30 Jul 05
1 edit
Assuming O is the center of the movement (where the pendulum is attached), A and C are the extremes, B is in the middle, and O is the intersection of AC and OB, I get (approximately):
a) 30 cm
b) 200 cm² (199.5...)
c) 300 cm²
d) 100 cm² (100.5..)
e) 5.4 cm
miscalcs are very realistic though.
01 Aug 05
Originally posted by ilywrinwow, that brings back memories, learned that one when I was in
Check this 🙂
http://www.bbc.co.uk/dna/h2g2/A827453
grade school at the First Lutheran School in El Monte California.
Wonder how you would do it for cube roots? can this method be
generalized for Nth roots?
01 Aug 05
Originally posted by sonhouseWell i just did a little googling and found this generalization 🙂
wow, that brings back memories, learned that one when I was in
grade school at the First Lutheran School in El Monte California.
Wonder how you would do it for cube roots? can this method be
generalized for Nth roots?
http://en.wikipedia.org/wiki/Shifting_nth-root_algorithm
02 Aug 05
Originally posted by LittleBearFor square root, I recommend using the "Divide and Average" algorithm. It has quadratic convergence (ie, it smokes), and it can be generalized for nth root. From what I see in the algorithms presented here, they seem to have linear convergence (ie, they're slow).
my rec.
If only I could remember how to get the square root of a number using only paper and pencil... :'(
Here's Divide and Average.
Say you want to find the square root of the number k.
Set X(0) to be a guess of the square root of k.
Now, set X(n+1) = (1/2)*(X(n)+(k/(X(n))).
Another good thing about Divide and Average is that if you make a mistake, your mistake gets smaller, and smaller with every iteration. With linear convergence methods, if you make a mistake, your results are worthless, regardless of how far you carry the calculation after your mistake. With Divide and Average, you can make a mistake at every step, and each mistake gets damped more and more, as you iterate, so you always approach the answer you are interested in.