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Exponential function 2^x-sum.

  1. Oct 1, 2011 #1
    I posted a related thread, in the same forum regarding matlab, but I want to discuss this here separately.

    Few minutes ago I became very confused baffled and surprised at the same time.

    consider function [itex]2^{x}[/itex].

    Now consider the sum of this function for n iterations.(from 0 to n)

    [itex]\sum_{x=0}^n 2^{x}[/itex].

    Using my humble knowledge of c++ I found that you need 19 iterations (including zero as iteration) for a number 524287.

    so 1+2+4+8+16...+262144=524287

    By case, found that this sum is 1 less than the [itex]2^{19}[/itex].

    so [itex]524287=2^{19}-1[/itex]

    Why? I am not best mathematician you will come across but this i found very interesting.
     
  2. jcsd
  3. Oct 1, 2011 #2
    The binomial theorem states that
    [tex]
    (x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}.
    [/tex]

    Now set, x=1 and y=1.
     
  4. Oct 1, 2011 #3
    I don't think I understand.

    General rule I got out of my numbers is:

    [itex]\sum_{i=0}^n 2^{i}=2^{n+1}-1[/itex]

    By putting zeros into my binomial equation i get:

    [itex]2^n =1+ \sum_{k=1}^n {n \choose k}1^{n-k}1^k=1+\sum_{k=1}^n {n \choose k}[/itex]

    How do I relate factorial formula to [itex]\sum_{i=0}^n 2^{i}[/itex]
     
  5. Oct 1, 2011 #4

    AlephZero

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    Homework Helper

    This is an example of a "geometric seriies" (google will find lots of web pages). It is a bit of a "special case" because you used 2 as the multiplying factor.

    In general, suppose you have a sum
    [tex]S = 1 + x + x^2 \cdots + x^{n-1}[/tex]
    You can multiply both sides by x to get
    [tex] Sx = x + x^2 + x^3 + \cdots + x^n[/tex]
    Now subtract the first sum from the second one. All the terms in the sums cancel out except for the first and last ones, and you get
    [tex]Sx - S = S(x-1) = x^n -1[/tex]
    When [itex]x = 2[/itex] the result is even simpler because [itex]x-1 = 1[/itex], and you get
    [tex] 1 + 2 + 2^2 ... + 2^{n-1} = 2^n - 1[/tex]

    BTW I don't really understand what the Binomial theorem has to do with this either.
     
  6. Oct 1, 2011 #5
    This makes sense. Thank you !!
     
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