Introduction
In a previous article, we discussed the Taylor expansion of a single-variable function $f(x)$.
So, how can we express the Taylor expansion of multivariable functions?
First, we will discuss the Taylor expansion of a two-variable function $f(x, y)$.
When a function $f(x, y)$ is infinitely differentiable in an interval containing $x = a, y = b$,
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 Does This Hold?
Let us express $x$ and $y$ of a two-variable function $z = f(x, y)$ using a parameter $t$ as follows:
x &=& x_0 + at \\
y &=& y_0 + bt
\end{eqnarray*}
Here, $x_0, y_0, a, b$ are constants. Then, since $z = f(x, y) = f(x_0 + at, y_0 + bt)$,
z = z(t)
\end{eqnarray*}we can regard $z$ as a single-variable function of $t$.
Now, let us find the derivative of $z(t)$ with respect to $t$. From the chain rule for differentiation, it can be expressed as follows:
\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,
\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 these, (1) becomes
\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}
Next, the second derivative of $z(t)$ with respect to $t$ is:
\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)より}{=}& \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)より}{=}& \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*}
It can be expressed as above.
Similarly, the third derivative… and the $n$-th derivative are:
\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 $z(t)$:
Here, from the Maclaurin expansion of $z(t)$,
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 $t = 1$ gives
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:
(☆)左辺: 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:
z(0) &=& f(x_0, y_0) \\
\\
\frac{\mathrm{d}z(0)}{\mathrm{d}t} &\underset{(2)より}{=}& \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)より}{=}& \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*}
Therefore,
(☆)右辺: &&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*}
Thus, (☆) becomes
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,
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*}
Thus, we have obtained the expression for the Taylor expansion of a two-variable function.
Taylor Expansion of Multivariable Functions
In general, the Taylor expansion of multivariable functions can be written as follows:
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*}
Therefore, the Taylor expansion of an $n$-variable function is as follows:
&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*}
