Definition:Real Function/Two Variables/Substitution for y

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Definition

Let $S, T \subseteq \R$ be subsets of the set of real numbers $\R$.


Let $f: S \times T \to \R$ be a (real) function of two variables:

$z = \map f {x, y}$

Then:

$\map f {x, a}$

means the real function of $x$ obtained by replacing the independent variable $y$ with $a$.

In this context, $a$ can be:

a real constant such that $a \in T$
a real function $\map g x$ whose range is a subset of $T$.


Examples

Example: $\map f {x, y} = x^2 + x y^2 + 5 y + 3$

Let $\map f {x, y}$ be the real function of $2$ variables defined as:

$\forall \tuple {x, y} \in \R^2: \map f {x, y} := x^2 + x y^2 + 5 y + 3$


Substituting $2$ for $y$

Let $2$ be substituted for $y$ in $\map f {x, y}$.

Then:

\(\ds \forall x \in \R: \, \) \(\ds \map f {x, 2}\) \(=\) \(\ds x^2 + x \times 2^2 + 5 \times 2 + 3\)
\(\ds \) \(=\) \(\ds x^2 + 4 x + 13\) simplifying


Substituting $a$ for $y$

Let $a$ be substituted for $y$ in $\map f {x, y}$.

Then:

\(\ds \forall x \in \R: \, \) \(\ds \map f {x, 2}\) \(=\) \(\ds x^2 + x \times a^2 + 5 \times a + 3\)
\(\ds \) \(=\) \(\ds x^2 + a^2 x + 5 a + 3\) simplifying


Substituting $\map g x$ for $y$

Let $\map g x$ be substituted for $y$ in $\map f {x, y}$, where $\map g x$ is a real function defined on all $\R$.

Then:

\(\ds \forall x \in \R: \, \) \(\ds \map f {x, \map g x}\) \(=\) \(\ds x^2 + x \times \paren {\map g x}^2 + 5 \map g x + 3\)


Example: $\map f {x, y} = x + y$

Let $\map f {x, y}$ be the real function of $2$ variables defined on the domain $S \times T$ as:

$\forall \tuple {x, y} \in S \times T: \map f {x, y} := x + y$

where $S$ and $T$ are the closed real intervals:

\(\ds S\) \(:=\) \(\ds \closedint {-1} 1\)
\(\ds T\) \(:=\) \(\ds \closedint 0 2\)


Substituting $\dfrac 1 2$ for $y$

Let $\dfrac 1 2$ be substituted for $y$ in $\map f {x, y}$.

Then:

\(\ds \forall x \in \closedint {-1} 1: \, \) \(\ds \map f {x, \dfrac 1 2}\) \(=\) \(\ds x + \dfrac 1 2\)


Substituting $3$ for $y$

Let $3$ be substituted for $y$ in $\map f {x, y}$.

Then $\map f {x, 3}$ is undefined, as $3 \notin \closedint 1 2$.


Sources

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