User:Egm6321.f10.team4.Yoon/HW4


 * (a) Using Method 1- Eq.2.2 and Eq.2.3


 * Taking derivative of Eq.2.8 yields:


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\begin{array}{*{20}l} {f_x = 2x} & {f_{xx}  = 2}\\ & {f_{xy} = 0}\\ & {f_{xp} = 0}\\ {f_y = 0}  & {f_{yy}  = 0}\\ & {f_{yp} = 0}\\ \end{array} $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * Similarly taking derivative of Eq.2.9 yields:


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\begin{array}{*{20}l} {g_x = p+2xy} & {g_{xp} = 1} & {{g}_{y}}={{x}^{2}}-{{\upsilon }^{2}} & {g_{yp}  = 0 }  \\ {g_p = x} & {g_{pp}  = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


 * Substituting the above values to Eq.2.3 (first relation) and Eq.2.4 (second relation):


 * {| style="width:100%" border="0" align="left"


 * $$2+2p\times 0+{{p}^{2}}\times 0=1+p\times 0-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }


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$$
 * $$\Rightarrow 2=1-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * $$\displaystyle (Eq.2.10)
 * $$\displaystyle (Eq.2.10)
 * }
 * }


 * So first relation for second exactness condition is not satisfied.


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$$
 * $$0+p\times 0+2\times 0=0\Rightarrow 0=0$$, second relation is satisfied
 * $$\displaystyle (Eq.2.11)
 * $$\displaystyle (Eq.2.11)
 * }
 * }


 * Since the first relation is not satisfied, the Eq.2.1 is not Exact