# Error approximation using mean value theorem for mv-function

1. Oct 24, 2016

### lep11

Obviously $\mathbb{R^2}$ is convex, that is, any points $a,b\in\mathbb{R^2}$ can be connected with a line segment. In addition, $f$ is differentiable as a composition of two differentiable functions. Thus, the conditions of mean value theorem for vector functions are satisfied. By applying the theorem we get

$$|\cos(0,3^2+(-0,3)^3)-1|$$
$$=|\cos(0,3^2-0,3^3)-\cos(0^2+0^3)|$$
$$=|Df(c)((0,3;-0,3)-(0,0))|$$
$$=|<\nabla{f(c)},(0,3;-0,3)>|$$
$$\leq\|\nabla{f(c)}\|\|(0,3;-0,3)\|$$(by C-S inequality)
$$=\|\nabla{f(c)}\|\sqrt{0,3^2+0,3^2},$$
$$=\sqrt{(-2c_1\sin(c_1^2+c_2^3)^2+(-3c_2^2\sin(c_1^2+c_2^3))^2}\sqrt{0,3^2+0,3^2},$$
$$\leq{\sqrt{(4c_1^2+9c_2^4}\sqrt{0,3^2+0,3^2}}$$
$$\leq...?$$

where $c=(c_1,c_2)$ lies on the line connecting $(0,3;-0,3)$ and the origin. Consequently $\|c\|=\sqrt{c_1^2+c_2^2}\leq\sqrt{0,3^2+0,3^2}.$ Further, $\nabla{f(c)}=(-2c_1\sin(c_1^2+c_2^3),-3c_2^2\sin(c_1^2+c_2^3)).$

How could one estimate $\|\nabla{f(c)}\|?$ It is a matrix (Frobenius) norm, right?

How to proceed?

Last edited by a moderator: Oct 24, 2016
2. Oct 24, 2016

### Staff: Mentor

Are you sure, you haven't lost too much space already? Why do you consider norms in the first place, rather than calculating $|<\nabla{f(c)},(0,3;-0,3)>|\;$? I also think, that the approximations of the sine values should be really tight. Is the usage of the mean value theorem mandatory?

3. Oct 25, 2016

### lep11

It is mandatory to use the mean value theorem. And now

$|<\nabla{f(c)},(0,3;-0,3)>|\;=|-0,6c_1\sin(c_1^2+c_2^3)+1,8c_2^2\sin(c_1^2+c_2^3))|=|-0,6c_1+1,8c_2^2||\sin(c_1^2+c_2^3)|\leq|-0,6c_1+1,8c_2^2|...?$

or alternatively
$|<\nabla{f(c)},(0,3;-0,3)>|\;=|-0,6c_1\sin(c_1^2+c_2^3)+1,8c_2^2\sin(c_1^2+c_2^3))|\approx|-0,6c_1(c_1^2+c_2^3)+1,8c_2^2(c_1^2+c_2^3))|\leq|-0,6c_1(c_1^2+c_2^3)|+|1,8c_2^2(c_1^2+c_2^3)|...?$

Neither seems to lead me in the right direction :/

Last edited: Oct 25, 2016
4. Oct 25, 2016

### Staff: Mentor

Shouldn't it be $0.9$ as coefficient instead of $1.8$? And you might have to take the signs of $c_i$ into account for an upper bound of $c_1^2+c_2^3$

5. Oct 25, 2016

### lep11

Yes it should. Btw, which one of the above is better estimation? I am still not getting the desired upper bound. :(

6. Oct 25, 2016

### Staff: Mentor

I would step in here $|<\nabla{f(c)},0,3;-0,3)>|\;=$
$=|-0,6c_1\sin(c_1^2+c_2^3)+0.9c_2^2\sin(c_1^2+c_2^3))|=|-0,6c_1+0.9c_2^2||\sin(c_1^2+c_2^3)|$.
With $c_1=c_2=0.3$ I get $31 \cdot 10^{-3}$, so it's too big.
But isn't $c_2 \leq 0$, which means $\sin (c_1^2+c_2^3) \leq \sin c_1^2 \leq 0.3^2 = 0.09$ ?

7. Oct 25, 2016

### lep11

$0,09\cdot|-0,6\cdot0.3+0,9\cdot0,3^2|\approx0,00891$ which is too big as well.
Next I tried
$|<\nabla{f(c)},(0,3;-0,3)>|\;=|-0,6c_1\sin(c_1^2+c_2^3)+0,9c_2^2\sin(c_1^2+c_2^3))| \leq|-0,6c_1+0,9c_2^2||\sin(c_1^2+c_2^3)|\leq|0,9c_2^2|\sin(c_1^2+c_2^3)\approx|0,9c_2^2||c_1^2+c_2^3|\approx5,1\cdot10^{-3}<7\cdot10^{-3}$with $c_1=0,3$ and $c_2=-0,3.$ Is this correct? $|-0,6c_1+0,9c_2^2|\leq|0,9c_2^2|?$

How can this be so hard?

Last edited: Oct 25, 2016
8. Oct 25, 2016

### Staff: Mentor

It's clear until $|<\nabla{f(c)},(0,3;-0,3)>|\; \leq |-0,6c_1+0,9c_2^2|\cdot |\sin(c_1^2+c_2^3)|$

Now $c_1 \in [0\, , \,0.3]$ and $c_2 \in [-0.3\, , \,0]$ which means the signs come into play and you lost me. We have to take the worst case here, i.e. the maximal possible value. To make it easier to look at, let's take positive values $a:=c_1 \geq 0$ and $b:=-c_2 \geq 0$.
Thus we have to maximize $m(a,b):=|-0.6 a + 0.9 b^2 \, |\,\cdot\,| \sin (a^2-b^3) \;|$.
Why should $|-0.6 a + 0.9 b^2 \, | \leq |0.9 b^2 \, |$ be the case? For $(a,b)=(0.3\, , \,0)$ this isn't true.

Because there isn't much space to be generous without further knowledge about $(a,b)=(c_1,-c_2)$.