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henpen
Group Title
\[ f(x,y)=f(x_0,y_0)+(xx_0)\frac{\partial f}{\partial x}+(yy_0)\frac{\partial f}{\partial y}+\frac{1}{2!} ( (xx_0)^2\frac{\partial^2 f}{\partial x^2} \]
\[ +2(yy_0)(xx_0)\frac{\partial^2f}{\partial xy}+(yy_0)^2\frac{\partial^2 f}{\partial y^2})+... \]
Why?
 one year ago
 one year ago
henpen Group Title
\[ f(x,y)=f(x_0,y_0)+(xx_0)\frac{\partial f}{\partial x}+(yy_0)\frac{\partial f}{\partial y}+\frac{1}{2!} ( (xx_0)^2\frac{\partial^2 f}{\partial x^2} \] \[ +2(yy_0)(xx_0)\frac{\partial^2f}{\partial xy}+(yy_0)^2\frac{\partial^2 f}{\partial y^2})+... \] Why?
 one year ago
 one year ago

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henpen Group TitleBest ResponseYou've already chosen the best response.0
I'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).
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1351261479700:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1351261612819:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
So all coefficients define by m,n derivatives of f(x,y)
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
I understand what it is saying, but why is the second assumption allowable? That is the essence of my question.
 one year ago
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