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Mathematical Induction: Chapter 1, Example 4 from WiM? by Courant

  1. Jan 8, 2010 #1
    This is not homework, per se, but I have recently started reading Courant and Robbins' What is Mathematics?

    1. The problem statement, all variables and given/known data
    Prove by mathematical induction.
    (1+q)(1+q2)(1+q4) ... (1+q2n) = (1-q2n+1) / (1-q)

    2. Relevant equations
    Only the problem itself.


    3. The attempt at a solution
    Using the method the book briefly introduced, I first established that A1 was true.

    A1 = (1+q)(1+q2) = 1+q+q2+q3

    Furthermore,

    (1-q21+1) / (1-q) = (1-q4) / (1-q) = (1-q2)(1+q2) / (1-q) = (1+q)(1+q2)

    Next, I attempted to show that Ar+1 follows from Ar, but this is where I ran into trouble.

    1. (1+q)(1+q2)(1+q4) ... (1+q2r) = (1-q2r+1) / (1-q)

    2. (1+q)(1+q2)(1+q4) ... (1+q2r)(1+q2r+1) = {(1-q2r+1) / (1-q)}(1+q2r+1)

    This is the point at which I hit a wall. I feel like it's something absurdly simple, but I've been stuck on this for hours; I think it's the variable raised to a power raised to a binomial power that is throwing me for a loop.

    I tried simply multiplying (1-q2r+1) and (1+q2r+1); that gives me another binomial that is the difference of two squares:

    {(1-(q2r+1)(q2r+1)} / (1-q)

    But again, stumped. I believe that if I multiply those two terms together I get (1-q2r+1+2r+1) / (1-q). If that's the case, this is the point at which I have completely lost it. However, I am not confident that I'm doing that properly.

    Just to see if I could work backwards, I took a look at what I think Ar+1 would look like:

    Ar+1 = (1-q2r+2) / (1-q)

    Any hints on where/what I'm doing wrong would be very much appreciated.

    P.S. - All of this is very new to me, so I apologize if it's not quite as clear or as polished as it should be.
     
  2. jcsd
  3. Jan 8, 2010 #2
    I don't understand your parenthesis.
    well..
    [tex]A_{r+1}=(1+q^{2^{r+1}})\frac{1-q^{2^{r+1}}}{1-q}[/tex]
    Just by focusing on the numerator,
    [tex](1+q^{2^{r+1}})(1-q^{2^{r+1}})[/tex]
    Gives us
    [tex]1-q^{2^{r+1}\times2}[/tex]


    I'll leave the rest to you.
    Whenever you are stuck, check your previous workings , then try other approach to that question.
     
  4. Jan 8, 2010 #3
    Sorry about the parenthesis and brackets, I'm not sure how to clearly represent fractions with the forum-code.

    And I'm afraid that 1-q2r+1x2 takes me right back to where I was stuck. Any additional hint to nudge me in the right direction? :)
     
  5. Jan 8, 2010 #4
    law of indices:

    [tex]a^b \times a^c[/tex]=[tex]a^{b+c}[/tex]

    Now:
    [tex]2^{r+1} \times 2^{1}[/tex] = ?
     
    Last edited: Jan 8, 2010
  6. Jan 8, 2010 #5
    Aaaaargh.

    It's not like I didn't just do that this morning on a previous problem. Nor is it like I wasn't staring at 1-q2(2r+1) while trying to solve this.

    2r+1 x 21 = 2r+1+1 = 2r+2

    Thanks very much for the help.
     
  7. Jan 8, 2010 #6
    Welcome! Never ever abandon your law of indices agn (=
     
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