# Matlab: Divide and Average Method

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Justin Lee 2021 年 2 月 17 日

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...

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### 回答 (1 件)

John D'Errico 2021 年 2 月 17 日

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
asqrt =
2.5
asqrt =
2.05
asqrt =
2.00060975609756
asqrt =
2.00000009292229
asqrt =
2
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.

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