A New Operation

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

With the caveat that domains need to be aligned, we have many operations on functions that make up an arithmetic on functions...

$\blacktriangleright$ Addition

The sum of two functions is again a function.

The additive identity is the zero function, $Zero(x) = 0$ for all real numbers.

For each function, $f$ , there is another function $-f$ , such that $f + (-f) = 0$ . $f$ and $-f$ are inverses of each other with respect to addition.

$\blacktriangleright$ Multiplication

The product of two functions is again a function.

The multiplicative identity is $One(x) = 1$ for all real numbers.

For each function, $f$ , there is another function $\frac {1}{f}$ , such that $f \cdot \left ( \frac {1}{f} \right ) = 1$ , again domain restrictions might be needed to avoid problems. $f$ and $\frac {1}{f}$ are inverses of each other with respect to multiplication. (If we were talking about numbers, then we might also use exponential notation $\frac {1}{n} = n^{-1}$ .

$\blacktriangleright$ Composition

Composition is a new operation on functions.

The composition of two functions is again a function.

The identity function, $Id(x) = x$ for all $x$ , is the identity element with respect to composition.

$f \circ Id = f \, \text { and } \, Id \circ f = f$

The inverse of $f$ with respect to composition is another function whose composition with $f$ produces the identity function.

Stealing the exponential notation from numbers, the symbol for the inverse of $f$ is $f^{-1}$ .

$f \circ f^{-1} = Id \, \text { and } \, f^{-1} \circ f = Id$

If we write a compopsition in function notation, $(f \circ g)(x) = f(g(x))$ , then we see that the algebra of composition is replacement.

Composition

Let $K(u) = \frac {u}{u-1}$ with its natural or implied domain: $(-\infty , 1) \cup (1, \infty )$ .
Let $T(w) = w^2 + 5w + 7$ with its natural or implied domain: $\mathbb {R}$ .

Form the composition $K \circ T$ .

$(K \circ T)(w) = \frac {w^2 + 5w + 7}{(w^2 + 5w + 7)-1} = \frac {w^2 + 5w + 7}{w^2 + 5w + 6}$

$K$ can accept any number, except $1$ . Therefore we need to find out when $T = 1$ and take out the domain numbers where $1$ occurs.

$\begin{align*} T(w) & = 1 \\ w^2 + 5w + 7 & = 1 \\ w^2 + 5w + 6 & = 0 \\ (w+2)\left( \answer{w+3} \right) & = 0 \end{align*}$

We need to remove $-2$ and $\answer {-3}$ from the domain of $K \circ T$ , because $T(-2)=1$ and $T(-3)=1$ and $1$ cannot be an input into $K$ .

The domain for $K \circ T$ is $(-\infty , -3) \cup (-3, -2) \cup (-2, \infty )$

$(K \circ T)(w) = K(T(w)) = \frac {w^2 + 5w + 7}{w^2 + 5w + 6} = \frac {w^2 + 5w + 7}{(w+2)(w+3)}$

The composition is a function.

This is a good reminder that a formula is not a function. There are always domain considerations. In the case of a composition, we need to restrict the domain of the inner function to avoid inner function values that are not in the domain of the outer function.

Composition

Let $g(x) = \frac {x-3}{x+1}$ with its natural or implied domain: $(-\infty , -1) \cup (-1. \infty )$ .
Let $H(t) = \frac {t+3}{1-t}$ with its natural or implied domain: $(-\infty , 1) \cup (1. \infty )$ .

$(g \circ H)(y) = g(H(y)) = \frac {\left ( \frac {y+3}{1-y} \right ) - 3}{\left ( \frac {y+3}{1-y}\right ) + 1} = \frac {y+3-3(1-y)}{y+3+(1-y)} = \frac {4y}{4} = y$

$g$ and $H$ are inverse functions.

Of course, that isn’t the whole story. As we said earlier, formulas are not functions.

$(g \circ H)(r) = r$ and so $g \circ H = Id$ , the identity function. But remember our caveat. There are always domain issues. In this case $(g \circ H)(r) = r$ for almost all of the real numbers.

We know that,

$(g \circ H)(r) = \frac {\left ( \frac {r+3}{1-r} \right ) - 3}{\left ( \frac {r+3}{1-r}\right ) + 1}$

This is equivalent to $r$ for almost all real numbers.

In this case, $(g \circ H)(r) = g(H(r))$ , which means that $r \ne 1$ , since $1$ is not in the domain of $H$ . Secondly, $-1$ is not in the domain of $g$ . We also cannot have values of $r$ that make $H(r) = -1$ .

$H(r) = \frac {r+3}{1-r} = -1$

$r + 3 = -(1-r) = -1 + r$

$3 = -1$

There are no such real numbers.

Wait. There is more to this story. We would like the other way as well: $(H \circ g)(z) = z$ and so $H \circ g = Id$ .

Thus, $z \ne -1$ . We must also avoid $g(z) = 1$ , because $1$ is not in the domain of $H$ .

$g(z) = \frac {z-3}{z+1} = 1$

$z - 3 = z + 1$

$-3 = 1$

There are no such real numbers.

$\blacktriangleright$ Therefore, $g$ and $H$ are inverse functions on $(-\infty , 1) \cup (-1, 1) \cup (1,\infty )$ .

1 The Other Half of the Story

We would like our function arithmetic to mimic our arithmetic for numbers. For numbers, the inverses are commutative.

$4 + (-4) = (-4) + 4 = 0$

$4 \cdot \frac {1}{4} = \frac {1}{4} \cdot 4 = 1$

In our example, we also want $(H \circ g)(z) = H(g(z)) = z$ . It does.

$(H \circ g)(z) = \frac {\left ( \frac {z-3}{z+1} \right ) + 3}{1 - \left ( \frac {z-3}{z+1}\right )} = \frac {z-3 + 3(z+1)}{z+1-(z-3)} = \frac {4z}{4} = z$

In this case, we cannot have $z = -1$

$\blacktriangleright$ Therefore, $g$ and $H$ are inverse functions on $(-\infty , -1) \cup (-1, 1) \cup (1, \infty )$ .

ooooo=-=-=-=-=-=-=-=-=-=-=-=-=ooOoo=-=-=-=-=-=-=-=-=-=-=-=-=ooooo
more examples can be found by following this link
More Examples of Function Algebra