SOME ILLUSTRATIONS ON COMPOSITION OF FUNCTIONS.
1) Let A = {
f(x) =
;
≠ 0 and 1} and consider the map f: A→A given by the formula
. We claim
a) f is a function from A to A. In fact, given x in A, since x ≠ 1 we see that 1−x is not zero, so its reciprocal
exists and is a fraction. But we need to verify that
(which is clear!) and
falls in A. That is, that
≠ 1. To see that this inequality holds, observe that
is not zero
= 1 only when 1−x
=1, so that, when x =0, which can not be the case since x (the input) was in A. Therefore, f(x) lands in A
for each x ∈ A, as claimed. Thus, f is a function from A to A.
b) f
is one-to-one. In fact, if a ≠ a’ then 1−a ≠ 1−a’ and so
≠
, that is
f(a) ≠ f(a’). We have proved that f is one-to-one.
Let us compose f with itself several times, that is, let us find f f, f f f, etc
If x ∈ A we have f f (x) = f ( f(x) ) = f (
)=
and so f f f (x) = f ( f f (x) ) = f (1− ) =
=
=
=
=1− ;
= x. Therefore f f f fixes every
x and so f f f = identity map. The cycle thus repeat….By the way, observe that f is invertible (i.e.,
bijective) with f = f f
2) Consider the map f:{1, 2,…, n,…}→ {1, 2,…, n,…}given by the formula
f(x) =
if x is even, and f(x) =
if x is odd.
a) Find f(x) for several values of x of your choice.
b) Next, take any x and apply f as many times as you want. For instance, start with x = 6. The effect of
applying f several times can be given in a compact way as: 6→3→5→8→4→2→1. If you begin with x =
9 you will obtain the sequence 9→14→7→11→17→26→13→20→10→5→8→4→2→1 Do the same
with some other x’s of your choice. If you do not make mistakes you will notice that no matter with
number you chose to start with, eventually you arrived to 1.(The number of steps to arrive down to 1 is
not fixed, though; look at the examples above) That this is always the case with this function is a
conjecture, an open, unresolved problem to date. What you have done is to verify the validity of the
conjecture only in some cases; you have not proved that the conjecture is true.
3) Remember: the order of the composition of functions is important. Therefore, gof means
“apply f first, then apply g”. That is, gof (x) = g(f(x))
Consider the following functions:
⎛ 30⎞
f(n) = ⎜ ⎟ defined for n = 0, 1, . . . , 30
⎝ n ⎠
g(k) = units digit of k
€ h = gof
H = fog
Compute
a) h(28), h(h(28)), h(h(h(28))), etc
b) H(28) and H(H(28)).
Also,
What is the maximum value (that is, the maximum output) taken
i)
by f?
ii)
by g?
iii)
by h?
iv)
by H?
4) Consider the functions defined by
f(n)= largest prime into the decomposition of n (for each n >1)
g(k) = 3k+5, for all k integer
h = gof
Compute h(74309125) and h(h(74309125))