Taylor's polynomial for a 2 variables function

**arbolis** · July 15th 2009, 09:15 PM

I must say the veracity of the affirmation : The second degree Taylor's polynomial of $f(x,y)=e^{xy}$ around $(0,1)$ is $p(x,y)=1+xy+\frac{y^2}{2}$ .
My attempt : I didn't use the $u=xy$ substitution. Rather I did it the hard way : $p(x,y)=f(0,1)+\frac{\partial f(0,1)}{\partial x}(x)+\frac{\partial f(0,1)}{\partial y}(y-1)=1+0+0=1$ .
Not sure I've done it right though. Could you second my answer?
( I just plotted the functions and $p(x,y)=1+xy+\frac{y^2}{2}$ seems to fit pretty well... meaning I made a mistake but I don't see where!)

**CaptainBlack** · July 15th 2009, 09:31 PM

Originally Posted by arbolis

I must say the veracity of the affirmation : The second degree Taylor's polynomial of $f(x,y)=e^{xy}$ around $(0,1)$ is $p(x,y)=1+xy+\frac{y^2}{2}$ .
My attempt : I didn't use the $u=xy$ substitution. Rather I did it the hard way : $p(x,y)=f(0,1)+\frac{\partial f(0,1)}{\partial x}(x)+\frac{\partial f(0,1)}{\partial y}(y-1)=1+0+0=1$ .
Not sure I've done it right though. Could you second my answer?
( I just plotted the functions and $p(x,y)=1+xy+\frac{y^2}{2}$ seems to fit pretty well... meaning I made a mistake but I don't see where!)

You need to include the second order terms (the second partial derivatives)

CB

**arbolis** · July 15th 2009, 09:51 PM

Originally Posted by CaptainBlack

You need to include the second order terms (the second partial derivatives, however all but the mixed partial derivative are zero)

CB

I see. I get $\frac{\partial f^2(x,y)}{\partial x ^2}=y^2 e^{xy}$ , so evaluated in $(0,1)$ and divided by $2$ and multiplied by $x^2$ gives $\frac{x^2}{2}$ .
However I get that $\frac{\partial f^2(x,y)}{\partial y ^2}=x^2 e^{xy}$ , evaluated in $(0,1)$ it gives $0$ . Hence $p(x,y)=1+\frac{x^2}{2}$ . ( I checked what you told me, that the mixed partial derivatives are 0, that is, $f_{xy}(0,1)=0$ )
I feel like I'm missing something obvious. I'll go to sleep and come back tomorrow. Thanks for your help CaptainBlack.

**CaptainBlack** · July 16th 2009, 07:14 AM

Originally Posted by arbolis

I see. I get $\frac{\partial f^2(x,y)}{\partial x ^2}=y^2 e^{xy}$ , so evaluated in $(0,1)$ and divided by $2$ and multiplied by $x^2$ gives $\frac{x^2}{2}$ .
However I get that $\frac{\partial f^2(x,y)}{\partial y ^2}=x^2 e^{xy}$ , evaluated in $(0,1)$ it gives $0$ . Hence $p(x,y)=1+\frac{x^2}{2}$ . ( I checked what you told me, that the mixed partial derivatives are 0, that is, $f_{xy}(0,1)=0$ )
I feel like I'm missing something obvious. I'll go to sleep and come back tomorrow. Thanks for your help CaptainBlack.

I think the mixed partiala are 1. (there is an error in may calcs. backing that post)

$\frac{\partial f}{\partial x}=ye^{xy}$

$\left.\frac{\partial f}{\partial x}\right|_{x=0,y=1}=1$

$\frac{\partial f}{\partial y}=xe^{xy}$

$\left.\frac{\partial f}{\partial y}\right|_{x=0,y=1}=0$

$\frac{\partial^2 f}{\partial y\partial x}=e^{xy}+yxe^{xy}$

so:

$\left. \frac{\partial^2 f}{\partial y\partial x}\right|_{x=0,y=1}=\left. \frac{\partial^2 f}{\partial x\partial y}\right|_{x=0,y=1}=1$

CB

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