Question

Use Taylor's formula to find a quadratic approximation of f(x, y) = cos cos y at the origin. Estimate the error in the approximation f|x| <= 0.1 and |y| <= 0.1

Answer 1

Given function is $f(x,y)=(cos \space x)(cos \space y)$

The quadratic approximation of f(X,y) at origin is

$f(X,y)=f(0,0)+xf_x(0,0)+yf_y(0,0)+\frac{1}{2}[x^2f_{xx}(0,0)+2xyf_{XY}(0,0)+y^2f_{yy}(0,0)]....(1)$

Compute the value of f(0,0) as follows

$f(X,y)=cos\space x\space cos\space y\\ f(0,0)=cos\space (0)\space cos\space (0)\\=(1)(1)\\=1$

Compute the value of $f_x(0,0)$ as follows

$f_x(X,y)=-sin(X)\\f_x(0,0)=-sin(0)\\=0$

Compute the value of $f_{xx}(0,0)$ as follows

$f_{xx}(X,y)=-cos(X)\\f_{xx}(0,0)=-cos(0)\\=0$

Compute the value of $f_y(0,0)$ as follows

$f_y(X,y)=-sin(y)\\f_y(0,0)=-sin(0)\\=0$

Compute the value of $f_{yy}(0,0)$ as follows

$f_{yy}(X,y)=-cos (y)\\f_{yy}(0,0)=-cos(0)\\=1$

Compute the value of $f_{xy}(0,0)$ as follows

$f_{0,0}(X,y)=sin(X)sin(y)\\f_{Xy}(0,0)=sin(0)sin(0)\\=0$

Now substitute the above values in equition 1

$f(X,y)=1+0+0+\frac{1}{2}(-x^2+0-y^2)\\=1-\frac{x^2}{2}-\frac{y^2}{2}$

Compute the error in approximation if $f|x| ≤ 0.1 \space and\space |y| ≤ 0.1$ as follows

$E(X,y)= \frac{1}{6}(x^3f_{xxx}+3x^2yf_{xxy}+3xy^2f_{xyy}+y^3f_{yyy})$

Since the partial derivatives are given f(X,y) are the multiples of sines and cosines function the absolute value of the derivatives will always be less than or equal to 1. Therefore

$E(X,y)≤ \frac{1}{6}((0.1)^3+3(0.1)^3+3(0.1)^3+(0.1)^3)≤ 0.00134$

Thus the error in the approximation is $E(X,y)≤ 0.00134$