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Inequality Question

  1. Apr 2, 2012 #1
    I want to prove
    cos2(x)/(n2 + 1) ≤ 1/(n2 + 1)

    I know this is an obvious inequality but I want to know if my reasoning is correct.

    For the expression cos2(x)/(n2 + 1) to be as large as possible the numerator must → ∞ but cos2(x) is bounded above by 1.

    cos2(x) = 1 for x = 2∏k where k ≥1 is an integer.

    cos2(2∏k)/((2∏k)2 + 1) = 1/ ((2∏k)2 + 1) for integers k ≥ 1.

    Now I want to compare n2 + 1 to (2∏k)2 + 1:

    n2 ≤ (2∏k)2 where n and k are consecutive integers from 1 to ∞.

    n2 + 1 ≤ (2∏k)2 + 1
    and
    1/(n2 + 1) ≥ 1/ ((2∏k)2 + 1 )

    so cos2(x)/(n2 + 1) ≤ 1/(n2 + 1)
     
  2. jcsd
  3. Apr 2, 2012 #2
    This is unnecessarily complicated, if you can assume the identity (cos(x))^2 ≤ 1, then just divide both sides by (n^2 + 1)
     
  4. Apr 2, 2012 #3
    Yes I know that, I just wanted to see if I was correct in what I was saying.
     
  5. Apr 2, 2012 #4

    Mark44

    Staff: Mentor

    The numerator can't approach infinity
    Do you mean that if n = 2, k = 3, and if n = 3, k = 4? I get the idea that what you said isn't what you meant.
    As already noted by poopsilon, what you have is much more complicated than what is needed, not to mention unclear. For any integer n (and for that matter any real number), -1 ≤ cos2(n) ≤ 1. This idea and the "squeeze" theorem are all you need to establish the inequality you started with.
     
  6. Apr 2, 2012 #5
    Mark44

    "The numerator can't approach infinity"


    Miike012
    I know this that is why I said...
    " but cos^2(x) is bounded above by 1."

    Mark44

    "Do you mean that if n = 2, k = 3, and if n = 3, k = 4? I get the idea that what you said isn't what you meant."


    Miike012
    "where n and k are consecutive integers from 1 to ∞."

    So if I have the following inequality
    n^2 ≤ (2∏k)^2

    Then if n and k are consecutive integers from 1 to ∞ what I am meaning to say is...

    when n is 1 k is 1
    when n is 2 k is 2
    n = 3, k = 3......

    so,
    (1)^2 ≤ (2∏(1))^2

    (2)^2 ≤ (2∏(2))^2

    (3)^2 ≤ (2∏(3))^2

    And so on...
     
  7. Apr 3, 2012 #6

    Mark44

    Staff: Mentor

    Since n and k are equal at each step, there's no need for two variables. What you wrote was very confusing. You could have said
    n2 ≤ (2##\pi n)^2##, for n = 1, 2, 3, ...
     
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