I never really liked math as a teen and I'm just now starting my journey into mathematics and things like this make me really appreciate beauty in it. Excellent job.
just came to say the same, I really miss the evenings where I went to the internet and started reading maths texts, first divulgation ones, then more complex (but still in the divulgation/not really formal field) and finally more formal ones (which i didn’t fully understand until i went to university). I like to think that math is like art, first the mathmatician fights to find the proof, then the rest of us admire the piece and finally we get to understand it. Still amazes me how even after that many years learning math it stills shows me there is more beauty hidden out there
@@pritamyadav17 I am doing my best. I do have about 200+ videos already up if you want to check the back catalog (but the older ones were when I was first learning so they could be updated perhaps).
I have seen the majority of them on olympics or challenges, and i finally discovering that it has some logix behind, the point that it isn't just uses to be in a random question, but the beauty of geometry.
विविध भौमितिक आकृत्या,त्यांचे अनंत विभाग करून त्यांची बेरीज ,प्रात्यक्षिकासह दर्शवल्यामुळे अनेक घटकांची माहिती मिळाली, यामुळे विविध कल्पना सुचतात, भूमिती मध्ये लपलेल्या सौंदर्याच दर्शन घडले धन्यवाद सर
For sure it is! Notice that the dissections for k=6 and k=7 can be generalized for any integer k. You can also prove the formula you have in a variety of ways for any k>1.
@@Goaw25511/0 does not approach anything. The limit of s to 0 for 1/s approaches infinity from above. From below for example it approaches negative infinity. If you take s=((-1)^n)/n you can take the limit of n to infinity and see that it doesnt converge at all. Thus we mathematicians dont like to talk about 1/0
The reason why i've always struggled with math is because i'm a very visual person.I feel like the difference between just knowing how to do math and knowing how math works and why things are the way they are adds a whole depth to the subject that is never taught in schools. A depth that allows you to understand the world around you better. Once you can look at a circle and fully understand what pi is, then you will never look at a circle the same way again. You will constantly have that teaching reinforced because it enhanced your understanding of the world. Which is why we invented math in the first place.
for the circle one, cant you prove the same thing with the hexagon one previously? can't you split it into an infinitely large number of segments and prove infinitely many geometric series?
For sure. Both n=6 and n=7 in this video can be generalized to any geometric series of the form 1/n where n is a positive integer. I even have another old video showing how you can use the circle idea (and so the polygon idea) to get some different series: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bxSCJR6RRxs.html
I think actually from (1/3)^i you can do all proofs on circle and perhaps there is general solution for 1/n+(1/n)²+(1/n)³+...=1/(n-1) I think so cuz you cut inside circle smaller one so that you can cut a bigger piece into n-1 parts and n-th part is circle where you repeat process for eg if you have series with 1/4^i you make 3 pieces on circle and there must be a smaller circle so that small circle is equal to each of one pieces from bigger one co you taking 1/4 of circle and going into small one and repeat process
@@nonameee0729 for sure! You can even use the circle for sums of 1/2, but it's a bit strange because you get an inner circle of 1/2 area and an annulus of 1/2 area... so the annuli just shrink in at various powers of 1/2.
I had no idea you had then all in one place. How beautiful and perfect that the infinite sum is the previous fraction. It can't help but be such and yet still.....❤
Thanks! I made them one at a time, but then I tried making a compilation video here (and I turned them into shorts). The compilation video did better than all my other long-form videos, so maybe I'll have to create more compilations.... Appreciate you watching them and commenting!
(1/k)^i for i:0 => infinity and k integer grater then 1 is 1/(1 - 1/k) which is k/(k-1). Now we can subtract fisrt item which is 1/k^0=1 and we get k/(k-1) - 1 = 1/(k-1)
4:05 I think this circle method can be used to prove the general situation since any circle can be dissected such that there are r-1 sections surrounding a central circle. So this is the general proof that the sum of any geometric sequence of Σ(1/r)^n is equal to 1/(r+1).
Beautiful video for a fascinating concept. I "sense" this has a deep meaning in our universe. I know their inversed, but it's like a sum of powers of an integer originates the next integer. Or maybe I'm just crazy. Probably the latter.
To make sense of an infinite sum we let it be the limit of the partial sums with n terms as n goes to infinity. So the partial sums get infinitely close to the infinite sum but the infinite sum is the limit so the infinite sum is the fraction shown.
In the geometric series of ½+¼+⅛ you wrote 1/2^i which is fine and correct, but not acceptable. The letter "i" is reserved for the root of -1. And what if you were to solve a problem with a triangle blocked inside a circle, would you mark one of the angles in the triangle with the Greek letter π? of course not. The spectators or students in the class will not understand what you mean when you say 2π. Twice the angles π, or twice pi. The same with the letter i. From the day it was established that i is the root of -1, this letter (i) should not be used for any variable in an equation. Besides of this, an amazing video, and teaches a lot. Graphics and illustration at a high level. BIG LIKE.
Thanks! In my experience, the letter "i" is often used as the index of a summation. So I think it is fairly standard, especially when complex numbers aren't involved. I agree that if I were using complex numbers in any way here, I wouldn't use i for the index . Thanks!
@@MathVisualProofs Thanks for your response. After 35 years as a math teacher, I can say that I have never called an angle the letter delta (even if there is no ∆x in the problem) nor pi (even if there is no π^2 or 2π in the problem) And for similar reasons I didn't call the variable e and more... For me these are "holy" letters or, as a student once told me, these are "married" letters... 😉 they are already taken. In any case, I mentioned at the beginning that it is correct and okay to use the letter i but... there is a "but" here I love your videos and I admit that something in this visual illustration of yours is new and fascinating to me. Thanks for the great videos.❤️👍
That is interesting! I usually use X for the root of -1. I mean someone can see X^2=-1. So (3x+2)^2=-5+6x. Jokes aside what I have written is Z[X]/(x^2+1) which is isomorph to C. I understand your point, but Math is not about symbols but rather the meaning behind them.
Can you explain why you can't just use any regular polygon for this? For finding 1/8, you use a heptagon with another regular heptagon inside that is equal to 1/8 of the larger heptagon's area. Why can you not also try using a hexagon with a smaller hexagon that is equal to 2/8 of the larger hexagon's area? If you do this, you shade in 1/8 of the larger hexagon every time, but then end up with the number 1/6. Why am I wrong?
@@MathVisualProofs If we have the inside circle be equal to 2/8 of the total circle and cut the donut into 6 pieces, then it becomes 1/6. If we have the inside circle be equal to 3/8 of the total circle and cut the donut into 5 pieces, then it becomes 1/5. And so on. So does the infinite sum of 1/8 become 1/7, 1/6, 1/5, or all of them? Or are visual proofs just not totally exact.
Idk if it's just a coincidence, but: The infinite sum of (1/n)^x, and for each sum x increases +1, equals 1/(n-1). This could be just a short and finite pattern, but it could be an infinite pattern too... So... Idk...
is there any proof that the parts of the pentagon are truly 1/6 each? not that I say they aren't but... how do you find the size of the inner pentagon?
You have to construct the pentagon with the appropriate radius. This is possible with straightedge and compass (though not straightforward). So you scale the radius of the inscribed circle down by the right value and then the outer ring will be evenly divided into 5 equal pieces.
@@MathVisualProofs that's the think I'm asking. by what value you scale the radius? how do you find that value? for the rest of the series it's easy to see how it's done but for pentagons it's a "scale to the right value" with no hint of how to find that value.
in 3:07 you take all sides to be 1/6 so does that mean the pentagon inside the pentagon has the same area as the other part of pentagon(Bigger one) or is it something else plz explain brother, Thanks
All the trapezoids have area 1/6 because I take the central pentagon to have area 1/6. The other space is 5/6 of the area and is spread equally among 5 trapezoids so they are 1/5 of 5/6 or 1/6 area too.
so you are practically done with geometric sums. I challenge you to try to give visual proves for certain harmonic progressions and arithmetic progressions. (I know I'm evil XD)
I have developed a proof of sum upto infinite powers of 1/2 which includes bisection of angles by extending the hypotenuse further into a base forming an infinite length base
Artistically, I can appreciate doing it with several different shapes but to demonstrate that it generalize to N partitions, I find it easier to show that you can do it all with just a circle.
Looks like it works for denominators that are not integers, too, as long as they are more than 1. For example, sum of powers of 1/pi = 1/(pi-1). I don't know how to show that, geometrically, though. Perhaps you can help me.
Here’s one way to do it in general : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-V7L5nsRj7CE.html . I have a playlist that contains others too : ru-vid.com/group/PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw
That's neat! Your work is a blessing! It is a healing for mathematical bruises and wounds. Your knowledge is sharp and as fulfilling as light! Your understanding and teaching abilities are divine! We are thankful to have you and RU-vid. Blessings to live a winning life!
Something you will discover, by using the same method as the proof at 3:04, the proof goes like this: For any number 'n' from 2 to infinity, the infinite sum of 1/n^i, where i = 1 to infinity, is equal to 1/(n-1). If n=1 then the resulting infinite sum is infinity. The geometric proof works for all whole numbers greater that or equal to 4 but breaks down lower than that. There is probably a way to use different geometric proofs for n=[1, 4) but I don't know them off the top of my head. Edit: The infinite sum of 1/3^i can be visually proven by using the approach at 3:51
Splendid approach of Jacques Lacan split from the 2 registers (except the 3rd the real) imaginary and symbolic. The symbolic difficulty (algebra) is touchable. It is the same difficulty to become alphabetized from orality for the childs !
so imagine applying this to one it will result with infinity being the sum of infinite one to the power of any value as 1^x is always 1, therefor approaching the limit value of having 0 as the divisor which is a neat idea
This probably counts for 1/0. There is so much proof that 1/0 is infinity: Since 1+(1/n)+((1/n)^2)+((1/n)^3)+((1/n)^4)+… = 1/(n-1), and 1/1 and 1^n = 1, then you can possibly make an infinite chain of ones with this method. Also, if 1/0.1 = 10, 1/0.001 = 1000, 1/0.000001 = 1 million, etc, then 1/0 should be infinity, no matter what. (Since there is an infinite amount of decimal places in 0) Thus, 1+1+1+1+1+… = 1/0
*_It is possible to apply certain areas that the opponent cannot in a short time be able to recognize. But this representation is only an approximation, because when approaching infinity if we use the device or solution to infinity, we will realize that the two sides are not equal. Anyone who thinks that they are equal is completely delusional._* *_however the video❤️ is still very nice!_*
The initial triangle is the whole (1). If you color half of it (1/2) then half of the rest (1/4), then half of the remaining (1/8), etc., you can see that eventually you fill the entire triangle with nothing left uncolored. So the grand total sum of successive halving is the whole you started with (1).
