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Newtons Square root method

  1. Dec 21, 2012 #1
    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?
     
    Last edited: Dec 21, 2012
  2. jcsd
  3. Dec 21, 2012 #2

    jedishrfu

    Staff: Mentor

    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
     
  4. Dec 21, 2012 #3
    Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|
     
  5. Dec 21, 2012 #4

    haruspex

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    Has that resolved your problem or are you still stuck?
     
  6. Dec 22, 2012 #5
    Very stuck PLEASE HELP
     
  7. Dec 22, 2012 #6

    jedishrfu

    Staff: Mentor

    Be more specific where exactly and why? sometimes in just explaining and thinking about it the answer will come.
     
  8. Dec 23, 2012 #7
    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 :)
     
  9. Dec 23, 2012 #8

    HallsofIvy

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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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