In this video, I show that the sum of differences between successive positive integer powers x^n for 0 ≤ x ≤ 1 can be written as an infinite series, and the resulting sum is equal to 1 or the unit ...
Transcript
Exercise 1 states graph the curves y equals x to the power of n from x is between 0 and 1, 4 n equals 0 1, 2, 3, 4, etc, just increasing an integer value on a common screen. Then it states by finding the areas between successive curves give a geometric demonstration of the fact, shown in example 6 that the series from n equals 1 to infinity of the summation 1 over n times n plus 1 equals to 1. So show it geometrically. So solution, graphing the curves using the amazing decimals graph and calculator reveals that the total area between the curves from x equals 0 to x equals 1 approaches the unit square of area 1. So here I've graphed this one here, n equals 0, so y equals x0, that's just equals to 1. This is just equals to 1, and that's just this line right here. This is just y equals 2, x0, or n is 1. So that's a line, and we're only going to go from 0 to 1, so that's a line from here to here, then y equals x1, that's just right here, y equals x to the 1. So the area between this curve right here, this is actually half a square, which is going to be 1 over 2. So, yeah, so you can highlight this one like that. Let's make this better. Yeah, so this is that arrow there. And then this one right here is a parabolic one, that's just y equals x squared. And as you see, it keeps going more and more, but then this area is going to be in between here. And as you can see here, it's getting smaller and smaller in between. And here I have y equals x50, is all the way here. This one's X10. This one's X100. And notice how it's pretty interesting. It goes like this. and almost spikes straight up. So it goes zero everywhere and spikes to infinity at high powers there, which is like a step function, which is quite fascinating. But notice that this total area right here, this is just a square. So this is just a perfect square like that. So that total area, area is just equal to, well, high times width or length times width equals to just one. So just a unit square. Also another note I want to make is a notice here for the odd power, so 1 and 3, goes like this, this is x to the 3, this is x to the 1, and this one goes negative here. All the even ones are just mere images, and when you have a negative value for x, it just becomes a positive, because positive or negative times negative, it's going to be positive. We have even numbers of numbers, and there it goes, it goes all positive like that, but for the odd ones. it's asymmetric and goes downwards, very, very fascinating. So yes, now going further, and let's write this out mathematically. I'm going to use integrals now. So the area between successive curves, Y equals X to the power of N minus 1, and Y equals X and 4, x is between 0 and 1 is. So again, this is between these two. Notice has the increase the power, it goes down further to the right there. So the area between X0 and X1, and hence, you could just basically find an integral and subtract. the difference you have between the y equals x n minus 1 or 0 and then y equals x1 like that or n so let's just uh give an example here just a generic one so let's say we have a x y axis like this y x and let's say we have and remember all these they could start from 0 up to 1 and then we put the height 1 like that so we get a square And notice here that every single one goes to this point, because if you have x is equal to 1, so 1 to the power of anything, it's going to be 1. So 1 times 1 times 1, for example, it's the power of 100. Yeah, you have 100 of these 1 multiplying by each other, it's still going to be 1. So they're all going to go to this point like that, at 1 1, at x is 1, and y is 1, it's right there. So now let's say we have a generic curve like this, and this is at, So this is at y equals to x to the power of n minus 1, like that. Now the area underneath this curve, recall this just an integral. So area underneath the curve from 0 to 1, that's just going to be an integral like that. I'll just highlight this like this. Actually, it'll make it zigzag. I'll make the zigzag but better like that. All right, so that's this area right here. I'll just put this over here, this area. I'll call this A n minus 1. So area underneath n minus 1 equals to you up to this x-axis right there is equal to remember, this is just integral of 0 to 1 of x-n minus 1, dx, like that. And now, and now what I'll do is just draw a second successive curve. Let's say it goes from like here. and it goes all the way up top there, and this is our y equals to xN, and now the area underneath this curve, it'll draw it zigzag this way, like that. This area is just going to be our a n, this equals to zero, the integral from zero to one of x n d x, like that. So then the difference here, This area across here, I'll draw this in blue, this one right here I'll call this is a difference now. This is the difference, and I'll call that just a... let's call us a i or this is the other difference so you just subtract a n minus a n minus 1 subtracted by a n like that or better yet we just write this as d for difference difference at the nth one from the previous one so then we have the same terminology there equals to a n minus 1 minus a n so that's the difference in areas there so we get the area between the curves Yeah, so let's just write this out and you can subtract the integrals using integral rules and anti-derivatives and so on. So let's go here, the difference equals to A-N minus 1 minus A-N. The difference in areas at any term N of this. or of these curves of y equals x n equals to now the integral from 0 to 1 and you can just add them all together so the difference is going to be you can just put them in the same bracket this equals to x n minus 1 like that minus x n then d x like that and then applying the anti-derivative of this one and derivative of x to the power of n minus 1 will we add a power there. So it's going to be x-n, and then divide this by n, so that if you take the derivative of x-n, it's going to be put the n down and minus power, so x-n minus 1, but we also have to consider the n there. So it becomes like that. And then this one here, x-n, same thing, we have to add one, x-n plus 1, and n plus 1. So if you take the derivative of x -n plus 1, We bring down n plus 1 and subtract the power, and then yes, we'll have xn, these will cancel. And this is the put brackets there, so from 0 to 1, like that. So from 0 to 1, this equals to plug in the 1 there, we get 1, the power of n is just 1, over n, minus 1 the power of n plus 1, that's just going to be 1, because 1 to the power of anything, just 1. then n plus 1 like that, and also we plug in 0, x0, power of n is still going to be 0, so yeah, minus 0, minus 0, for completeness, put this in bracket like that, so we have this, those just vanish. So now what we're going to do here is multiplied by these common denominators, just to get the common denominator, to multiply to top and bottom of both of these accordingly, and again, that's just a partial fraction decomposition that we had before, and that's just going to be our example 6 series. So this is going to be 1 over n times it by n plus 1 over n plus 1 to get the same common denominator. Minus here is going to be n times, yeah, n over n times it by 1 over n plus 1. So this equals to the same common denominator, n plus, yeah, n times n plus 1. Like that, and then we have, yeah, so 1 times n plus 1 is just n plus 1, minus n times 1 is just n. So the n is canceled. Results with the 1 there. So this equals to, and this is equals to 1 over n plus 1. So the difference in area between any consecutive curves is just 1 over, let's write, dn equals 2, yeah, the n equals to 1 over n times n plus 1, like that. So circle this. Yeah, so that's what we have. So we can now, yeah, we can see from the above diagram that as n approaches infinity, the sum of the areas between the successive curves approaches the area of the unit circle, that is 1. So this is going to approach 0 between here. So this area between here is getting smaller and smaller. So notice how it's getting smaller and smaller going downwards this way. So that's approaching zero, but the total area of summing all of them together we get is 1. So this full area is going to be 1 like that. So we could write this as a series. So the summation of the difference from n equals 1 to infinity is the series. So 1 over n plus 1. And we already calculate this one in example, in example 6, that's equals to 1. as expected, which is quite amazing. So let's put a hashtag amazing. Yeah, so this one is exactly one, and you can just find it using integrals or a summation series, or you're just using example 6, the series we had there.