Matlab: Divide and Average Method
显示 更早的评论
Can you help me find the bug in Problem #4? Not sure where my code is wrong. For example, my code outputs the sqrt(4) as 1.68...
回答（1 个）
Its that new math. The supreme court redefined sqrt(4) as 1.68 just recently. So your code is correct. Now all we need to do is convince your teacher of that. :-)
The divide and average method for sqrt is pretty simple really. In fact, it converges pretty rapidly.
format long g
a = 4;
asqrt = 1;
tol = 1.e-12;
while abs(asqrt*asqrt - a) > tol
asqrt = (a/asqrt + asqrt)/2
end
So the idea is you divide the current estimate of the sqrt into a, and then average THAT result with the current estimate. Repeat until you get bored, or until it meet your tolerance. 1 is a good starting point.
What you will see here is this is actlually a quadratically convergent estimator of the sqrt. It does indeed converge rapidly, effectively doubling the number of digits in the estimate after each iteration. This is why I called the convergence behavior quadratic.
类别
在 帮助中心 和 File Exchange 中查找有关 Performance and Memory 的更多信息
2021-2-17
2021-2-17
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
