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 very exciting problem-solving lesson with iteration.

We're gonna use everything we've learned about iteration to explore the concept and to solve some problems. With that in mind, you definitely want your calculator with you today.

You might even want access to some graphing software if you've got that to hand.

If you want a reminder as to what iteration is, pause the video and read through that now.

So we're gonna start by exploring, using different values of x0 in the same iterative process.

And in the second part of the lesson we're gonna explore using different iterative formulas to solve the same equation.

Right, let's get started.

We can sometimes choose iteration to find solutions or approximate solutions to quadratics.

It's quadratics that we're gonna focus on today.

So how could we solve this quadratic equation? Pause the video.

What would you do? Right, the easiest way is gonna be to factorise, and this does factorise nicely.

Yeah, x + 5 x - 3 = 0.

So we know we've got solutions, x = negative 5, and x = 3.

Now let's see what happens if we try and find these solutions using iteration.

Of course that wouldn't be the way that we would find these solutions.

We would find these by factorising, and we've just seen how easy it is.

Let's just explore what happens if we try and use iteration.

Right, so here's one iterative formula we could try.

We know that 3 is an exact solution, so when we use x0 = 3, we should get an output of 3.

So pause the video and have a go.

Lovely, hopefully you should get an output of three.

Three balances our equation because it is an exact solution.

Okay, what happens if we use x0 = 2? So that isn't a solution.

If we put it into our iterative formula, what happens? Try this out.

There we go.

So I'm gonna generate a few iterations, and keep going.

If you keep generating iterations, it should eventually get to three.

So we are able to use iteration to find that solution.

Okay, I'd like you to try x0 = 5 and x0 = 7.

What happens this time? Right, hopefully you found that they both converge on 3.

So, so far it doesn't seem to have mattered what value of x0 we start on, we get to that solution of 3.

Izzy says, "I wonder if this will happen for any value of x0?" Try this one.

Oh well, our first iteration requires us to square root -1.

This leads to an error.

Our calculator cannot evaluate the square root of a negative.

This isn't gonna work.

And let's just think about this.

If 15 - 2x = 0, then x is gonna be 7.

5.

So if that value in the square root is 0, then x would have to be 7.

5.

So any value of x0 greater than 7.

5 is not going to work.

Okay, what happens if x0 is negative itself? So I'd like you to try x0 = negative 10, and x0 = negative 21.

What happens? Right well, when x0 is negative 10, the iterations converge on 3 again, even though it was negative 10, they converge on the solution 3.

When x0 is negative 21, then the first iteration is 7.

54, which then causes an error for the next iteration, 'cause remember we said x can't be bigger than 7.

5, because that leads to us trying to square root a negative.

And this is gonna be the case for any value smaller than negative 21 as well.

Izzy says "All the values that worked gave the solution of 3, not negative 5." Yeah, there didn't seem to be a value that we could use which gave us the solution of negative 5 apart from using negative 5 itself, which would give you negative 5 on the first iteration, 'cause it's the solution.

When an iterative formula is used to find a solution, sometimes different values of x0 converge to the same solution.

Sometimes certain values of x0 cause an error, and it's likely that even if one solution can be found, the other solution can not be found using the same iterative formula.

So what we're seeing here is iteration can be useful to check values, find approximate values, especially for tricky equations that we don't know how to solve another way.

However, it does have limitations.

Right, how could we solve this quadratic? What would you do? Right, it does not factorise, so you're gonna wanna use the quadratic formula or if you've got the equation feature on your calculator you could use that.

That's what I've done here and I've got two solutions.

Now again, let's see what happens when we try to use iteration to find the same solutions.

So here's an iterative formula.

If we use x0 is 1, or x0 is 2, this converges on the solution x = 0.

3819, and so on.

If you want to pause the video and try that out, you could do that now.

What happens if we use x0 = 3? Give this a go.

Right, the iterations keep getting bigger and they don't converge.

Try x0 is negative 3, what happens this time? Again on the first iteration, negative 3 squared is the same as 3 squared, so we have the same problem, it doesn't converge.

So we've managed to find one solution by using x0 is 1 or 2, but it doesn't work for x0 is 3 or negative 3.

What we can do is we can use graphs to investigate what is happening here.

Now, this is quite a subtle idea, but it's really interesting.

So if you've got time to play around with this, I suggest you get some graphing software out, and have a play around yourself.

So what I've plotted here is I've plotted the line with equation y = x, and that's because if you look at our iterative formula, the left hand side is essentially x.

So I plotted y = x, and then I've also plotted the parabola x squared + 1 all over 3.

And what I'm doing, I'm trying to find when those are equal to each other, 'cause that will be a solution.

Now there's two solutions, and that's where the graphs intersect.

Now I'm gonna zoom in, and I'm gonna see what happens when I use the value which is not the exact solution, which is what we did with the iteration.

We used x0 = 2 first.

So if I find x is 2, and I go up to my parabola, and I look across, that gave me a value of 1.

67.

So our first output was 1.

67.

Now what I need to do now is find 1.

67 on my x-axis, 'cause that's gonna be my next input.

The line y = x is gonna be really helpful here, because x and y values are the same.

So if I just track that line back down, that will find 1.

67.

I'm not gonna be able to be this accurate all the time.

So I'm gonna try and do this as accurately as I can using the lines.

So, 1.

67 was my first output, so that's gonna become my next input.

So if I track that up to the curve, I then got an output of 1.

259.

Again, that's gonna become my next input, so let's find it on the x-axis.

That's gonna be my next input.

So let's see what output that gave me.

I'm not gonna write that down this time, but we can use my y = x line to find it on the x-axis and input it again.

And you can see if we keep doing this, it is gonna converge on that solution.

Right, well now let's see what happens when we used x0 = 3 instead.

So before we used x0 = 2 and it converged on that lower solution.

Let's try x0 = 3.

So I need to go up to my curve, that's my first iteration.

Find it on the x-axis.

Okay, now that's gonna be my next input.

Find it on the x-axis.

Oh, hang on, we can see that our outputs are getting further away from both solutions.

I keep going.

There we go.

And these iterations are gonna keep generating bigger and bigger values, and they're not going to converge.

So it's the curvature of the graph in relation to y = x, which showed us how some values of x0 did not converge.

Now this is not something you're gonna need to be able to do yourself.

You don't need to be able to guess when certain values are gonna work and not work, but it is useful to see that there is a reason behind why some values of x0 work and why some don't.

This also showed us how all the values that did work converged to one solution but not the other.

There was not a way of getting it to converge to that larger solution.

All values less than the solution converged to the lower solution, and all values greater didn't converge at all.

As I said, just something really fun to investigate, and that's what we're doing today.

We're improving our mathematical skills by investigating these things.

Now, picking values of x0 close to the solution increases the chance that an iterative formula will converge.

However, sometimes even that will not work.

So here are the solutions to this equation.

I've rounded them, which is why they're approximate solutions, and I've chosen an iterative formula.

Right, pause the video.

What happens if we use x0 = 3? Right, well the real solution is 2.

89, so 3 is not far away.

Something really strange happens this time.

I hope you had time to play around.

The values seem to oscillate between different values, but they don't seem to converge on a solution.

You can keep pressing that EXE or that equals button, we don't seem to be getting anywhere.

Try x0 = 2.

9 and x0 = is negative 6.

9.

These are really close to the solutions.

Do they manage to find the value? No, neither of those worked as well.

So we tried values really close to our solutions and none of them seemed to produce a solution, so that iterative formula is not gonna be of any use to us.

Let's have a look at it on a graph.

Okay, so I've plotted y = x, and I've plotted y = 20 - x squared all over 4.

So there's a value quite close to the solution, you can see how close I've started to that intersection point, and I track it, and I find the value on the x-axis, and then I find the next value.

I find that on the x-axis, I input that value.

I find that on the x-axis, and I keep going, and you can see that all we do is we jump round and round the axes, but it is not converging on a solution.

Right, quick check.

Here's an iterative formula for this equation.

Which values of x0 converge on the solution x = 3? Try this out.

Right, x0 = 1, it causes an error, because we need to square root a negative for the first iteration, but the other two do converge on the solution, x = 3, and that's an exact solution.

We could have just solved that equation by factorising, couldn't we? That would've been the easiest way.

If we have a look at our graphs.

If I input 10, I get that output just less than 7.

If I put that in, I get that output, and so on.

You can see that we're converging on that solution.

We can also see from the graph that inputting any values less than about 1.

2 is not going to work.

Okay, time for practise.

You're gonna start by actually finding the real solutions.

You can factorise, it's probably the easiest way.

What happens when you substitute those solutions into this iterative formula? And then play around, which of the following values of x0 converge on one of the solutions? Give it a go.

Right, you've got another one to play around with.

Try these different values of x0 for this iterative formula, and explain what happens.

For Question Three then, I want you to make a prediction about which values of x0 will converge on a solution.

Doesn't matter if you're right or wrong, just making a bit of a guess.

Then I'd like you to test your prediction and find a solution to two significant figures.

Can you use that iterative formula to find a solution? Then, by substituting the upper and the lower bounds and that's into the original quadratic, I'd like you to show your solution is correct to two significant figures.

So you're looking for that change in sign rule between the upper and the lower bounds to test your solution.

Give that one a go.

Right, this is another iterative formula for the same equation.

I've drawn the graphs for you.

I'd like you to make a prediction about which values of x0 will converge on a solution.

So use the graphs.

Can you track some values of x0, and try and decide which ones are gonna converge? Test your answer, and can you find a solution to three significant figures? Use that iterative process with one of the values of x0.

Let's have a look.

So our solution should be x = 3, and x = 2.

If you substitute those in, the input and the output are the same, they're exact solutions, so they should balance our equation.

Thankfully they do.

So, testing these out, x0 = 0 and x0 = 1.

Both converge on 2.

x0 = 4 does not converge.

Okay, this is a different equation, and a different iterative formula.

We're just testing it out.

You should get an error on the first iteration, 'cause the calculator cannot evaluate the square root of a negative.

Now for x0 = 1, this one took ages to converge.

At first it doesn't seem to be converging at all.

So I can understand if you said, "No, it doesn't converge." However, the difference between iterations are decreasing, if only slightly, and after a very long time it does converge on a solution.

But I'd understand if you gave up and thought that that wasn't going to converge, because it did take a very long time.

x0 = negative 1, you get an error on the second iteration, 'cause the first iteration has an output greater than 3, so that gives us a negative value to try and square root, which the calculator won't evaluate.

For Question Three, it doesn't matter what predictions you made, but you get an error if x0 = 0, and that makes sense, because you're doing 0 subtract 3 which is negative 3, and then try and square root it.

When x0 = 1 or 10, this converges on a solution, and it should be x = 4.

3 to two significant figures.

Well, I dunno if you found that.

And then we can test using our lower bound and our upper bound.

There's a change of sign, so there's a solution between those two values.

If there's a solution between those two values, then the solution will round to 4.

3.

Right, well then if you guessed x0 = 3 and x0 = 4, it doesn't matter if you didn't, 'cause you'd be able to find this out by trying it, and you get an approximate solution of 0.

697 to three significant figures, and we can see that on the graph as well.

Not that accurately, obviously, but we can see that there's a solution between zero and one.

Well done, I hope you enjoyed playing around with that.

So now what we're gonna do is explore different iterative formulas for the same equations.

So we've seen how changing the value of x0 can affect whether iterations lead to a solution.

It's also possible to write different iterative formulae for the same equation.

So let's look at this equation.

To rearrange this into an iterative formula, we need the output to be x and the input to contain x as a term.

How would you start to rearrange this equation? Pause the video.

What would you do first? Right, there's many answers to this.

Here's one way of doing it.

I added one to both sides, then subtracted 3x and then square rooted.

Izzy says, "Don't forget the negative root." Now, what this means is we can actually write a different iterative formula with a negative root.

So I've got two equations here which I can turn into iterative formulas in a minute.

One with a positive root of 1 - 3x, one with a negative root of 1 - 3x.

So now I can write them as iterative formulas, x subscript n + 1 = the square root of 1 - 3xn, or you can do the same with the negative square root.

Right, are there other ways we can rearrange this initial equation? What else could we do? I wonder if it came out with this way.

We can still add one to both sides This time subtract x squared and divide by 3.

Again we need to turn it into an iterative formula using our subscripts.

Or here's another way.

We can factorise, and that isolates an x.

So we've got x bracket x + 3, and then we can divide by x + 3, so, x = 1 over x + 3.

And again, I can turn that into an iterative formula.

Right, we now have four different iterative formulae for this equation.

And this is not necessarily an exhaustive list, there might be others that you found.

Izzy says, "I wonder if they all work to find a solution." Now we've had the most luck previously using x0 values close to the real solutions.

So what I want you to do, is I want you to try x0 = 0.

3, and negative 3.

3 with that first formula, and see what happens.

Pause the video, give yourself time to explore.

Right, because of the nature of the square root, x0 = 0.

3 causes an error on the fourth iteration, and x0 = negative 3.

3 causes an error on the second iteration.

This iterative format doesn't work to find solutions.

The values I gave you are quite close to the real solutions, and it didn't work.

And you can try lots of values of x0.

This formula is not gonna be particularly successful.

Okay, well, let's try it with a negative root, 'cause that changes what happens here.

Once we square root, multiply by negative, put it back in, it might not cause the same problem with square rooting.

Give it a go.

Right, that was more successful.

It takes a bit of time, but eventually this converges on the solution x = negative 3.

303, that's to three decimal places.

You need lots of iterations to make sure you are accurate to that degree of accuracy.

Let's try it with the other input, so with negative 3.

3.

Now our other value of x0 converged on the same solution.

It gets there a bit quicker, probably because it was closer to that solution to start with.

So both of those inputs converged on negative 3.

3 using that iterative formula.

Izzy says, "I wonder if there's a way to find the other solution." Well, we've got some other formulas, so why don't you try those inputs, but with this formula.

What happens this time? Give it a go.

Right, hopefully you found that both values of x0 converge on the other solution, the 0.

303 to three decimal places.

Okay, just for completion sake, what about that fourth formula? What happens this time? x0 = 0.

3 converges on the solution 0.

303 again.

But the other input leads to an error.

This was quite an interesting one, because inputting negative 3.

3 gave an output of negative 3.

The problem with negative 3 here, is negative 3 add 3 is 0, so that gives you a denominator of 0, and divided by 0 is undefined, so that causes us an error as well.

So a bit of a summary.

Different iterative formulas can find different solutions, but they may also lead to errors, or they may lead to processes that don't converge as well.

So it's not easy to know straight away whether an iterative formula is going to work.

And again, studying the graphs can help us see what's happening here.

So that was our positive root.

We can see the inputting anything greater than a third is not gonna work.

And even if we pick an x0 value, which works for the first iteration.

There you go, we get an output of 2, we track 2 onto the x-axis.

We then can't input that.

Okay, let's look at our negative root.

This was more successful, we can see that with the curvature of the graph, can't we? So we know that we can't input anything greater than a third, but it's likely to work for some other values.

If you'd like to pause the video, try following some paths of different values, and see what happens.

We did try some different values.

You should see that they converge on that negative solution.

We can't actually see the other solution on this graph, because we've only taken the negative root.

Right, this was our third one.

Both solutions are visible.

But try following the path of different values to see which solution they converge on.

Right, this was interesting.

If you try a value less than about negative 3.

4 or greater than 3.

4, you'll see that it won't converge.

It's to do with where the line y and x is in comparison to the curve.

You'll see that it just disappears down into the negative numbers.

If you try a value between those roughly, you'll see they converge on the solution x = 0.

303.

And this was our last one, the 1 over x + 3.

Now this has an asymptote at x = negative 3.

Most other values converge on the solution x = 0.

303.

Not quite all, but most of them.

If you follow the path of one of those, you'll see converge on that positive solution.

Interestingly, even if you start with a negative value, when you track the path, you'll see it jumps up to that positive section as well, which is quite interesting.

Once you finish playing around, we'll have a look at the task.

Right, which of these could be the iterative formula for that equation? So, quick check.

Do you know how to make an iterative formula? Well done if you spotted that either B or C could be iterative formulas for this equation.

The top one should have been xn + 3 in the square root.

Be careful with those positives and negatives.

Which of the following iterations converge on a solution when x0 = 2? So try it out with your calculator, which converge on a solution? So, the first one converges on that solution.

The second one does something weird, it oscillates between negative 2 and 1.

The third one converges on a different solution, and the fourth one converges on the same solution as the first one.

So the different formulas did manage to find us two different solutions.

Right, time to put that into practise.

You've done lots of hard work already.

You really explored how to do this.

So let's see if you can have a go at these questions.

When you're happy with Question One, come back and we'll look at Question Two.

Right, Question two, you've got a new equation.

I've told you what the approximate solutions are to three significant figures, and then I'd like you to have a go at these questions.

Off you go.

Question Three, again I've given you a new equation, and I've told you two approximate solutions.

I'd like you to write your own iterative formulas this time.

Then see if any of them, and it might be that none of them do, but see if any of them converge on one of the solutions, and then see if any of them converge on the other solution.

You have to pick your own values for x0, so it might take a little bit of trial and improvement and playing around.

Give this a go.

Let's have a look.

So, I've turned my three equations into iterative formulas, so check that you've written those correctly.

And then if we use x0 = 4, the first one causes an error, because we get a denominator of zero.

The second one does not converge at all.

And the third one converges on the approximate solution roughly 3.

732.

Right, if we change x0 = 3, we now do get an approximate solution with the first formula, the other solution, which is 0.

2679.

The second formula converges on the same solution.

The third formula converges on the other solution.

If we change x0 = 0.

2, the first formula converges on the same solution, again, 0.

2679.

The second formula also converges on that solution again.

But the third formula causes an error this time, 'cause we end up having to square root a negative number.

So check that you've shown that these can be rearranged correctly.

There's my working out for the first one.

And the second one, you're gonna need to factorise and divide through by the bracket.

Right, when x0 = 1, the first formula converged to an approximate solution, but the second formula caused an error.

When x0 was negative 10, the first formula converged to the same solution, and the second formula still caused an error.

So we haven't found a formula that converges to that other solution yet.

I wonder if you found one.

If you took the negative square root, that does converge on the solution as long as you use values of x, not less than 1.

6 recurring.

If you use that formula there, it seems to converge on the other solution 1.

44 again.

It is possible that you found other iterative formulas as well, but that negative root was the one that I found that did get you to that solution.

For Question Three, different formulas that you could have written.

I've written four there.

You might wanna pause the video, and check that you've got some of the same ones as me.

For B, there was so much to explore here.

I hope you had a lot of fun playing around.

I found that the positive root formula will converge on x = 9.

58 when x0 was 9.

There's other values of x0 as well that that worked, and there's potentially other formulas that that works for as well.

I found that to get the other solution, I could use those two formulas shown there, so the ones with the division, one where I'm dividing by 10, and the one that I'm dividing by 10 - x.

And again I used x0 = 9 and they worked well.

There'll be other values of x0 that work as well.

Fantastic, there was so much going on in that task, so, well done for persevering with it.

And, again, we're just having a bit of fun playing around with these ideas and seeing what happens.

Hopefully what you've come away with, is that iteration is a really useful tool, but it has limitations.

There are ways of seeing whether iterative formulas are likely to work, and we've seen that different values of x0 change what happens with the formula, but there's loads going on with these ideas, and this is something that you can carry on exploring as a mathematician into the future.

Thank you for joining us today, I hope you enjoyed that, and I hope you had a little bit of fun playing around with those problems. If you want to read through what we've learned today, pause the video and do that now.

Otherwise, I really hope that you'll join us again for another lesson.