I have to solve the following exercise:
Find the secondary approach of $f(x,y)=\sin x\sin y$ near the point $(0,0)$. How accurate is the approach if $|x|\leq 0{,}1$ and $|y|\leq0{,}1$?
(I have to prove that $|E|\leq {\frac{8}{6}}(0{,}1)^3$ )
$-$ I found :
the secondary approach is for $n=2$
\begin{align}f(x,y)= &f(0,0)+xf_x+yf_y+\frac{1}{2}(x^2f_{xx}+2xyf_{xy}+y^2f_{yy})\\&+ \frac{1}{6}(x^3f_{xxx}+3x^2yf_{xxy} +3y^2xf_{xyy}+y^3f_{yyy})(φ,n)\end{align}
$f(0,0)=0$
$f_x(0,0)=\cos0\sin0=0$, $f_y(0,0)=\sin0\cos0=0$
$f_{xx}(0,0)=-\sin 0\sin0=0$, $f_{yy}(0,0)=-\sin0\sin0=0$
$f_{xy}(0,0)=\cos0\cos0=1$
therefore $f(x,y)\simeq 0+0+0+\frac{1}{2}(x^20+2xy+y^20)=xy$,
the error estimation is $|E(x,y)|=\frac{1}{6}(x^3f_{xxx}+\dots+y^3f_{yyy})(φ,n)$
$-$How am I going to prove that $|E|\leq \frac{8}{6}(0{,}1)^3?$ I don't know what I have to do. Are the things that I've done above correct?
Best Answer
I suppose that in $0{,}1$, the comma is a decimal comma.
Observe that all partial derivatives at any order are the products of a sine or a cosine in $x$ and a sine or a cosine in $y$, up to a $-$ sign. So the absolute value of these derivatives are bounded by $1$, and by the triangle inequality, $$|E(x,y)|=\frac{1}{6}|(x^3f_{xxx}+\dots+y^3f_{yyy})(φ,n)|\le \frac{1}{6}|(|x|^3 +3x^2|y|+3|x|y^2+|y|^3)$$ which is no more than $\;\dfrac{1}{6}(0.1+0.1)^3=\dfrac8610^{-3}\;$ on $\;[-0.1,0.1]\times[-0.1,0.1]$.