y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}
From your observation
y^2-y-x=0
y=\frac{1}{2}(1 \pm \sqrt{1+4x})
as x>0,y>0
y=\frac{1}{2}(1+\sqrt{1+4x})
But from the first formula, y(x=0) should be zero. How can we get value of y(x=1) from it which does not show us initial value ? I am afraid this formula is not defined well enough.
Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.
Hello !
I derived equations of stress tensor 2D transformation.
Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture.
I want to obtain expression that connects tensor for case 1 and tensor for case 2.
My attempt:
Are these equations correct? Is there more easier expression for stress tensor...