# Functions and Sets?

1. Jun 17, 2008

### Mol_Bolom

Could functions be placed into sets?

Such as

L = f(x), g(x), h(x)

Lets say we have some properties which tie these functions together...

f(x) = ax+b
g(x) = cx+d
h(x) = ex+f

f(x)=g(x)=h(x)
This would be several functions which are also constant. Such as a polynomial.
(ax+b)(cx+d)(ex+f), there are three linnear functions which all equal 0.

Another way which I think would be plausible, though I am not sure...

f(x) = ax + b
g(x) = ax + b+1
h(x) = ax + b+2

Or

f(x) = ax + b
g(x) = ax2 + bx
h(x) = ax3 + bx2

Or

f(x) = ax + b
g(x) = a(x-1) + b
h(x) = a(x-2) + b

The way I think this would work...

Of a polynomial...

L = (3x+4), (5x+7), (8x-2)

120x3 + 298x2 + 142x - 56 = L1-1(x) * L2-1(x) * L3-1(x)

Before anyone gets angry about this question, note I am learning about this on my own, and I really don't like how that was done...But it's just a wild guess...Also, it has been bugging me for a bit so I figured I ought to ask about it before I take it too far...

The only thing that makes me think it is plausible is if every function (or if it is a better description, then equation) is tied with every other equation in some manner.

Here is another example of what I mean...

L = x-1, x2-x, x3-x2,,,

This function list could be tied by two methods (If methods isn't the right word, please tell me which word is correct). Each function in the list is multiplied by x. The second is that as x increases, the next element is used.

Let the domain be the natural numbers, 1,2,3,,,.

Lx(x)

L2(2) = 4 - 2 = 2. Since the second element is x2 - x.

Again, this is just a curiosity...

2. Jun 17, 2008

### gunch

Yes.

What exactly do you want here? You can define the set of the three functions f,g and h. If you want to impose that f(x)=g(x)=h(x) for all x, then we must have a=c=e and b=d=f. The polynomial f(x)*g(x)*h(x) have no relation to the actual set, but you can define it. For instance let,
$$f(A) = \prod_{g\in A}g(x)$$
Then f maps $$\{f,g,h\}$$ to (ax+b)(cx+d)(ex+f).

Yes. As long as you can describe them you can construct them (assuming of course that your description doesn't contain any contradictions)

Again a set doesn't "work", it just is.

I don't really see what the -1 are doing, as you aren't using the inverse functions, but yes (here I assume you have multiplied the polynomials properly):
$$f(L) = 120x^3 + 298x^2 + 142x - 56$$

If you're able to describe how they're tied together, then you can make a set of them.

This is an example of an inductive definition (sometimes called recursive). You basically say: L is the smallest set that satisfy the following rules:
1. The function x-1 is in L.
2. If $$f(x) \in L$$, then $$f(x)\cdot x$$ is in L.
This results in an infinite, but well-defined set of functions.

A set isn't ordered. x-1 doesn't come before $$x^2-x$$ just because you mentioned that first. What you're probably looking for is a sequence, which is like a set, but ordered. This sequence could be defined as:
$$L_1(x) = x-1$$
$$L_{n+1}(x) = L_n(x)\cdot x\qquad\textrm{for positive integers n}$$

3. Jun 17, 2008

### Mol_Bolom

I want to see if I understand $$f(A) = \prod _{g\in A} g(x)$$ correctly.

L = q(x), r(x), s(x)

$$f(L) = \prod _{g\in L} g(x)$$

Or is the f and g you used in the equation the same f and g in the set?

4. Jun 17, 2008

### Mol_Bolom

I have been pondering over this as well as a few other things and I would like some help as to parts of functions...
(I think I might have written out some of the details in the first thread incorrectly because of this problem)...

There is the definition of the function...
$$f(x)=ax+b$$

X is the domain of f
Y is the range of f

$$Y = f(X)$$

Would it be more appropriate to lable a variable/set which represents the range before the equal sign when writing a function out?
And when writing a function out where the variable comes after the equal sign such as,
$$f(x) = y$$
would mean then that the function f would equal y at x?

The reason is when I had mentioned before that f(x) = g(x) = h(x) was because in a polynomial each linnear equation/function would return the same answer. f(x) = 0, g(x) = 0, h(x) = 0...Which is what I was implying with f(x)=g(x)=h(x)...Although, I do know that this would be rather incorrect, but oddly enough correct...

$$(ax+b)(cx+d)(ex+f) = q_3 x^3 + q_2 x^2 + q_1 x^1 + q_0 x^0$$
x could either equal $$-b / a, -d / c, -f / e$$...

Thanks again...

Last edited: Jun 17, 2008
5. Jun 18, 2008

### gunch

Sorry, I assumed you was familiar with the symbol which is simply the product. So f(L) is the product of all functions in L (g isn't a specific function).

So you want to find the roots of f(x)g(x)h(x)? Then it would be incorrect to state f(x) = g(x) = h(x), consider f(x) = g(x) = h(x) = 1, and what if $$f(x)\not= g(x)$$, but f(x)=0? You should simply state that one of them is 0.

6. Jun 18, 2008

### Mol_Bolom

If we have a set of three functions...

L = f(x), g(x), h(x)

f(x) = (3x+5)
g(x) = (2x-1)
h(x) = (x-7)

For any $$L_n = 0$$ then,

$$\prod _{g \in L} g(x) = 0$$

For any element in L that returns 0 then,
the product of all the elements of L multiplied by that x would equal 0.

Does this make sense?

I do know that 0 times any number is 0, what I am wondering is if I wrote all that out correctly, or if I wrote it out badly...I am still working on the correct terminology to use when explaining things, also...

Last edited: Jun 18, 2008
7. Jun 18, 2008

### matt grime

It is common place to write brackets around things to indicate they are sets:

L={f,g,h}

Now, where did L_n come from?

No, it doesn't.

First, what are the domains of those functions? The domain is part of the definition of a function. What you've just written are expressions, or elements of the polynomial ring Z[x] (the set of all polynomials in x with integer coefficients).

What do you mean by an element in L 'returning 0'? What does 'that x' refer to?

8. Jun 18, 2008

### Mol_Bolom

Ah...After reviewing my Calculus book, I guess I mixed up different parts together...

How about it being defined as?...

For any set of linnear equations such that any element at one point of x returns 0 then the product of that set is equal to 0.

9. Jun 18, 2008

### matt grime

Uh, what does it mean for 'an element at one point of x to return 0'? What is x? How does it have points? Whatever that means, do you then mean that some product of functions is equal to the 0 function? None of that makes sense, I'm afraid.

Whatever you're getting at the normal sense of product of functions means that if I have finitely many polynomials, then their product is a polynomial (and is only the zero polynomial if one of the set of polys is the zero polynomial). Assuming everything is to be viewed as an element in R[x], the polynomials in one variable - x - with real coefficients.

10. Jun 18, 2008

### Mol_Bolom

This is getting wierd...

I'm going to back track here...

f(x) = 3x+5
X is the set of real numbers in which x can be. The domain of f?
Y is the set of numbers which is returned by f. The range of f?

If x = 5, then f(5) = 20
If x = 9, then f(x) = 32
if f(x) = 15, then x = 10/3
if f(x) = 24, then x = 19/3

Wouldn't the values of x exist in the set X, and wouldn't the values of f(X) exist in set Y?

Could there exist a single element in X which could be 7?
could there exist at a single element 15/3000000000 in X?
And wouldn't the same hold true for Y?

Last edited: Jun 18, 2008
11. Jun 18, 2008

### Diffy

Ok go for it.

Yup.
Yup.

yup. yup. yup. yup.

Yes elements of a set exist in said set.

In this specific case, yes, but not always. Say f(x) = Sqrt(-x). Then 7 is not in your domain (assuming we are working in real numbers).

In this specific case, yes, but not always. Say f(x) = Sqrt(-x). Then 7 is not in your domain (assuming we are working in real numbers).

Again, in your specific case where f(x) = 3x + 5 yes the same holds true for Y. This is because you have chosen a very nice function that happens to be a bijection on the reals.

I recommend you do some reading on set theory and review definitions for function, inverse, injection, surjection, bijection etc.

12. Jun 18, 2008

### matt grime

Diffy, you should bear in mind that it doesn't make sense to refer to sqrt(x) as being a function of the real numbers and then to exclude the negative ones as you just did. Despite what is taught in many courses, questions such as 'determine the domain of sqrt(x)' are nonsensical: the domain of a function is part of its definition. Such elementary mistakes in the teaching of mathematics are a real pain. I don't go as far as to define a function from X to Y as a subset of XxY satisfying certain properties, though.

13. Jun 18, 2008

### Mol_Bolom

"do some reading on set theory"

Thank you diffy...

However, I do know that certain numbers blow up in certain kinds of functions...Which is why I defined each function to be linnear...I know common linnear functions will never blow up with any number that is entered into them...At least I thought it would have been the easiest to start with...But anyway, I'll sure read up on set theory, wado, thanks, takk, Kolaval...

By the way, which books would you suggest for studying set theory on my own?

14. Jun 18, 2008

### matt grime

You have a finite number of functions from R to R. Since that is how you've defined them, then they cannot 'blow up' by *definition* of each being a function from R to R.

As far as I can tell is "if one of the functions f(x) is such that (without loss of generality) f(0)=0, then is the product of all the functions 0 (i.e. the function g(x) such that g(x)=0 for all x)?" Obviously, the answer is no: just take the simplest case of having one function in your collection (and f(x)=x being that single function if you want to be even more concrete). I don't know what made you make the conjecture, though.

15. Jun 18, 2008

### Mol_Bolom

Hmm...After reading about what you wrote matt...I've come to the conclusion that what I was thinking does not exist...

The set, say it be L, would be a set of linnear functions. Each element in each of these functions would be the value of x as the specific function returns 0...

L = {(3x+4), (5x+2), (8x-3)}
Thus...
L-1 = {-4/3, -2/5, 3/8}

But I know that L-1 is wrong, and I happen to have forgotten how to pull out the value of x in f(x) = 0.

Perhaps if I made two sets...
L is the set of functions.
S is the set of x where each function in L returns 0...
Therefore

S = {-4/3, -2/5, 3/8}

16. Jun 18, 2008

### Mol_Bolom

Out of curiosity, how does one show a series of numbers or a list in order?

Lets say we use two numerical systems for the same number...

1563 Using the decimal numerical system.
3,18,3 Using the best representation of the Mayan numerical system, or vigesimal, for the same number (1,563).

And if we write them out in quadratic equations.
1x3 + 52 + 61 + 3x0
3x2 + 18x1 + 3x0

The first one has four numbers 1, 5, 6, and 3
The Mayan one has 3 numbers 3, 18, and 3.

I am looking for something, such as a function, that I could use in another equation...

17. Jun 18, 2008

### matt grime

Linear has one n.

Now, what do *you* mean when you say a function has an element?

You are simply asking for the set of zeroes of the functions.

That almost makes perfect sense.

Given a set of functions L, you can define another set S:={ x : f(x)=0 for some f in L}. This is the set you wrote down.

You could also define a set T:={ x: f(x)=0 for all f in L}, and when you do that you're getting on to algebraic geometry. That is the set you described when you said you wanted the x where 'each function returns zero'.

Last edited: Jun 18, 2008
18. Jun 18, 2008

### matt grime

However one likes, so long as one makes clear what they are doing. I have no idea, from that description, what you want to do, so can be no more help.

19. Jun 18, 2008

### Mol_Bolom

Eh, I've gotten a lot further in a month, so who knows what another month will bring...But for now that will do...Thanks for all your help...

What I am trying to do is understand how things are written and said. I've been looking at mayan numbers as well as some others, and when I compair them to quadratic equations, I really see no difference...That's to say if x in a quadratic is -8, thus the sqare root of a number such as 1,0,0 would be -8. x^2 = 64, but x in itself would be -8. However, the square root of 1,0,0,0,0 would be 64, not -64.

I'm still not ready to discuss this yet, but I feel that I am getting closer...Just a lot more things to learn...

Still, Thanks for all your help...

20. Jun 19, 2008

### Diffy

Thanks Matt,

Just trying to help, but I guess I am in over my head! For each question I ask on this forum, I try to answer other's questions but I am no expert so I appreciate your correction.