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...?

 

 

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, 201312 yr by 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.

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!

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?