Litov's mean value theorem

Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...

Age

11 to 14

Challenge level

Secondary

STATISTICS AND PROBABILITY

Statistics

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

Litov's Mean Value Theorem printable sheet

Start with two numbers, say 8 and 2.

Let's generate a sequence where the next number is the mean of the previous two numbers.

So the next number is half of $(8 + 2)$, and the sequence continues: $8, 2, 5$

The next number is half of $(2 + 5)$, and the sequence continues: $8, 2, 5, 3.5$

What would happen if you continued this process indefinitely?

Choose a few pairs of starting numbers and repeat the process.

Each time, your sequence should get closer and closer to a value which we call the limit.

Can you find a relationship between your starting numbers and the limit of the sequence they generate?

Can you explain why this happens?

Now start with three numbers.

This time, we can generate a sequence where the next number is the mean of the last three numbers.

Check you agree that if we start with $4, 1, 10$, the next number is 5, and the number after that is $\frac{16}{3}$.

What would happen if you continued this process indefinitely?

Choose some more sets of three starting numbers.

Can you find a relationship between your starting numbers and the limit of the sequence they generate?

Can you explain why this happens?

Extension

Explore what happens when you have $n$ starting numbers and you generate a sequence where the next number is the mean of the last $n$ numbers.

Student Solutions

Well done to Ken from Blessed Robert Johnson Catholic School who sent us his solution to this problem:

After a while of playing with the numbers on a spreadsheet I have discovered that the formula to find the "limiting value" for $2$ starting numbers is: $$\frac{x+2y}{3}$$ where $x$ is the first number chosen and $y$ is the second number chosen.

The formula to find the "limiting value" for 3 starting numbers will be: $$\frac{x+2y+3z}{6}$$

Some older students also worked on this problem. Terence from Brumby Engineering College has made a prediction for what the limit will be when there are $n$ starting numbers. He thinks it will be

$$\frac{a_1+2a_2+3a_3+4a_4+\ldots+(n-1)a_{n-1}+n a_n}{\frac{1}{2}n(n+1)}$$ where $a_1$, $a_2$, $\ldots$, $a_{n-1}$ and $a_n$ are the $n$ starting numbers.

Can you see why Terence has made this prediction?

As mathematicians, we like to prove our answers, rather than just predict. Here is an outline of how we might prove the formula when there are two starting numbers.

Suppose that the first two numbers are $a$ and $b$. Let's work out the next few terms of the sequence. We have

$$a,\:b,\:\frac{a+b}{2},\:\frac{a+3b}{4},\:\frac{3a+5b}{8},\: \frac{5a+11b}{16},\:\frac{11a+21b}{32},\:\frac{21a+43b}{64}$$ and so on. We'd like to show that in the limit the coefficient of $b$ is twice the coefficient of $a$, and also that the sum of these coefficients is the denominator. Can you see why this will prove the formula?

The $n^{\textrm{th}}$ term in the sequence is of the form $$\frac{ \alpha_n a+\beta_n b}{2^{n-2}}$$ (for $n\geq 2$). We can work out $\alpha_n$ in terms of $\alpha_{n-1}$ and $\alpha_{n-2}$. Can you see how?

We have $\alpha_n=\alpha_{n-1}+2\alpha_{n-2}$ (for $n\geq 4$). Similarly, we have $\beta_n=\beta_{n-1}+2\beta_{n-2}$ (for $n\geq 4$).

It turns out that we can use these formulae to prove that $\beta_n=2\alpha_n +(-1)^n$ (for $n\geq 2$). Test this formula on the above examples to see that it works for these cases.

We'd like to show that as $n$ tends to infinity, $\beta_n$ becomes twice $\alpha_n$. Let's think about $\beta_n/\alpha_n$. From the above formula, this is $2+(-1)^n/\alpha_n$. But $\alpha_n$ is getting bigger and bigger, so the $(-1)^n/\alpha_n$ part of this gets smaller and smaller. In fact, it tends to $0$, and so the limit of the whole expression is $2$.

It is also possible to use the formulae for $\alpha_n$ and $\beta_n$ to show that $\alpha_n+\beta_n=2^{n-2}$ (for $n\geq 3$) (that is, the coefficients of $a$ and $b$ sum to the number on the bottom of the fraction). Again, test this formula on the examples above.

Combining the last two paragraphs, we find that the limit of the sequence is $$\frac{a+2b}{3}$$ as predicted.

Teachers' Resources

Why do this problem?

This problem provides a context for genuine discovery and student directed research/investigation. It is an ideal situation for students to work out how to use ICT to speed up the calculations and support them in their quest to rediscover Litov's Theorem.

Possible approach

"I found this problem called Litov's Mean Value Theorem. I'm hoping we can work out what the theorem is. We can start with any two numbers, say $8$ and $2$. These are the start of a sequence. The rule is that the next number in the sequence is the average of the last two numbers. So what comes next? Why? And then what?...

Invite students to choose their own pair of starting numbers, to calculate the sequence and find its limit. Students could use calculators for this activity. Giving students free choice can result in a lot of information being collected in a short space of time. While this is going on and the results are appearing on the blackboard, ask some students to think about how these calculations could be done on a spreadsheet.

Bring the class together and ask for observations, comments, suggestions and predictions. Demonstrate the use of a spreadsheet for testing these predictions quickly. The use of the computer makes it possible to operate at a new level and the computer shows the limiting process clearly.

"Given all that information would anyone like to check a result or predict what will happen to any pair of numbers?"

There's a chance to discuss whether these sequences will ever actually reach their limits.

Students could test their hypotheses working on paper, or everyone could be given access to spreadsheets. When students are convinced that they know how to find these limits, challenge them to suggest some reasons why the limits behave as they do.

Students could then move on to working on these:

What would happen if sequences were generated from three initial values by:

averaging the last three numbers (i.e. $(a+b+c)/3$)?

adding the last three numbers and dividing by 2 (i.e. $(a+b+c)/2$)?

Key questions

Can you tell where these numbers are heading?

Does it matter if I swap the two starting numbers around?

What do these long decimals mean? How big is that number, roughly?

Possible support

This problem is a good context for work on organisation skills and calculator competence with opportunities for making conjectures, and refining conjectures.

Possible extension

What happens when you have $n$ start numbers and the rule for working out the next number changes to finding the average of the last $n$ numbers?

Laurinda Brown (1983) wrote about using this problem in the classroom: in Mathematics...with a Micro 1, pp.22-25, Waddingham, Jo (ed), Bristol, County of Avon, Resources for Learning Development Unit. The lesson notes above are adapted from her descriptions of using the problem.

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