Two Questions Regarding a Corollary about Uniqueness

[Science Student]

Here's the corollary: Let X be a nonempty set and f : N×X → X be a function. For each x ∈ X there exists a unique sequence x(1), x(2), ... such that x(1) = x and x(n+1) = f(n,x(n)).

 

Question 1: What exactly does x(1) = x mean? In the same context, what would x(2) equal?

 

Question 2: What does the second part of this equation mean x(n+1) = f(n,x(n))? Is it an ordered pair for an x-y graph or...?

 

 

[John]

I'm assuming these are questions you're asking for yourself, and not questions you have to answer for an assignment. Mods forgive me if I'm incorrect.

I'm assuming N is supposed to be the natural numbers. We have some function f that has the Cartesian product of N and X (i.e. all the ordered pairs (n, x) for n in N and x in X) as its domain and X as its codomain.

 

The corollary states that if we take any x in X, then there exists a unique sequence starting with x (this is what x(1) = x means) and defined by the recurrence relation x(n + 1) = f(n, x(n)), or alternatively x(n) = f(n - 1, x(n - 1)). Thus, letting x(1) = x, we have that x(2) = f(1, x(1)), x(3) = f(2, x(2)), etc. The precise form of this sequence will depend on the definition of f. All we know is that each term in the sequence must be an element of X (since X is the codomain of f).

As an example, let X be the rational numbers, and let f be defined such that f(n, x) = nx. Then, taking x = 3/5, we have that

 

x(1) = 3/5,

x(2) = f(1, x(1)) = (1)(3/5) = 3/5,

x(3) = f(2, x(2)) = (2)(3/5) = 6/5,

x(4) = f(3, x(3)) = (3)(6/5) = 18/5,

x(5) = f(4, x(4)) = (4)(18/5) = 72/5,

 

and so on.

Edited December 16, 2013 by John

[Science Student]

I'm assuming these are questions you're asking for yourself, and not questions you have to answer for an assignment. Mods forgive me if I'm incorrect.

 

I'm assuming N is supposed to be the natural numbers. We have some function f that has the Cartesian product of N and X (i.e. all the ordered pairs (n, x) for n in N and x in X) as its domain and X as its codomain.

 

The corollary states that if we take any x in X, then there exists a unique sequence starting with x (this is what x(1) = x means) and defined by the recurrence relation x(n + 1) = f(n, x(n)), or alternatively x(n) = f(n - 1, x(n - 1)). Thus, letting x(1) = x, we have that x(2) = f(1, x(1)), x(3) = f(2, x(2)), etc. The precise form of this sequence will depend on the definition of f. All we know is that each term in the sequence must be an element of X (since X is the codomain of f).

 

As an example, let X be the rational numbers, and let f be defined such that f(n, x) = nx. Then, taking x = 3/5, we have that

 

x(1) = 3/5,

x(2) = f(1, x(1)) = (1)(3/5) = 3/5,

x(3) = f(2, x(2)) = (2)(3/5) = 6/5,

x(4) = f(3, x(3)) = (3)(6/5) = 18/5,

x(5) = f(4, x(4)) = (4)(18/5) = 72/5,

 

and so on.

These are my own questions to make sense of the corollary. Unfortunately, my professor does not seem to think that additional information is needed for the not-so-genius math students. Thank-you so much!

[Olinguito]

Here's the corollary: Let X be a nonempty set and f : N×X → X be a function. For each x ∈ X there exists a unique sequence x(1), x(2), ... such that x(1) = x and x(n+1) = f(n,x(n)).

 

For each [latex]x\in X[/latex], let [latex]x_1=x[/latex]. Then let [latex]x_2=f(1,x_1)[/latex]. Then let [latex]x_3=f(2,x_2)[/latex], [latex]x_4=f(3,x_3)[/latex], [latex]x_5=f(4,x_4)[/latex], … and so on. This is what’s called a recursively defined sequence. The existence of such a sequence is obvious; the corollary asserts that it is the only one possible.

 

By the way, could you state the theorem or proposition to which this result is a corollary?