Is it possible to find the $n$th digit of $\pi$ (in base $10$)? Is it possible that there exists some function $f:\mathbb N_1\to \{0,1,2,3,4,5,6,7,8,9\}$, where $$f(1)=\color{red}1, f(2)=\color{red}4, f(3)=\color{red}1, f(4)=\color{red}5, f(5)=\color{red}9, f(6)=\color{red}2, \ldots \quad ?$$
I know that such a function, if it exists, cannot be a polynomial, since all polynomials are finite-dimensional, whereas the sequence containing the digits of $\pi$ has infinitely many elements, but could it be that there is some function (even if we cannot explicitly state it) which gives the $n$th  term of the sequence containing the digits of $\pi$?
If this is true, and such a function does exist, can the same be said for all transcendental numbers?

Edit: to be absolutely clear, I am looking for the explicit function $f$ such that $f(n)$ is the $n$th digit of $\pi$ (in base $10$).
 A: Yes, of course that function exists. You have just defined it.
If you want a symbolic-looking expression for your function, then how about
$$ f(n) = \lfloor 10^n\pi \rfloor \bmod 10 $$

The larger point, however, seems to be that you're using the word "function" in a much narrower sense than (contemporary) mathematics does. It looks like you want a "function" to be a symbolic expression, but in mathematics a function is simply anything which associates one and only one output value for each input value in its domain. It is immaterial how this association is done, as long as we're sure that it gives an output for every input, and that the output depends only on what the input is.
If you like set-theoretic formalizations, then a function $X\to Y$, where $X$ and $Y$ are sets, means a set of pairs with the first component being a member of $X$ and the second component being a member of $Y$ (that is, $f\subset X\times Y$), satisfying the conditions that:


*

*for every $x\in X$ there is an $y\in Y$ such that $(x,y)\in f$ (every input has an output).

*for every $x,y,z$, if $(x,y)\in f$ and $z\ne y$, then $(x,z)\notin f$ (there is only one output for each input).


It doesn't where we got the set from. In your case the function $f:\mathbb N\to\{0,1,2,3,4,5,6,7,8,9\}$ is
$$ f = \{ (n,d) \mid \text{The }n^{\text{th}}\text{ decimal of }\pi\text{ is }d\} $$
a perfectly cromulent function.
A: What exactly means explicit? For a given $n$, there is certainly an algorithm that computes the $n$'th digit of $\pi$. 
But the answer is no if you ask about some other real numbers. There are non-computable numbers that encode the halting problem, for example. There is no chance to have an algorithm computing the digits of some numbers in general -- and hence not any reasonable "formula" composed of common basic operations for those numbers.
