# Which of these claims are true?

Consider the sequences of numbers $$\left\{0, 1, 2\right\}$$ with length $$n$$. There are $$3^n$$ such sequences. I define each sequence like a function. If a function consists of {0,1,2} elements of the length $$«n»$$, let's consider this function $$\phi (n).$$

Because, we can deduce all sequences for only finite number $$n$$, If we accept any infinity sequence equal to a specific function, problematic points occur. Because, we have infinitely number of function, which is we can not deduce all functions.

Definition: In mathematics, a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., nth root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but usually no limit.

For infinity sequence , which consist of elements $$\left\{0,1,2\right\}$$, if we have a specific $$n-$$th term any formula, we have a "specific mathematical function" (algebraic closed form expression or non-elementary function), which is we are looking for.

I can choose an infinitely number of functions that are Closed-Form Expression in term of a $$"n".$$

Examples: let $$n\in\mathbb{Z^{+}} \bigcup \left\{0\right\}$$

$$\phi(n)=n+2-3\lfloor \frac{n+2}{3}\rfloor=\left\{0,1,2,0,1,2\cdots \right\}$$

$$\phi(n)=n+1-3\lfloor \frac{n+1}{3}\rfloor=\left\{ 2,0,1,2,0,1\cdots \right\}$$

$$\phi(n)=n-3\lfloor \frac{n}{3}\rfloor=\left\{ 1,2,0,1,2,0\cdots\right\}$$

$$\phi(n)=n^n-3\lfloor \frac{n^n}{3}\rfloor=\left\{1,1,0,1,2,0,1,1,0,1,2,0\cdots \right\}$$

$$\phi(n)=n^n+n-3\lfloor \frac{n^n+n}{3}\rfloor=\left\{2,0,0,2,1,0,2,0,0,2,1,0\cdots \right\}$$

$$\phi(n)=\lfloor 10^n \pi \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \pi \rfloor }{3}}\rfloor$$

$$\phi(n)=\lfloor 10^n e \rfloor - 3 {\lfloor \frac{ \lfloor 10^n e \rfloor }{3}}\rfloor$$

$$\phi(n)=\lfloor 10^n \sqrt2 \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \sqrt2 \rfloor }{3}}\rfloor$$

$$\phi(n)=\lfloor 10^n \log \pi \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \log \pi \rfloor }{3}}\rfloor$$

$$\cdots \cdots \cdots \cdots \cdots$$

For these periodic and non-periodic sequences there are exist $$n-$$th term "closed-form expression."

Then, we can define an infinitely number of "specific mathematical functions", (non-elementary, non-algebraic) which is non-periodic.

Example:

$$\phi(n)=\lfloor 10^n \displaystyle\int_0^\infty e^{-x^n}dx \rfloor - 3 {\lfloor \frac{ \lfloor 10^n \displaystyle\int_0^\infty e^{-x^n}dx \rfloor }{3}}\rfloor$$

$$\cdots \cdots \cdots \cdots \cdots$$

Claims: (only for infinitely number sequences)

A) There exist infinitely number of sequences that, consist of elements $$\left\{0,1,2\right\}$$, which is can not express by the any "closed-form expression" or any "specific mathematical function".

B) There exist infinitely number of sequences that, consist of elements $$\left\{0,1,2\right\}$$, which is can express by the any closed-form expression.

C) There exist infinitely number of sequences that, consist of elements $$\left\{0,1,2\right\}$$, which is can express by the any "specific mathematical function" (non-elementary, non-algebraic).

Which of these claims are true? I'm looking for a proof that confirms or denies the claims.

Thank you!

• It is hard to understand your notations since it seemed inconsistent: at the beginning you define $\phi_k(n)$ as a sequence of length $n$, and then at some point it becomes an infinite sequence. If you want to ask whether there is an infinite sequence of $\{0,1,2\}$ which cannot be represented as closed form, I will say yes: Such sequence is uncountable, while closed form expressions of finite length is countable. – Hw Chu Apr 23 at 19:49
• What previous comments are saying is that $3^n$ applies only to finite sequences. As soon as you start with the "$\cdots$" or talking about non-periodic sequences you are talking about infinite sequences and we throw the $3^n$ out the window. You are better off never to mention it at all. – David K Apr 23 at 19:59
• Assuming infinite sequences, for Claim 1, you've already given a mathematical formula for any sequence, provided that in each case that $\alpha$ itself has a formula. Claim 2 is true because there are infinitely many irrational numbers that have no formulas to express the numbers themselves. – David K Apr 23 at 20:02
• I gave up after the first couple of paragraphs. The question is incomprehensible. If it didn't have a bounty, I'd vote to close as unclear. – Gerry Myerson Apr 26 at 5:07
• It's not a question of mistakes – what you have posted is a word salad. – Gerry Myerson Apr 26 at 7:59

## 4 Answers

Others have already addressed the claims. Here I discuss a bit about finite vs infinite sequences, as you seem to be a bit confused.

• For any fixed $$n$$, there are $$3^n$$ sequences of length $$n$$ with symbols from $$S = \{0,1,2\}.$$

• For any fixed $$n$$, there are $$3^n$$ functions from $$\{1,2,...,n\}$$ to $$S$$. Note that the function domain (possible input values) is only the set $$\{1,2,...,n\}$$, not all $$\mathbb{N}$$ (all positive integers). E.g. if $$n=4$$ then $$f(7)$$ is undefined, since the input value is invalid for this $$f$$. These $$3^n$$ functions are in bijective mapping with the $$3^n$$ sequences: the sequence $$(f(1), f(2), ..., f(n))$$ fully specifies $$f$$.

• If you union together all the sets of all finite sequences (of any length), the resulting set is of course infinite. It is countably infinite.

• Similarly, if you union together all the sets of all functions from a finite domain $$\{1,2,...,n\}$$ to $$S$$, the resulting set is again countably infinite, and again in bijective mapping with the set of finite sequences.

Nothing above involve infinite sequences. Here they come now:

• Any infinite sequence is $$(s_1, s_2, s_3, ...)$$ where each $$s_i \in S$$. The set of all infinite sequences is of course infinite, but it is uncountably infinite.

• The set of functions $$f: \mathbb{N} \rightarrow S$$ is in bijective mapping with the set of infinite sequences, because again $$(f(1), f(2), ...)$$ fully specifies the function. This set of function is again uncountably infinite.

So, the status of your claims depends on whether you mean finite or infinite sequences, or equivalently, functions with finite domains (of the form $$\{1,2,...,n\}$$ for some $$n$$) or functions with domain $$\mathbb{N}$$.

If you include infinite sequences, then the answers are (A) Yes, (B) Yes, (C) Yes -- see the answer by Ross Milikan, for example.

If you mean only finite sequences (remember: there are still countably infinite such finite sequences), then, IMHO every such finite sequence has a closed form expression because IMHO the usual written form of the finite sequence itself is a closed form expression -- it is made from these five symbols: $$\color{red}{0,1,2,(}$$ and $$\color{red}{)}$$. Thus the answers would be (A) No, (B) Yes, (C) Yes.

• @Student - you're welcome! one key idea is that a function is just a mapping from input to output. there does not need to be an explicit formula (by any definition of "explicit"). and as Austin Mohr (and more briefly, Ross Milikan) pointed out, there are only countably-infinite explicit formulas. this is because any reasonable definition of "explicit" or "closed form" or "specific function (elementary or non-elementary)" must include "can be described by a finite set of mathematical and/or natural-language symbols in a certain arrangement" – antkam Apr 26 at 14:22
• @Student - then the other key idea is that, if $\mathbb{N}$ is function domain (equivalently: if each sequence is infinitely long), the number of functions (no. of infinitely-long sequences) is uncountably-infinite, which is strictly bigger than countably-infinite. this is the famous diagonal argument. from these you can derive answers to (A), (B), (C) – antkam Apr 26 at 14:26
• @Student - Personally I differentiate between a beginner who is rude / not trying (e.g. many many posters who just want answers to their homework) vs a beginner who is trying hard, even if not very successfully (e.g. you). And I would never downvote the latter. Anyway, glad you found all these useful! – antkam Apr 26 at 16:21
• I dont understand your first question. The second question: answer is no. Putting some technicalities aside, $\{0,1\}^\infty$ is "the same" as the real numbers in the interval $[0,1]$: map such a real number to its binary expansion. Similarly, $\{0,1,2\}^\infty$ can be a real number's ternary (base $3$) expansion. – antkam Apr 28 at 15:33
• If you are interested in these issues, google Cantor and Cardinal numbers and diagonal argument. or maybe start with this famous hotel – antkam Apr 28 at 15:35

Claim A is true because there are uncountably many sequences but only countably many closed formulas.

Claim B is true because there are infinitely many closed form expressions. One class of them is the base $$3$$ representations of the natural numbers.

I can't understand claim C.

• Is it possible answer me this question in a comment..? For infinity sequence: consist of only elements $\left\{0,1\right\}$ A) if the length of all each sets is $n$, then we have $2^n$ uncountable sets B) if the length of all each sets is $2 ^ n$, then we have $2^{2^n}$ uncountable sets . C) if the length of all each sets is $2 n$, then we have $4^n$ uncountable sets. Then can we say the set B is more uncountable than sets A. And the sets C is more uncountable than sets A ? Thank you very much ... – Elvin May 19 at 15:36
• You first talk about infinite sequences, then you talk about length $n$. These are inconsistent. Yes, if the length is $n$ there are $2^n$ sequences, but that is finite if $n$ is finite, not uncountable. If you have sequences of $\{0,1,2,3,\}$ of length $n$, yes there are $4^n$. If you have countably infinite sequences, there are $\mathfrak c=2^{\aleph_0}$ of them whether they are sequences of $2, 4,$ or even countably many symbols. – Ross Millikan May 19 at 16:21

If it concerns the enumeration of the decimals of irrationals, Claim 1 is false.

Indeed

$$\phi_6(n)=\lfloor\pi\cdot10^n\rfloor\bmod3$$ fulfills the specification and is a "mathematical formula". Same holds for other constants.

• Is Claim 2 is also wrong? – Elvin Apr 23 at 20:21
• @Student: this was answered by Hw Chu. – Yves Daoust Apr 23 at 20:24
• How can we calculate $(\pi×10^5)\mod3$ ? – Elvin Apr 23 at 20:31
• @Student: oooops, made a typo. Now fixed. In case the floor function is "forbidden", it can be "emulated" using $\arctan(\tan(x))$. – Yves Daoust Apr 23 at 20:37
• In short answer, "is it possible to express any infinity sequence which is consist of only $\left\{0,1,2\right\}$ with a closed form formula?" Answer is NO. Is it correct? – Elvin Apr 24 at 16:28

You may know this already, but there are two important types of infinite sets: countable and uncountable. For our purposes here, it's enough to know that countable sets are much, much smaller than uncountable ones. One way to put it is that any countable set is $$0\%$$ as large as any uncountable one.

As others have mentioned, there are uncountably-many sequences using the characters $$0$$, $$1$$, and $$2$$. (In fact, there are uncountably-many sequences using just $$0$$ and $$1$$.)

Now, pick your favorite language in which to express mathematical truths subject to the following two rules:

1. The language must contain only finitely-many different kinds of symbols (so I can comprehend the language with my finite brain).
2. Any expression you make with the language must have finite length (so I have enough time to read the expression with my finite life).

For example, my favorite language includes every mathematical symbol I've ever encountered and every English character (it's the only tongue I speak). From this language, I can produce not only the closed-form expressions and mathematical functions you ask about, but I can also write computer programs to produce sequences or construct English sentences to describe sequences.

Once you have decided on a language, define the describable sequences to be the set of all sequences that you can describe with your language. The set of describable sequences is countable! Not only is it impossible to describe all the sequences, but we can only describe $$0\%$$ of them.

• I've written all this and possibly misunderstood a crucial fact: Are your sequences finite in length or not? – Austin Mohr Apr 26 at 0:38
• We have infinity sequences. (0...,1.. ,2...) – Elvin Apr 26 at 4:17
• In that case, my answer is relevant: There are uncountably-many infinite-length sequences on $\{0, 1, 2\}$, but there are only countably-many possible descriptions. – Austin Mohr Apr 26 at 17:23
• I'm so grateful for the answer. Because I learned a lot from this answer. Right now, I'm surprised at who I'm gonna give a bounty to. I got great answers. – Elvin Apr 27 at 9:16