Understanding church numerals I am learning lambda calculus as my previous questions indicate . But in a different book , namely , The structure and interpretation of  computer program I cam across the concept of "church numerals " and the stumbled upon the following link about church numerals :

https://www.cs.rice.edu/~javaplt/311/Readings/supplemental.pdf

It goes on stating the following in the first page :

The Church numerals that follow just have additional applications of the successor function:

$$\begin{align}
C_2 &amp;= \lambda f . \lambda x . f(fx) \\ 
C_3 &amp;= \lambda f . \lambda x . f(f(fx))\\
C_4 &amp;= \lambda f . \lambda x . f(f(f(fx)))\\
&amp;\vdots\\
C_n &amp;= \lambda f . \lambda x . f nx
\end{align}
$$

and then he mentions that had we been having a little more powerful language then we could define the following two functions :
$$\begin{align}
S &amp;= \lambda r.\text{“ring the small bell (ding) and apply r”}\\
Z &amp;= \lambda r.\text{“ring the big bell (dong)”}
\end{align}
$$
and then says that the application of  

$C_4$ to  $S$ and $Z$ would yield "ding ding ding ding dong"

How to explain the above result without even knowing what "f" is ?

POST EDIT : 
After I got the first answer , I tried computing the sequence by the help of clue provided . 

By applying step by step I will get (lambda r.ding r)^4 (lambda r.dong) = (lambda r.ding r)^4 dong .

Now each evaluation of (lambda r.ding r) leads to "ding r " .

So the overall evaluation becomes "ding ding ding ding r dong ". 

Am I doing it right ?If yes then , how to account for omission of r in the expression?

P.S :
For some trivia information , has there been any known application of this idea of church numerals to any field of mathematics ?
Using your definition for Church Numerals and trying to see what's going on:

$$ \begin{align} 0 &amp;= \lambda f. \lambda x. x \\
                 1 &amp;= \lambda f. \lambda x. f x\\
                 2 &amp;= \lambda f. \lambda x. f f x\\
                 3 &amp;= \lambda f. \lambda x. f f f x\\
&amp;\vdots
\end{align} $$

Now let's say have a function $g$ and some $a$, try to use $g$ and $a$ as arguments to $0$, that is:

$$ \begin{align} 0 ~ g ~ a &amp;= (\lambda fx. x)~g~a\\
\quad &amp;\rhd_{1\beta} ([g/f] (\lambda x.x)) a = (\lambda x.x) a \\
\quad &amp;\rhd_{1\beta} [a/x]x = a
\end{align} $$

So $g$ applied $0$ times to $a$ returns $a$.

Now try to do the same with $1$.

$$ \begin{align} 1 ~ g ~ a &amp;= (\lambda fx. f~x)~g~a\\
\quad &amp;\rhd_{1\beta} ([g/f] (\lambda x.f~x)) a = (\lambda x.g~x) a \\
\quad &amp;\rhd_{1\beta} [a/x]g~x = g ~ a
\end{align} $$

So $g$ applied $1$ time to $a$ is indeed $g(a)$.

Your problem is $4 ~ g ~ a$, where $g$ is $S$ and $a$ is $Z$.

$$ \begin{align}
S &amp;= \lambda r.\text{“ring the small bell (ding) and apply r”}\\
Z &amp;= \lambda r .\text{“ring the big bell (dong)}”
\end{align}
$$

Doing $1 ~ S ~ Z$ we already know that evaluates to $S(Z)$. And that translates to "print 'Ding' then apply $r$ ", where $r$ just prints 'Dong'. So it outputs 'Ding Dong'.