# Thread: Taylor's polynomial for a 2 variables function

1. ## Taylor's polynomial for a 2 variables function

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!)

2. 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

3. 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.

4. 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