A New Operation

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composition

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 $(-\infty , 1) \cup (-1, 1) \cup (1,\infty )$.

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 $(-\infty , -1) \cup (-1, 1) \cup (1, \infty )$.

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more examples can be found by following this link
More Examples of Function Algebra

2025-01-07 01:05:07