# Newtons Square root method

by Bonaparte
Tags: method, newtons, root, square
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 P: 26 1. The problem statement, all variables and given/known data Let e be the number close to sqrt(a) by Newtons Method (That is picking a number, diving a by it, and taking their average, divide a by average, get a number, find their average, so on). Using |e
 P: 2,996 e =>0 and e0 right? if e == sqr(a) then you'd have e*e = a and hence a < a which is wrong
 P: 26 Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|
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P: 9,863
Newtons Square root method

 Quote by Bonaparte Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|
Has that resolved your problem or are you still stuck?
 P: 26 Very stuck PLEASE HELP
P: 2,996
 Quote by Bonaparte Very stuck PLEASE HELP
Be more specific where exactly and why? sometimes in just explaining and thinking about it the answer will come.
 P: 26 So were trying to prove the second one smaller then first (I think), that is: |sqrt(a)-e|<|a/e-e| sqrt(a)-e
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,556 To find $\sqrt{a}$, we choose some starting value, e, and calculate a/e. There are three possibilities: 1) $e= \sqrt{a}$. Then $e^2= a$ so that $e= a/e$. We get the same number again and so know that we are done. 2) $e< \sqrt{a}$. Then multiplying both sides by $\sqrt{a}$, $e\sqrt{a}< a$ and $\sqrt{a}< a/e$. That is, $e< \sqrt{a}< a/e$. The square root of a is somewhere e and a/e and the midpoint, (e+ a/e)/2, is as good a place to look as any. 3) $e> \sqrt{a}$. Then multiplying both sides by $\sqrt{a}$, $e\sqrt{a}> a$ and $\sqrt{a}> a/e$. That is, $e> \sqrt{a}> a/e$. The square root of a is somewhere e and a/e and the midpoint, (e+ a/e)/2, is as good a place to look as any.

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