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anonymous
 3 years ago
\[ f(x,y)=f(x_0,y_0)+(xx_0)\frac{\partial f}{\partial x}+(yy_0)\frac{\partial f}{\partial y}+\frac{1}{2!} ( (xx_0)^2\frac{\partial^2 f}{\partial x^2} \]
\[ +2(yy_0)(xx_0)\frac{\partial^2f}{\partial xy}+(yy_0)^2\frac{\partial^2 f}{\partial y^2})+... \]
Why?
anonymous
 3 years ago
\[ f(x,y)=f(x_0,y_0)+(xx_0)\frac{\partial f}{\partial x}+(yy_0)\frac{\partial f}{\partial y}+\frac{1}{2!} ( (xx_0)^2\frac{\partial^2 f}{\partial x^2} \] \[ +2(yy_0)(xx_0)\frac{\partial^2f}{\partial xy}+(yy_0)^2\frac{\partial^2 f}{\partial y^2})+... \] Why?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm having trouble with getting the intuition behind the Taylor series when the function is multivariabled. I know that you can prove the Taylor series with integration by parts, but I'm not sure how you would use that here (if at all). I assume the answer to this question will be closely associated with the total differential of a multiplication variable to a high degree of approximation (I mean with dx^n dy^m lying about the place, where n>1 and/or m>1).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351261479700:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351261612819:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So all coefficients define by m,n derivatives of f(x,y)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I understand what it is saying, but why is the second assumption allowable? That is the essence of my question.
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