Some Thoughts on a Real-Valued Function

Problem

Spending so much time to work out the locale support for dates and valid HTML code in my second blog which was powered by Jekyll-Bootstrap, I couldn’t recall this fact.

Suppose that $\mathcal{D}$ is a subset of $\R^2$ that contains an $\varepsilon$-neighbourhood of a point $(x_0,y_0)$. If

1. $f: \mathcal{D} \to \R$ is has first-order partial derivatives in the $\varepsilon$-neighbourhood of $(x_0,y_0)$.
2. The first-order partial derivatives of $f$ are continuous at $(x_0,y_0)$.

Then we can write

\begin{equation} \Delta f = \frac{\partial f}{\partial x} (x_0,y_0) \Delta x + \frac{\partial f}{\partial y} (x_0,y_0) \Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y, \label{fact} \end{equation}

where $\Delta f := f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0), \varepsilon_1$ and $\varepsilon_2 \to 0$ as $\Delta x$ and $\Delta y \to 0$.

The First-Order Approximation For a Differentiable Function

At first, I thought that equation \eqref{fact} followed from formula \eqref{thm}.

\begin{equation} \lim_{\vect{h} \to \zeros} \frac{f(\vect{x}+\vect{h})-[f(\vect{x})+\langle\nabla f(\vect{x}),\vect{h}\rangle]}{\norm{\vect{h}}} = 0 \label{thm} \end{equation}

I forgot that this statement assumed the continuous differentiability on an open set. However, we only know that the first-order partial derivatives are continuous at one point $(x_0,y_0)$.

Solve an easier problem first

This question should be much easier and much more intuitive if the domain of the function $f$ in \eqref{fact} is one-dimensional. Then, by drawing a curve and sketching its tangent line at a point, one can intuitively realise that the geometric meaning of $\varepsilon$.

In fact, one constructs

\begin{equation} \varepsilon (\Delta x) := \begin{cases} \frac{\Delta f}{\Delta x}-f'(x_0) &\text{if }\Delta x \ne 0,\\ 0 &\text{if } \Delta x = 0, \end{cases} \label{def1} \end{equation}

where $\Delta f := f(x_0+\Delta x)-f(x_0)$ in this section since $f$ is now one-dimensional.

Then one can make use of the differentiability of $f$ at $(x_0)$ to say that $\varepsilon \to 0$ as $\Delta x \to 0$.

To get the one-dimensional version of \eqref{fact}, we get rid of the denominator by multiplying both sides by $\Delta x$ in the case of $\Delta x \ne 0$.

\begin{equation} \begin{aligned} \varepsilon \Delta x &= \Delta f - f' (x_0) \Delta x\\ \Delta f &= f' (x_0) \Delta x + \varepsilon \Delta x \end{aligned} \label{mult1} \end{equation}

Observe that equality \eqref{mult1} also holds when $\Delta x = 0$.

Back to the problem

Make use of the previous section

From \eqref{def1}, we observe that $\varepsilon$ is defined as the difference between a difference quotient between two points $x_0$ and $x_0+\Delta x$ and the derivative $f'(x_0)$. We can define $\varepsilon_1$ and $\varepsilon_2$ in a similar way.

\begin{align} &\begin{split} & \varepsilon_1(\Delta x,\Delta y) :=\\ &\begin{cases} \frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x}-\frac{\partial f}{\partial x}(x_0,y_0) &{\small \text{if } \Delta x \ne 0},\\ 0 &\text{if } \Delta x = 0, \end{cases} \end{split} \label{def2a}\\ &\begin{split} & \varepsilon_2(\Delta x,\Delta y) :=\\ &\begin{cases} \frac{f(x_0+\Delta x,y_0+\Delta y) - f(x_0+\Delta x,y_0)}{\Delta y} - \frac{\partial f}{\partial y}(x_0+\Delta x,y_0) &\text{if } \Delta y \ne 0,\\ 0 &\text{if } \Delta y = 0, \end{cases} \end{split} \label{def2b} \end{align}

We then multiply \eqref{def2a} and \eqref{def2b} by $\Delta x$ and $\Delta y$ respectively, just like what we’ve done in \eqref{mult1}.

\begin{align} &\begin{split} \varepsilon_1(\Delta x,\Delta y) \Delta x =& f(x_0+\Delta x,y_0) - f(x_0,y_0)\\ &- \frac{\partial f}{\partial x}(x_0,y_0) \Delta x \end{split} \label{mult2a}\\ &\begin{split} \varepsilon_2(\Delta x,\Delta y) \Delta y =& f(x_0+\Delta x,y_0+\Delta y) - f(x_0+\Delta x,y_0)\\ &- \frac{\partial f}{\partial y}(x_0+\Delta x,y_0) \Delta y \end{split} \label{mult2b} \end{align}

The trick is to change the independent variables one-by-one since we can only make use of partial derivatives. After defining \eqref{def2a}, there’s no need to scratch our head for \eqref{def2b} to fit the pizzle. We may first try to use \eqref{def2a} to get \eqref{mult2a}. Then by observing the term $\Delta f$ in \eqref{fact}, we realize that we should add the term $f(x_0+\Delta x,y_0+\Delta y)$ and remove $f(x_0+\Delta x,y_0)$ so as to get $\Delta f$ in \eqref{fact}.

Remaining problem

In \eqref{mult2b}, the partial derivative with respect to $y$ is taken at $(x_0+\Delta x,y_0)$, which shouldn’t appear in \eqref{fact}. Therefore, we need a way to get rid of the $\Delta x$ inside the bracket. That leads us to the one unused condition—the second given condition in \eqref{fact}.

By the continuity of the first-order partial derivatives at $(x_0,y_0)$,

\begin{equation} \lim_{\Delta x \to 0} \frac{\partial f}{\partial y}(x_0+\Delta x,y_0) = \frac{\partial f}{\partial y}(x_0,y_0)\\ \lim_{\Delta x \to 0} \left ( \frac{\partial f}{\partial y}(x_0+\Delta x,y_0) - \frac{\partial f}{\partial y}(x_0,y_0) \right ) = 0 \label{cts1} \end{equation}

Thus, we define

\begin{equation} \varepsilon_3(\Delta x) := \frac{\partial f}{\partial y}(x_0+\Delta x,y_0) - \frac{\partial f}{\partial y}(x_0,y_0) \label{cts2} \end{equation}

so that $\varepsilon_3 \to 0$ as $\Delta x \to 0$. With \eqref{cts2}, we can replace the first-order partial derivative with respect to $y$ at $(x_0+\Delta x,y_0)$ in \eqref{mult2b} by the one at $(x_0,y_0)$.

\begin{equation} \begin{split} \varepsilon_2(\Delta x,\Delta y) \Delta y =& f(x_0+\Delta x,y_0+\Delta y) - f(x_0+\Delta x,y_0)\\ &- \left ( \frac{\partial f}{\partial y}(x_0,y_0) + \varepsilon_3(\Delta x) \right ) \Delta y \end{split} \label{b4repl} \end{equation}

Therefore, from \eqref{b4repl}, we see that it’s legitimate for us to rename $\varepsilon_2-\varepsilon_3$ as \varepsilon_2 to get an equation which looks more similar to \eqref{mult2a} than \eqref{mult2b} does.

\begin{gather} \begin{split} &\varepsilon_2(\Delta x,\Delta y) \Delta y\\ =& f(x_0+\Delta x,y_0+\Delta y) - f(x_0+\Delta x,y_0) - \frac{\partial f}{\partial y}(x_0,y_0) \Delta y \end{split} \label{mult2bb}\\ \begin{split} &\varepsilon_1(\Delta x,\Delta y) + \varepsilon_2(\Delta x,\Delta y)\\ =& f(x_0+\Delta x, y_0+\Delta y) - f(x_0,y_0)\\ &- \frac{\partial f}{\partial x}(x_0,y_0) \Delta x - \frac{\partial f}{\partial y}(x_0,y_0) \Delta y \quad \text{(\eqref{mult2a}+\eqref{mult2bb})} \end{split}\\ \begin{split} &f(x_0+\Delta x, y_0+\Delta y) - f(x_0,y_0)\\ =& \frac{\partial f}{\partial x}(x_0,y_0) \Delta x + \frac{\partial f}{\partial y}(x_0,y_0) \Delta y + \varepsilon_1(\Delta x,\Delta y) + \varepsilon_2(\Delta x,\Delta y) \end{split} \label{result} \end{gather}

The result \eqref{result} is what we desired in \eqref{fact}.

Generalisation to $n$-dimension

By reusing the trick of changing the variables once at a time from \eqref{def2a} to \eqref{mult2b}, and a suitable renaming and rearrangment of terms, one can generalise the result in \eqref{fact} to a function $f:\mathcal{D} \to \R$ defined on a subset $\mathcal{D}$ of $\R^n$ containing a $\varepsilon$-neighbourhood of a point $\vect{x}_0 \in \R^n$.

\begin{equation} \Delta f = \langle \nabla f(\vect{x_0}) + \vect{\varepsilon},\Delta \vect{x} \rangle, \label{generalisation} \end{equation}

where $\Delta f := f(\vect{x_0}+\Delta \vect{x})-f(\vect{x_0})$ and $\vect{\varepsilon} \to \zeros$ as $\Delta \vect{x} \to \zeros$.

As you can see in \eqref{generalisation}, writing the statement in its vector form is more concise than writing out each partial derivative in \eqref{fact}.