Taylor's Formula
Write taylor's formula for $F(x,y)= \sin(x)\sin(y)$ using $a=0$, $b=0$, and $n=2$.
$$\sin(h)\sin(k)=hk−\frac 16h(h^2+3k^2)\cos\theta h\sin\theta k−\frac 16 k(3h^2+k^2)\sin\theta h\cos\theta k$$ That's the answer, I don't get how to get it though.
I don't understand how to use the taylor's formula to find this answer. I've tried and failed, I have no idea where $h$,$k$, and $\theta$ comes from. Is anyone able to help me out?
so the formula you want to apply reads \[ F(a+h, b+k) = \sum_{i,j=0}^{n} \frac 1{i!j!}\cdot \frac{\partial^{i+j}F}{\partial x^i\partial y^j}(a,b)h^ik^j + \sum_{i+j = n+1} \frac 1{i!j!} \cdot \frac{\partial^2F}{\partial x^i\partial y^j}(a+\theta h, b + \theta k)h^ik^j \] and tells you how $F$ behaves in a neighbourhood of $(a,b)$. The corresponding theorem says the for each $(h,k)$ there is some $\theta \in [0,1]$ such that this formula holds.
Here we have \begin{align*} F(x,y) &= \sin x \sin y\\\ F(0,0) &= 0\\\ \frac{\partial F}{\partial x}(x,y) &= \cos x \sin y\\\ \frac{\partial F}{\partial x}(0,0) &= 0\\\ \frac{\partial F}{\partial y}(x,y) &= \sin x \cos y\\\ \frac{\partial F}{\partial y}(0,0) &= 0\\\ \frac{\partial^2 F}{\partial x^2}(x,y) &= -\sin x \sin y\\\ \frac{\partial^2 F}{\partial x^2}(0,0) &= 0\\\ \frac{\partial^2 F}{\partial x\partial y}(x,y) &= \cos x \cos y\\\ \frac{\partial^2 F}{\partial x\partial y}(0,0) &= 1\\\ \frac{\partial^2 F}{\partial y^2}(x,y) &= -\sin x \sin y\\\ \frac{\partial^2 F}{\partial y^2}(0,0) &= 0\\\ \end{align*} So in the first sum above all but one Term are zero, the non-vanishing term is \[ \frac 1{1!1!}\cdot \frac{\partial^2 F}{\partial x\partial y}(0,0)hk = hk\] Now the third partial derivatives \begin{align*} \frac{\partial^3 F}{\partial x^3}(x,y) &= -\cos x \sin y\\\ \frac{\partial^3 F}{\partial x^2\partial y}(x,y) &= -\sin x \cos y\\\ \frac{\partial^3 F}{\partial x\partial y^2}(x,y) &= -\cos x \sin y\\\ \frac{\partial^3 F}{\partial y^3}(x,y) &= -\sin x \cos y\\\ \end{align*} Pluging in, we get \begin{align*} \sum_{i+j = n+1} &\frac 1{i!j!} \cdot \frac{\partial^2F}{\partial x^i\partial y^j}(a+\theta h, b + \theta k)h^ik^j \\ &= -\frac 16\cdot\cos\theta h\sin\theta k\cdot h^3 - \frac 12\cdot \sin\theta h \cos \theta k \cdot h^2k\\\ &\quad{}- \frac 12\cos\theta h \sin \theta k\cdot hk^2 - \frac 16 \cdot \sin\theta h \cos \theta k \cdot k^3\\\ &= -\frac 16 h(h^2 + 3k^2)\cos\theta h\sin \theta k -\frac 16 k(3h^2 +k^2) \sin\theta h\cos\theta k. \end{align*} .. as we wanted.
It might be a slight overkill to use the general Taylor expansion in two variables for this particular example. In fact, since the function factors in two independent parts, you can proceed as follows:
$$\sin\left(h+\theta h\right)\sin\left(k+\theta k\right)=\left(\sin h\cos\theta h+\sin\theta h\cos h\right)\left(\sin k\cos\theta k+\sin\theta k\cos k\right)$$ $$=\left(\left(h-\frac{h^{3}}{6}+O\left(h^{5}\right)\right)\cos\theta h+\sin\theta h\left(1-\frac{h^{2}}{2}+O\left(h^{4}\right)\right)\right)\left(\left(k-\frac{k^{3}}{6}+O\left(k^{4}\right)\right)\cos\theta k+\sin\theta k\left(1-\frac{h^{2}}{2}+O\left(k^{4}\right)\right)\right)$$ Opening the brackets and gathering terms you should arrive at the same result. If this is not the kind of solution expected from you, it is still a good means of checking the answer
