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Induction proof please help

  1. Feb 8, 2007 #1
    Induction proof please help!!!!!!!

    1. The problem statement, all variables and given/known data

    Let x be any real number in the interval (0,1). Prove that for any natural number n greater or equal to 2 it is true that
    (1+x^2)^n is greater or equal to (1+x^n)^2
    (^ means exponent)

    2. Relevant equations


    3. The attempt at a solution

    Ok well I have been attempting to do this problem for a couple days and it has come to a point when I am not getting any further.:grumpy:

    Here is what I have so far:
    I believe that I should be doing a proof by Induction.

    Let n be a natural number greater than or equal to 2. Let 0 < x <1 and let p(n): (1+x^2) >(or equal to) (1+x^n)^2

    Base case: WWTS p(2) is true.
    p(2): (1+x^2)^2 >(or equal to) (1+x^2)^2. As (1+x^2)^2 = (1+x^2)^2, p(2) is true.

    Inductive step: Assume p(n): (1+x^2)^n >(or equal to) (1+x^n)^2 is true. WWTS p(n+1): (1+x^2)^(n+1) >(or equal to) (1+x^(n+1))^2 is true.

    I guess this is basically where I get stuck. I know that I have to use my assumption p(n) to try to get to p(n+1). I have tried adding different things to each side, multiplying each side by different things but nothing seems to give me what I want.
    I recognize that to get the left sides of the equations to be the same I need to multiply (1+x^2)^n by (1+x^2). And in order to get the right sides to look the same I need to first multiply each out to get rid of the ^2 (just so it is easier to see what is needed)....
    But I am seriously stuck. Any help would be greatly appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 8, 2007 #2


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    Hi, I did not check your work carefully. It is possible that it is a dead end, maybe not. In any case, have you tried casting the problem in terms of binomial series?

    With use of the famous identity on the binomial coefficients, this problem should be easy to tame:http://en.wikipedia.org/wiki/Pascal's_rule
  4. Feb 9, 2007 #3
    I should work using binomial coefficients.

    But I first would examine the relationship between
    [tex] x^n [/tex]
    [tex] x^{n+1} [/tex]

    The fact that x belongs to (0,1) should matter somewhere... ;-) (it's easy to check with some examples that it doesn't work if that condition is not verified)
  5. Feb 9, 2007 #4
    I am not sure how to use binomial coefficients can you please explain more?
    I realize that x^n times X gives me x^(n+1)
    and I have realized that inorder to get the left sides to look the same it is necessary to multiply (1+x^2)^n by (1+x^2) but when I do this, I know that I also have to multiply the right side by this. But,
    (1+x^n)^2 * (1+x^2) =
    (1 +2x^n+x^2n) * (1+x^2)=
    (1 +x^2+2x^n+ 2x^n*x^2+ x^(2n+2))=
    (1 + x^2 +2^n*x^(n+2)+ x^(2n+2))

    But I am trying to get (1+x^(n+1))^2
    which is
    (1+2x^(n+1)+(x^(n+1))^2) =
    (1+2^(n+1)*x^(n+1) +x^(2n+2)

    and these are not the same..I dont think....
    What am I doing wrong?
  6. Feb 9, 2007 #5
    Your equations are fine, but you have to prove an inequality. Have you noticed
    [tex] x^n > x^{n+1} [/tex] ?
    (this doesn't hold if you let x have any value :wink:)
    If it works using that, it may be easier than using binomial coeff. There is a link above pointing to an explanation of the subject... in short words, they are used for writing the general expansion of
    for any n (also called Newton's binomy IIRC)

    BTW:My natural language is Spanish. I suspect that I'm doing some grammar incorrectly, so please forgive me and ignore the misspelling. o:)
  7. Feb 10, 2007 #6
    Ok here is what I have come up with. please tell me if there is a flaw in my reasoning.

    I have
    (1+x^2)^n *(1+x^2) {call this equation A} >/= (greater than or eqaul to) (1+2(x^n)+x^(2n)+x^2+2(x^n)*x^2) {call this equation B}
    I want (1+x^2)^n *(1+x^2) {A} >/= (1 + 2(x^n+1) + x^(2n+2){call this C}
    I am thinking of the transitive property if A >/= B and B >/= C then A >/= C

    Now, comparing equations B and C we can cancel like terms ending up with
    2(x^n)+x^(2n)+x^2+2(x^(n+2)) for B and
    2(x^(n+1)) for C.

    Now since x^n > x^(n+1), 2(x^n) > 2(x^(n+1))
    Therefore, B > C
    Using the transitive property A >/= C

    Am I right?!? Please say yes o:)
  8. Feb 12, 2007 #7
    I looks good to me!
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