Functions

[Ostronomos]

Let S = R and T = R and f(x) = x^2. This is clearly a function which assigns to each real number another real number. Note that it is also a parabolic function. x ∈ R since the domain is in the reals and it assigns another real to the range.

Can anyone tell me if my explanation of this function is at fault or if anything is missing?

 

[Jedster]

Let S = R and T = R and f(x) = x^2. This is clearly a function which assigns to each real number another real number. Note that it is also a parabolic function. x ∈ R since the domain is in the reals and it assigns another real to the range.

Can anyone tell me if my explanation of this function is at fault or if anything is missing?

Parabolic because f(x)=f(-x)
Seems fine to me although I don't understand what S & T refer to.

 

[Polymath257]

Let S = R and T = R and f(x) = x^2. This is clearly a function which assigns to each real number another real number. Note that it is also a parabolic function. x ∈ R since the domain is in the reals and it assigns another real to the range.

Can anyone tell me if my explanation of this function is at fault or if anything is missing?

I would say that f:S->T is a function. Depending on your level and amount of detail, you might want to say that f={(x,x^2 ):x\in S} is a function f:S->T.

I would say the graph is a parabola, not that the function is parabolic.

Also, depending on your level, you may need to prove that x^2 makes sense and is in T for every x in S. This may depend on the specific definition of the set of real numbers you are using.

 

[Polymath257]

Parabolic because f(x)=f(-x)
Seems fine to me although I don't understand what S & T refer to.

Nope, f(x)=f(-x) means it is an even function. S and T are the domain and co-domain.

 

[Jedster]

Nope, f(x)=f(-x) means it is an even function. S and T are the domain and co-domain.

Thanks.

 

[Yerda]

Let S = R and T = R and f(x) = x^2. This is clearly a function which assigns to each real number another real number. Note that it is also a parabolic function. x ∈ R since the domain is in the reals and it assigns another real to the range.

Can anyone tell me if my explanation of this function is at fault or if anything is missing?

Sometimes in maths classes you'll hear stuff about one-to-one and onto (or surjective). f isn't one-to-one here, but if you limit the domain you can produce two (or more) one-to-one functions. I'm not sure if f is onto, as it only maps the positive real numbers, but every positive real number is the image under f of an element (two infact) of the domain (so if the codomain is the positive reals it is onto).

I don't think I've ever came across these ideas anywhere except in a pure maths setting in 1st year.

@Polymath257 are these concepts ever useful?

 

[Ostronomos]

I recently read the rigorous mathematical proof of limits and I thought I would reiterate it here:

Suppose f is a function whose domain contains two neighboring intervals: f: (a,c)∪(c,b)->R. We wish to consider the behavior of f as x approaches c. If f approaches a particular finite value l as x approaches c, then we say that the function f(x) has limit l as x approaches c. We write:

Lim x->c F(x) = L.

The proof of this is as follows:

Let a<c<b be the domain of f. If δ<0 such that ε<0 then:

|f(x)-l|<δ

where 0<|x-c|<ε.

 

[Yerda]

I recently read the rigorous mathematical proof of limits and I thought I would reiterate it here:

Suppose f is a function whose domain contains two neighboring intervals: f: (a,c)∪(c,b)->R. We wish to consider the behavior of f as x approaches c. If f approaches a particular finite value l as x approaches c, then we say that the function f(x) has limit l as x approaches c. We write:

Lim x->c F(x) = L.

The proof of this is as follows:

Let a<c<b be the domain of f. If δ<0 such that ε<0 then:

|f(x)-l|<δ

where 0<|x-c|<ε.

The epsilon-delta stuff is more of a definition that you can use to prove a function has a limit at a given argument. Also delta and epsilon are small but positive (i.e. e,d>0) and the distance from f(x) to the limit should be smaller than epsilon when the distance from x to the point c on the interval (a,b) is smaller than delta.

This picture helps:

If you take a value of x in the green strip (so within distance delta from c) and you take the function value at x then it should be within the red strip (within distance epsilon from f) if f has a limit at c.

Are you taking a calculus class?

 

[Polymath257]

I recently read the rigorous mathematical proof of limits and I thought I would reiterate it here:

Suppose f is a function whose domain contains two neighboring intervals: f: (a,c)∪(c,b)->R. We wish to consider the behavior of f as x approaches c. If f approaches a particular finite value l as x approaches c, then we say that the function f(x) has limit l as x approaches c. We write:

Lim x->c F(x) = L.

The proof of this is as follows:

Let a<c<b be the domain of f. If δ<0 such that ε<0 then:

|f(x)-l|<δ

where 0<|x-c|<ε.

Nope. You clearly attempted the epsilon, delta *definition* of the limit of a function, but got it very messed up.

1. it is the definition of the concept of a limit, not a proof. You can use the definition in a proof, of course.

2. you want epsilon>0 and delta>0. You have them negative, not positive.

3. The quantifiers are important!

The actual definition of 'f(x) has a limit of L as x approaches c' is:

for every epsilon>0,
there exists a delta>0 such that
whenever 0<|x-c|<delta,
we have |f(x)-L|<epsilon.

In what you wrote, there is no clear connection between epsilon and delta. In the actual definition, delta is a function of epsilon: for every epsilon, there is a delta.

Next, we want |f(x)-L|<epsilon *whenever* 0<|x-c|<delta.

You had the epsilon and delta interchanged and forgot an important quantifier on the x.

(Quantifiers state existence of something or that something always happens: for every epsilon, there exists a delta, such that for every x with 0<|x-c|<delta, we have |f(x)-L|<epsilon)

This is a typical place where students have difficulty. Usually, students are bad at basic logic, and, in particular, have not had to deal with nested quantifiers much before this definition. This makes the epsilon/delta definition a right of passage. Once you have learned 'epsilonics', you have made it past the *first* hurdle to real mathematics.

An important exercise: using the epsilon/delta definition, show the function defined by f(x)=1 for x>0 and f(x)=-1 for x<0 has no limit as x-->0.

Hint: you need to find an epsilon that 'works' for every delta. What is it? More specifically, what is the *largest* epsilon that works to show the limit does not exist?