Finding Taylor Polynomial of Degree 2 for f(x,y,z)

[AKG]

I have a question that asks me to find the Taylor polynomial of degree 2 of the function:

$$f(x, y, z) = (x^2 + 2xy + y^2)e^z$$

at (1, 2, 0). I have Taylor's Theorem given as follows:

If $f : V \to \mathbb{R}$, V is open, $V \subseteq \mathbb{R}^n$, $c \in V$, $h \in \mathbb{R}^n$, f is of class $C^{k + 1}$, and $c + th \in V$ if $0 \leq t \leq 1$, then:

$$f(c + h) = \sum _{l = 0} ^{k} \left ( \sum _{\{\alpha \in \mathbb{Z}_+ ^n : |\alpha | = l\} } \frac{D_1 ^{\alpha _1} \dots D_n ^{\alpha _n}f(c)}{l!}(h_1 ^{\alpha _1}, \dots , h_n ^{\alpha _n})\right ) + \sum _{\{\alpha \in \mathbb{Z}_+ ^n : |\alpha | = k + 1\}}\left ( \int _0 ^1 \frac{(1 - t)^k}{k!}D_1 ^{\alpha _1} \dots D_n ^{\alpha _n} f(c + th)(h_1 ^{\alpha _1}, \dots , h_n ^{\alpha _n})dt\right )$$

Is this ugly thing above the thing I'm supposed to be working with? And if I'm asked for the polynomial of degree 2, then should my value for k be 1 or 2 (or something else)? Thanks.

 

[Nate810]

Yes, the ugly thing above is Taylor's Theorem which you are supposed to be working with. For the polynomial of degree 2, your value for k should be 2.

 

[wais]

Yes, the expression above is the formula for Taylor's Theorem. In this case, since you are asked to find the Taylor polynomial of degree 2, your value for k should be 2. This means that your polynomial will have terms up to the second degree. To find the polynomial, you will need to calculate the derivatives of f(x,y,z) up to order 2 and evaluate them at the point (1,2,0). Then, using the formula, you can find the coefficients for each term in the polynomial. Remember that the polynomial will have the form:

P(x,y,z) = f(1,2,0) + a(x-1) + b(y-2) + c(z-0) + d(x-1)^2 + e(x-1)(y-2) + f(y-2)^2 + g(z-0)^2

where a,b,c,d,e,f,g are the coefficients you will calculate. Hope this helps!