For positive reals a and g, the two sequences ai and bi, with a0 =a g0 =g ai+1

For positive reals a and g, the two sequences ai and bi,...

Key Concepts

Arithmetic Mean

The arithmetic mean is a type of average computed as the sum of two numbers divided by two. It plays a key role in iterative methods that update sequences based on the current estimates, serving as a simple and direct way of blending values towards convergence.

Geometric Mean

The geometric mean is defined as the square root of the product of two nonnegative numbers. It provides a multiplicatively balanced measure and is particularly effective in iterative processes where scale-invariant behavior is desired.

Arithmetic-Geometric Mean Iteration

This is an iterative process that repeatedly computes the arithmetic and geometric means of two sequences. Starting from two initial positive numbers, the two sequences converge to a common limit known as the arithmetic-geometric mean (AGM), a constant with applications in areas such as elliptic integrals and number theory.

Convergence of Iterative Sequences

Iterative methods like the AGM process rely on the idea that continually applying certain functions (in this case, the arithmetic and geometric means) to a pair of values will lead to convergence to a fixed point. This convergence is utilized both in theoretical contexts and in practical numerical algorithms.

Numerical Implementation using a While Loop

Implementing the AGM computation in a numerical setting typically involves a while loop that repeats the iterative updating until the difference between successive terms is less than a specified tolerance. This approach ensures that the algorithm stops when the computed values are sufficiently close to the common limit.

Transcript

00:01 Okay, so this is the inequality that we would like to show for part one or for part a.

00:06 Is that for all n we have that an is greater than an plus 1 is greater than bn plus 1 is greater than bn great, so to prove that we're actually going to prove something that's a little simpler and then we'll see that that follows from this so we want to show is that a -n is greater than bn for all n.

00:31 Great, okay, so let's prove it.

00:32 Of that.

00:34 So we're going to use induction.

00:40 So for the induction, so the base case is going to be n equals 1.

00:45 So we want to show for the base case is that a1 is greater than b1.

00:52 So to show that, so we'll probably, well, so let's see what that means.

00:59 Well, what we want to show.

01:01 So we want to show that that a plus b over two is greater than square root of a b well okay we can see that this is true if and only if um you see this is true if and only if um well let's square both sides so a squared plus two a b plus b squared over four is greater than ab okay and then that line is true if and only if well let's multiply both sides by four so we get a squared plus 2ab plus b squared is greater than 4ab great okay and then uh let's subtract 4 ab from both sides so this line is true if and only if a squared minus 2ab plus b squared is greater than 0 and then now we can factor so that line is true if and only if a minus b squared is greater than 0 well remember that a is greater than b so a minus b is greater than zero so a minus b squared is greater than zero so this line is true so following our chain of um if and only of spec this tells us um that uh that this is that this is true so since a minus b squared is greater than zero what we've shown is that a um a plus b over 2 is greater than square of ab where this is uh this is a 1 and this is uh b1 okay so then so then we need to do the inductive step.

02:37 So assume that this is true, that a .n is greater than bn for some n greater than equal to 1.

02:49 And then we want to show a .n plus 1, it's greater than bn plus 1.

02:57 But this is actually quite easy now.

03:00 So we've actually already done all the hard work in the base case.

03:06 So right, so what does this mean? so a and plus 1 greater than bn plus 1.

03:11 Well, that just means that i want to show that a, n plus bn over 2 is greater than squared of a and bn.

03:22 But replacing this a .n with a and bn with b, and this an with a, and b, well, these have just replaced those roles in the base case that our a and our gn are g in our g.

03:36 B's had.

03:37 The only assumptions that we had on a and b in the base case was that a was greater than b.

03:42 Well, now we've got that a .n is greater than bn.

03:46 So it's just the exact same thing.

03:48 So the exact same proof that we did for the base case shows that this inequality is true since a .n is greater than bn.

03:59 So right.

04:00 So we've shown, so by same proof, and by, i mean, the exact same proof.

04:08 By same proof, a .n.

04:11 Plus 1 is greater than bn plus 1.

04:17 So by induction, the result follows.

04:24 So a .n is greater than bn for all in.

04:28 Great, okay.

04:29 And so that finishes the proof of this proposition that a .n is greater than bn for all n.

04:34 So we want to see.

04:35 But remember, so for part a, we wanted to show, was that, so we want to, for part a, we want to show that a, n is greater than an plus 1 is greater than bn plus 1 is greater than bn for all n.

04:56 This means for all.

04:58 So for all n.

05:00 Great.

05:01 Okay.

05:01 So we need to do a little bit more work, but not a whole lot of work.

05:04 So, well, we've got, and again, we have, just what we just proved is that a, n is greater than bn for all n.

05:16 So, well, what does this tell us? so this tells us that, so what we can compare is that a .n is greater than, well, it's observed, just that a .n is greater than a .n plus bn over two.

05:30 And why is this true? well, since a .n is greater than bn, this is just the average of a .n with a smaller number.

05:38 So this is going to make, this quotient is going to be smaller than a n because it's the quote it's it's uh the average of a n with a number um and this is of course equal to a n plus one so average let's see so by average of a .n uh with smaller number um so that one was easy so we've actually got that for all n a n is greater than a n plus one um and again that's implying this uh that we know that a n is greater than b for all in.

06:14 Okay? and then we also want to show, of course, that a bn plus one is greater than bn.

06:21 Well, this is true if and only if, well, let's let's let's repeat.

06:26 We have an expression for bn plus one.

06:28 So bn plus one is squared of a n, bn.

06:32 So, so this top one is true if and only if square of a and bn is greater than bn.

06:39 Well, let's square both sides.

06:41 Then we have that a and bn.

06:43 That's, that that's true if and only if a and bn is greater than bn squared.

06:48 Well, let's just divide both sides by bn squared.

06:51 We know that bn is not zero, or rather let's divide both sides by bn, i mean.

06:56 So we know that bn is not zero, so we can divide both sides by bn.

06:59 And we also know that each bn is positive.

07:02 Think about why we know that each bn is positive.

07:05 But right, so just dividing both sides by bn, we get that this is true if and only if a n is greater than bn.

07:12 But we know.

07:13 That a .n is greater than bn for all n.

07:15 That's what we showed with this proposition that we started with.

07:19 So we've confirmed following this back that bn plus 1 is greater than bn for all n...

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