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Newtons Square root method

by Bonaparte
Tags: method, newtons, root, square
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Bonaparte
#1
Dec21-12, 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/e-e|<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/e-e|
sqrt(a)-e<a/e-e (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?
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jedishrfu
#2
Dec21-12, 08:43 AM
P: 2,969
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
Bonaparte
#3
Dec21-12, 08:58 AM
P: 26
Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|

haruspex
#4
Dec21-12, 06:57 PM
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Newtons Square root method

Quote Quote by Bonaparte View Post
Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|
Has that resolved your problem or are you still stuck?
Bonaparte
#5
Dec22-12, 10:33 PM
P: 26
Very stuck PLEASE HELP
jedishrfu
#6
Dec22-12, 10:53 PM
P: 2,969
Quote Quote by Bonaparte View Post
Very stuck PLEASE HELP
Be more specific where exactly and why? sometimes in just explaining and thinking about it the answer will come.
Bonaparte
#7
Dec23-12, 06:35 AM
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<a/e-e (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 :)
HallsofIvy
#8
Dec23-12, 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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