Linear Dependence with Zero Vector: A Simple Solution?

eyehategod
Messages
82
Reaction score
0
I have to prove that a set that has the zero vector is linearly dependent. Can anyone help me out?
 
Physics news on Phys.org
As usual a set {x1,x2...} is linearly dependent if you can find constants (not all zero) such that c1*x1+c2*x2+...=0. What happens if there is only one vector? Doesn't that make it easy?
 
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