The following integral was on the qualifying test:
$\displaystyle \int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx $
Any ideas? I'm stumped.
How about the following:
let $\displaystyle y = \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} $
then $\displaystyle y^{2} = 1+x\sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} = 1+xy$
solving for y
$\displaystyle y = \frac{x}{2} \pm \frac{\sqrt{x^{2}+4}}{2} $
then using the positive root since we're integrating over positive values of x
$\displaystyle \int^{1}_{0} \sqrt{1+x \ \sqrt{1+x \ \sqrt{1+x \ \sqrt{...}}}} \ dx \ = \int^{1}_{0} \Big(\frac{x}{2} + \frac{\sqrt{x^{2}+4}}{2} \Big) \ dx $ ?
No no! That is correct. I didn't realize it was supposed to be an infinite surd . Yes, that method works...if you can prove that the surd converges .
P.S. You can choose the $\displaystyle \pm$ according to the fact that $\displaystyle y>0$ for $\displaystyle x\in[0,1]$