hi all
help me plz
proof that if f(x) is a function form natural numbers to natural numbers and f(f(f(x))) is a strictly increasing polynomial f(x) is also a stricly increasing polynomial

hi all
help me plz
proof that if f(x) is a function form natural numbers to natural numbers and f(f(f(x))) is a strictly increasing polynomial f(x) is also a stricly increasing polynomial
This looks like homework. Is it?
If so, you can expect no help here. Just a hint with notation, though: f(x) is NOT a function, it is the image of the object x in the domain under the "action" of the function f . i.e. f(x) is an object in the codmain, not a function.
yeah wrong notation sorry but it's no homework because it's still summer holidays i just got this from a book and i'm trying to solve it
i can tell you how far i got:
i know that f is injective since f^3 (the composition) is injective and now i'm trying to prove that f has to be stricly increasing but since f is probably not continous i had no working idea how
Since the domain and rainge of f are the natural numbers, discrete topological spaces, it s a dead nuts certainty that f is continuous.Originally Posted by memmeth
That will not help much.
However, if one looks at the question from the point of view of abstract algebra one would quite naturally blook at f(x) as a an abstract polynomial in x and hence the notation is quite natural.Originally Posted by Guitarist
Well, notation is arbitrary, as they say, but it is certainly useful to distinguish between a polynomial form and a polynomial function ; in the former case, is a polynomial in the indeterminate whereas in the second, is an element in the domain, and is an element in the codomain.
The distinction arises as follows:
Writing as the domain of a polynomial function, as the codomain, every polynomial form determines a unique polynomial function , but the converse need not be true.
Like, suppose that over the polynomial functions are equal as functions () and and yet and are distinct forms (), and yet, since both these forms solve to zero. As a matter of fact this last equality is true for any prime, by a theorem of Fermat
In other words, equality has a different meaning for forms and functions
For that reason, I think it preferable to use different notations.
You are being overly pedantic and quite silly for the application at hand.Originally Posted by Guitarist
We are not working over fields of characteristic other than zero here.
There is no need to confuse the OP with such subtleties that are not germane to the problem at hand.
Please try to avoid derogatory personal remarks in this forumOriginally Posted by DrRocket
As it happens, my post was entirely and exclusively to do with notation, which I admitted in my first sentence was arbitrary.There is no need to confuse the OP with such subtleties that are not germane to the problem at hand.
Let the OP decide whether or not my post was in any way helpful for their original question. If not, maybe (s)he might have nonetheless found it mildly interesting and want to discuss it further. Or, for that matter, maybe any one else might. Let them decide.
Please do not try to direct the way in which discussions develop on this forum.
Please do not put your own spin on my comments, or misinterepret them to fit your own agenda.Originally Posted by Guitarist
The major difficulty in proving this assertion is that it is false.Originally Posted by memmeth
Consider the following function f.
On {1,2,3} f is defined as follows: f(1)=2, f(2)=3, f(3)=1.
Then f(f(f(x))) =x for x=1,2,3
For the remainder of the natural numbers let f(x)=x for x> 3.
f(f(f)x)) = x for all natural numbers, f^3 is therefore strictly increasing and a polynomial but f is neither a polynomial nor strictly increasing.
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