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If a>-1, show that (1+a)^n>=1+na

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]a> -1, (1+a)^n \geq 1+na[/tex]


    2. Relevant equations
    [tex]a>-1[/tex]
    [tex]a+1> 0[/tex]

    3. The attempt at a solution
    If I let n=0, and then n=1 i get that 1≥1, and a+1≥ a+1.

    Add to each side n+1
    [tex](1+a)^{n+1}+(1+a)^{n} \geq 1+na+(1+a(n+1))[/tex]
    Then...? Perhaps subtract the original expressions on each side and then evaluate if i can show that n+1-n is ≥ n+1-n
    [tex](1+a)^{n+1}-(1+a)^n \geq 1+a(n+1)-(1+an)[/tex]
    [tex]a(a+1)^n \geq a[/tex]

    However, I don't really know what I'v done. for a>-1, is it obvious that a(a+1)^n≥a?
    Or have I attacked the problem in a wrong way? I'm new to induction, and completely green when it comes to bigger than, geq, etc. instead of =
     
  2. jcsd
  3. Oct 5, 2011 #2

    Mark44

    Staff: Mentor

    Either of these serves as your base case. I would use n = 1, though.
    ???
    You started off sort of OK by establishing a base case (in fact, two of them).
    There are two more parts in an induction problem.
    1) The induction hypothesis, which you assume is true. For your problem, assume that the proposition is true for n = k.
    That is, assume that (1 + a)k >= 1 + ka
    2) The induction step. Use the induction hypothesis of the previous step to show that the proposition is true for n = k + 1. IOW, show that (1 + a)k + 1 >= 1 + (k + 1)a is true.
     
  4. Oct 5, 2011 #3
    If I set n=k+1 I can write the expression like this


    [tex](a+1)^k(a+1)\geq 1+(k+1)a[/tex]
    [tex](a+1)^k(a+1)\geq 1+ak+a[/tex]
    Since we assumed that (1 + a)^k ≥ 1 + ka, can I then look at
    [tex](a+1)^k(a+1)-(a+1)^k\geq 1+ak+a-(1+ak)[/tex]
    [tex]a(a+1)^k\geq a[/tex]

    Since k is at least 0, which gives me a ≥ a.

    This whole thing feels very unintuitive, I dont really know when I have proven something.
     
    Last edited: Oct 5, 2011
  5. Oct 5, 2011 #4

    Mark44

    Staff: Mentor

    No, you can't do this (above). You have to show this, not start off with it.

    Start with (a + 1)k + 1 = (a + 1)*(a + 1)k
    Now, what can you do with the (a + 1)k expression?
     
  6. Oct 5, 2011 #5
    In words:

    The original problem, if I set up a base(n=0 already tested and proven) n=1[n=n-(n-1)], I get the statement to be true. Then I will try to see what happens if n=k[n=n-(n-k)], 1<k<n.

    Then I will get (a+1)^k ≥ 1+ak. If this is true, then k-1 will also be true.
    n=k-1[n=n-(n-(k+1))

    (a+1)^(k-1) ≥ 1+a(k-1)
    =
    (a+1)*(a+1)^(k-2) ≥ 1+ak-a
    Since k is a positive integer bigger than 1, it is at least 2. Lets check it for k=2
    (a+1)(1) ≥ 1+a, it holds!

    I will try to see if it holds for n=k

    (a+1)^k ≥ 1+ak, and if k=2

    a^2+2a+1 ≥ 1+2a, and since a^2 is positive this holds aswell!

    n=k+1

    (a+1)*(a+1)^k ≥ 1+(k+1)a
    =
    (a+1)*(a+1)^k ≥ 1+ak+a

    If k=2

    a^3+3a^2+3a+1 ≥ 1+3a, and since if a is negative it is smaller than -1 and that says that a^3 will be smaller than 3a^2 so again it holds.


    If my statements are correct, how can they be improved. If not, where did I go wrong?
     
  7. Oct 5, 2011 #6

    SammyS

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    a2 ≥ 0

    ∴ 1 + (k+1)a + a2 ≥ 1 + (k+1)a
     
  8. Oct 5, 2011 #7
    Ahh, I had completely forgotten that I could simplify on both sides of the ≥. Thanks both of you.
     
  9. Oct 5, 2011 #8

    Mark44

    Staff: Mentor

    Why do you have the stuff in the brackets? n=1[n=n-(n-1)] and n=k[n=n-(n-k)]


    No, this is not right. You are assuming what you need to prove.

    Also, two inequalities are not equal, so don't connect them with =.
    It's hard to follow what you are doing, but it looks to me like you are just establishing some more base cases.

    There's a certain template that you should follow.
    1) Establish that the proposition is true for some base case (such as n = 1). You have done that.
    2) Assume that the proposition is true for n = k.
    IOW, assume that (1 + a)k >= 1 + ka.
    3) Using the assumption in #2, show that the proposition is true for n = k + 1.
    IOW, show that (1 + a)k + 1 >= 1 + (k + 1)a
    Take a look at what I did in post #4.
     
  10. Oct 5, 2011 #9
    Ok, I'm off to sleep now but I will do another take on it tomorrow. The brackets were to show that 0<1<k<n<n+1.
    Thanks so far though.
     
  11. Oct 5, 2011 #10

    Mark44

    Staff: Mentor

    You don't need that.
     
  12. Oct 6, 2011 #11
    Ok, I will go from my base case, n=1 is true.

    Now I assume n=k is true. From that assumption I show that n=k+1 is true.

    (a+1)^(k+1) ≥ 1+a(k+1)

    (a+1)*(a+1)^k ≥ 1+ak+a

    (a+1) ≥ (1+ak+a)/(a+1)^k

    (a+1) ≥ (1+ak+a)(a+1)^-k

    or perhaps

    (a+1)^(k+1) ≥ 1+a(k+1)

    (a+1)*(a+1)^k ≥ 1+ak+a

    -ak-a+(a+1)*(a+1)^k ≥ 1

    Are my induction-process-assumptions correct? So I know that I can focus on algebraic operations on my "(a+1)^(k+1) ≥ 1+a(k+1)" expression to show that it is true for n=k+1.
     
  13. Oct 6, 2011 #12

    Mark44

    Staff: Mentor

    Don't say that n = 1 is true and that n = k is true. Instead, say that that the proposition is true for n = 1, and that the proposition is true for n = k. Your job is to show that the proposition is true for n = k + 1.
    When you write the above, you are tacitly assuming that it is true when you apply operations to both sides. You can't do that - you need to provide the work that shows it is true.

    Your work should look something like this:
    (a + 1)k+1 = (a + 1)*(a + 1)k = ...<stuff>... ≥ ...<other stuff>... ≥ 1 + a(k + 1)

    In that way, you will have shown that (a + 1)k+1 ≥ 1 + a(k + 1). You need to fill in where I have <stuff> and <other stuff>.

    Something you need to fill in is what you can exchange for (a + 1)k, based on your induction hypothesis.

     
  14. Oct 6, 2011 #13
    I think i got it earlier today. Im the phone atm, so I cant supply any details, but I ended up comparing (1+ak)(a+1) \geq (1+ka+a)
     
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