Topic 8: Functional Equations

Topic 8: Functional EquationsDr J Frost ([email protected])

Last modified: 21st August 2013

Key to question types:

SMC Senior Maths Challenge

BMO British Maths Olympiad

The level, 1 being the easiest, 5 the hardest, will be indicated.

Those with high scores in the SMC qualify for the BMO Round 1. The top hundred students from this go through to BMO Round 2.Questions in these slides will have their round indicated.

Frost A Frosty SpecialQuestions from the deep dark recesses of my head.

Uni University InterviewQuestions used in university interviews (possibly Oxbridge).

Classic ClassicWell known problems in maths.

MAT Maths Aptitude TestAdmissions test for those applying for Maths and/or Computer Science at Oxford University.

STEP STEP ExamExam used as a condition for offers to universities such as Cambridge and Bath.

Slide Guidance

?Any box with a ? can be clicked to reveal the answer (this works particularly well with interactive whiteboards!).Make sure you’re viewing the slides in slideshow mode.

A: London B: Paris C: Madrid

For multiple choice questions (e.g. SMC), click your choice to reveal the answer (try below!)

Question: The capital of Spain is:

Slide Guidance

What are functional equations?

𝑓 (𝑥 )=𝑥+1You should be familiar with the concept of a function: something which takes an input and produces an output.

Now suppose we used this function to form an equation:

𝑓 (𝑥 ) 𝑓 (𝑦 )= (𝑥+1 ) (𝑦+1 )?

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What are functional equations?

We’ve now defined the function in terms of itself rather than just in terms of its input. When a function makes reference to itself, it’s known as a functional equation.

We’ve established the above is true when .But what if we just had the above equation, and didn’t know what the original function was. How could we work them out? Trial and error?

And even if we guessed and found it satisfied the functional equation, how would we know (i.e. how would we have proved) that this is the only possible function that satisfies it?

𝑓 (𝑥 ) 𝑓 (𝑦 )= 𝑓 (𝑥𝑦 )+ 𝑓 (𝑥 )+ 𝑓 (𝑦 )−2

Types of questionsThe types of questions you might see on functional equations often fall into one of the two categories:

1 Given functional equation(s), what are the possible original (i.e. ‘explicit’) functions?

2 Given functional equation(s), can we work out a particular output of the function, e.g. what is, without even having the explicit definition of the function?

Terminology

(where is a constant)

What is the name of these functions?

Constant Function

𝑓 (𝑥 )=𝑥 Identity Function

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Convention: We often use and for real numbered variables and for integer variables.

The key is to specialise, that is, trying some specific values of and and see if they yield any clues about the function.

e.g. Let . What can we conclude?

Then:

Thus for all values of .?

Find all functions that satisfy for all real .

Strategy #1 – Specialising

What would be a good starting point?

Let . Then .Thus and so .

Now given that we know what is, is there any way we can set and such that we can relate and ?

Let . Then:

Thinking about the Algebra slides, what must we conclude?

Squared terms must be positive, and the sum is 0, so , and thus for all

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Find all functions that satisfy for all real .

Strategy #1 – Specialising

Now let’s return to the previous functional equation, but restrict the domain a little more…

We could try . This would allow us to conclude that:

Thus

But it would be difficult to proceed further, because we’re now no longer allowed to use or , since 0 is outside the domain of the function (i.e. positive reals)

But since we know . Could we let and be something such that we can relate and ?

Let , then:

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Find all functions that satisfy for all positive real .

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But this approach doesn’t always work…

But this approach doesn’t always work…Find all functions that satisfy for all positive real .

We’ve managed to establish:

Proceeding further by specialising won’t get us far.For those who’ve done bits of C2, could you guess a function that satisfies the above?

(where and are constants)

Proving this is the only family of solutions* though isn’t possible using this technique. One method of proving this is using partial differentiation, which is covered at the end of these slides.

* I say ‘family of solutions’ because we can choose different values for constants and , but the underlying form of the function remains the same.

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Round 2

Round 1

BMO

Olympiad ExampleQuestion: Find all functions , defined on the real numbers and taking real values, which satisfy the equation for all real values and .

Using , , thus .

We can use so that .This gives us Now if , then Thus or

Case 1:. Then letting so that we get rid of the

Which is equivalent to

Case 2:. Then letting for the same reason:

𝑓 (0 )=1𝑓 (1 )𝑜𝑟 𝑓 (−1 )=0

See if you can work out these on your own using the specialisation method we’ve seen:

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Full Solution:

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Summary so farMake clever substitutions. Common ones are:

or Often helps us determine , or simplify our functional equation.

which lets us get rid of the . Although if is in the functional equation, then we obviously need to know what is.

A good idea when is in our functional equation. But we’d need to know what is.

A good idea when is in our functional equation. But we’d need to know what is.

It often helps to try and establish what and is via substitution, in order that we can make further substitutions!

A function is self-inverse, when applying it twice gets you back to the original input, i.e.

𝒇 (𝒙 )=𝟏𝒙 𝒇 (𝒙 )=𝒌−𝒙 𝒇 (𝒙 )= 𝒙

𝒙 −𝟏?

Examples:

Strategy #2 – Exploiting Self-Inverses

Now since is a self-inverse function, what two substitutions might be sensible to yield two equations?

Letting :

Letting

We now have simultaneous equations. We could eliminate by appropriate elimination. This gives us our solution:

𝑓 (𝑥 )=4−2𝑥2

3 𝑥?

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Strategy #2 – Exploiting Self-Inverses

Strategy #3 – InductionSometimes we can spot the function that satisfies the functional equations:

For all integers :

What is ?

𝒇 (𝒏 )=𝟐𝒏−𝟏?

Just looking at the functional equations above, we can see that we have the arithmetic sequence 1, 3, 5, 7, … as increases, which obviously has the formula

We could formally show this by induction.Suppose is true for . Then in our inductive step:

And in the base case, where , we have that .

Strategy #4 – Using a function substitutionWe’ve so far approached functional equations by substituting the input, e.g. or . But sometimes we must want to replace the function with a completely different one to make the functional equation easier to solve.

There will sometimes be clues in the equation with regards to the substitution we can make.

Round 2

Round 1

BMO

The function is defined on the set of positive integers by , and for all . (i) Prove that is always an integer. (ii) …

Notice that the appears both within the function brackets and outside (as does ). Can you think therefore what might be a sensible substitution?

𝑔 (𝑛 )= 𝑓 (𝑛 )𝑛?

Strategy #4 – Using a function substitution


Notice also that the arguments of are and . What’s significant about using these as inputs? They represent even and odd integers.

This suggests we should do the same for . Use the definition of above to simplify these:

Wow, that substitution really did give us simpler functional equations!

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Strategy #4 – Using a function substitution


We’ve got to prove is always an integer. But since , it’s sufficient to prove is an integer, as will then be.I smell induction!

We can use the second type of induction: supposing that up to are all integers, and thus that is:

Base case: , and 1 is an integer.Inductive case: Suppose to are integers. Then if is even, . Now , and , so by induction, is an integer. Thus is an integer.If is odd, then . Now . Again, so is an integer, and clearly then is.?

Strategy #5 – Using a different number base


One more exotic approach (which I won’t cover in detail here), which particularly seems to crop up in more difficult questions, is to define a new function in terms of the binary representation of the input. e.g. This is often a viable approach when we find and defined in terms of .

Suppose we let “the count of 1s in the binary representation of ” (e.g. ) and , then it turns out that we find that satisfies all the functional equations stated in the question! (I’ll leave this as an exercise…)Because the count of ones is always an integer, then is always an integer as required by the question.

The challenge then is to find a function in terms of the binary representation of the input that will satisfy the functional equations given.Note that this approach isn’t really applicable for types of questions where we’re asked to find all which satisfies the functional equations, because any valid binary function we found is just one solution: it doesn’t prove we can’t find others.

Finding specific outputsSometimes, given some functional equations, we don’t actually need to work out what is.We might just be required to find “the possible values of ”.

Let be a function where (i.e. the domain is non-negative integers) and

Suppose the three functional equations hold:

i)

ii) if ends in a 3.

Question (a): Find .

Oxford interview question (candidates were given a day in advance)

Hint: 17 ends with a 7. What could we do with 17 to make it end with a 3?You don’t need condition (ii) for this question.

Finding specific outputs

i)

ii) if ends in a 3.

Question (a): Find .


Note that . We multiplied by 9 to get a number ending in 3.i.e. . At this point it seems sensible to use (i) since we have a product inside the function brackets.

From (i) we have that Thus

A solution

Finding specific outputs

Let be a function where (i.e. the domain is non-negative integers) and

Suppose the three functional equations hold:

i)

ii) if ends in a 3.


By using (i), Note that . But since for all , must be 0 or 1.

Thus could either be 2 or 3.?

Question (b): What are the possible values of ?

Common functional equations to remember

When domain: 𝑓 (𝑥 )=𝑎 log (𝑥 )+𝑏

𝑓 (𝑥+𝑦 )= 𝑓 (𝑥 ) 𝑓 (𝑦 ) 𝑓 (𝑥 )=𝑎𝑥

Known as Cauchy’s Functional Equation

For integers/rationals:

But additional more complicated solutions for reals.

Cauchy’s Functional Equation

Solving this functional equation has been used as a topic for discussion in an Oxford maths interview.

𝑓 (𝑥+𝑦 )= 𝑓 (𝑥 )+ 𝑓 (𝑦 )

Let :

Prove that over the integers.

But we could keep adding extra ’s:

By continuing this (‘inductively’) we have that:

where is an integer, but is any real( is at the moment restricted to integers, because in the workings above we had ‘ lots’ of , and we can’t have a fractional number of things)

where is an integer, but is any real


Prove that over the integers.

Let and

Then i.e. We’ve proved it over the integers (since is an integer).

Prove that over the rational numbers.

We know when is an integer. Observe that where is an integer.

Then which we were allowed to do because is an integer and we know that when is an integer, as at the top of the slide)

Thus dividing by : , i.e. where , i.e. is rational.

So far:


Prove that over the reals if is an increasing function.

Suppose we wanted to show that .If is increasing, then if and only if .Then we can bound the value between two rationals:

But since we showed that for rational :

We could make these bounds arbitrarily tighter:

Thus in the limit,

because is increasing

ζEpilogueTopic 8 – Functional Equations

A proof using partial differentiation that the family of solutions to is .

Proof of solutions to

We want to prove that if:

then when :

To do so, we can use partial differentiation. Here’s a very quick intro:(but I’m assuming knowledge of the chain rule and product rule for differentiation, which is covered in the C3 A Level module)

Partial Differentiation

By the chain rule, and conventional differentiation, we have for example:

But in partial differentiation, all variables except the ones we’re differentiating are treated as constants. Thus:

Notice the curly d.

Proof of solutions to

(1)(2) Partially differentiating (1) with respect to . Note

that disappears because it doesn’t use , so everything is a constant.

(3) Letting

(4) Letting and rearranging.

(5) Integrating w.r.t.

This therefore provides us with an alternative way to solve functional equations: by partially differentiating.

for

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Topic 8: Functional Equations