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1. Simplify an expression

1. Homework Statement I have to prove that the expression $$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$ is equal to $$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$ where $\omega_r= \frac{1}{\sqrt{LC}}$...
2. Least squares problem: am I solving it correctly?

Right, thanks! I corrected that and now I have as a solution $$(\frac{-1}{6}, \frac{1}{3}, \frac{1}{6})$$ is this correct now?
3. Least squares problem: am I solving it correctly?

So what I did was: I took the square of the length: $$(x_1)^2 + \frac {2}{9} + \frac {2}{3} x_1$$ And then I calculated the 1st derivative of this expression and zero gave me value for x_1: -8/3 I made a substitution of this value in my original expression and I got my solution
4. Least squares problem: am I solving it correctly?

I already explained in my first post !
5. Least squares problem: am I solving it correctly?

Then how do you think I should do it? I'm not understanding...
6. Least squares problem: am I solving it correctly?

Thanks for the reply! So in that case I need to minimize the square of the length of (x_1, \frac{1}{3}, \frac{1}{3} + x_1) right? Because if I minimize the square length of (x_1, 0, x_1) I reach to a zero solution, right? Than I can write that the family of the least squares solution is...
7. Least squares problem: am I solving it correctly?

1. Homework Statement In R^3 with inner product calculate all the least square solutions, and choose the one with shorter length, of the system: x + y + z = 1 x + z = 0 y = 0 2. The attempt at a solution So I applied the formula A^T A x = A^T b with A as being the matrix with row 1...