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Find all values such that the inequality is true

  1. Sep 7, 2016 #1
    Mod note: Moved from a technical math section, so missing the homework template.
    This is for an Intro to Analysis course. It's been a very long time since I've taken a math course, so I do not remember much of anything.

    =============
    Here is the problem:


    For the inequality below, find all values n ∈ N such that the inequality is true:

    (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025.


    ============================
    Here is my attempt at the problem:


    Looking at the following set

    {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


    We want to find the lower bound of this set.


    Suppose A denotes the above set, then we have


    A= {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


    Since the above rational function can be reduced to 1/(2n+1) we have


    1/(2n+1) <0.025


    Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3.... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20.


    ==========
    If you can only just tell me what topics I need to review to answer this correctly, I would appreciate it.
     
    Last edited by a moderator: Sep 8, 2016
  2. jcsd
  3. Sep 7, 2016 #2

    ChrisVer

    User Avatar
    Gold Member

    what is n2,n3?
    for example if n=1, n2=1 and n3=170000 you can make the inequality true....
    [itex]\frac{1+2+3}{2 \times 170000 + 5 + 8 +3} = \frac{6}{340016}= 0.00001764622 < 0.025[/itex]
    and so n=1 is keeping the inequality
     
  4. Sep 7, 2016 #3
    Sorry. I didn't check to see if the format changed once I copied and pasted my problem:
    ==============================================================

    Here is the problem:


    For the inequality below, find all values n ∈ N such that the inequality is true:

    (n2 + 2n +3) / (2n3 + 5n2+ 8n + 3) < 0.025.


    ============================
    Here is my attempt at the problem:


    Looking at the following set

    {n ∈ N: (n2+ 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


    We want to find the lower bound of this set.


    Suppose A denotes the above set, then we have


    A= {n ∈ N: (n2 + 2n +3) / (2n3 + 5n2 + 8n + 3) < 0.025}


    Since the above rational function can be reduced to 1/(2n+1) we have


    1/(2n+1) <0.025


    Where we get n>19.5. Since the lower bounds of the set are 19.5, 19.4, 19.3.... And so on, this set is bound below by the natural number n=19. Therefore, the above inequality holds true for all n ∈ N greater than or equal to 20. Thus A= {20, 21, 22, 23,...}
     
  5. Sep 9, 2016 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I don't see the purpose of all those "steps" before reducing the rational function (which you should probably show in more detail).
    Okay (showing the steps wouldn't hurt).
    What is the relevance of those numbers?

    Once you have n>19.5 you can directly conclude that n>=20, and write down the set of solutions.
     
  6. Sep 9, 2016 #5

    RUber

    User Avatar
    Homework Helper

    One method that might help with the simplification is to flip the inequality.
    ##\frac{ n^2 +2n+3}{2n^3 + 5n^2 + 8n+3} < .025 \equiv \frac{2n^3 + 5n^2 + 8n+3}{ n^2 +2n+3}>40##
    Which as you pointed out can be written as
    ##2n+1 > 40.##
    And you already have the solution.
     
  7. Sep 9, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Be careful: to do that you have to check that the two sides cannot be negative. This is easy to do here, but it is a necessary step.
     
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