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Bernoulli inequality proof

  1. Nov 9, 2005 #1
    Hi I'm doing a small induction proof for bernoullis inequailty:
    Proof:
    Given the inequality [tex]A(n) = (1+x) ^n \geq 1+nx[/tex]
    [tex]r \geq -1[/tex], [tex]n \in \mathbb{N}[/tex]
    Initial step:
    A(n=1) is true cause [tex](1+x) \geq 1 + x[/tex] is true.
    Induction step:
    A(n) is true is since n = 1 and [tex]r \geq -1[/tex] so
    [tex]0 \geq 0[/tex]
    Therefore by the rules of induction
    A(n+1) is true.
    q.e.d.
    Is my proof sufficient ??
    Best Regards,
    Bob
     
    Last edited: Nov 9, 2005
  2. jcsd
  3. Nov 9, 2005 #2
    No.

    In the inductive step, n is supposed to be an /arbitrary/ natural number for which A(n) is true.
     
  4. Nov 9, 2005 #3
    Okay and thank you for your answer,
    What do I need to add under the induction step to complete the proof?
    Do I need to show that A(n+1) is true?
    By that I mean

    [tex] A(n+1) = (1+x) ^{n} (1+x) \geq (1 + nx) (1+x) [/tex]
    [tex] A(n+1) = (1+x) ^{n} (1+x) \geq (1 + x + nx^2)[/tex]

    Is this correct to way to show that A(n+1) is true ?
    Sincerley
    Bob
     
    Last edited: Nov 9, 2005
  5. Nov 9, 2005 #4
    Yes. Given that A(n) is true for some (arbitrary) n, show that A(n + 1) is true.
     
  6. Nov 9, 2005 #5
    You could also prove it by saying:

    [tex]f\left(x\right)=\left(1+x\right)^{n}-1-nx,[/tex]

    then use derivatives to show that it is greater than zero and increasing on a certain interval.
     
  7. Nov 9, 2005 #6
    Correct me if I'm wrong:

    d/dx = n x ^ (n-1)

    Since [tex]n \in \mathbb{N}[/tex] then if n > 0, A(n+1) groves and therefore its true.

    Is that what you mean ?

    best regards,
    Bob
     
  8. Nov 10, 2005 #7

    VietDao29

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    Homework Helper

    ???
    I just wondered what r is in your first post?? :confused:
    -----------------
    There are 3 steps in proof by induction:
    (1) Test if the statement's true for n = 0.
    (2) Assume the statement is true for n = k.
    (3) Prove the statement is true for n = k + 1 using the induction hypothesis (2).
    ----------
    (1) So you have shown that for n = 1, the equality is true. Or you can even show that the inequality is true for n = 0.
    For n = 0, you'll have:
    (1 + x)0 ≥ 1 + 0x. And that's true!
    (2)Then assume the inequality is true for n = k, ie:
    (1 + x)k ≥ 1 + kx.
    (3)Now let's prove it's true for n = k + 1. That means, you have to prove:
    (1 + x)k + 1 ≥ 1 + (k + 1)x.
    So (1 + x)k + 1 = (1 + x)k (1 + x)
    Since x ≥ -1, so 1 + x ≥ 0.
    Using the induction hypothesis, you have:
    (1 + x)k ≥ 1 + kx
    <=> (1 + x)k (1 + x) ≥ (1 + kx) (1 + x) (Note that: 1 + x ≥ 0).
    <=> (1 + x)k (1 + x) ≥ 1 + x + kx + kx2. From there, can you show that (1 + x)k + 1 ≥ 1 + (k + 1)x?
    ---------------------
    You don't have to prove it by taking the derivatives of f(x). But by the way, your f'(x) is wrong.
    f(x) = (1 + x)n - 1 - nx, then
    f'(x) = (1 + x)n' - 1' - nx' = n(1 + x)n - 1 - n.
     
    Last edited: Nov 10, 2005
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