I'm just computing the 2nd order taylor polynomial for $f(x,y) = tan(x + 3y + \frac{\pi}{4})$ centered at (3,-1) and wondering if I have done this correctly or if anyone has any suggestions on how I can improve my answer (I've never done this before with mulitvariables so just want to be sure I am on the right track):
I have that: $$f(3,-1) = tan(\frac{\pi}{4}) = 1$$ $$f_{x}(3,-1) = sec^2(\frac{\pi}{4}) = 2$$ $$f_{y}(3,-1) = 3sec^2(\frac{\pi}{4}) = 6$$
So the gradient vector for $f$ is $(2,6)$
The letting $u = x + 3y + \frac{\pi}{4}$, I get that $\frac{du}{dx}= 1 $ and $\frac{dy}{dx}= 3 $
So, $$f_{x}(u) = sec^2(u)$$ $$f_{xx}(u) = 2tan(u)sec^2(u)\frac{du}{dx}$$ $$f_{xy}(u) = 2tan(u)sec^2(u)\frac{du}{dy}$$ and similarly,
$$f_{y}(u) = 3sec^2(u)$$ $$f_{yx}(u) = 6tan(u)sec^2(u)\frac{du}{dx}$$ $$f_{yy}(u) = 6tan(u)sec^2(u)\frac{du}{dy}$$
Substituting back for $u$ and plugging in $(3,-1)$ I get:
$$f_{xx}(3,-1) = 2tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot1 = 4$$ $$f_{xy}(3,-1) = 2tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot3 = 12$$ and similarly,
$$f_{yx}(3,-1) = 6tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot1 = 12$$ $$f_{yy}(3,-1) = 6tan(\frac{\pi}{4})sec^2(\frac{\pi}{4})\cdot3 = 36$$
So,
$$T_{2}((x,y),(3,-1)) = 1+(2,6) \cdot (x-3,y+1) + \frac{1}{2} \bigl( \begin{smallmatrix} x-3 & y+1\\ \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} 4 & 12\\ 12 & 36 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x-3 \\ y+1 \end{smallmatrix} \bigr)$$
Performing the calculations I get:
$$= 1+2x-6+6y+6+\frac{1}{2}[(x-3)(4x-12+12y+12)+(y+1)(12x-36+36y+36)]$$
$$ = 2x^2 + 18y^2 + 12xy + 2x + 6y + 1$$
Is this correct and have I done the steps correctly? Many thanks in advance!
For what it's worth after two years, yes, you did everything correctly.
To back that up, the Mathematica expression for Taylor Series
yields indeed
$$1+2 x+2 x^2+6 y+12 x y+18 y^2.$$
At least that's one unanswered question less.