Results 1 to 3 of 3

Math Help - Double Induction for (1+t)^t > 1+nt

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    8

    Double Induction for (1+t)^t > 1+nt

    Hi, I have the following problem which I think I have solved correctly (it would be great if you could check it), but I don't know how to answer to "simple" questions about my proof... any help would be very much appreciated, thanks!

    The problem says:
    "Let t>-1 be given with t≠0. Show by induction that for n≥2 we have
    (1+t)^t>1+nt."

    And here are the two questions that I don't know:

    "Where is the condition t>-1 used in your proof? Is the result true if instead we had -2<t<-1?"

    My solution to the first bit is:

    Let F(n)=(1+t)^n>1+nt.

    Proof by induction on n.

    Base n=0. Then (1+t)^n=(1+t)^0=1=1+0t=1+nt

    Induction. Suppose that there is a number k (k≥2) such thaqt (1+t)^k>1+kt for any t>-1 (t≠0). We need to show that (1+t)^{k+1}>1+(k+1)t for any t>-1, t≠0.

    (1+t)^{k+1}=(1+t)^k*(1+t). By the induction hypothesis, (1+t)^k>1+kt, so we have:
    (1+t)^k*(1+t) \geq (1+kt)*(1+t)=1+kt+t+kt^2=1+(1+k)t+kt^2.

    Since t^2>0 and k \geq 2, 1+(1+k)t+kt^2 \geq 1+(1+k)t.

    So (1+t)^{k+1} \geq 1+(k+1)t

    And so the result holds for n=0 and hence for all n≥2 and t>-1,t≠0.
    Last edited by jmgilbert; March 10th 2011 at 12:38 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Everything looks good, except for one thing: you skipped the n = 1 case! It's a small technicality, but it does actually render your proof invalid. You definitely have the right idea with your proof. I would clean up your notation a bit. Note: it's \geq or \ge, not /geq. That'll make your post a lot more intelligible.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Two comments. First, as Ackbeet says, you must start the induction at n=2, not n=0, if you want to get the strict inequality (for t\ne0).

    The second comment is that in the inductive step you multiplied both sides of an inequality by (1+t). Now you ought to know that an inequality is preserved if you multiply both sides by a positive number, but not if you multiply both sides by a negative number. That should help you to answer the "simple" questions about the proof.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Strong induction vs. structural induction?
    Posted in the Discrete Math Forum
    Replies: 13
    Last Post: April 21st 2011, 12:36 AM
  2. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  3. Mathemtical Induction Proof (Stuck on induction)
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 8th 2009, 09:33 PM
  4. Double Abs. Value
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 19th 2008, 02:18 PM
  5. Double or Nothing?
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: July 27th 2008, 04:31 AM

Search Tags


/mathhelpforum @mathhelpforum