Newtons Square root method

Bonaparte

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?

jedishrfu

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

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

haruspex

Has that resolved your problem or are you still stuck?

Bonaparte

Very stuck PLEASE HELP

jedishrfu

Be more specific where exactly and why? sometimes in just explaining and thinking about it the answer will come.

Bonaparte

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

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.