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Expanding 2^(ab)-1

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to understand a proof for Mersenne number theorem, but I'm having trouble understanding the expansion from 2ab-1 into (2a-1)((2a)b-1+(2a)b-2+....+2a+1)).
    2. Relevant equations
    Not sure

    3. The attempt at a solution
    I don't know where to start really and I can't seem to find any proof that explains what is going on in this step.
     
  2. jcsd
  3. Oct 21, 2015 #2

    Mark44

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    They are factoring ##2^{ab} - 1## into the two factors you show. Note that ##(a^m)^n = a^{mn}##.
     
  4. Oct 21, 2015 #3
    Sorry, I forgot to include that part. I understand that (am)n = amn. I don't see how see how (2a)b-1 helps with factoring this though. To be honest, before I saw this proof, I thought there were no other ways to factor it. Is there some factoring rule that I am missing?
     
  5. Oct 21, 2015 #4

    Mark44

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    It's not a rule that you would learn in an algebra class, I don't believe. Try it out with some specific numbers. For example, with a = 3 and b = 2, you have ##2^6 - 1 = (2^3 - 1)(2^3 + 1)##.
    With a = 3 and b = 3, you have ##2^9 - 1 = (2^3 - 1)(2^6 + 2^3 + 1)##
     
  6. Oct 21, 2015 #5
    I was trying it out with (22)3 and it simplifies to the same result. I just really wanted to know how that equivalence was derived.
     
  7. Oct 21, 2015 #6

    Dick

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    Look at the formula for summing a geometric series.
     
  8. Oct 21, 2015 #7
    Would that make it: ((2ab(1-2ab)/(1-2)) -1 = -22ab -1 ? Am I doing something wrong here? Ignoring the -1 right now, the first term is 2ab, n = ab, and the common ratio is 2.
     
  9. Oct 21, 2015 #8

    Ray Vickson

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    Have you ever seen the expansion
    [tex] x^n - 1= (x-1)(1+x+x^2+ \cdots x^{n-1}) [/tex]
    for any ##x## and positive integer ##n##? Just apply it to ##x = 2^a## and ##n = b##, provided that ##b## is a positive integer.
     
  10. Oct 21, 2015 #9
    I don't know how that expansion is derived. I understand how to get expression on the right side by dividing the original polynomial by (x-1), but I don't know how (x-1) was factored with begin with.
     
  11. Oct 21, 2015 #10

    Ray Vickson

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    Sorry, I assumed everybody had seen that result before, as it is a part of elementary algebra. You can derive it from the formula for the sum
    [tex] 1 + x + x^2 + \cdots + x^k [/tex]
    Google 'geometric series'.
     
  12. Oct 21, 2015 #11
    Would't that result in = 1(1-xk)/(1-x) = (1-x)k-1?
     
  13. Oct 21, 2015 #12

    Ray Vickson

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    No.

    Anyway, why would you think that ##(1-x^k)/(1-x) = (1-x)^{k-1}##? That is false. Try it for yourself: put ##x = 1/2## and ##k = 3## into both sides and see what happens.

    Have you done what I suggested (Google 'geometric series')?
     
  14. Oct 21, 2015 #13

    epenguin

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    It looks like you know everything but don't know your know it.
    You know how to get from left to right and right to left.
    If the question is how did anyone hit on the fact (x - 1) is a factor of (x n - 1) without someone telling them, or the examples, they could have noticedn that x = 1 makes (x nn - 1) be 0 and after a bit they said

    And you should have met it in another form, see #6, though not always the relation is pointed out.
     
    Last edited by a moderator: Apr 28, 2017
  15. Oct 21, 2015 #14
    Yeah, I did google geometric series. I looked at the sum of geometric series formula when the other guy posted about it. And I got:
    1 = first element
    x = common ratio
    k = nth element
     
  16. Oct 21, 2015 #15

    Ray Vickson

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    And what formula did they give you for the finite sum?
     
  17. Oct 21, 2015 #16
    That makes sense, but how does 2a-1 factor out of (2a)b-1? 2 raised to a power minus 1 will not be zero anymore unless the exponent is 0.
     
    Last edited by a moderator: May 7, 2017
  18. Oct 21, 2015 #17
    (a1)(1-rn/1-r) = (1)(1-xk/1-x) = (1-xk)/(1-x)
     
  19. Oct 21, 2015 #18

    epenguin

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    :oldtongue:Re #16 Just look at formula in #8 for instance.

    2a is x and b is n.b

    Note that since 2ab = (2a)b = (2b)a you can get another valid formula just exchanging a with b everywhere, there is nothing to distinguish between a and b algebraically, though extra-mathematically you might have decided on them standing for particular different things.

    But maybe the question was a different one: you did not realise that nothing was being raised to power -1: the guy was just using correct notation (for a subtraction of 1 from (2a)b) which a lot of people don't bother with here. :oldtongue:
     
    Last edited: Oct 21, 2015
  20. Oct 21, 2015 #19

    Mark44

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    No, these aren't right.
    ##2^{ab} = (2^a)^b \neq 2^{a^b}##
    For example, ##2^{2\cdot 3} = (2^2)^3 = 2^6 = 64##, but ##2^{2^3} = 2^8 = 256##
     
  21. Oct 21, 2015 #20

    epenguin

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    Thank you typo corrected
     
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