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Finding expression for ε

  1. Oct 27, 2012 #1
    Hello,

    Is there a better way to find n such that for every n>n0 |sqrt(n^2+3)-sqrt(n^2-1)|<ε?
    I have tried reformulating the expression on the left side so that only one of the square roots remains, to not much avail. The only way I could solve it was by squaring both sides. I did get the correct expression for ε for every ε<1, but it does not work for larger ε's, hence, alas, it is wrong.
    Any word of advice, please?
     
  2. jcsd
  3. Oct 27, 2012 #2

    Mark44

    Staff: Mentor

    I don't see why you would need to worry about ##\epsilon## ≥ 1. In fact, you should probably be able to assume that ##\epsilon## < 1 - the whole goal of this exercise is to show that you can choose n large enough so that the difference of the two square roots is arbitrarily small.
     
  4. Oct 27, 2012 #3
    Thank you very much for replying.
    n0 >= sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon)
    Which not only seems very messy but also yields wrong results for epsilon greater than 1. The question states that epsilon is positive, so I presume my expression must hold for any epsilon, even if greater than 1.
    I know that the right answer should probably be 1/(2epsilon), but how may I derive it?
     
  5. Oct 27, 2012 #4
    My bad, I meant that the right answer should probably be 2/epsilon!
     
  6. Oct 27, 2012 #5

    Mark44

    Staff: Mentor

    If ##\epsilon## is suitably small, then ##\epsilon^4## and ##\epsilon^2## can be neglected. This makes the expression on the right side approximately equal to √16/##2\epsilon##, which gives you the (corrected) result below.
     
  7. Oct 27, 2012 #6
    But what if epsilon were equal to 40? In that case epsilon^4 and epsilon^2 may not be neglected and 2/epsilon still yields the right value for n0, whereas sqrt(epsilon^4-4epsilon^2+16)*1/(2epsilon) doesn't!
     
  8. Oct 27, 2012 #7

    Mark44

    Staff: Mentor

    Why would anyone care? ##\epsilon## is almost always used to represent small (close to 0) numbers.
     
  9. Oct 27, 2012 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since that is a decreasing function of n, and is 2 when n= 1, as soon as [itex]\epsilon[/itex] is larger than 2, you can take [itex]n_0= 1[/itex].
     
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