i guess Im asking the wrong place but does any of you know a method to log back into an instagram account?? I somehow lost my password. I would appreciate any tips you can give me!
@Benedict Uriah I really appreciate your reply. I found the site through google and I'm trying it out atm. Takes a while so I will get back to you later with my results.
Thank you very much for such wonderfully done videos. It is indeed great to see the pictures and the intuitive ideas behind those great formulas. You do make it much easier to understand them, though I needed to watch the (n+1/2) for the height a few times to realize where it came from ;))))
That formula is also the formula for counting all visible squares in a square divided by smaller equal squares with n sides. ;-) I've also derived that formula when I was high school student after our Mathematics teacher taught us how to derive an equation when x and y values are given! ;-) I've come up to derivation of this formula when my brother challenged me to count the number of squares in a 3 x 3 square using matchsticks. Obviously for a young fellow like me, I answered 9 which was wrong because of the fact that there are four 4x4 squares plus the one 3x3 square plus the 9 smallest squares which total is 14 visible squares. Because of that, I wondered how to count squares in a big number of side like in chessboard. ;-)
Do you happen to recall where you came across it? Not saying I’d get it, but a crack at the harder stuff is never a bad thing so long as someone isn’t hard on themselves if they don’t understand it.
@@drakesmith471 I don't remember, but I think there is a video in blackpenredpen explaining it. If not, I'm sure that if you look something like "sum of first n integers" in UA-cam (or google) you'll get the explanation.
@@facilvenir it’s funny you mentioned the black pen red pen video, I got it suggested to me but dismissed it thinking it would be geometric by nature and thus repeat of this. Thank you for setting me on the next leg of the journey.
Hi. I'm just wondering if 1² + 2² + 3²... +24² = 4900, and the square root of 4900 is 70, is it possible to perfectly arrange all the 1x1, 2x2, 3x3 up to 24x24 squares inside the big square of 70x70 without space or overlapping areas?
+Daniel Vieiralves Though, I would't be able to comprehend it. Speaking about the sum of powers.... x + (x+1) + (x+2)... x^2 + (x+1)^2... Was this the intuition for Bernoulli numbers? I'm very interested in them because they appear in the expansions of trig functions, such as tan(x). All I know is that there is a first and second set of Bernoulli numbers which differ from just one number.
+Daniel Vieiralves Eu não posso entender a tua língua matemática e estratégia. É muito complexa quando pode ser mais fácil. I still prefer Sal's method. Easier to see the derivation. When you figure about Bernoulli's numbers please tell me. Obrigado de novo.
I was able to reason out the sum of natural numbers to n on my own. I was trying to do similar with this one but I couldn't visualize how to make a predictable solid out of the pyramid. I especially couldn't visualize how those pyramids would stack together. But this is really great. Is there a general way to prove these formulas?
misteratoz Yes, but the one I've seen uses algebra. I tried deriving the case for cubes with geometry and I got pretty close, but it's substantially harder than this one, and I found a much easier way without geometry.
Actually, it's really hard because you need to do this in the 4th dimension, you can't make flat surfaces with cubes, only more cubes... Like 3^3=27, but 27 can't make a square, only a cube (a 3x3 cube). But maybe there is a way in the third dimension.
Tyrone Lannister sorry for bringing you back to this video after so long, but I just found it and hope to explain it if you haven't already learnt yourself Proof by induction is not how you get the formula, it's a way to prove it right for all values of n in a completely algebraic not geometric way. first, your prove it true for n=1. Then, make the assumption it is true for n=k, and you get k(k+1)(2k+1)/6 and then you find the arithmetic progression expansion (1+4+9+16...+k^2). then you find the expansion for n=k+1, in this case is (k+1)(k+2)(2k+3)/6 then, you also find the arithmetic progression expansion (1+4+9+16...+k^2+(k+1)^2) and you equate it to the expansion for k. you should see that this expansion is simply the same as the expansion for k + (k+1)^2. thus you can equate the two right hand sides and see if they are equal. From this you get k(k+1)(2k+2)/6+k^2=(k+1)(k+2)(2k+3)/6. if your algebra is correct then these should be equal. Thus, you've proved a ink between subsequent integers. Plugging in k=1, you can show that as this is true from the first part, then it must be true to the subsequent integers 2, and if true for 2, it is true for 3 and so on and hence your proved it is true for all positive integers. This is proof by induction.
Even though the video is about squared integers, the visual explanation works with cubes ... Isn't it supposed to be explained with squares? But it wouldn't work!
This explanation is saying 1 is represented by one cube, 2 squared is being represented by 4 cubes, 3 squared is being represented by 8 cubes so each individual cube is just being represented as one so the aim is to find out the total number of cubes so we can get 1+4+9+16…+n.
if you take it as squares, you get n(n+1)/2 when you do that, for n(n+1)/2 to be= n(n+1)(2n+1)/6 , n has to be 1 , so either your wrong or im missing something , either way this is bad video
This is real mathematics. No memorization, only understanding.
i guess Im asking the wrong place but does any of you know a method to log back into an instagram account??
I somehow lost my password. I would appreciate any tips you can give me!
@Elijah Briar Instablaster ;)
@Benedict Uriah I really appreciate your reply. I found the site through google and I'm trying it out atm.
Takes a while so I will get back to you later with my results.
@Benedict Uriah It worked and I actually got access to my account again. I am so happy!
Thank you so much, you really help me out!
@Elijah Briar happy to help =)
THAT WAS BEAUTIFUL
Amazing channel. I really love that !
I never ever memorize any law or equation unless I don't know where it really came from.
Thank you very much for such wonderfully done videos. It is indeed great to see the pictures and the intuitive ideas behind those great formulas. You do make it much easier to understand them, though I needed to watch the (n+1/2) for the height a few times to realize where it came from ;))))
You should do this one next time! (1+2+3+ ... +n)^2 = 1^3 + 2^3 + 3^3 + ... + n^3 = {[n(n+1)]/2}^2
It is cool!!! This is missing part from my junior and senior middle school. This makes all algebra lively sense and meaningful.
I'm just simply amazed... That was a beautiful explanation about the Sum of integers squared. You are awesome. Keep it up! :D
So concise and so precise. Elegant!
Excellent presentation of the subject. Dr Rahul RohtakIndia.
Geometric explanations really are the best way to undestand these things huh :P
Wow that was so much better than my textbook! THANK YOU
Awesome! the reason why I love mathematics
Great explanation. i loved the animation :)
Awesome explanation! ;) Out of curiosity...what software did you use for the graphics?
Valerio Varricchio i m commenting for ans
The visuals really help thank you so so very much I subbed :)
Thanks for the sub!
What ! only 17k subscribers ??
those are all the future doctors I think, good luck.
That formula is also the formula for counting all visible squares in a square divided by smaller equal squares with n sides. ;-)
I've also derived that formula when I was high school student after our Mathematics teacher taught us how to derive an equation when x and y values are given! ;-)
I've come up to derivation of this formula when my brother challenged me to count the number of squares in a 3 x 3 square using matchsticks. Obviously for a young fellow like me, I answered 9 which was wrong because of the fact that there are four 4x4 squares plus the one 3x3 square plus the 9 smallest squares which total is 14 visible squares.
Because of that, I wondered how to count squares in a big number of side like in chessboard. ;-)
Very nicely done!
wow... beautiful animation i'm subbed
1:15 What is the reasoning behind that step? Creating two other copies?
Thank you. But why do we have to do the steps after 2:36?
thanks for sharing, what animation software did you use ?
wow! great video and explanation
which software did you use to 3D animate the cubes?
Math is God's Language! This is awesome! Thank you sir!
There's also an algebraic explanation for this. It's much more difficult to understand but more academic.
Do you happen to recall where you came across it? Not saying I’d get it, but a crack at the harder stuff is never a bad thing so long as someone isn’t hard on themselves if they don’t understand it.
@@drakesmith471 I don't remember, but I think there is a video in blackpenredpen explaining it. If not, I'm sure that if you look something like "sum of first n integers" in UA-cam (or google) you'll get the explanation.
@@facilvenir it’s funny you mentioned the black pen red pen video, I got it suggested to me but dismissed it thinking it would be geometric by nature and thus repeat of this. Thank you for setting me on the next leg of the journey.
This was really help ful!!
Okay this image of the three stacked squares vaguely puts me in mind of the fact that the derivative of x cubed is 3 (x squared).
Your comment was more interesting than the video!
🙄🤔🤔😮👍👍🙏
Exceptional
Thanks very nice vid
pls explain integration and differentiation
thank you
How long has the equation been around?
At least 5 years
Why 1/2? Didnt get that part
because the last layer is missing half
if it were complete it would be one more
Beautiful! :D
For the first time I enjoyed mathematics.
Is everything really solvable with geometry?
Wocannasei Lai possibly not.
Wocannasei Lai possibly not.
it's a good question ;)
thank you man
i love this one
very thanx for video!
Nice
Thnkss
starting at 2:37 is wrong
how could
n(n+1)(n+1/2)=
=[n(n+1)(2n+1)]÷2?
shouldn't it be [2n(2n+2)(2n+1)]÷2?
so it could cancel out on all of the ecuation?
2:39
please help
2n(2n+2)(2n+1)=[n(n+1)(n+0.5)]*8
8=2x2x2
8=2x2x(4/2)
8=(2x2x4)/2
does this make sense to you
Chua Yew Hui
Yesss thanks
(10*11*21)/6=(2310)/6=385 clever !
I checked this on my calculator !
Hi. I'm just wondering if 1² + 2² + 3²... +24² = 4900, and the square root of 4900 is 70, is it possible to perfectly arrange all the 1x1, 2x2, 3x3 up to 24x24 squares inside the big square of 70x70 without space or overlapping areas?
But sir we need cubes intsread of squares.
@@AmanSingh-or6yf No..? He wants to arrange squares inside a big square. Not sure whether things won't overlap but you don't need cubes...
how to derive this formula algebraically ?(i mean without the help of geometry?)
Khan Academy has an explanation. It required system of equations.
Something like that.
+Daniel Vieiralves
Interesting....
+Daniel Vieiralves
Though, I would't be able to comprehend it.
Speaking about the sum of powers....
x + (x+1) + (x+2)...
x^2 + (x+1)^2...
Was this the intuition for Bernoulli numbers?
I'm very interested in them because they appear in the expansions of trig functions, such as tan(x).
All I know is that there is a first and second set of Bernoulli numbers which differ from just one number.
+Daniel Vieiralves
Eu não posso entender a tua língua matemática e estratégia.
É muito complexa quando pode ser mais fácil.
I still prefer Sal's method. Easier to see the derivation.
When you figure about Bernoulli's numbers please tell me.
Obrigado de novo.
+Daniel Vieiralves
+Daniel Vieiralves
P.s.
Your English is PERFECT.
The only thing was the "wasn't".
*Which part weren't you able to grasp.
:)
didn't know the formula but amusing
You can use 6 copies to make a perfect cube of n x (n+1) x (2n+1), which would be much better than this 1/2 "half height" proof.
Chounoki It will not be a cube, but a rectangular prism
+Chounoki you could also slice the green thing at the top in half and (vertically) and put the cut off half on the red area
👏👏👏
Could you accept the subtitles for Brazil please ?
wow amazing :)
I was able to reason out the sum of natural numbers to n on my own. I was trying to do similar with this one but I couldn't visualize how to make a predictable solid out of the pyramid. I especially couldn't visualize how those pyramids would stack together. But this is really great. Is there a general way to prove these formulas?
misteratoz Yes, but the one I've seen uses algebra. I tried deriving the case for cubes with geometry and I got pretty close, but it's substantially harder than this one, and I found a much easier way without geometry.
AMAZING)
make one with 1^3+2^3 etc..
It's possible I assume, but harder to put the pieces together.
Actually, it's really hard because you need to do this in the 4th dimension, you can't make flat surfaces with cubes, only more cubes... Like 3^3=27, but 27 can't make a square, only a cube (a 3x3 cube). But maybe there is a way in the third dimension.
Much better than #3blue1brown
Just use induction.
+Josh Liu pls tell me how this is done.
Tyrone Lannister sorry for bringing you back to this video after so long, but I just found it and hope to explain it if you haven't already learnt yourself
Proof by induction is not how you get the formula, it's a way to prove it right for all values of n in a completely algebraic not geometric way.
first, your prove it true for n=1. Then, make the assumption it is true for n=k, and you get k(k+1)(2k+1)/6 and then you find the arithmetic progression expansion (1+4+9+16...+k^2). then you find the expansion for n=k+1, in this case is (k+1)(k+2)(2k+3)/6 then, you also find the arithmetic progression expansion (1+4+9+16...+k^2+(k+1)^2) and you equate it to the expansion for k. you should see that this expansion is simply the same as the expansion for k + (k+1)^2. thus you can equate the two right hand sides and see if they are equal. From this you get k(k+1)(2k+2)/6+k^2=(k+1)(k+2)(2k+3)/6. if your algebra is correct then these should be equal. Thus, you've proved a ink between subsequent integers. Plugging in k=1, you can show that as this is true from the first part, then it must be true to the subsequent integers 2, and if true for 2, it is true for 3 and so on and hence your proved it is true for all positive integers. This is proof by induction.
Induction doesn't help with understanding. It only helps with proofs.
Guarda mio video
Geometria numerata
Superficie area cerchio
I cannot understand that half height thing. That's it 😣
Damn
Even though the video is about squared integers, the visual explanation works with cubes ... Isn't it supposed to be explained with squares? But it wouldn't work!
This explanation is saying 1 is represented by one cube, 2 squared is being represented by 4 cubes, 3 squared is being represented by 8 cubes so each individual cube is just being represented as one so the aim is to find out the total number of cubes so we can get 1+4+9+16…+n.
Very nice but with 6 of these pyramids no algebra was necessary!
if you take it as squares, you get n(n+1)/2 when you do that, for n(n+1)/2 to be= n(n+1)(2n+1)/6 , n has to be 1 , so either your wrong or im missing something , either way this is bad video
nvm, i missed something
To calculate the sum of the first _n_ *squares,* let us talk about *cubes,* first…
Sorcery
Nước ngoài ngta học bản chất ở vn chơi học thuộc, mẹo :))
what the hell math is crazy
I CAN’T UNDERSTAND!!! AM I DUMB????