Question

expand sin(xy)in power of (x-1)and (y-π/2) upto second degree terms by using Taylor's series
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Answer 1

Answer:

...excosy=1+xcosy+(xcosy)22!+(xcosy)33!+... Now use the fact that cosy=sin(π/ ...

Answer 2

Expand the function using Taylor's series up to second degree  

Explanation:

1. given a function in 'x' and 'y' than for expanding it up to second degree using Taylor's theorem for a point $x=a,\ y=b$ is given by,                                                                                                                    [tex]\\ \\ f(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\frac{f_{xx}(a,b)}{2}(x-a)^{2}+f_{xy}(a,b)(x-a)(y-b)+\frac{f_{yy}(a,b)}{2}(y-b)^2[/tex]   here, $f_x,\ f_{xx$ - partial derivatives at the given point.
2. here , we have $f(x,y)=sin(xy)$   and   $(a,b)->\ (1,\frac{\pi }{2} )$  hence we have,                                [tex]f(a,b)= f(1,\frac{\pi }{2} )=sin(\frac{\pi }{2} )=1\\ f_x= ycos(xy)\ \ \ \ \ \ \ \ ->f_x(1,\frac{\pi }{2} )=\frac{\pi }{2} cos(\frac{\pi }{2} )=0\\ f_y= xcos(xy)\ \ \ \ \ \ \ \ ->f_y(1,\frac{\pi }{2} )= cos(\frac{\pi }{2} )=0\\ [/tex]                                                        
3. next we have second derivatives,                                                                             [tex]f_{xx}=-y^2sin(xy)\ \ \ \ \ \ \ \ \ \ \ \ \ \ ->f_{xx}(1,\frac{\pi }{2} )=-\frac{\pi ^2}{4}sin(\frac{\pi }{2} )=-\frac{\pi ^2}{4}\\ f_{yy}=-x^2sin(xy)\ \ \ \ \ \ \ \ \ \ \ \ \ \ ->f_{yy}(1,\frac{\pi }{2} )=-sin(\frac{\pi }{2} )=-1\\ f_{xy}=cos(xy)-xysin(xy)\ \ \ ->f_{xy}(1,\frac{\pi }{2} )=cos(\frac{\pi }{2})-\frac{\pi }{2}sin(\frac{\pi }{2} )=-\frac{\pi}{2}\\[/tex]      
4. putting these values in the expansion expression we get,                                    $sin(xy)=1+0(x-1)+0(y-\frac{\pi }{2})-\frac{\pi ^2}{8}(x-1)^2-\frac{\pi }{2}(x-1)(y-\frac{\pi }{2}) -\frac{1}{2}(y-\frac{\pi }{2} )^2$  
5. hence the expansion of the function up to second degree terms in power of $(x-1),\ \ (y-\frac{\pi }{2} )$ is ,                                                                        $sin(xy)=1-\frac{\pi^2(x-1) ^2}{8}-\frac{\pi(x-1)(y-\frac{\pi }{2}) }{2} -\frac{(y-\frac{\pi }{2} )^2 }{2}$   ------>ANSWER