Items tagged with stopping

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What are the stopping criteria for fsolve?
I cannot find anything in the help page and there seems to be no way of adding an optional argument to fsolve about errors.

I was initially surprised by the results of the first two fsolve commands below:

restart;
infolevel[fsolve]:=2:
fsolve([x->1,x->3],[0.4,8]);
fsolve([x->1,x->3],[0..7,8..9]);
fsolve(x->1,0.4); #OK, returns unevaluated
fsolve([1,1],{x,y}); #OK, returns NULL

I assume that in the first two examples the criterion used is that at some point in the process the iterates [x(n+1),y(n+1)] and [x(n),y(n)] are close enough together and the difference between results from the two is small enough (clearly 0).

 

An intersection in my neighbourhood, currently controlled by a 2-way stop, is under consideration to become a 4-way stop.  This means the traffic that currently has the right-of-way will be required to come to a complete stop, wheras previously they could have coasted down the hill, and accelerated up the other side.   Politics aside, I was curious to explore the following question:

I was wondering if anyone could explain the theoretical reason (ie derive the
expression for the probability) why the optimal solution for the secretary problem
with n=100 is equal to 100/exp(1) and the probability is (1/exp(1))*100

Robert Israel tend to be good at these kind of things....:-)

The problem can be stated as follows (wikipedia):

 1. There is a single secretarial position to fill.
 2. There are n applicants for the position, and the value of n is known.

I have a problem understanding how the below reasoning works
for Bruss article “Sum the Odds to One and Stop,”

i) I would like to understand the die problem and more specificaly
why the probability is k(1/6)^1*(5/6)^(k-1). I have a hard time understanding this

ii) How can I apply this reasonong to a coin toss example?


I have managed...

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