1.8 Taylor’s Formula For Functions of Several Variables
Recall: \(n^{\text {th}}\) degree Taylor polynomial \[f(x) \approx f(a) + \frac {f'(a)(x-a)}{1!}+ \frac {f''(a)(x-a)^2}{2!}+ \frac {f'''(a)(x-a)^3}{3!}+ \cdots \cdots \cdots \cdots + \frac {f^n(a)(x-a)^n}{n!}\]
Two variables \[F(x,y) = F(x_1,y_1) + dF(x_1,y_1,x-x_1,y-y_1) + \frac {d^n}{n!}F(x_1,y_1,x-x_1,y-y_1)\]
For \(n = 1\) \[F(x,y) = F(x_1,y_1) + dF(x_1,y_1,x-x_1,y-y_1)\]
This is known as the mean value theorem.
If \(f(x,y)\) and its partial derivatives up to the \(n^{\text {th}}\) order are finite and continuous for all points \((x,y)\) where \(a\leq x\leq a+h\hspace {0.2cm},\hspace {0.2cm} b\leq y \leq b + k\). Then
\[f(a+h,b+k) = f(a,b) + \bigg (h\frac {\partial }{\partial x} + k \frac {\partial }{\partial y}\bigg ) f + \frac {\bigg (h\dfrac {\partial }{\partial x} + k \dfrac {\partial }{\partial y}\bigg )^2f}{2!} + \frac {\bigg (h\dfrac {\partial }{\partial x} + k \dfrac {\partial }{\partial y}\bigg )^3 f}{3!} + \cdots \cdots \cdots \]
Let \(a=0,\hspace {0.3cm} b = 0,\hspace {0.3cm} h=x,\hspace {0.3cm} k = y\)
\begin {align*} f(x,y) = & f(0,0) + \frac {1}{1!}\bigg (x\frac {\partial f}{\partial x} + y \frac {\partial f}{\partial y}\bigg ) +\frac {1}{2!}\bigg (x^2\frac {\partial ^2 f}{\partial x^2} + y^2\frac {\partial ^2f}{\partial y^2} +2xy\frac {\partial ^2f}{\partial x\partial y}\bigg )\\\\ & + \frac {1}{3!} \bigg (x^3\frac {\partial ^3f}{\partial x^3} + 3x^2y \frac {\partial ^3f}{\partial x^2\partial y} + 3xy^2\frac {\partial ^3f}{\partial x\partial y^2} + y^3 \frac {\partial ^3f}{\partial y^3}\bigg ) +\cdots \cdots \cdots \cdots \\ \end {align*}
Expand \(f(x,y) = e^x \sin y\) about \((0,0)\) in powers of \(x\) and \(y\) up to terms of third degree.
Solution.
\(f(x,y) = e^x\sin y\)
\(f_x = e^x\sin y\hspace {1cm} f_x(0,0) = 0\)
\(f_y = e^x \sin y\hspace {1cm} f_y(0,0) = 1\)
\(f_{xx} = e^x\sin y \hspace {1cm} f_{xx} (0,0) = 0\)
\(f_{xy} = e^x \cos y \hspace {1cm} f_{xy} (0,0) = 1\)
\(f_{yy} = -e^x \sin y \hspace {1cm} f_{yy} (0,0) = 0\)
\(f_{xxx} = e^x \sin y \hspace {1cm} f_{xxx} (0,0) = 0\)
\(f_{xxy} = e^x\cos y \hspace {1cm} f_{xxy} (0,0) = 1\)
\(f_{xyy} = -e^x\sin y \hspace {1cm} f_{xyy} (0,0) = 0\)
\(f_{yyy} = -e^x \cos y \hspace {1cm} f_{yyy} (0,0) = -1\)
By Taylor’s theorem \begin {align*} f(x,y) = & f(0,0) + xf_x(0,0) + yf_y(0,0) + \frac {1}{2!}\bigg (x^2f_{xx}(0,0) + 2xyf_{xy}(0,0) +y^2f_{yy}(0,0)\bigg )\\ + &\frac {1}{3!}\bigg (x^3f_{xxx}(0,0) + 3x^2yf_{xxy}(0,0) + 3xy^2f_{xyy}(0,0) + y^3f_{yyy}(0,0)\bigg ) + \cdots \cdots \cdots \\ \end {align*}
\[\implies \hspace {1cm} f(x,y) = y + xy + \frac {x^2y}{2} - \frac {y^3}{6} + \cdots \cdots \cdots \cdots \]
Solution.
\(f(x,y) = x^2y + 3y -2\)
\(f_x = 2xy\hspace {1cm} f_x(1,-2) = -4\)
\(f_y = x^2 + 3\hspace {1cm} f_y (1,-2) = 4\)
\(f_{xx} = 2y \hspace {1cm} f_{xx}(1,-2) = -4\)
\(f_{yy} = 0 \hspace {1cm} f_{yy} (1,-2) = 0\)
\(f_{xy} = 2x\hspace {1cm} f_{xy} (1,-2) = -4\)
\(f_{xxx} = 0\)
\(f_{xxy} = 2\)
\(f_{xyy} = 0\)
\(f_{yyy} = 0\)
\begin {align*} f(x,y) = & f(1,-2) + (x-1)f_x(1,-2) + (y+2)f_y(1,-2)\\ & + \frac {1}{2!}\big [(x-1)^2 f_{xx}(1,-2) + 2(x-1)(y+2)f_{xy}f(1,-2) + (y+2)^2f_{yy}(1,-2)\big ]\\ & + \frac {1}{3!}\big [3(x-1)^2(y+1)f_{xxy}(1,-2)\big ] \end {align*}
\begin {align*} f(x,y) = & -10 - 4(x-1) + 4(y+2) + \frac {1}{2!}\big (-4(x-1)^2 +4(x - 1) (y+2)\big ) + \frac {1}{3!}\big (6(x-1)^2(y+2)\big )\\\\ \end {align*}
Find the expansion for \(\cos x \cos y\) in powers of \(x\) and \(y\) up to the fourth order terms.
\[\textbf {Ans:}\hspace {1cm} f(x,y) = 1 - \frac {x^2}{2}-\frac {y^2}{2} + \frac {x^4}{24} + \frac {x^2y^2}{4} + \frac {y^4}{24}+\cdots \cdots \cdots \cdots \]
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