# Family of functions satisfying a particular criterion

• issacnewton
In summary: So yeah, you were on the right track.@fresh_42 There are solutions of the form ##f(x) = 2x + c##, too.
issacnewton
Homework Statement
Let ##f:\mathbb{Z}\to \mathbb{Z}## be a family of functions following the criterion. $$f(2a) + 2 f(b) = f(f(a+b))$$. Find the family of functions
Relevant Equations
definition of function
Since, both the domain and the range is set of integers, we must have just operations of addition and multiplication only in the function. That means, function should be some kind of a polynomial. Plugging ##a=0## and ##b=0##, I can deduce that ##3 f(0) = f(f(0))##. Also I can deduce that ##3 f(2a) = f(f(3a))##. But after this, I am stuck. So, any guidance will be helpful. This is a math Olympiad question.

Thanks

Since ##1## generates the integers, it is certainly useful to consider ##f(1)##. E.g. compare ##f^2(0+1)## and ##f^2(1+0)##.

Ok, we would get the following equations
$$f(2) + 2 f(0) = f(f(1))$$
$$f(0) + 2 f(1) = f(f(1))$$
And subtracting one from the another, we would get
$$f(2) + f(0) = 2 f(1)$$

I would also try multiplication: Is ##x \stackrel{f}{\longmapsto} c\cdot x## a solution? The tricky part is probably to get rid of ##f^2##. So what happens if you consider ##f^2(0)=:c## or ##f^2(1)=:c\,?## Or consider ##f(-1)##.

I don't see the solution, but I think deriving other rules as the one you just got is helpful. E.g. what happens if you repeat the calculation which led to ##f(2)+f(0)=f(1)##, i.e. ##f^2(2)=f(2)+2f(1)=f(4)+2f(0)=f(0)+2f(2)## and so on. Once you have found a recursion, you will be done.

IssacNewton said:
Since, both the domain and the range is set of integers, we must have just operations of addition and multiplication only in the function.

It may turn out that the answer obeys those conditions, but there are many types of functions from ##\mathbb{Z}## to ##\mathbb{Z}## that are not polynomials.

I am also able to prove that ## f(f(0)) = 3 f(0) ##, ##f(2f(0)) = 5 f(0)##, ## f(3f(0)) = 7 f(0)##. So, we can prove with induction that if ##n \in \mathbb{N}##
$$f(n\cdot f(0)) = (2n+1) f(0)$$

IssacNewton said:
$$f(2a) + 2 f(b) = f(f(a+b))$$.

One observation is that the right hand side of the equation remains the same if you swap ##a## and ##b##.

IssacNewton said:
$$f(2a) + 2 f(b) = f(f(a+b))$$. Find the family of functions

I see how to guess a family of functions informally. I don't yet see how to deduce it.

For a given number ##c##, if we want a function that can gives the same value in a calculation that involves applying it to each pair ##a,b## such that ##a+b = c##, nothing comes to mind except ##f(x) = k x ##. Making that guess, we have

##k2a + 2kb = k(k(a+b))##
##2k(a+b) = k^2(a+b)##
##2 = k## or ##0 = k##

Hmm, how about this. Setting ##b=0##, we get:
$$f(2a)+2f(0)=f(f(a))$$
Setting ##a=0## and substituting ##b## for aa, we get:
$$f(0)+2f(a)=f(f(a))$$
Right hand sides are equal, so we can identify them:
$$f(2a)=2f(a)−f(0)$$
This looks like a useful relation.
Let's try to put another value in the initial equation, for example ##a=1##. We get:
$$f(2)+2f(x)=f(f(x+1))=f(0)+2f(x+1)$$
Where we used the second equation from the top to simplify the right hand side.
From there we get:
$$f(x+1)−f(x)=const.$$
So we have an arithmetic progression, and therefore ##f## is linear. Finally, we substitute ##f(x)=ax+b## for some constants ##a## and ##b##. Solving, we get the solution.

EDIT: Oops, I just realized this was in the homework section, so I cut the solution down. This should be enough for a hint.

Last edited:
Delta2
How about considering the cases a=b and a=-b?

Well sure, you can do that, but those relations won't contribute much, I think. The problem becomes relatively easy once it's concluded that the function is linear. Generally in these types of problems, substituting 0, 1, -1, usually helps acquire some good relations to start off. Either that or switching places of ##a## and ##b## looks interesting since the right side is symmetrical with respect to them.

Last edited:
##f(x)=2x## is obviously a solution, and by symmetry we have ##f(2x)=2f(x)-f(0)##. If I haven't made a mistake, we also have from ##f^2(3x)##
\begin{align*}
f(2x)&=2f(x)-f(0)\\
f(4x)&=4f(x)-3f(0)\\
f(6x)&=6f(x)-f(0)\\
f(8x)&=8f(x)-7f(0)
\end{align*}
which is a bit strange. The questions now are whether ##f(0)\neq 0## is possible, and whether ##f(x)=2x## is the only solution.

@fresh_42 Actually, I cut the solution down, since it's homework, but after plugging in the second condition, that it is linear, you obtain precisely whether ##f(x) = 2x## is the only solution(turns out it isn't, actually, although at first I thought that it is). Solving you get that solutions have the form of ##f(x) = 2x + c##, for arbitrary ##c##.

Last edited:

## What is a family of functions?

A family of functions is a group of functions that share similar characteristics or properties. These functions can be represented by a general equation or formula, with different values plugged in to create a specific function within the family.

## What is a criterion for a family of functions?

A criterion for a family of functions is a condition or rule that the functions within the family must satisfy. This could be a specific property, pattern, or behavior that all functions in the family must exhibit.

## How do you determine if a function belongs to a particular family?

To determine if a function belongs to a particular family, you must check if it satisfies the criterion set for that family. This can be done by plugging in values and observing if the function exhibits the desired properties or behaviors.

## Can a function belong to more than one family?

Yes, a function can belong to more than one family if it satisfies the criteria for multiple families. This often occurs when the families have overlapping or similar criteria.

## Why are families of functions useful in mathematics?

Families of functions are useful in mathematics because they provide a broader understanding of a group of related functions. They can help identify patterns and relationships between different functions, and can be used to solve a variety of problems and equations.

### Similar threads

• Calculus and Beyond Homework Help
Replies
8
Views
590
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• General Math
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
15
Views
914
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
2K