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Newtons Square root method 
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#1
Dec2112, 08:23 AM

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<sqrt(a)+e prove that if a/ee<1/10 then sqrt(a)e<1/10 Note that e is using the Newtons method a few times, not necessarily infinity, for any number of times. Also this is about positive integers, and 0 only, root and a. 2. Relevant equations 3. The attempt at a solution So were trying to prove the second one smaller then first (I think), that is: sqrt(a)e<a/ee sqrt(a)e<a/ee (since both are positive, as using the given inequality subtract e from both sides) so sqrt(a)<a/e e*sqrt(a)<a, but e is not necessarily smaller then sqrt a, what am I missing? 


#2
Dec2112, 08:43 AM

P: 3,097

e =>0 and e<sqr(a) with a>0 right?
if e == sqr(a) then you'd have e*e = a and hence a < a which is wrong 


#3
Dec2112, 08:58 AM

P: 26

Sorry, i had a typo, its a +, not a , e< sqrt(a)+e



#4
Dec2112, 06:57 PM

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Newtons Square root method



#5
Dec2212, 10:33 PM

P: 26

Very stuck PLEASE HELP



#6
Dec2212, 10:53 PM

P: 3,097




#7
Dec2312, 06:35 AM

P: 26

So were trying to prove the second one smaller then first (I think), that is:
sqrt(a)e<a/ee sqrt(a)e<a/ee (since both are positive, as using the given inequality subtract e from both sides) so sqrt(a)<a/e e*sqrt(a)<a, but e is not necessarily smaller then sqrt a, what am I missing? There :) 


#8
Dec2312, 06:55 AM

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To find [itex]\sqrt{a}[/itex], we choose some starting value, e, and calculate a/e. There are three possibilities:
1) [itex]e= \sqrt{a}[/itex]. Then [itex]e^2= a[/itex] so that [itex]e= a/e[/itex]. We get the same number again and so know that we are done. 2) [itex]e< \sqrt{a}[/itex]. Then multiplying both sides by [itex]\sqrt{a}[/itex], [itex]e\sqrt{a}< a[/itex] and [itex]\sqrt{a}< a/e[/itex]. That is, [itex]e< \sqrt{a}< a/e[/itex]. 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) [itex]e> \sqrt{a}[/itex]. Then multiplying both sides by [itex]\sqrt{a}[/itex], [itex]e\sqrt{a}> a[/itex] and [itex]\sqrt{a}> a/e[/itex]. That is, [itex]e> \sqrt{a}> a/e[/itex]. 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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