Taylor Expansions Of Two And Multivariable Functions, Simple Proofs

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Introduction
Why is it possible?
Taylor expansion of multivariable functions

Introduction

In a previous article, we discussed the Taylor expansion of the univariate function $f(x)$.

Explaining The Theory Of Taylor Expansion

Derive the Taylor expansion. The beauty of the Taylor expansion is that it allows us to express functions in polynomial form.

So how can the Taylor expansion of a multivariable function be expressed?

First, we discuss the Taylor expansion of the bivariate function $f(x, y)$.

Taylor expansion of a two-variable function

If the function $f(x, y)$ is infinitely differentiable on the interval containing $x = a, y = b$, then

\begin{eqnarray*}
f(x + a, y + b) = \sum_{k = 0}^{\infty} \frac{1}{k!} \left(a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y} \right)^k f(x, y)
\end{eqnarray*}

Why is it possible?

Let $x, y$ of the bivariate function $z = f(x, y)$ be expressed in terms of the mediating variable $t$ as follows

\begin{eqnarray*}
x &=& x_0 + at \\
y &=& y_0 + bt
\end{eqnarray*}

Let $x_0, y_0, a, b$ be constants. Then $z = f(x, y) = f(x_0 + at, y_0 + bt)$, so

\begin{eqnarray*}
z = z(t)
\end{eqnarray*}Thus, $z$ can be regarded as a one-variable function of $t$.

Now, let us find the derivative of $z(t)$ with $t$. From the derivative of the composite function, we can express it as follows

\begin{eqnarray}
\frac{\mathrm{d}z}{\mathrm{d}t}=\frac{\partial z}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial z}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}t}
\end{eqnarray}

Here,

\begin{eqnarray*}
\frac{\mathrm{d}x}{\mathrm{d}t} &=& \frac{\mathrm{d}}{\mathrm{d}t}(x_0 + at) = a \\
\frac{\mathrm{d}y}{\mathrm{d}t} &=& \frac{\mathrm{d}}{\mathrm{d}t}(y_0 + bt) = b
\end{eqnarray*}

Substituting this, (1) becomes

\begin{eqnarray}
\begin{split}
\frac{\mathrm{d}z}{\mathrm{d}t} &=& a \frac{\partial z}{\partial x} +b \frac{\partial z}{\partial y} \\
&=& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) z
\end{split}
\end{eqnarray}

Then the second-order derivative of $z(t)$ by $t$ is

\begin{eqnarray*}
\frac{\mathrm{d}^2z}{\mathrm{d}t^2} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}z}{\mathrm{d}t} \right) &\underset{(2)}{=}& \frac{\mathrm{d}}{\mathrm{d}t} \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) z \\
&=& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) \frac{\mathrm{d}z}{\mathrm{d}t} \\
&\underset{(2)}{=}& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) z \\&=& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 z \tag{3}
\end{eqnarray*}

Similarly, the third-order derivative and … $n$-order derivative is

\begin{eqnarray*}
\frac{\mathrm{d}^3z}{\mathrm{d}t^3} &=& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^3 z \\
& \vdots & \\
\frac{\mathrm{d}^n z}{\mathrm{d}t^n} &=& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^n z \\
\end{eqnarray*}

Now, from the Maclaurin expansion of $z(t)$, we have

\begin{eqnarray*}
z(t) = z(0) + \frac{\mathrm{d}z(0)}{\mathrm{d}t}\, t + \frac{1}{2!}\frac{\mathrm{d^2}z(0)}{\mathrm{d}t^2}\, t^2 + \cdots
\end{eqnarray*}

and substituting $t = 1$, we obtain

\begin{eqnarray*}
z(1) = z(0) + \frac{\mathrm{d}z(0)}{\mathrm{d}t} + \frac{1}{2!}\frac{\mathrm{d^2}z(0)}{\mathrm{d}t^2} + \cdots \tag{☆}
\end{eqnarray*}

Now, since $z(t) = f(x_0 + at,\ y_0 + bt)$, The left-hand side of (☆) is

\begin{eqnarray*}
(☆) LHS: z(1) &=& f(x_0 + a\cdot 1,\ y_0 + b\cdot 1) \\
&=& f(x_0 + a, y_0 + b)\end{eqnarray*}

The right-hand side of (☆) is

\begin{eqnarray*}
z(0) &=& f(x_0, y_0) \\
\\
\frac{\mathrm{d}z(0)}{\mathrm{d}t} &\underset{(2)}{=}& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) z(0) = \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) f(x_0, y_0) \\
\\
\frac{\mathrm{d^2}z(0)}{\mathrm{d}t^2} &\underset{(3)}{=}& \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 z(0) = \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 f(x_0, y_0) \\
&\vdots&
\end{eqnarray*}

and so it will be

\begin{eqnarray*}
(☆)RHS: &&z(0) + \frac{\mathrm{d}z(0)}{\mathrm{d}t} + \frac{1}{2!}\frac{\mathrm{d^2}z(0)}{\mathrm{d}t^2} + \cdots\\
&=&f(x_0,\ y_0) + \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) f(x_0, y_0) + \frac{1}{2!} \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 f(x_0, y_0) +\cdots
\end{eqnarray*}

Therefore, (☆) is

\begin{eqnarray}
f(x_0+a,\ y_0+b) = f(x_0,\ y_0) + \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) f(x_0, y_0) + \frac{1}{2!} \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 f(x_0, y_0) +\cdots
\end{eqnarray}

Replacing $x_0, y_0$ with $x, y$ respectively

\begin{eqnarray*}
f(x+a,\ y+b) &=& f(x,\ y) + \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right) f(x, y) + \frac{1}{2!} \left( a \frac{\partial }{\partial x} +b \frac{\partial }{\partial y}\right)^2 f(x, y) +\cdots \\
\\
&=&\sum_{k = 0}^{\infty} \frac{1}{k!} \left(a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y} \right)^k f(x,\ y)
\end{eqnarray*}

We have obtained a tabular expression for the Taylor expansion of a two-variable function.

Taylor expansion of multivariable functions

In general, the Taylor expansion of a multivariable function can be written as follows

\begin{eqnarray*}
f(x+a) &=& \sum_{k = 0}^{\infty} \frac{1}{k!} \left(a \frac{\partial}{\partial x} \right)^k f(x) \\
\\
f(x+a,\ y+b) &=& \sum_{k = 0}^{\infty} \frac{1}{k!} \left(a \frac{\partial}{\partial x} + b
\frac{\partial}{\partial y}\right)^k f(x,\ y) \\
\\
f(x+a,\ y+b,\ z+c) &=& \sum_{k = 0}^{\infty} \frac{1}{k!} \left(a \frac{\partial}{\partial x} + b
\frac{\partial}{\partial y} + c \frac{\partial}{\partial z} \right)^k f(x,\ y,\ z) \\
&\vdots&
\end{eqnarray*}

Thus, the Taylor expansion of the $n$ variable function is as follows

Taylor expansion of $n$ variable functions

\begin{align*}
&f(x_1+a_1,\ x_1+a_2, \dots, x_n+a_n) \\
&\ \ \ \ \ \ \ \ \ \ \ \ = \sum_{k = 0}^{\infty} \frac{1}{k!} \left(a_1 \frac{\partial}{\partial x_1} + a_2
\frac{\partial}{\partial x_2} + \cdots + a_n \frac{\partial}{\partial x_n} \right)^k f(x_1,\ x_2, \dots, x_n)
\end{align*}