Taylor series for function of two variables In my textbook taylor series for function of two variable has been written like this:
$$ f(a+h,b+k)=f(a,b)+f_x(a,b)h+f_y(a,b)k+\frac 1 2 (f_{xx}(a,b)h^2+ 2hkf_{xy}(a,b)+f_{yy}(a,b)k^2)+h^2+k^2)^{\frac{3}{2}}B(h,k) $$
in which $B(h,k)$ is a bounded function around the center.
My question is where did $(h^2+k^2)^{\frac{3}{2}}$ come from ?
why is $B(h,k)$ bounded and what does it mean in this context that $B(h,k)$ is bounded?
in my textbook doesn't mention anything about these.
Can someone please explain these
 A: For a single variable function $f$, the Taylor polynomial $T_nf$ of degree $n$ has the property that
$f(x_0+h)=T_nf(x_0;h)+\vert h\vert^{n+1} B_n(h)$ with $B_n$ bounded. This essentially says that $f$ is approximately equal to the Taylor polynomial plus some remainder which goes to 0 very quickly as $h\to0$. To be specific, it goes to $0$ quicker than $h^n$. This is guaranteed by the boundedness of $B_n$: The remainder is $h^{n+1}$ (so of higher order than $h^n$) multiplied by something bounded, so it doesn't slow down the speed with which the remainder goes to 0.
This can be generalized to multi-variable functions by replacing $h$ with the vector $\vec h=(h_1,\dots,h_n)$ (or in 2d, you could also name the components $(h,k)$) and $\vert h\vert$ with the norm $\vert \vec h\vert=\sqrt{h_1^2+\dots+h_n^2}$, or in 2d with $\sqrt{h^2+k^2}$. What it should say to actually capture the intuition is something like this:
$f(\vec a+\vec h)=T_2f(\vec a;\vec h)+\vert\vec h\vert^{2+1} B_2(\vec h)$, where $\vec a=(a,b),~\vec h(=h,k)$ and where $T_2f$ is just all the terms I've suppressed in this notation (so the actual Taylor polynomial itself). Now just plug in $\vert \vec h\vert=\sqrt{h^2+k^2}$ and you'll see where the $(\dots)^{\frac{3}{2}}$ comes from.
