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How can I find a bound on the error of approximation of $f(x,y) = sin(xy)$ by its Taylor polynomial $L(x,y)$ of degree 1 on the disc $B_{1/2}(0,0)$?

I know:

$E_L = f(x,y) - L(x,y)$

$M_L \ge max\{\mid f_{xx}(x,y)\mid, \mid f_{xy}(x,y)\mid, \mid f_{yy}(x,y) \mid \}$

$\text{for } (x,y) \in $ $B_d(a,b) = \{(x,y)\mid (x-a)^2+(y-b)^2 \le d\}$

$\text{then,}$

$|E_L(x,y)\mid \le2M_Ld^2 \text{ for all } (x,y) \in B_d(a,b)$

In my case I imagine that will have something like:

$f_{xx} = -y^2\,sin(xy), f_{xy} = cos(xy)-xy\,sin(xy) \text{ and }f_{yy} = -x^2\,sin(xy)$

And

$B_{1/2}(0,0) = \left\lbrace(x,y)\mid x^2+y^2 \le \left(\frac12\right)^2 \right\rbrace$

So my question is, how can I calculate $M_L$ and, therefore $E_L$?

Thanks in advance

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1 Answer

up vote 0 down vote accepted

Let $B = B_{1/2}(0,0)$ to simplify notation a litte. Start estimating the second order derivatives using the triangle inequality and obvious bounds:

\begin{align} \max_B |f_{xx}| &\le \max_B y^2 \le \frac14 \\ \max_B |f_{xy}| &\le \max_B (1 + |xy|) \le \frac54 \\ \max_B |f_{yy}| &\le \max_B x^2 \le \frac14. \end{align}

Hence, you can take $M_L = \dfrac54$.

(With a little more effort, you can get a sharper bound on $|f_{xy}|$, using the fact that $t \mapsto \cos t - t\sin t$ is even and decreasing in $t$ for the relevant values f $t$. Consequently, $\max_B |f_{xy}| \le 1$ and hence you can take $M_L = 1$.)

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Nice. I want only validate if I understood the entire process to find each max. I imagine that because $\mid cos(xy) \mid \le 1$ and $\mid sin(xy) \mid \le 1$, $f_{xx}$ and $f_{yy}$ became $y^2$ and $x^2$. But I not understood why $\mid cos(xy) - xy\,cos(xy)\mid$ became $(1 + \mid xy \mid)$ and not $(1 - \mid xy \mid)$ for $f_{xy}$. After that I only need to replace $x$ and $y$ for $\frac12$ to find the max for each derivative. Isn't it? –  user78723 Jun 6 '13 at 22:39
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@user78723 Yes, that's it. For the estimate on $f_{xy}$, $|a-b| = |a+(-b)| \le |a|+|-b| = |a|+|b|$ by the triangle inequality, but in general it's not true that $|a-b| \le |a|-|b|$. (Take for example $a=2$, $b=-1$.) –  mrf Jun 6 '13 at 22:43
    
Perfect. Thanks! –  user78723 Jun 6 '13 at 22:48
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