallet with USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How should I go about transferring them to Binance?
nurd...
First comment/ Second view
First idiotic comment too!
As usual, you are not proving anything. You take something that works, plot it and say "look, it works". As a consequence, you induce people who do not know maths into thinking they have a proof, when the only thing they have is the observation of something true, which had no way to be "not working" because, guess what, you were not trying to prove false things. So far, the only videos that could be claimed as "visual proofs" in this channel are maybe infinite sum ones, under the condition that all steps were simple enough to be self obvious, and not just the observation of something that could not be not working.
01:38 - 05:30 🤷♂️
With visual intuition you can try to write down a rigorous proof.
This is specially for all of you:
In the new year eve, i thought , as a math enthusiast, why don't I give my math friends a math problem and wish them new year through that?And then , working for about an hour , I made a crazy problem myself. I sent it to some of my friends and some solved it too and were really impressed. This is a finalized version for you. Try it out yourselves.
Let f(x) = 45.e^(-x/4).[cos (2x)/{(4/65).(e^(-π/8) - e^(π/8)}]
Now find the integral of this function over the interval [-π/2 , π/2] , square the answer and see what you get!