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oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0$$f(x,y)=xy^2+\cos xy\\f_x(x,y)=y^2y\sin xy\\f_y(x,y)=2xyx\sin xy\\f_{xx}(x,y)=y^2\cos xy\\f_{yy}(x,y)=2xx^2\cos xy\\f_{xy}(x,y)=2y\sin xyxy\cos xy\\f_{xxx}(x,y)=y^3\sin xy\\f_{yyy}(x,y)=x^3\sin xy\\f_{xxy}(x,y)=2y\cos xy+xy^2\sin xy\\f_{xyy}(x,y)=22x\cos xy+x^2y\sin xy\\f_{xxxx}(x,y)=y^4\cos xy\\f_{yyyy}(x,y)=x^4\cos xy\\f_{xxyy}(x,y)=x^2y^2\cos xy+4xy\sin xy2\cos xy\\f_{xxxy}(x,y)=xy^3\cos xy+3y^2\sin xy\\f_{xyyy}(x,y)=3x^2\sin xy+x^3y\cos xy$$ Notice we take advantage of symmetry of mixed partials. Observe so far that at \((0,0)\)$$f=1\\f_x=0\\f_y=0\\f_{xx}=0\\f_{xy}=0\\f_{yy}=0\\f_{xxx}=0\\f_{xxy}=0\\f_{xyy}=2\\f_{yyy}=0\\f_{xxxx}=0\\f_{xxxy}=0\\f_{xxyy}=2\\f_{xyyy}=0\\f_{yyyy}=0$$Now let's work on our power series expansion: $$\begin{align*}f(x,y)\approx &f\\+&(xf_x+yf_y)\\+&\frac12\left(x^2f_{xx}+2xyf_{xy}+y^2f_{yy}\right)\\+&\frac16\left(x^3f_{xxx}+3x^2yf_{xxy}+3xy^2f_{xyy}+y^3f_{yyy}\right)\\+&\frac1{24}\left(x^4f_{xxxx}+4x^3yf_{xxxy}+6x^2y^2f_{xxyy}+4xy^3f_{xyyyy}+y^4f_{yyyy}\right)\end{align*}$$Plugging in our values, we find:$$f(x,y)\approx1+xy^2\frac12x^2y^2$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0The easy way, however, is just to use our wellknown expansions:$$\cos xy=1\frac12(xy)^2+\frac1{24}(xy)^4\dots\approx1\frac12x^2y^2$$Then we just add to our polynomial (which is its own power series expansion):$$f(x,y)\approx1+xy^2\frac12x^2y^2$$
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