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Proof by Induction

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex](1+x)^{k}[/tex][tex]\geq[/tex]1+kx

    2. Relevant equations



    3. The attempt at a solution
    I want to show for P(k+1)
    (1+x)^(k+1)[tex]\geq[/tex]1+kx+x
    (1+x)^k*(1+x)[tex]\geq[/tex]1+kx+x
     
  2. jcsd
  3. Dec 7, 2008 #2
    Use this:

     
  4. Dec 7, 2008 #3
    I don't really understand how I use that...
     
  5. Dec 7, 2008 #4
    You want to use an inequality involving [tex](1 + x)^k[/tex] to derive an inequality involving [tex](1+x)^k (1+x)[/tex], which is a multiple of it.

    An inequality still holds if you multiply both sides by a positive number. If you multiply both sides of an inequality by a negative number, then you have to flip the sign from [tex]\le[/tex] to [tex]\ge[/tex] or vice versa.
     
  6. Dec 7, 2008 #5
    Ok, so:
    (1+x)^k(1+x)>(1+kx)(1+x)
    >(1+2kx+x)
     
  7. Dec 7, 2008 #6
    Right idea, but there is an error in your expansion.

    Does the question give any restrictions on x? Is what you wrote still true of 1 + x < 0?
     
  8. Dec 7, 2008 #7
    No restrictions.
    1+kx+x+kx^2
     
  9. Dec 7, 2008 #8
    When k = 3 and x = -4,

    [tex](1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx[/tex]

    which makes the statement false.
     
  10. Dec 7, 2008 #9
    So, we have to assume 1+x>0
     
  11. Dec 7, 2008 #10
    so, x>-1
     
  12. Dec 8, 2008 #11

    Mark44

    Staff: Mentor

    I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.
     
  13. Dec 8, 2008 #12
    Once I get here I'm unsure where to go
     
  14. Dec 8, 2008 #13
    Look at your first post; you need to show that that is [tex]\ge 1 + kx + x[/tex].
     
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