Lesson video

Well done for taking the first step to learning by choosing to load this video today.

My name is Ms. Davies and I'm gonna be helping you throughout this lesson.

Iteration is a fantastic topic with lots to explore and play around with.

There's some really fascinating results that's gonna come out of today, so I hope you have fun exploring things.

With that in mind, make sure you've got your calculator handy.

You will need that today and we'll talk about how to use that efficiently throughout this lesson.

Let's get started then.

Welcome to this lesson when we're approximating solutions to equations.

By the end of this lesson, you'll be able to use iteration to find approximate solutions to lots of different equations.

Iteration is the repeated application of a function or process in which the output of each iteration is used as the input for the next iteration.

We're gonna explore that idea today, and we're gonna start by looking at convergent iterations.

So you have seen many iterative processes which produce ever-increasing values.

If I was applying compound interest of 5% to a value of 100 pounds, if I start with X0 is 100, then I could do 100 times 1.

05 to get 105, and that's the value after one application of 5%.

That then becomes X1.

I then input that into my function, which is multiplying by 1.

05, 'cause that increases by 5%.

So I'm using a calculator, I can use my answer button times 1.

05 and that'll give me X2.

Again, if I've got this set up in my calculator now with answer times 1.

05, I just need to press EXE or equals again and keep going.

And you'll see that because I'm applying 5% interest to the new value each time, each iteration is larger than the previous.

And more importantly, the difference between successive iterations is increasing.

They're getting bigger but by more each time.

So some iterative processes will produce iterations which converge on a set value.

Rather than getting bigger and bigger and bigger or smaller and smaller and smaller, they're gonna get closer to a set value.

Here's one for you to try out.

Try the formula Xn + 1 = the square root of Xn where X0 is 81.

Can you find the first four iterations of that one, what's happening? So what hopefully you notice about these iterations is each iteration leads to a value which is smaller than the previous value, so we're not getting bigger and bigger this time.

We are getting smaller.

But more importantly, the amount they're decreasing by is decreasing.

The differences between successive iterations are decreasing.

First we decrease by 72, then only by 6, then only by 1.

26, then only by 0.

4159.

So the gaps between the iterations are getting smaller each time.

It's that fact that means if we carry on this process, they're gonna converge on a set value.

So if you're looking for a convergent iteration, you're looking for the differences between each iteration to be decreasing and that's whether they're increasing each time or decreasing each time or look at some later where they increase, then decrease, and alternate between the two.

But we're looking for the differences between iterations to be decreasing.

Okay, I'd like to make a prediction now.

What value do you think these iterations are getting closer to? Pause the video.

What do you think? Okay, so there's X5.

For this investigation, we do not need to keep track of which iteration we are calculating, so you don't need to write down each iteration.

You can just keep pressing the EXE button or the = button and see what happens to the values.

So I wonder whether you said that the iterations seem to be converging on the value one.

If you generate some more iterations, you can see if this value gets even closer to one.

There you go.

That's the 31st iteration and you can see that it's very, very close to one now.

Right, your turn.

Using the same iterative formula we had before, but this time with X0 as 5, what is the value of X5? Off you go.

Brilliant, hopefully we're reminding ourself of that efficient calculator use.

So we've got X1, X2, X3, X4, then X5 is 1.

051581198, depending on how many digits your calculator display gives you.

Okay, so how do we know that this formula is converging? Can you remember what we said? Right, we need to look at the differences between iterations and see that the differences are decreasing each time.

What value do you think they are converging on? If you've got an idea, test it out with a few more iterations.

Okay, this is interesting.

They converge on one again.

So changing the value of X0 did not change the value that the iterations converge on.

And you'll find for many examples that is the case, but for other examples that's not the case.

So can we use any value of X0 to get a process which convergences on one? What do you think? Right, well done if you said no, 'cause X0 has to be positive.

We cannot evaluate the square root of a negative.

We have to start on a positive value.

Izzy says, I wonder what will happen if we start on a positive value less than one? Try the same iterative process, but with X0 as 0.

25.

What happens this time? Let's have a look.

There's your First four iterations.

And you can see they're converging, 'cause the differences are getting smaller each time.

Even though the iterations are increasing this time, it's the differences between iterations that's decreasing.

Again, they will converge on one.

That's the value of X25 and it's not 0.

9999999, and we can see that it's getting closer and closer to one.

Sometimes different values of X0 will converge on the same number and sometimes they will not.

When an iterative process does converge on a value, it's useful to write that value to a specific degree of accuracy, and that's what we're gonna focus on now.

So here's a new iterative formula.

The next value is found by doing 6 - the current value squared all over 4.

We're gonna start with X0 as 0.

If you want to do this at the same time as me, please do or pause the video and have a go yourself.

So there's X1, X2, X3, X4.

This time the iterations are decreasing, then increasing alternately.

However, the differences are decreasing, so they are converging.

What value do you think they might converge on? Make a prediction.

Right, well it's impossible to tell at the moment exactly what value they are converging on, but we can approximate to one significant figure.

Now several iterations in a row round to one, to one significant figure.

And we said for this example that we're increasing, then decreasing, then increasing, then decreasing ultimately, then we know that the value we are converging on must be between successive iterations.

So the value will be between 1.

28 and 1.

09.

All values between those round to one to one significant figure.

So we know that this is gonna round to one to one significant figure.

Now we don't know what this is gonna round to to two significant figures.

We're gonna need more iterations.

Okay, there's another few.

And we can see there's gonna be a value between 1.

176 and 1.

1542.

And it's because in this example, we're increasing then decreasing alternately.

So we're getting closer and closer to the value.

We're gonna be a little bit above it, then a little bit below it each time until we get to that value that it's converging on.

So because it's between those values, it's gonna round to 1.

2, 'cause every number between those two rounds to 1.

2.

The more digits you determine, the more confident you can be about the rounded value.

Aisha has calculated the seventh iteration and says to three significant figures, this converges on 1.

18, to four, 1.

176, and to five, 1.

1761.

Is she correct? Right, and this is the key point.

We cannot tell with this information.

We need to look at successive outputs and see whether they're rounding to the same thing to a certain degree of accuracy.

We can't just calculate the seventh iteration and round it.

We need to know that those different digits are fixed to know that we've got a solution to the right degree of accuracy.

I'll show you what I mean, there's X7, there's X8, there's X9.

We can still only determine this to two significant figures.

X7 rounds to 1.

18.

X8 rounds to 1.

15.

But X9 would round to 1.

17.

So we don't know which of those at the moment.

This is gonna converge onto two significant figures.

Let's do a few more iterations.

After 11 iterations, we can say this converges on 1.

16, 'cause we can see the X10 and X11 both round to 1.

16.

And if you're not sure, just generate some more iterations.

If you use your calculator efficiency, keep pressing that EXE or the = button and then you can be more confident that you've got it rounding to the right degree of accuracy.

Okay, I'd like you to try that now.

Keep generating iterations.

What value does this converge on to four significant figures? Well done if you said 1.

162 Now there are gonna be other values of X0 which will converge on the same value.

This will not always work for all values of X0.

It all depends on what iterative formula you're using.

I'd like you to try X0 = 6 and see what happens.

Let's have a look.

(mouse clicks) Right, the differences between the iterations are increasing and they're increasing rapidly.

This is not gonna converge on our value this time.

So the value of X0 does matter.

So we've seen an example where the values alternate between being slightly bigger and slightly smaller than the true value.

Let's see what happens with this example.

(mouse clicks) There's our first four iterations.

If you look carefully, you'll see that the values are increasing, but the differences are decreasing each time.

Now we have to be a bit more careful with our degrees of accuracy here.

It's not as easy to be confident that we've got it to the right degree of accuracy.

If you're ever unsure, just generate more iterations, and that'll help you see when a digit is fixed.

Okay, so we're fairly confident now that this is gonna round to 2.

4.

You might wanna do a few more iterations to make sure.

Even though each iteration's getting bigger, doesn't seem to be the case that it's gonna reach 2.

5.

If you keep generating iterations and they're all fixed on 2.

41, then you know that that's definitely gonna be the case.

Right, here are four consecutive iterations using this formula.

Which statements are definitely true? What do you reckon? Right, let's look at what type of example we have where you can see that the values are increasing, then decreasing alternately.

Well, that helps us out here, 'cause we know that these are all definitely gonna round to two to one significant figure and we can see that they're definitely going to converge on 2.

3 to two significant figures.

Now, well, I dunno if you spotted that this does not converge on 2.

31.

X8, X9, and X10 all rounds to 2.

32.

And well done if you also said that we don't have enough iterations to determine this to four significant figures yet.

If you look at X9.

that rounds to 2.

316, X10 rounds to 2.

317.

So we need more iterations to know which of those it's going to converge on.

Right, which of these iterative processes are converging? What do you think? You may have noticed it's the same formula.

We're using different values of X0.

The first one is converging.

The differences between successive iterations are decreasing.

That one is not converging.

You'll see that the differences between successive iterations are increasing.

And we've got another one that's converging.

Although this time the iterations are increasing each time, the differences are still decreasing.

Right, time for a practise then.

Read through these iterative formulas and find the required iterations.

Off you go.

Right, question three, I've written the first eight iterations of this formula.

Have a go at answering those questions.

And for question four, you've got a new formula.

What value does it converge onto two significant figures, four significant figures? And then you're gonna change the value of X0 and see what happens.

Off you go.

And finally, your last formula to try.

Can you work out the value it's converging on? And I want that accurate to three significant figures.

And then what happens if we change the initial input to X0 is 6? Off you go.

Let's have a look.

Pause the video and check you've got the correct iterations.

For the first one, yes, the difference between successive iterations is decreasing, so it is converging.

And the same is true for that second formula.

When you're happy with your answers, we'll have a look at the next one.

All right, question three it's gonna round to 0.

7.

If you look at X3, all values after X3 round to 0.

7.

B, we know it's gonna round to 0.

74.

Again, we know that by the third iteration, 'cause all success of iterations also round to 0.

74 For C, we need a few more iterations to work this one out.

It's gonna be 0.

742.

X6, X7, and X8 all round to 0.

742, and we know that the value's gonna be between X7 and X8 because this is one of those examples again where we are increasing and decreasing, ultimately getting closer and closer to the right answer.

For D, we can't determine it yet because X7 rounds to 0.

7416, whereas X8 rounds to 0.

7417.

We need more iterations to know which of those it's going to converge on.

For question four, well done if you found that it was gonna be 1.

8, 1.

796, and changing the value of X0 did not change the value we were converging on.

It's also 1.

8.

Finally for five, this is gonna converge on 1.

32.

And for B, this is not gonna converge this time, so changing the value of X0 did affect what was happening, so we're not gonna converge on a value for X0 = 6.

Right, let's now see if we can use this to solve some equations.

So let's try and find a solution to this equation.

X cubed + 4x - 21 = 0.

Izzy says, "I don't know how to solve a cubic equation like this." This is quite a tricky cubic equation to solve.

So what we're gonna do is we're gonna try some values and see if we get close.

So I've tried X is 1, 2, 3, and 4.

We can see that 2 is the closest value 'cause I wanted this equal to 0, so 2 is the closest value.

What value do you think we could try next? Well, if you suggested something between two and three, probably closer to two, but you might've said 2.

5, it's in the middle.

Let's try 2.

5 and we get 4.

625.

That's still too big.

And we could keep doing this.

We could keep doing this trial and improvement method, but it could take a while, and it's still only gonna be at an approximate value.

So we need something a little bit quicker.

And that's what's useful about this iterative method.

Like Aisha says, it's the fact that we could just keep pressing that EXE or that = button and we didn't need to keep typing in the input each time, and that made it a really quick process.

So that's what we're gonna do.

We're gonna write our equation as an iterative formula.

Now to be an iterative formula, the current output must become the next input.

So we need to write our equation so it's self-referring with X as the output.

So then once we get an output of X, that goes back in to our formula and becomes the next input, and so on.

You'll see this when we play around with it.

So there are many ways to rearrange this equation.

Let's look at one way.

If I add 21 to both sides, subtract 4x and then cube root.

So you'll see now that we have X as the output and X as our input.

So when we get a value of X out, we can put it straight back into the formula, get a new value of X.

What we're doing is we're looking for a value of X which balances this.

We want to write this as an iterative formula, so we can write that as X subscript n + 1 = the cubed root of 21 - 4x subscript n.

Now if this formula converges, then each time we calculate the next iteration, we'll get closer to a solution which balances the equation.

All we need to do is pick a value for X0.

Now it's possible that some values may work, some may not, so it's safest to pick a value reasonably close to the solution.

So we already estimated that two was gonna be reasonably close.

So let's go with X0 = 2, so we put it into our iterative formula.

You get X1 as 2.

351.

Now that value then, X = 2 did not balance our equation because if our equation was balanced, X would be the same on both sides of the equation.

But we put X in as 2 and it came out as 2.

351.

Right, let's try the next iteration.

So we put in 2.

351.

We get out 2.

2633.

So again it's not perfectly balanced.

If we use that Answer button, we can be efficient and we can generate some more values.

Now these values are converging 'cause the difference between iterations is decreasing.

So we can carry on.

It's one of those examples again where we're increasing, then decreasing ultimately.

So we know the value must be between 2.

2802 and 2.

2816.

Both of those rounds are 2.

28, so that's gonna be a solution, but it's an approximate solution to our original equation 'cause it's been rounded to three significant figures.

Now if you've got a calculator that can solve cubics, you might wanna check your answer now.

If we go into the equation tool, select Polynomial.

We need to be a cubic, so it's the second one down on my calculator, and then we can input the values.

It's 1x cubed, 0x squared, 4x - 21.

And there's a solution, 2.

281410263.

And we got 2.

28 to three significant figures, so we were correct.

Now there are many other solutions as well, but this is the one that we have found.

Right, so I'm gonna do an example on the left and you're gonna try one on the right.

We're gonna show that this equation can be rearranged in this way.

So I need to look carefully at what I'm aiming for.

I've got x cubed on one side and x on the other.

So let's start by adding 5x.

So that's x cubed + 3 = 5x.

Now I need to divide by 5, x cubed + 3 over 5 = x.

Remember there's loads of different ways to rearrange a formula into an iterative formula.

Right, I'd like you to have a go at this one.

It's the same starting equation.

Can you show it can be rearranged this way? So again, we're gonna start by adding 5x.

Then we're also gonna subtract three and then cube root.

Right, so I'm gonna turn that into an iterative formula and using X0 = 1, see if I can find a solution.

I'm gonna put that into my iterative formula, so I'm gonna use answer cubed + 3 over 5.

(mouse clicking) And after five iterations, I can be fairly confident now that's gonna round to 0.

7 to one significant figure.

It is getting smaller each time, so I do need to be careful to check whether it's gonna round to 0.

7 or whether it's gonna get small enough to round to 0.

6.

And always do more iterations if you're not sure, Right, you'll go use this iterative formula, find a solution to two significant figures.

Off you go.

So we have an approximate solution as 1.

8 to two significant figures.

We can't tell yet if this is gonna be 1.

84 or 1.

83 'cause again these are decreasing each time, so we're not sure just how small it's going to get.

Solving it on a calculator with an equation function, we get those three solutions.

And those are the two solutions we just approximated.

There's the one you approximated in the middle and there's the one I did on the right-hand side.

Right, time for a practise.

Which of these could be an iterative formula to find an approximate solution to that equation? And you've got two to do there.

Come back when you've got your answer.

Right, working the other way this time.

So this is the iterative formula.

Which equation could that be used to solve? Be careful with your positive and negative signs.

Got two there to do.

Right, question five, I'd like you to show that that equation can be rearranged in that way and then I've given you an iterative formula and a value for X0, And I'd like you to find X1, X2, and X3.

Then find X6 and I want an approximate solution to two significant figures.

Give that one a go.

Right, another one for you to try, exactly the same idea.

For C, you can generate as many iterations as you need to to be confident that you've got a solution to two significant figures, off you go.

And finally, slightly different type of formula this time, but gonna do exactly the same thing.

Can you show that it can be arranged in that way? Then use the formula to find the first few iterations.

Then I want an approximate solution to three significant figures, and make sure you generate enough iterations to be confident to three significant figures.

Come back when you're ready for the answers.

Let's have a look.

So that third one could be an iterative formula.

If we subtract two from both sides, add 4x to both sides, and then square root, and then we've just turned it into an iterative formula using our subscript.

For question two, it's gonna be a.

Again, we're gonna subtract 2 from both sides.

We're gonna add 10x to both sides and cube root this time.

For question three, the equation that that could be used to find a solution to is B.

For four you're looking at a.

And then you put it all into practise.

So there's our steps to rearrange.

I've given you X1, X2, and X3.

X5 was 1.

3283, and so on.

And X6 was 1.

2704, and so on.

Because we're increasing and decreasing ultimately, I know the actual value is gonna be between X5 and X6 and both of those round to 1.

3, so all values between them is gonna round to 1.

3.

So I know that there's a solution x = 1.

3 correct to two significant figures.

And there's the same for six.

Pause the video and check your answers to a and b.

For c, because successive iterations were decreasing each time, you need quite a lot of iterations to be sure if this is gonna be 3.

6 or 3.

5.

If you keep using your calculator efficiently, you will see that this converges on 3.

5.

And finally, factorising is gonna help us rearrange this one.

Well, I dunno if you spotted that.

So we can add one to both sides of the equation.

Then factorise the left-hand side to X bracket X squared + 4 and then we can divide both sides by that bracket.

So you've got X = 1 over x squared + 4.

That was quite a tricky rearrangement.

Well done if you did that.

So there's X1, X2, and X3.

And then again if we use our calculator efficiently, and this does converge on 0.

246 to three significant figures.

Just a reminder though, what you've done there is you've found an approximate solution to that original equation.

The x cubed + 4x - 1 = 0 equation.

Well done for your hard work today.

As always, when you're doing something new for the first time, you do need to practise it.

So feel free to go back through, have a look at those examples and have a bit more of a practise on using your calculator.

This is a really interesting idea to explore, and there's all sorts of interesting results that come out as to when this work, when this doesn't work, the different equations you can form, the different values of X0 you can use.

There's so much to explore mathematically here.

But you've got the general idea of how the iterative processes work and how to know you found a solution to a certain degree of accuracy.

So well done and I hope to see you back to explore this in more detail.