Taylor's theorem for multivariate functions

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Taylor's theorem

As we have seen in a previous post, the essence of differentiation is a linear approximation. 

See also: Partial and total differentiation of multivariate functions

Given a totally differentiable function \(z = f(x,y)\), we have in the neighbor of \(P = (a,b)\),

\[f(x,y) = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b) + o(\|X - P\|)\]

where \(X = (x,y).\) By discarding the higher order terms \(o(\|X - P\|)\), we have the equation of the tangent plane at \(P\):

\[z = f(a,b) + f_x(a,b)(x - a) + f_y(a,b)(y - b)\]

that gives the linear approximation of \(z = f(x,y)\) around the point \(P\). But we may obtain better approximations by looking into the details of \(o(\|X - P\|)\), that is, by incorporating higher-order derivatives. Just as in the univariate case, we have a multivariate version of Taylor's theorem, which can be stated concisely and proved easily by using differential operators.

See also: Differential operators

Theorem (Taylor's theorem)

Let \(f(x,y)\) be a function of class \(C^n\) on an open region \(U \subset \mathbb{R}^2\), and \((a,b) \in U\). If the line segment between \((x,y)\) and \((a,b)\) is in \(U\), the following holds:

\[\begin{eqnarray}f(x,y) &=& \sum_{m=0}^{n-1}\frac{1}{m!}\left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^{m}f(a,b)\\ && + \frac{1}{n!}\left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^{n}f(a+\theta h, b + \theta k)\tag{Eq:Taylor} \end{eqnarray}\]

where \(h = x - a\), \(k = y - b\), and \(\theta \in (0, 1)\). 

Remark. In (Eq:Taylor), \(h\) and \(k\) in the differential operators are regarded as constants. That is, we mean

\[\left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^{m}f(a,b)=\sum_{l=1}^{m}\binom{m}{l}\left[\frac{\partial^m f}{\partial x^{l}\partial y^{m-l}}(a,b)\right](x-a)^l (y-b)^{m-l}.\]

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Remark. Eq. (Eq:Taylor) is called the finite Taylor expansion of \(f(x,y)\), and its last term (that involving \(\theta\)) is called the remainder or residual. In particular, when \((a,b) = (0,0)\), the finite Taylor expansion is also called the finite Maclaurin expansion. □

Proof. Let us define the function \(g(t) = f(a + ht, b + kt)\). Since the line segment between \((x,y)\) and \((a,b)\) is in \(U\), \(g(t)\) is a function of class \(C^n\) defined on an open interval containing \([0, 1]\), and

\[\begin{eqnarray*} g(0) &=& f(a,b),\\ g(1) &=& f(x,y). \end{eqnarray*}\]

By the finite (univariate) Maclaurin expansion of \(g(t)\), we have

\[g(t) = \sum_{m=0}^{n-1}\frac{1}{m!}g^{(m)}(0)t^m + \frac{1}{n!}g^{(n)}(\theta t)t^n\tag{Eq:gMac}\]

where \(0 < \theta < 1\). \(g^{(m)}(t)\) (the \(m\)-th derivative of \(g(t)\)) is given as

\[\begin{eqnarray*} g^{(m)}(t) &=& \frac{d^m}{dt^m}f(a + ht, b + kt)\\ &=&\left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^{m}f(a + ht, b + kt). \end{eqnarray*}\]

where we have used the "total differential operator" \(\frac{d}{dt}\) defined by using the chain rule:

\[\frac{d}{dt}f(x(t), y(t)) = \left(\frac{d x}{dt} \frac{\partial}{\partial x} + \frac{dy}{dt}\frac{\partial}{\partial y}\right)f(x(t),y(t)).\]

By setting \(t = 0\), we have

\[g^{(m)}(0) = \left(h\frac{\partial}{\partial x} + k\frac{\partial}{\partial y}\right)^{m}f(a, b).\]

Substituting these into (Eq:gMac) and setting \(t = 1\), we have (Eq:Taylor). ■

Example. Let's find the Maclaurin expansion of \(f(x,y) = e^{2x - 3y}\) at \((x,y) = (0,0)\) with \(n =2\).

\[\begin{eqnarray} f_x(x,y) &=& 2e^{2x-3y},\\ f_y(x,y) &=& -3e^{2x-3y},\\ f_{xx}(x,y) &=& 4e^{2x-3y},\\ f_{xy}(x,y) &=& -6e^{2x-3y},\\ f_{yy}(x,y) &=& 9e^{2x-3y}. \end{eqnarray}\] 

The finite Maclaurin expansion with \(n = 2\) is given by 

 \[\begin{eqnarray} f(x,y) &=& f(0,0) + f_x(0,0)x + f_y(0,0)y\\ &&+ \frac{1}{2}\{f_{xx}(\theta x, \theta y)x^2 + 2f_{xy}(\theta x, \theta y)xy + f_{yy}(\theta x, \theta y)y^2\} \end{eqnarray}\]

where \(0 < \theta < 1\). Thus, we have

\[e^{2x - 3y} = 1 + 2x - 3y + \frac{1}{2}(4x^2 - 12xy + 9y^2)e^{2\theta x - 3\theta y}.\]

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Asymptotic expansion

Corollary (Second-order asymptotic expansion)

Let \(f(x,y)\) be a function of class \(C^{2}\)on an open region \(U\subset \mathbb{R}^2\), and \(P=(a,b), X=(x,y)\in U\). Suppose that the line segment between \(X\) and \(P\) is in \(U\). Then,

\[\begin{eqnarray} f(x,y)&=&f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\\ &&+\frac{1}{2}\left[f_{xx}(a,b)(x-a)^2 + 2f_{xy}(a,b)(x-a)(y-b) + f_{yy}(a,b)(y-b)^2\right] \\ && + o(\|X-P\|^2) \end{eqnarray}\]

as \(X \to P\).

Proof. Let \(h = x -a\) , \(k = y - b\), then \(\|X-P\|^2 = h^2 + k^2\). Let \(A = f_{xx}(a,b)\), \(B = f_{xy}(a,b)\), \(C = f_{yy}(a,b)\). By Taylor's theorem, 

\[\begin{eqnarray*} f(x,y)&=& f(a,b) + f_{x}(a,b)h + f_{y}(a,b)k\\ && + \frac{1}{2}[Ah^2 + 2Bhk + Ck^2] + \frac{1}{2}r(h,k) \end{eqnarray*}\]

where

\[\begin{eqnarray*} r(h,k) &=& (A' - A)h^2 + 2(B' - B)hk + (C' - C)k^2\\ A' &=& f_{xx}(a + \theta h, b + \theta k),\\ B' &=& f_{xy}(a + \theta h, b + \theta k),\\ C' &=& f_{yy}(a + \theta h, b + \theta k) \end{eqnarray*}\]

for some \(\theta \in (0, 1)\). Since \(f(x,y)\) is of class \(C^2\), all the second derivatives are continuous. Therefore, \(A'-A\), \(B' - B\) and \(C' - C\) all converge to 0 as \(X \to P\). Hence

\[(A' - A)\frac{h^2}{h^2 + k^2}, ~ (B' - B)\frac{hk}{h^2 + k^2}, ~ (C' - C)\frac{k^2}{h^2 + k^2}\]

all converge to 0 as \(X \to P\). In fact,

\[\begin{eqnarray*} \left|(A' - A)\frac{h^2}{h^2 + k^2}\right| &\leq& |A' - A| \to 0,\\ \left|(B' - B)\frac{hk}{h^2 + k^2}\right| &\leq& |B' - B| \to 0,\\ \left|(C' - C)\frac{k^2}{h^2 + k^2}\right| &\leq& |C' - C| \to 0. \end{eqnarray*}\]

Thus, \(r(h,k) = o(\|X - P\|^2)\). ■

Example. The second-order asymptotic expansion of \(f(x,y) = e^{2x - 3y}\) at \((x,y) = (0,0)\) is

\[e^{2x - 3y} = 1 + 2x - 3y + \frac{1}{2}(4x^2 - 12xy + 9y^2) + o(x^2 + y^2).\]

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