Great video, I think a basic proof similar to others shown is missing: linking 1 with n, 2 with n-1, etc, each with a half-circle arc, giving a diagram that looks like a rainbow or a Menorah, then the sum of each pair is n+1 and you have n/2 pairs (need to split into even/odd cases). This is very similar to the proof where you duplicated the sum and flipped it, but all of the proofs are similar so I think this also counts as different
I like Gauss': (first plus last) times (half the number of terms) Works for any arithmetic sequence. Works also for arithmetic sequences missing terms in a symmetric way like say, the sum of numbers in a square on a calendar.
This is amazing. I’ve always wanted to find a UA-cam video that proves the same thing in many many different ways. I think in the future you could do a similar video on the connection between binomial expansion, Pascal’s triangle, and the number of ways to choose from a collection.
If f(x) is the generating function for a sequence {a_n}, then f(x)/(1-x) is the generating function for the partial sums of a_n. Using this fact, we see that the generating function for 1, 3, 6, 10, ..., is 1/(1-x)^3. So the sum 1+2+...+n is the x^(n-1) coefficient in the series expansion of 1/(1-x)^3, which we can evaluate as 1/(n-1)! d^(n-1)/dx^(n-1) [1/(1-x)^3] = 1/(n-1)! 3*4*...*n*(n+1) / (1-0)^(n+2) = n(n+1)/2.
This is really well done. Lots of beautiful connections (some I'd never seen before), all explained clearly. I'm always a fan of recurrence relations (and the hidden generating functions), but I think Pick's Theorem is my favorite.
Thank you for the comment! And check the link in the description for even more proofs :) I think I agree with you about Pick's theorem. Especially because it ties together area and counting proofs :)
Great job, Tom! And good luck with this impressive #SoME2 submission. I'm still working on mine. It will be an opener for the Visual Group Theory series.
I have only put up some of the code: github.com/Tom-Edgar/MVPS . I have been learning a lot over the past almost two years, but much of the code is quite messy. Those ones are some of the best (though none are documented).
Here's one that came to me after I couldn't stop thinking about this video: Consider the function f(x) = 1 + x + ... + x^n = (x^(n+1) - 1) / (x - 1). Evaluating f'(1) in the first expression gives the sum of the first n positive integers. Starting from the right expression and computing lim x->1 f'(x), we can use l'Hopital's rule to get the desired formula! edit: I see I wasn't the first to come up with this!!
Hah! Thanks. I still use his software and my videos are typically only short. Not long explainers. I have a long way to go. Still appreciate the sentiment :)
all I know about math is that 2+2=5. the rest of it I don´t know much.. great videos it gives insights to make us start loving what the universe of numbers is all about!
very cool. I started to get lost during the combinometric proofs, but the water one got me back. Maybe one day I'll return to this and see it more deeply.
Very nice video! I appreciate the work that went into this. I also didn't know some of the proofs, I really liked the one using Euler's formula! However, we used that the number of faces inside of the graph is n². This is equivalent to the summation formula we want to prove so it feels like we hide something at that step, right? Here is one proof I just thought of: Consider the expression f(x)=1+x+x²+...+xⁿ If we calculate f'(1), we can either use the summation rule to obtain the sum 1+2+...+n. On the other hand we can first use the geometric formula for this expression which gives f(x)=(xⁿ⁺¹ - 1)/(x-1) (for x≠1) Now we use the quotient rule we obtain f'(x)=((n+1)xⁿ *(x-1) - (xⁿ⁺¹ -1))/(x-1)² We can't just plug in x=1 because this is undefined. But we can use L'Hôpital twice to obtain f'(1)=((n+1)(n(n+1)-n(n-1)) - (n(n+1)))/2=n(n+1)/2
Thanks! I love that proof too! ( it is in the linked article but I couldn’t figure out how best to visualize it for this). As to the Euler’s proof. I think you can get the number of faces by a scaling argument too but yes it is also equivalent to the sum formula so maybe a bit of a stretch. Still fun to see that theorem in this context :)
@1:14 Ah yes.. the classic "use divination to find the answer then prove it's true". It's like a swimming school throwing somebody in the ocean and if they swim it's proof that they learnt to swim. (at some point... also don't be distracted by the dead bodies on the ocean floor)
Now, I admit I'm not familiar with even the term linear recurrence, let alone whatever field it's drawing from, but I feel like I'm missing a LOT of steps in that "proof." I'm certain these formulas come from somewhere, but I can't immediately tell where.
Yes. That one requires some knowledge of the theory of linear recurrences. I don’t think this video was the place. Here is a video showing how to deal with two term linear recurrences : ua-cam.com/video/YZCP_MhYqQk/v-deo.html. But it also only gives the idea and doesn’t delve into repeated roots too much. The idea in this video is to inspire you to learn about that area of mathematics now :)
I was thinking of a few before I watched and only had a few. The water flow is my favourite! Such a nice one which I didn't know about and totally for me as I had the center of mass in my head all video long. I wondered: Is there a way the Pythagorean theorem is right around the corner in this, seen as there as it triangle galore all over?
Great thought. I don’t know one that uses PT explicitly but perhaps it’s there. You’d want a n by square root of n triangle. Can you fit twice the sum of integers in there nicely?
14:50 I'm confused about how you got the result that that triangle has n^2 faces, without already having a proof of the sum of integers After all, the faces are made up of the upward facing triangles, and the downward facing triangles, meaning that the number of faces is just T_n + T_(n-1) Which is equal to T_(n-1)+n+T_(n-1)=2T_(n-1)+n This means that knowing that that Triangle has n^2 faces seems equivalent to knowing the sum of integer formula already, making the proof a circular Argument in a way
Awesome stuff! I love how esoteric and field-expanding some of the proofs got. My favorite proof is this: > Let’s assume we already know 1 + 3 + 5 + … + 2n - 1 = n^2 > Let’s add 1 to each term in the sum on the left. There are n such terms, so to keep equality, we should add n 1’s on the right: 2 + 4 + 6 + … + 2n = n^2 + n > This looks like our target result. We just have to divide the left side by 2, and what does this give on the right? (n^2 + n)/2 Thoughts: Pretty simple and cute. I always love manipulating equations in cool ways, so this has a soft spot for me ever since 7th grade. Try thinking of a good visual animation of this too… it’s not anything mind blowing, but it’s still fun.
Before shading the area below the line, y= x + 0.5, at 05:50, it would have been interesting to see that line drawn on the grid at 05:00. It was not immediately obvious to me that lifting *_both_* ends of the sloped line at 05:00 by 0.5 does slice the top squares in such a way that the top portions of those squares can be rotated and shifted to fill the valleys below the raised line.
I've read numerical proof for the sum of n integers but the proof never really stick in my head. The moment you arrange the coin in a triangle however it clicks for me that it has something to do with the area of the triangle.
When I was in precalc I got bored as I couldn't really understand what the professor was saying. So I ended up inventing that formula using what we had just learned about differential equations. Although mine was (N^2+N)/2 Most useful equation ever when playing a lot of card/board games. Now I could computer arbitrarily large sums of ascending numbers in my head instantly and I looked like a genius when I could very quickly add up scores in my head.
Sort of because it works out! It’s interesting that the area under that curve is n^2/2 +n/2. That’s kind of expected because the derivative of x^2/2+x/2 is x+1/2.
@@MathVisualProofs nvm, I get it now. The integral is calculated in two ways, both of which are equal. The integrand is really chosen just because it works
Try to proof with telescopic, multiply with 2/2, first term multiply with (2-0)/2, second term (3-1)/2, third term with (4-2)/2 until n^th term with ((n+1)-(n-1))/2 , we get telescopic form
I'm a year late, but I just found another proof: The sequence of sums of the first n positive integers is 1, 3, 6, 10, 15, 21, 28, 36, etc. For our purposes, we're going to start the sequence with a 0 so it looks like 0, 1, 3, 6, 10 etc. and considering the 0 to be the term we get for n=0. When we take this sequence's difference (meaning we do f(n + 1) - f(n) for each term), we get 1, 2, 3, 4, 5, 6, 7, 8, 9, ... because the next number in the sequence is always the last one plus the next greatest positive integer. When we take the difference again, we end up with a row of all 1's. What we are doing here is basically calculus with sequences. Now that we have our original sequence with an added 0 at the start, the first difference, and the second difference, we can use the Gregory-Newton interpolation formula to come up with a formula for our sequence. This is where we take the n=0 terms of our sequence and its differences, multiply each by the binomial coefficient n choose k where k is which difference each term represents (k=0 for our original sequence), and add them all together to get a formula for our original sequence. For the original sequence we get 0, for the first difference we get 1 x n, and for the second difference we get 1 x n(n - 1)/2. Now we just need a bit of algebraic manipulation to finish the proof. Distributing the second difference term gives us (n^2 - n)/2. Let's also manipulate the first difference term to give us 2n/2. Now we can combine both fractions into (n^2 - n + 2n)/2 which can then be condensed into (n^2 + n)/2 aka n(n + 1)/2. And that's the proof!
Used it because it works! That's the trickiest part of that proof: knowing which curve to study. But if you differentiate (x^2+x)/2 (the eventual sum), then you get x+1/2 as needed.
Many many textbooks use this as a classic first proof by induction. Second, the rectangular array visualization (which was known to the Greeks) is a literal translation of “Gauss’s trick” to a visual proof
I don’t know of such an ebook. Roger Nelsens three Proofs without words books will have most of the visualizations. I am working my way through them : ua-cam.com/play/PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw.html
Here is a video I did that gives some idea about where that comes from : ua-cam.com/video/YZCP_MhYqQk/v-deo.html. But that is only two terms and doesn’t really explain why we handle the repeated roots the way we do. I suggest searching for “solving linear recurrences” to find some more in depth notes.
@@MathVisualProofs haha I liked it as a surprise how long it kept going, really amazing seeing how everything is connected! and I have explored around some of these approaches before! Gives me Mathologer vibes how it ties things together ^_^
I think once you know what’s going on you eventually see them as the same proof - the interesting bit is how they can be reframed in different places and are then adjacent to or include so many ideas and techniques.
The fundamental idea is that if you put n*n + n markers on a table, you always have the same number of markers if you just shuffle them around and regroup them however and never add or delete any. Ditto for an area cut into squares, with some squares cut along a diagonal as needed.
@@MathVisualProofs I see what you're saying. It just sounds a little like you're saying that your proof doesn't depend on induction. That first there is proof and then induction depends on the proof, which is not how thought works.
@@MathVisualProofs Because no matter how many times you find something to be "right", you'll never be sure that next time you do it it's going to be "right" again. That's why everything in physics is a "right" theory until proven wrong. In math instead, we define what is "right" and what isn't (e.g. 1+1=2). With induction you can't prove anything, because for example no matter how many times you see a stone fall on the ground, there's no way to determine for sure what would happen if you did the "experiment" once more. Physics is all about sensible guesses, which we assume to be laws of nature until proven wrong. Math is all about truths we define ourselves.
Tell me which of the 12 is your favorite! Or maybe you have a favorite I didn't include?
I like the triangle area proof. It's so elegant
@@michaelmam1490 triangular numbers from triangular area-makes so much sense! Thanks!
Great video, I think a basic proof similar to others shown is missing: linking 1 with n, 2 with n-1, etc, each with a half-circle arc, giving a diagram that looks like a rainbow or a Menorah, then the sum of each pair is n+1 and you have n/2 pairs (need to split into even/odd cases).
This is very similar to the proof where you duplicated the sum and flipped it, but all of the proofs are similar so I think this also counts as different
@@AssemblyWizard for sure. That’s a nice one. Thanks for sharing!
I like Gauss': (first plus last) times (half the number of terms)
Works for any arithmetic sequence. Works also for arithmetic sequences missing terms in a symmetric way like say, the sum of numbers in a square on a calendar.
This is amazing. I’ve always wanted to find a UA-cam video that proves the same thing in many many different ways.
I think in the future you could do a similar video on the connection between binomial expansion, Pascal’s triangle, and the number of ways to choose from a collection.
Thanks! Definite a good suggestion for the future - I will put it on the list :)
If f(x) is the generating function for a sequence {a_n}, then f(x)/(1-x) is the generating function for the partial sums of a_n. Using this fact, we see that the generating function for 1, 3, 6, 10, ..., is 1/(1-x)^3.
So the sum 1+2+...+n is the x^(n-1) coefficient in the series expansion of 1/(1-x)^3, which we can evaluate as 1/(n-1)! d^(n-1)/dx^(n-1) [1/(1-x)^3] = 1/(n-1)! 3*4*...*n*(n+1) / (1-0)^(n+2) = n(n+1)/2.
Excellent one for sure! I wasn’t sure how to do visualization of gen funs, so I didn’t include it. But I love it! Thanks!
You have a great reading voice, I could easily see you having success in voice acting/audiobooks
Thanks! Probably not in my future, but good to have options :)
The center of mass proof is amazing! All of them are, but that one is the best one
Thanks! I definitely like that proof.
I haven’t seen all your videos-you have many-but I’ve seen a lot.. and this is my favorite one. It offers a ton in a short amount of time. Excellent
Thanks! This was fun to make :)
This is really well done. Lots of beautiful connections (some I'd never seen before), all explained clearly. I'm always a fan of recurrence relations (and the hidden generating functions), but I think Pick's Theorem is my favorite.
Thank you for the comment! And check the link in the description for even more proofs :) I think I agree with you about Pick's theorem. Especially because it ties together area and counting proofs :)
Great job, Tom! And good luck with this impressive #SoME2 submission. I'm still working on mine. It will be an opener for the Visual Group Theory series.
Thanks! Oh that's great! I look forward to seeing what you do for that. I'll keep my eyes out for it.
I came here from UA-cam shorts and I was wondering if you had a public repository with all the manim scripts. That would be very helpful!
I have only put up some of the code: github.com/Tom-Edgar/MVPS . I have been learning a lot over the past almost two years, but much of the code is quite messy. Those ones are some of the best (though none are documented).
@@MathVisualProofs Thank you so much!!!
Here's one that came to me after I couldn't stop thinking about this video:
Consider the function f(x) = 1 + x + ... + x^n = (x^(n+1) - 1) / (x - 1). Evaluating f'(1) in the first expression gives the sum of the first n positive integers. Starting from the right expression and computing lim x->1 f'(x), we can use l'Hopital's rule to get the desired formula!
edit: I see I wasn't the first to come up with this!!
Yes! This one is excellent (love the geo series formula). It’s in the linked paper but I couldn’t come up with a great visualization for it. Thanks!
This guy right here is the real MVP of Mathematics on UA-cam. 3blue1brown? Who is that?
Hah! Thanks. I still use his software and my videos are typically only short. Not long explainers. I have a long way to go. Still appreciate the sentiment :)
all I know about math is that 2+2=5. the rest of it I don´t know much.. great videos it gives insights to make us start loving what the universe of numbers is all about!
Great job with this collection, kudos! 👏
Thanks!
very cool. I started to get lost during the combinometric proofs, but the water one got me back. Maybe one day I'll return to this and see it more deeply.
The combinatorial one is a challenge for sure. Thanks for checking it out!
9:45 This bijection has had an aha-effect on me! So easy - and so brilliant.
Glad you liked it!
Hey man, wonderful video here ! Animations are gorgeous.
Do you have a github in order to see the Manim code you used to animate this video ?
The code for this video is not up there yet. I have a small repo with code from some videos: github.com/Tom-Edgar/MVPS
Very nice video! I appreciate the work that went into this. I also didn't know some of the proofs, I really liked the one using Euler's formula! However, we used that the number of faces inside of the graph is n². This is equivalent to the summation formula we want to prove so it feels like we hide something at that step, right?
Here is one proof I just thought of: Consider the expression
f(x)=1+x+x²+...+xⁿ
If we calculate f'(1), we can either use the summation rule to obtain the sum 1+2+...+n. On the other hand we can first use the geometric formula for this expression which gives
f(x)=(xⁿ⁺¹ - 1)/(x-1) (for x≠1)
Now we use the quotient rule we obtain
f'(x)=((n+1)xⁿ *(x-1) - (xⁿ⁺¹ -1))/(x-1)²
We can't just plug in x=1 because this is undefined. But we can use L'Hôpital twice to obtain
f'(1)=((n+1)(n(n+1)-n(n-1)) - (n(n+1)))/2=n(n+1)/2
Thanks! I love that proof too! ( it is in the linked article but I couldn’t figure out how best to visualize it for this). As to the Euler’s proof. I think you can get the number of faces by a scaling argument too but yes it is also equivalent to the sum formula so maybe a bit of a stretch. Still fun to see that theorem in this context :)
Very nice. And nice to see you're getting some traction!
Thanks! This has been the best traction yet... still a long way to go :)
A few years ago, I used the formula “(n(n+1)) / 2” to make another formula, where instead of counting up by “1”, you count up by “x”.
(n(n+x)) / (2x)
Lol don't think this is correct.... If you know what an Arithmetic Progression means 😁
awesome keep it up
Thanks
Thanks for checking it out!
@1:14 Ah yes.. the classic "use divination to find the answer then prove it's true". It's like a swimming school throwing somebody in the ocean and if they swim it's proof that they learnt to swim. (at some point... also don't be distracted by the dead bodies on the ocean floor)
Haha. It’s unfortunate that that appears to be the most common way people encounter the formula…
Well I'm convinced.
😀
Now, I admit I'm not familiar with even the term linear recurrence, let alone whatever field it's drawing from, but I feel like I'm missing a LOT of steps in that "proof." I'm certain these formulas come from somewhere, but I can't immediately tell where.
Yes. That one requires some knowledge of the theory of linear recurrences. I don’t think this video was the place. Here is a video showing how to deal with two term linear recurrences : ua-cam.com/video/YZCP_MhYqQk/v-deo.html. But it also only gives the idea and doesn’t delve into repeated roots too much. The idea in this video is to inspire you to learn about that area of mathematics now :)
I was thinking of a few before I watched and only had a few. The water flow is my favourite! Such a nice one which I didn't know about and totally for me as I had the center of mass in my head all video long.
I wondered: Is there a way the Pythagorean theorem is right around the corner in this, seen as there as it triangle galore all over?
Great thought. I don’t know one that uses PT explicitly but perhaps it’s there. You’d want a n by square root of n triangle. Can you fit twice the sum of integers in there nicely?
14:50 I'm confused about how you got the result that that triangle has n^2 faces, without already having a proof of the sum of integers
After all, the faces are made up of the upward facing triangles, and the downward facing triangles, meaning that
the number of faces is just T_n + T_(n-1)
Which is equal to T_(n-1)+n+T_(n-1)=2T_(n-1)+n
This means that knowing that that Triangle has n^2 faces seems equivalent to knowing the sum of integer formula already, making the proof a circular Argument in a way
Is a good point but I think you can get the argument by scaling.
In my opinion this sum equals to -1/12
Hah! Only if you never stop ;)
This channel is hidden gem!
Thanks for thinking so! Appreciate it :)
I agree. I hope we can help make it less hidden!
i love your chanel
Thanks! Appreciate the feedback :)
Awesome stuff! I love how esoteric and field-expanding some of the proofs got. My favorite proof is this:
> Let’s assume we already know 1 + 3 + 5 + … + 2n - 1 = n^2
> Let’s add 1 to each term in the sum on the left. There are n such terms, so to keep equality, we should add n 1’s on the right: 2 + 4 + 6 + … + 2n = n^2 + n
> This looks like our target result. We just have to divide the left side by 2, and what does this give on the right? (n^2 + n)/2
Thoughts: Pretty simple and cute. I always love manipulating equations in cool ways, so this has a soft spot for me ever since 7th grade. Try thinking of a good visual animation of this too… it’s not anything mind blowing, but it’s still fun.
Excellent! Nice one. Thanks!
I think u mean "n" not "1" but I love this!!
@@MonkOrMan Fixed! Thanks for seeing it
@@myrus5722 where did you fix.... It's still '1' has to be 'n'
@@050138 Oh I had it wrong in two places maybe? Should be completely fixed now
فيديو مذهل، أقدّر هذا المجهود الكبير وأحييك على إتقانك
أتساءل إن كان هناك برهان باستخدام علم المثلثات
Good question! I would guess yes :)
Before shading the area below the line, y= x + 0.5, at 05:50, it would have been interesting to see that line drawn on the grid at 05:00. It was not immediately obvious to me that lifting *_both_* ends of the sloped line at 05:00 by 0.5 does slice the top squares in such a way that the top portions of those squares can be rotated and shifted to fill the valleys below the raised line.
That’s a good one! Wish I had thought of it.
I've read numerical proof for the sum of n integers but the proof never really stick in my head. The moment you arrange the coin in a triangle however it clicks for me that it has something to do with the area of the triangle.
😀👍
When I was in precalc I got bored as I couldn't really understand what the professor was saying. So I ended up inventing that formula using what we had just learned about differential equations. Although mine was (N^2+N)/2
Most useful equation ever when playing a lot of card/board games. Now I could computer arbitrarily large sums of ascending numbers in my head instantly and I looked like a genius when I could very quickly add up scores in my head.
Very cool! One of my top fave formulas.
One small tip: Please consider also international viewers and don't use sizes like gallons. But otherwise, a very interesting video!
Yes of course :) Luckily it is really just a problem about rates, so you can assume liters per minute. Thanks!
Had to do a double take there. YTs algorithm threw an ad with grifter and con deluxe Elon Musk at me before the video.
Weird. I don’t understand the algorithm at all.
amazing lesson
Thanks! 😃 I'm glad you liked it
I don’t need sleep. I NEED ANSWERS
How do you know so many diverse things, do they teach these topics in ug math?
I’ve been around a long time and have thought a lot about math. You pick things up over time.
hey! i'm confused why, in the FTC proof, the function y = x + 1/2 was used
Sort of because it works out! It’s interesting that the area under that curve is n^2/2 +n/2. That’s kind of expected because the derivative of x^2/2+x/2 is x+1/2.
@@MathVisualProofs nvm, I get it now. The integral is calculated in two ways, both of which are equal. The integrand is really chosen just because it works
subscribed before i even started watching :D
great video :D
Thanks! Glad you enjoyed it :)
Reminds me of Philip Ording's 99 Variations on a Proof. Great job!
An excellent book! Thanks :)
Try to proof with telescopic, multiply with 2/2, first term multiply with (2-0)/2, second term (3-1)/2, third term with (4-2)/2 until n^th term with ((n+1)-(n-1))/2 , we get telescopic form
great video
Thank you!
1:05. The triangle area...
I'm a year late, but I just found another proof:
The sequence of sums of the first n positive integers is 1, 3, 6, 10, 15, 21, 28, 36, etc. For our purposes, we're going to start the sequence with a 0 so it looks like 0, 1, 3, 6, 10 etc. and considering the 0 to be the term we get for n=0. When we take this sequence's difference (meaning we do f(n + 1) - f(n) for each term), we get 1, 2, 3, 4, 5, 6, 7, 8, 9, ... because the next number in the sequence is always the last one plus the next greatest positive integer. When we take the difference again, we end up with a row of all 1's. What we are doing here is basically calculus with sequences. Now that we have our original sequence with an added 0 at the start, the first difference, and the second difference, we can use the Gregory-Newton interpolation formula to come up with a formula for our sequence. This is where we take the n=0 terms of our sequence and its differences, multiply each by the binomial coefficient n choose k where k is which difference each term represents (k=0 for our original sequence), and add them all together to get a formula for our original sequence. For the original sequence we get 0, for the first difference we get 1 x n, and for the second difference we get 1 x n(n - 1)/2. Now we just need a bit of algebraic manipulation to finish the proof. Distributing the second difference term gives us (n^2 - n)/2. Let's also manipulate the first difference term to give us 2n/2. Now we can combine both fractions into (n^2 - n + 2n)/2 which can then be condensed into (n^2 + n)/2 aka n(n + 1)/2. And that's the proof!
My favorite was the bijective proof.
Suck a cool one, right?
This video is great thank you! And your thumb’s nail is beautiful!
Thanks for watching!
Good video! What manim version did you use?
Thanks! This is done with manimgl. I haven’t moved to ce yet.
Where did the y = x + 1/2 came from in the FTC proof?
Used it because it works! That's the trickiest part of that proof: knowing which curve to study. But if you differentiate (x^2+x)/2 (the eventual sum), then you get x+1/2 as needed.
You can also do it the way mathematicians have always done it when the problem is too hard to solve. Call it an axiom.
Hah! Yes. I suppose that’s one way.
How about thinking simple about deep things
-1/12
At 14:52 how did you determine the number of faces? Don't we first need to know (or prove) that the sum of the first n odd integers is equal to n*n?
One way to get it is to use the formula. Another way is to use a scaling argument.
First, no one learns this formula by induction.
Second, Gauss' proof is 2*S=(1+n)+(2+n--1)+(3+n-2)+....+ (n+1)=n*(n+1)
Many many textbooks use this as a classic first proof by induction. Second, the rectangular array visualization (which was known to the Greeks) is a literal translation of “Gauss’s trick” to a visual proof
Good Topic
Thanks!
amazing content! keep on it
Thank you! I’ll see what I can do :)
Interesante las diversas interpretaciones, a lo que yo considero la inducción, especialmente la de n(n+1)(0.5)
Thanks for checking it out!
Can u give us a link to ebook of sequences and series , that give visual explaination and ulstration
I don’t know of such an ebook. Roger Nelsens three Proofs without words books will have most of the visualizations. I am working my way through them : ua-cam.com/play/PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw.html
I am really impressed by this video, just beautiful....
Thank you! It was fun to make!
I love this
Thanks for checking it out!
Thanks!
Woah! Thank you. I appreciate it :)
Thanks!
Thanks for checking it out.
Can someone explain how the equation at 11:30 arises? I understand where the coefficients come from but not what x is supposed to represent.
Here is a video I did that gives some idea about where that comes from : ua-cam.com/video/YZCP_MhYqQk/v-deo.html. But that is only two terms and doesn’t really explain why we handle the repeated roots the way we do. I suggest searching for “solving linear recurrences” to find some more in depth notes.
@@MathVisualProofs Thanks for the prompt reply! Enjoyed the video
@@lukeinvictus thanks!
Langland’s dream …
(n²+n)/2
But here's twelve different ways to get it!
Reminds me a Christmas song 😋
Should have phrased it that way maybe 😀
@@MathVisualProofs haha I liked it as a surprise how long it kept going, really amazing seeing how everything is connected! and I have explored around some of these approaches before!
Gives me Mathologer vibes how it ties things together ^_^
@@oncedidactic his videos are the best.
This is amazing! I subbed
Thanks!
Very Inserting!
Thanks!
Here is a proof using telescoping sum:
a(n) = n or n[ a(n) -a(n-1) ]
Let S(n) = a(1) + a(2) +...+a(n).
a(1)= 1[ a(1) - a(0) ]
a(2)= 2[ a(2) - a(1) ]
a(3)= 3[ a(3) - a(2) ]
..... ....................
..... ....................
a(n) = n[ a(n) -a(n-1)]
Summing all equations:
S(n) = -S(n-1) + n^2
S(n) + S(n-1) + n = n^2 + n. (Adding n on both side)
S(n) = (n^2 + n)/2.
The late Twentieth Century
no. I don't use induction.
aren't so much of these the same proofs?
I think once you know what’s going on you eventually see them as the same proof - the interesting bit is how they can be reframed in different places and are then adjacent to or include so many ideas and techniques.
The fundamental idea is that if you put n*n + n markers on a table, you always have the same number of markers if you just shuffle them around and regroup them however and never add or delete any. Ditto for an area cut into squares, with some squares cut along a diagonal as needed.
-1÷12
only if you let n go to infinity ;) So n(n+1)/2 - > -1/12 as n goes to infinity I guess...
generate all sets of pyhtagoran triangle integer solutions (up to certain number like 100k), all the possible variations, integer combinations
I am not sure I quite understand this one... can explain?
@@MathVisualProofs meh, its just an algorithm description suggestion
@@MathVisualProofs what you use it for is another thing
@@MathVisualProofs maybe find the 3d box shape that has only integers as the sides and the connecting lines
This is really cool. I was blown up by the bijective proof.
Very engaging indeed!
Thanks! Yes. That bijective proof is totally amazing - the kind of thing I wished I had thought of myself :)
great to know! thanks
👍
👍
😀
Unrelated, but what is the theorem that 2 + 2 = 4, 2 × 2 = 4, 2 ^ 2 = 4, etc. called?
I’m not sure I know if that’s a theorem. It’s just a coincidence?
@@MathVisualProofs: What is it called though?
@@MathVisualProofs: There are very few posts online about it.
Its not a theorem, why would it?
@@rafiihsanalfathin9479 : What is the 'rule' called then?
Induction CAN NOT be our last resort because ALL DEDUCTION depends on prior induction.
I think the visual proof of stacks of squares should be first. Induction is too dry and requires that you already know a formula.
@@MathVisualProofs I see what you're saying. It just sounds a little like you're saying that your proof doesn't depend on induction. That first there is proof and then induction depends on the proof, which is not how thought works.
You're confusing philosophical induction and mathematical induction
@@schweinmachtbree1013 thanks!
@@schweinmachtbree1013 mathematical induction also depends on philosophical induction
Induction is not a proof btw
Why not?
@@MathVisualProofs Because no matter how many times you find something to be "right", you'll never be sure that next time you do it it's going to be "right" again. That's why everything in physics is a "right" theory until proven wrong. In math instead, we define what is "right" and what isn't (e.g. 1+1=2).
With induction you can't prove anything, because for example no matter how many times you see a stone fall on the ground, there's no way to determine for sure what would happen if you did the "experiment" once more.
Physics is all about sensible guesses, which we assume to be laws of nature until proven wrong. Math is all about truths we define ourselves.