# Taylor's polynomial for a 2 variables function

• Jul 15th 2009, 09:15 PM
arbolis
Taylor's polynomial for a 2 variables function
I must say the veracity of the affirmation : The second degree Taylor's polynomial of $\displaystyle f(x,y)=e^{xy}$ around $\displaystyle (0,1)$ is $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$.
My attempt : I didn't use the $\displaystyle u=xy$ substitution. Rather I did it the hard way : $\displaystyle p(x,y)=f(0,1)+\frac{\partial f(0,1)}{\partial x}(x)+\frac{\partial f(0,1)}{\partial y}(y-1)=1+0+0=1$.
Not sure I've done it right though. Could you second my answer?
( I just plotted the functions and $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$ seems to fit pretty well... meaning I made a mistake but I don't see where!)
• Jul 15th 2009, 09:31 PM
CaptainBlack
Quote:

Originally Posted by arbolis
I must say the veracity of the affirmation : The second degree Taylor's polynomial of $\displaystyle f(x,y)=e^{xy}$ around $\displaystyle (0,1)$ is $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$.
My attempt : I didn't use the $\displaystyle u=xy$ substitution. Rather I did it the hard way : $\displaystyle p(x,y)=f(0,1)+\frac{\partial f(0,1)}{\partial x}(x)+\frac{\partial f(0,1)}{\partial y}(y-1)=1+0+0=1$.
Not sure I've done it right though. Could you second my answer?
( I just plotted the functions and $\displaystyle p(x,y)=1+xy+\frac{y^2}{2}$ 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
• Jul 15th 2009, 09:51 PM
arbolis
Quote:

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 $\displaystyle \frac{\partial f^2(x,y)}{\partial x ^2}=y^2 e^{xy}$, so evaluated in $\displaystyle (0,1)$ and divided by $\displaystyle 2$ and multiplied by $\displaystyle x^2$ gives $\displaystyle \frac{x^2}{2}$.
However I get that $\displaystyle \frac{\partial f^2(x,y)}{\partial y ^2}=x^2 e^{xy}$, evaluated in $\displaystyle (0,1)$ it gives $\displaystyle 0$. Hence $\displaystyle p(x,y)=1+\frac{x^2}{2}$. ( I checked what you told me, that the mixed partial derivatives are 0, that is, $\displaystyle f_{xy}(0,1)=0$)
I feel like I'm missing something obvious. I'll go to sleep and come back tomorrow. Thanks for your help CaptainBlack.
• Jul 16th 2009, 07:14 AM
CaptainBlack
Quote:

Originally Posted by arbolis
I see. I get $\displaystyle \frac{\partial f^2(x,y)}{\partial x ^2}=y^2 e^{xy}$, so evaluated in $\displaystyle (0,1)$ and divided by $\displaystyle 2$ and multiplied by $\displaystyle x^2$ gives $\displaystyle \frac{x^2}{2}$.
However I get that $\displaystyle \frac{\partial f^2(x,y)}{\partial y ^2}=x^2 e^{xy}$, evaluated in $\displaystyle (0,1)$ it gives $\displaystyle 0$. Hence $\displaystyle p(x,y)=1+\frac{x^2}{2}$. ( I checked what you told me, that the mixed partial derivatives are 0, that is, $\displaystyle f_{xy}(0,1)=0$)
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)

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

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

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

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

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

so:

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

CB