# some functions, ploting them and defining positive and negative areas

I have four following functions. I need to plot them and define positive and negative defined area. I have done most of them. But, i want you to check them, and correct them if false.

(1) $F(x,y)=x^2+2y^2$

for $x^2+2y^2=1$ , the graph is that

$C=\{(x,y)| -1\lt x \lt 1\ and -1/\sqrt{2} \lt y \lt 1/\sqrt {2}\}$

i.e the set C refers to inside of the ellipse

in this set C, the function is positive defined.

(2) $F(x,y)=xy$

for $xy=1$, the graph is that

this is hyperbol.

$C=\{(x,y)| x \lt 0 \ and\ y\lt 0,\ x\gt 0 \ and \ y\gt 0 \}$

in this set C, the function is positive defined.

(3) $F(x,y)=x+2y^2$

for $x+2y^2=0$ the graph is that

well, I cannot the set C which shows the positive and negative defined areas of the function.

(4) $F(x,y)=x^2+3xy+3y^2$

I can neither graph this function nor define the set C (positive and negatife defined areas).

Especailly, I cannot do part (3) and (4).

Hopefully help me doing these two parts. And please check other two parts (1) and(2). Thank you so much.

• your $C$ in (1) describes a rectangle, and, not the bounded part of the ellipse. Further that $F$ is positive almost everywhere (i.e. $\Bbb R^2-(0,0)$) because is a sum of squares, being zero only at $(0,0)$. – janmarqz Oct 25 '15 at 0:58
• Okay, now, I see. Well, how can I describe what you said as a set? Can you show this? @janmarqz thank you. And also, can you answer part(3) and (4)? – user315 Oct 25 '15 at 1:01
• the bounded part of your ellipse (without the border) is $C=\{(x,y)\mid -1< x< 1\ \& \ -\frac{1}{2}\sqrt{1-x^2}<y<\frac{1}{2}\sqrt{1-x^2}\}$... the other items in a while – janmarqz Oct 25 '15 at 1:08
• meanwhile, for (4) use wolframalpha.com/input/… to get an idea. Later I'll abound. – janmarqz Oct 25 '15 at 1:37
• well my friend, at juanmarqz.wordpress.com/2012/02/16/… you could see an example of how to find the new components of a vector when a change of basis happens... the answer that I'm gonna post uses those ideas – janmarqz Oct 25 '15 at 14:23

(4)

The function $F(x,y)=x^2+3xy+3y^2$ can be thought as the quadratic form $$F(x,y)=(x,y)\left(\begin{array}{cc}1&1\\2&3\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right).\qquad (*)$$

Let us abbreviate $p=\left(\begin{array}{c}x\\y\end{array}\right)$ and $Q=\left(\begin{array}{cc}1&1\\2&3\end{array}\right)$.

Now $Q$ isn't symmetric but can be decomposed as $$Q=\frac{1}{2}(Q+Q^{\top})+\frac{1}{2}(Q-Q^{\top}).$$ This decomposition gives us the matrices $$S=\frac{1}{2}(Q+Q^{\top})\quad \mbox{and}\quad A=\frac{1}{2}(Q-Q^{\top}),$$ which are $S$ symmetric and $A$ antisymmetric. The matrix $S$ is $$\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right).$$

Then ours expression for $F$ is $$F(p)=p^{\top}Qp=p^{\top}(S+A)p=p^{\top}Sp,\qquad (**)$$ because it is easy to compute $p^{\top}Ap=0$.

Also note that $$F(x,y)=(x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right)=x^2+3xy+3y^2.$$ Then the sign of $F$ will depend on the symmetric quadratic form $p^{\top}Sp$.

To decide the sign of this function, knowing that the level curve $x^2+3xy+3y^2=1$ is a tilted ellipse, one should try to calculate a new basis for the vectorspace $\Bbb R^2$. This change gives a pair of axes that one gets from the eigenvectors of the matrix $S$.

The eigenvectors are $$v_1=-\frac{2+\sqrt{13}}{3}e_1+e_2,$$ and $$v_2=\frac{-2+\sqrt{13}}{2}e_1+e_2.$$

With these, one can take the matrix $$B= \left(\begin{array}{cc} -\frac{2+\sqrt{13}}{3}&-\frac{2-\sqrt{13}}{3}\\ 1&1 \end{array}\right),$$ and to get the interpretation of the quadratic form in $(**)$ with the basis $\{v_1,v_2\}$, this will be through the formula $B^{\top}QB$ (this is dubbed orthogonal diagonalization for $Q$).

So the quadratic form for $S$ is $$p^{\top}Sp=({BB^{-1}p})^{\top}SBB^{-1}p=(B^{-1}p)^{\top}(B^{\top}SB)B^{-1}p$$

Hence any can compute that the resulting expression is \begin{eqnarray*} (x,y)\left(\begin{array}{cc}1&3/2\\3/2&3\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right) &=& (s,t)B^{\top}SB \left(\begin{array}{c}s\\t\end{array}\right)\\ &=& (s,t) \left(\begin{array}{cc} \frac{26-5\sqrt{13}}{9}&0\\ 0&\frac{26+5\sqrt{13}}{9}\end{array}\right) \left(\begin{array}{c}s\\t\end{array}\right)\\ &=& \frac{26-5\sqrt{13}}{9}s^2+\frac{26+5\sqrt{13}}{9}t^2 \end{eqnarray*}

Since both $\frac{26-5\sqrt{13}}{9},\frac{26+5\sqrt{13}}{9}$ are positive, then $F(x,y)>0$ as well as $x,y$ aren't zero simultaneously.

We have $\left(\begin{array}{c}s\\t\end{array}\right)=B^{-1}p$ to compute the new components of any vector $p=\left(\begin{array}{c}x\\y\end{array}\right)$.

• to check some of the claims that I made you can use wolframalpha.com/input/… – janmarqz Oct 25 '15 at 14:56
• For (3) the function $F(x,y)=x+2y^2$ has a level curve the parabola $x+2y^2=0$ , with she, you have a partition of the plane into three zones... you will decide how is the sign of $F$ by choosing some points: in the concave part choose, let say $(−1,0)$ and for the right-most zone choose $(1,0)$... How is the sign of $F$ on them? – janmarqz Oct 25 '15 at 17:50
• The sign of F changes by the value of x. – user315 Oct 25 '15 at 19:03
• that is a sample of $x+2y^2<0$ in concave part and $x+2y^2>0$ in the other – janmarqz Oct 25 '15 at 20:59