apologies beforehand if I still lack the pertinent "mathematical" language. I'm a law major having only recently switched to maths, and as much as it's embarrassing to admit, I'm very, very new to anything beyond high-school mathematics.
I'm reading Riley's book, and, in section 5.7, there is a formula for two-variable Taylor series, namely:
$$f(x,y) = f(x_0,y_0) + f_x\Delta{x} + f_y\Delta{y} + \frac{1}{2}f_{xx}(\Delta{x})^2 + f_{xy}\Delta{x}\Delta{y} + \frac{1}{2}f_{yy}(\Delta{y})^2 + ...$$
I've been trying to prove Taylor's theorem "my own way"; first I set:
$$f(x,y) \approx f(x_0,y_0) + \int_{x_0}^{x}\!f_x(x,y)\mathrm{d}x +\int_{y_0}^{y}\!f_y(x,y)\mathrm{d}y $$
then set:
$$f_x(x,y) \approx f_x(x_0,y_0) + \int_{x_0}^{x}\!f_{xx}(x,y)\mathrm{d}x +\int_{y_0}^{y}\!f_{xy}(x,y)\mathrm{d}y $$
along with:
$$f_y(x,y) \approx f_y(x_0,y_0) + \int_{x_0}^{x}\!f_{xy}(x,y)\mathrm{d}x +\int_{y_0}^{y}\!f_{yy}(x,y)\mathrm{d}y $$
now the two previous formulas for $f_x$ and $f_y$, when replaced in the original formula for $f$, would yield:
$$f(x,y) \approx f(x_0,y_0) + \\ \int_{x_0}^{x}\![f_x(x_0,y_0) + \int_{x_0}^{x}\!f_{xx}(x,y)\mathrm{d}x +\int_{y_0}^{y}\!f_{xy}(x,y)\mathrm{d}y]\mathrm{d}x + \\ \int_{y_0}^{y}\![f_y(x_0,y_0) + \int_{x_0}^{x}\!f_{xy}(x,y)\mathrm{d}x +\int_{y_0}^{y}\!f_{yy}(x,y)\mathrm{d}y]\mathrm{d}y $$
taking $f(x_0,y_0)$, $f_x(x_0,y_0)$, and $f_y(x_0,y_0)$, respectively, equal to $f(x,y)$, $f_x(x,y)$, and $f_y(x,y)$ in the vicinity of $(x_0,y_0)$, and simplifying further, gives:
$$f(x,y) \approx f(x_0,y_0) + f_x(x,y)\Delta(x) + f_y(x,y)\Delta{y} + \frac{1}{2}f_{xx}(x,y)(\Delta{x})^2 + 2f_{xy}(x,y)\Delta{x}\Delta{y} + \frac{1}{2}f_{yy}(x,y)(\Delta{y})^2$$
which, while very much close to the actual formula, has got an extra $f_{xy}(x,y)\Delta{x}\Delta{y}$ in it; I have been struggling to find out what exactly I have missed during the process but I have not made it anywhere; not yet, at least.
Could I ask what exactly I'm doing wrong? Thanks a lot!
