# Taylor Polynomial and Remainder

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• Dec 9th 2010, 04:47 PM
AKTilted
Taylor Polynomial and Remainder
Hi,

I'm not sure how to present an answer to this kind of question. It states:

Present the 3rd-Degree Taylor Polynomial $\displaystyle P_{a,2}(h)$ as well as the Lagrange remainder for the function $\displaystyle f(x,y) = x^3y + sin(xy)$.

Then let $\displaystyle a = (1, \pi)$ and write this polynomial in matrix form.

Thanks a lot!
• Dec 9th 2010, 05:17 PM
FernandoRevilla
What kind of difficulties have you had?.

Regards.

Fernando Revilla
• Dec 9th 2010, 05:23 PM
AKTilted
Well, I don't know how to apply the definitions to find an answer and I haven't read anything on the matrix forms for the taylor polynomial.
• Dec 10th 2010, 03:05 AM
HallsofIvy
The third degree Taylor polynomial for f(x, y), at $\displaystyle (1, \pi)$ is
$\displaystyle f(1, \pi)+ \frac{\partial f(1, \pi)}{\partial x}(x- 1)+ \frac{\partial f(1, \pi)}{\partial y}(y- \pi)+ \frac{1}{2}\frac{\partial^2 f(1, \pi)}{\partial x^2}(x- a)^2+$$\displaystyle \frac{1}{2}\frac{\partial^2 f(1, \pi)}{\partial x\partial y}(x- a)(y-\pi)$$\displaystyle + \frac{1}{2}\frac{\partial^2 f(1,\pi)}{\partial y^2}(y- \pi)^2+ $$\displaystyle \frac{1}{6}\frac{\partial^3 f(1,\pi)}{\partial x^3}(x- 1)^3+ \frac{1}{6}\frac{\partial^3 f(1, \pi)}{\partial x^2\partial y}(x-1)^2)(y- \pi)+$$\displaystyle \frac{1}{6}\frac{\partial^3 f(1,\pi)}{\partial x\partial y^2}(x-1)(y-\pi)^2+ \frac{1}{6}\frac{\partial f(1,\pi)}{\partial y^3}(y- \pi)^3$.