Finding Supremums and Infimums for Real Numbers | Homework

  • Thread starter Thread starter kmeado07
  • Start date Start date
kmeado07
Messages
40
Reaction score
0

Homework Statement



Show that for all x,y in R (real numbers), sup{x,y}=1/2(x+Y+|X-Y|), and inf{x,y}=1/2(X+Y-|x-y|)


Homework Equations





The Attempt at a Solution



i know that the supremum is the lowest upper bound and that the infimum is the largest lower bound. However i really don't know how to apply that to solving this question.
 
Physics news on Phys.org
If you have two numbers, one has to be at least as big (greater-than-or-equal-to) the other. Assume that x \le y. Then

\sup\{x,y\} = y <br />

and

\inf\{x,y\} = <br /> x


What do the two right-hand-sides simplify to?
 
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