Lesson video

Hello, Mr. Robson here.

Welcome to maths.

Today, we're evaluating iterative formulas.

Iterative formulas are really, really useful in maths and in the real world around us, so let's take a look.

Our learning outcome is, I'll be able to evaluate and interpret iterative formula for various real-life situations.

Some keywords we're gonna hear throughout the lesson.

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.

You're gonna hear those words a lot.

Two parts to our learning today.

Let's begin by looking at the inputs and outputs in iterative formulas.

We can evaluate a composite function composed of repeated iterations of the same function.

For example, f of x equals three x minus one and we're asked to evaluate f of f of f of f of two.

We begin by evaluating f of two.

When we substitute two into our function, we get five.

Five is the output of f of two and it becomes our next input when we got to evaluate f of f of two.

We input the five and we get an output of 14.

This output becomes our next input when we evaluate f of f of f of two.

When we input 14, we get an output of 41 and that output becomes the next input and we reach 122.

F of f of f of f of two equals 122.

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.

You just saw an iterative process in action.

The output of five became the input into the next iteration.

The output of 14 became the input into the next iteration.

The output of 14 became the input into the next iteration.

This case was fine, but what if we had been asked to evaluate f of f of f of f of f of f of f of f of four.

Can you imagine how much working out would be required for that one? I'm sure you can imagine it and I'm also sure you don't want to do it all.

So is there a more efficient way? We can simplify this process by writing an iterative formula and efficiently using our calculator.

Our iterative formula looks like this.

X subscript n is the existing x value.

X subscript n plus one is the next x value.

You could think of x subscript n as the input, that's what we're putting into the function, three lots of the existing x value minus one.

X subscript n plus one is the output.

X zero equals four, that means it's our first input.

Once we have an iterative formula and our x zero value, the first step is to input this initial value of four and press EXE.

EXE is short for execute.

So I type in four and press execute.

If you have a lesser calculator than a Casio ClassWiz, that might be the equals button for you, but hopefully you're in possession of a Casio ClassWiz, in which case EXE is your button.

Our calculator now has an answer of four stored in it and this is ready to be used as our first input.

We now apply this first input to our iterative formula using the Ans button on our calculator.

Ans being short for answer.

We know that our calculator has an answer of four programmed into it, so when we do three lots of ans minus one, the calculator does three lots of four minus one.

That gives us our first iteration of x one value is 11.

Our calculator now has this answer of 11 stored in it ready to be used as the next input.

Pressing EXE again automatically inputs 11 into the same function.

We get an output of 32.

Pressing EXE again automatically inputs 32 into the same function, output of 95.

When recording our outputs after each iteration, we can just write Ans in our working.

We show in the first iteration, x one, that we input the actual x zero value, but after that, we don't need to write any values as our inputs.

We keep going with this process, pressing EXE again and again.

We get our fourth iteration, our fifth iteration, our sixth iteration, seventh iteration, eighth iteration.

Eight f's means we perform eight iterations.

We've successfully evaluated our composite function.

Certain statements we can make now.

After three iterations, the output is 95.

After five iterations, the output is 851.

The input into the sixth iteration is 851.

F of f of f of f of f of f of f of four is 7655.

That's the x seven value, the seventh iteration.

Seven f's there.

Let's check that you can make statements just like that.

Can you fill in these blanks? Pause and try this now.

Welcome back.

Hopefully, our first statement read, "The input into the third iteration is 32." After six iterations, the output is 2552.

And finally, f of f of f of f of f of f of f of f of four is 22,964, the eighth iteration.

An iterative formula could require the input to be used in more than one place.

For example, f of x equals x squared minus two x plus three.

There's two variable terms in that function.

Our iterative formula's going to look a little different.

It's going to look like this.

X subscript n, the existing x value, is now input in two positions.

We're gonna start this one by typing one and pressing EXE, making one our first input.

Use the Ans button to perform the first iteration.

Our calculator display will look like that.

We can see that input of one going in twice into our iterative function.

And after one iteration, we get two.

Remember to show that you've input the x zero value in that first iteration.

Pressing EXE repeatedly puts our output back into our function.

We get the second iteration, the third iteration, the fourth iteration, the fifth iteration and the sixth iteration.

F of f of f of f of f of f of one equals 458,331.

Whilst this is an efficient way of evaluating composite functions, we're more likely to see problems written like this.

Using this iterative formula and an x zero value equal to one, find the values of x one, x two, x three and x four.

When presented with a problem like this, we begin with our x zero value.

X zero in this case is equal to one, so I'm gonna press one on my calculator, the execute button, one is now ready to be input into our iterative formula.

When I go to do that, you'll notice on my calculator display the Ans button has been used twice, because we're substituting into this iterative formula the existing x term in two different positions.

We now have our first iteration.

You'll notice that I've demonstrated by putting one in the bracket that I've inputted the correct x zero value.

Pressing EXE again gives us the second iteration and again for the third iteration and again for the fourth iteration.

It wasn't specified what we were to round to in this case, so I've left those numbers exact and use truncation in my communication.

Quick check you can do that.

Using this iterative formula and an x zero value of two, can you find the values of x one, x two, x three and x four? Pause and give this a go now.

Welcome back, well done for having a go at that problem.

Let's see how we did.

Should've begun by having an input of two in your calculator.

Then we go to put that into our iterative formula, which will look like so.

That's the first iteration when we go to record it and make sure we write bracket two to show that we've used the correct x zero value.

Then pressing execute again gives us the second iteration.

We can just use bracket ans in our written communication now.

Then EXE again for the third iteration, EXE again for the fourth iteration and we don't need to record all of those digits.

We can truncate that answer to make our communication a little more concise.

Jacob and Sam are working on this problem.

Using the iterative formula, the next x term equals the root of the current x term and x zero equals nine, find the values of x one, x two, x three and x four to two decimal places.

They've been asked to round their answers.

Jacob's method.

The first iteration equals 3.

00, the second iteration, 1.

73, the third iteration, 1.

31, the fourth iteration, 1.

14.

Sam's method.

It's the exact same for the first iteration and then Sam's working looks like this.

Who do you think is right, Jacob or Sam? Pause, study their work.

Who's right? Welcome back.

Hopefully you said Sam's right and Jacob's gone slightly awry here.

Sam is right, whilst we're ask to round answers to two decimal places, it's important to continue to input exact answers from the previous iteration and the Ans button accurately does that.

Inputting any rounded answers at any stage costs accuracy in the next iteration.

You can see the point at which Jacob has gone wrong.

Quick check you've got that.

Using this iterative formula and x zero equals three, Jacob's asked to find the values of x one, x two, x three and x four to two decimal places.

You can see that Jacob's wrong.

Write a sentence to explain what is wrong in his method and another sentence about how he can improve his work.

Pause and write those sentences now.

Welcome back.

You might've written, "Don't input rounded answers, this costs you accuracy." Secondly, as advice on how to improve his work, you might've said, "Use the Ans button to ensure an exact value is input each time." Practise time now.

Question one.

I'd like you to use the iterative formula to find f of f of f of f of f of f of f of f of f of nine.

Pause and do this now.

For question two.

Part A, I'd like you to this iterative formula, then x zero value of three to find the values of x one, x two, x three and x four.

For part B, I'd like you to use this iterative formula, then x zero value of 13 to find values of x one, x two and x three.

Pause and do this now.

Question three.

You're gonna use this iterative formula with an x zero value of one to find the value of x 38,695.

My tip for this one, don't show all of your working for this problem.

I don't expect you to write out 38,695 iterations.

I'd like you to just state your answer and include a sentence to justify it.

Pause, give this one a go, good luck.

Welcome back, feedback time.

Let's go through question one and see how you did.

The first iteration should look like so.

Second iteration, third iteration, fourth, fifth, sixth, seventh, eighth, ninth iteration.

Nine f's means we want the ninth iteration.

That composite function evaluates to negative 127,937.

For part two.

Our first iteration should look like so.

Our second iteration like so.

Third iteration and fourth iteration.

For part B.

Our first iteration, second iteration, and third iteration.

For question three, I hope you enjoyed this one.

It was interesting.

X 38,695 equals zero.

Why is that so? Well, the first iteration, an input of one, gives an output of zero.

So in the second iteration, zero becomes the input, giving us an output of one.

In the third iteration, one is the input, giving us an output of zero.

You can see where this is going.

Once you've spotted this pattern in the first handful of iterations, you notice that every odd iteration has an output of zero.

X 38,695 is an odd iteration, therefore it is equal to zero.

You needed that sentence to justify your answer.

On to the second half of our learning now where we're going to do some modelling with iterative formulas.

Iteration has a lot of practical applications.

For example, we can use it in this context.

An organisation purchases 250,000 pounds of shares in a space exploration firm.

These shares are forecast to grow with a rate of 7.

5% per anum.

The accountants want a forecasted value for these shares each year for the next five years.

We can write an iterative formula and use an iterative process to quickly calculate these amounts.

There's our same problem.

The iterative formula for such a problem will look like so.

We need to know the first five years of results.

We need to know y one, y two, y three, y four and y five.

We begin with our y zero input of 250,000.

We'll type Ans times 1.

075 into our calculator and that'll give us the first iteration.

Pressing EXE gives us the second iteration, EXE again for the third iteration, the fourth iteration and the fifth iteration.

Notice I have rounded these values.

We're talking about pounds and pence here, so I'm gonna round to two decimal places or the nearest penny.

Quick check you can do a similar example.

An investor purchases a small business for 800,000 pounds.

The value of the business is forecast to grow at a rate of 12% per anum.

The accountants want a forecasted value for this business each year for the next six years.

Use an iterative process to quickly report these values.

Pause and try this problem now.

Welcome back.

You should have used the iterative formula, the next y value equals the existing y value multiplied by 1.

12 with y zero equaling 800,000 and then found the first six iterations.

You start with 800,000 in our calculator and then the first iteration, the second iteration, third iteration, fourth, fifth, sixth iterations.

So remarkably, the small business, after six years, is worth one million 579,058 pounds and 15 pence.

Seems like a good investment to me.

Iteration can be used to create mathematical models, for example, this population model.

There are 14,000 fish in a lake.

2000 are fished and removed each year.

The remaining population reproduces and grows by 18% per anum.

The iterative formula looks like this.

The next year's population is in brackets, the current year's population minus 2000, that's the 2000 that are fished and removed.

And then close brackets, multiplied by 1.

18 to show the growth by 18% using a p zero value of 14,000.

That's the initial population.

Our calculator will quickly generate the year-on-year projected population numbers.

When using this model, we're gonna start with an input of 14,000 in our calculators.

That's going into our iterative formula to give us our first iteration.

After one year, the population's 14,160.

Press EXE again.

After two years, the population's 14,349.

You will notice 14,349 is not the exact answer on my calculator display.

When we get decimal answers, non-whole number answers in this case, we round to the nearest whole number with it being a population model.

We can have 0.

8 of a fish.

Well, you can, it's just not very good for the fish.

You press EXE again to get the third iteration, EXE again for the fourth iteration, the fifth iteration, the sixth iteration.

We're pressing button on our calculator to get the next year's population each time.

Using an iterative process allows us to quickly generate the projected year-on-year population.

Quick check you've got this.

Alex is solving this modelling problem using an iterative process.

There are 4000 fish in a lake.

500 are fished and removed each year.

The remaining population reproduces and grows by 23% per anum.

In what year will the population exceed 5000? You can see Alex's iterative formula, p one, p two, p three, p four values and Alex's conclusion that the population will exceed 5000 in the fourth year.

There are errors in Alex's work.

Spot them and advise Alex as to what they should do better.

Pause and do this now.

Welcome back.

Hopefully you said the iterative formula is fine, but there's a problem with that value.

P one equals 4000? Well, that's not right.

So if our p one value is not right, Alex's conclusion is not right.

You might have said, "4000 is the initial population, the p zero value." Also, you might've advised Alex, "You should show your method at each iteration." This might've helped Alex to spot their error.

When we write the first iteration out like so, we can see that we've inserted correctly the right initial population.

The correct p zero value of 4000 has gone into our iterative formula.

That's the first iteration, the second iteration, the third iteration.

We can now see that actually the population exceeds 5000 in the third year.

Practise time now.

Question one.

You purchase 30,000 pounds worth of shares in a solar panel firm.

They are forecast to grow at a rate of 8% per anum.

Your accountant needs a forecasted value for these shares for each of the next six years.

Use an iterative formula to calculate these values.

For question two, there are approximately 25,000 polar bears left on Earth.

Each year, an estimated 800 polar bears die of unnatural causes.

The remaining population grows at a rate of 1.

3% per anum.

Use an iterative process to model the next 10 years of population numbers.

Pause and try these problems now.

Question three.

There are 16,000 fish in a lake.

2500 are fished and removed each year.

The remaining population reproduces and grows by 20% per anum.

A new executive joins the fishery.

They want to change the limit on the number of fish removed each year to 5000.

Assuming that the rate at which the remaining population increases stays at 20%, use iterative modelling to show why the limit should not be increased to 5000.

Tricky problem this, but I think you'll enjoy it, so let's give it a go.

Pause and do that now.

Welcome back, feedback time.

Question one.

We are buying shares in a solar power firm.

The iterative formula should look like so.

Your first iteration should look like that, second iteration, third iteration, fourth, fifth, sixth.

Well, this is impressive.

After six years, your shares are worth 47,606 and 23 pence.

Nice investment.

For question two we were population modelling with the polar bears.

Your iterative formula should look like so.

Your first iteration like that, your second iteration, third, fourth, fifth, sixth, seventh, eighth, ninth and 10th iterations.

Look at that number on the 10th iteration, 19,852.

There's approximately 25,000 polar bears left on Earth and in 10 years' time, there's gonna be less than 20,000.

In fact, the polar bears haven't got that long left.

Maybe you'll want to go and do some research into why this is happening.

Question three.

We're working at the fishery and we got a new executive who've joined and they want to raise the limit on the number of fish they can remove each year.

We're trying to present why it should not happen.

In order to give this person feedback, you might divide your working into two scenarios.

Let's model scenario one where 2500 fish are removed per year and let's model scenario two separately alongside where 5000 fish are removed each year.

There's the two iterative formulas, the only difference being inside the bracket, the existing population minus 2500 in scenario one, the existing population minus 5000 in scenario two.

The first five iterations for scenario one look like so.

We can see the population is slowly growing.

What about scenario two? Well, there's the first iteration, the second, the third, the fourth.

When we go to do the fifth, you actually get a negative result in your calculator.

This is a population model so that can't happen.

What has happened, is that the population has been fished to exhaustion.

We don't have any fish left.

That's not good for a fishery.

This is how you're gonna convince the executive that that limit can't be raised.

Maybe one day in the future you'll be using mathematical modelling like this in your own organisation.

We're at the end of the lesson now, sadly, but we have learned we can use and evaluate iterative formulas to simplify a problem.

We can also evaluate and interpret iterative formulas in a variety of real-life situations from investments to mathematical modelling in contexts such as population change.

Hope you enjoyed the maths in today's lesson as much as I did and I look forward to seeing you again soon for more mathematics.

Goodbye for now.