2

If I have the following equation:

$$f = 2\gamma_2 \sigma_{2,1}^2 a^2 + 4 \gamma_2\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2$$

I can write it as:

$$f = 2\gamma_2(\sigma_{2,1}a + \sigma_{2,2}x_2)^2$$

But now I have:

$$g = 2\gamma_1 \sigma_{2,1}^2 a^2 + 2 (\gamma_1 + \gamma_2)\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2$$

Now I'd like to complete the squares, but I don't see how.

Michael Hardy
  • 1
Pietair
  • 383
  • 1
    What are you trying to do that you need to complete the square? Solve for $x_2$, or something else? – John Aug 07 '14 at 16:51
  • I want to proof (if possible) that $g \geq 0$ for $\gamma_1 \geq 0$ and $\gamma_2 \geq 0$. – Pietair Aug 07 '14 at 17:02
  • I suspect there's some meaning to the $\sigma$'s and perhaps the $\gamma$'s and $a$, that would help? What do they represent? – John Aug 07 '14 at 18:40

1 Answers1

0

\begin{align} g &= 2\gamma_1 \sigma_{2,1}^2 a^2 + 2 (\gamma_1 + \gamma_2)\sigma_{2,1}\sigma_{2,2}ax_2 + 2\gamma_2 \sigma_{2,2}^2x_2^2 \\ &= 2(\gamma_1 \sigma_{2,1} a + \gamma_2 \sigma_{2,2} x_2) ( \sigma_{2,1} a + \sigma_{2,2} x_2) \end{align}

Things are occurring in groups. Acknowledging them might clear away some of the noise.

Let $A = \sigma_{2,1} a$ and let $B = \sigma_{2,2} x_2$.

Then $g = 2(\gamma_1 A + \gamma_2 B)(A + B)$

You said you wanted to prove, if possible, that $\gamma_1 \ge 0$ and $\gamma_2 \ge 0$ implies $g \ge 0$. I'm going to have to assume that $A$ and $B$ can be any real number.

But if $\gamma_1 > 0$ and $\gamma_2 = 0$, then you get $g = 2 \gamma_1 A(A + B)$ which you can make negative if you make $A$ positive and $A+B$ negative.

So it seems that your hypothesis is false; unless there are things about $\sigma_{2,1} a$ and $\sigma_{2,2} x_2$ that you haven't told us about.

Steven Alexis Gregory
  • 26,769