Problem reducing quadratic to diagonal form

  • Thread starter Thread starter lep11
  • Start date Start date
  • Tags Tags
    Form Quadratic
lep11
Messages
380
Reaction score
7

Homework Statement


Reduce ##xy+zy## to diagonal form.

Homework Equations


The desired diagonal form is ##Q(\vec{x})=(\alpha_1(\vec{x}))^2+...+(\alpha_k(\vec{x}))^2-(\alpha_{k+1}(\vec{x}))^2-...-(\alpha_{k+l}(\vec{x}))^2,## where ##\alpha_i## are linearly independent linear functions. Also known as 'changing the variables in quadratic form'.

The Attempt at a Solution


##xy+zy=y^2+yx+yz+(\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2-((\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2)-y^2=(y+\frac{x}{2}+\frac{z}{2})^2+(\frac{x}{2}+\frac{z}{2})^2-y^2##, but now
##\alpha_1=y+\frac{x}{2}+\frac{z}{2}##
##\alpha_2=\frac{x}{2}+\frac{z}{2}##
##\alpha_3=y##
are linearly dependent. I have tried several different substitutions without success. There's also the matrix method which I am not familiar with. This rather simple problem is giving me headache.
 
Last edited:
Physics news on Phys.org
hint
$$xy=\left(\frac{x}{2}+\frac{y}{2}\right)^2-\left(\frac{x}{2}-\frac{y}{2}\right)^2$$
 
lurflurf said:
hint
$$xy=\left(\frac{x}{2}+\frac{y}{2}\right)^2-\left(\frac{x}{2}-\frac{y}{2}\right)^2$$
Using that lead to situation where the ##\alpha##'s weren't linearly independent.
 
sorry try
$$\left( \frac{x}{2} \pm \frac{y}{\sqrt{2}}+\frac{z}{2} \right)^2$$
As the two
 
lep11 said:

Homework Statement


Reduce ##xy+zy## to diagonal form.

Homework Equations


The desired diagonal form is ##Q(\vec{x})=(\alpha_1(\vec{x}))^2+...+(\alpha_k(\vec{x}))^2-(\alpha_{k+1}(\vec{x}))^2-...-(\alpha_{k+l}(\vec{x}))^2,## where ##\alpha_i## are linearly independent linear functions. Also known as 'changing the variables in quadratic form'.

The Attempt at a Solution


##xy+zy=y^2+yx+yz+(\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2-((\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2)-y^2=(y+\frac{x}{2}+\frac{z}{2})^2+(\frac{x}{2}+\frac{z}{2})^2-y^2##, but now
##\alpha_1=y+\frac{x}{2}+\frac{z}{2}##
##\alpha_2=\frac{x}{2}+\frac{z}{2}##
##\alpha_3=y##
are linearly dependent. I have tried several different substitutions without success. There's also the matrix method which I am not familiar with. This rather simple problem is giving me headache.

What happens if you set u= x + z so that Q = uy? Doesn't the identity in post #2 then work?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top