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ParthKohli Group Title

Notation of a sequence: What do we mean by \(n \ge 1\) in \(\{ 2^n\}_{n \ge 1}\)? Is it all the terms of the sequence only have \(n = 1\) or more? So basically \(n\) iterates to the positive numbers.

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    So, the sequence is \(\{2,4,8,16\cdots \}\).

    • 2 years ago
  2. ParthKohli Group Title
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    Am I correct?

    • 2 years ago
  3. ParthKohli Group Title
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    @agentx5 I need a genius for this.

    • 2 years ago
  4. agentx5 Group Title
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    Powers of two, while n is greater than or equal to 1. Yep! You're good

    • 2 years ago
  5. agentx5 Group Title
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    Infinite sequence in fact, right? (i.e.: what's the limit as n goes to infinity for 2\(^n\) ? ) ;-)

    • 2 years ago
  6. ParthKohli Group Title
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    Ah! Great stuff! \(\{ 1,2,3,4,5,6,7\} = \{n \} _{7 \ge n \ge 1}\)

    • 2 years ago
  7. ParthKohli Group Title
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    It never reaches anywhere, so it's infinity. That was an oral question :P

    • 2 years ago
  8. agentx5 Group Title
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    This is how you typically see these though, as series (sums of the terms in a sequence) \(\sum_{n \ge 1}^{\infty} 2^n\) = \(\infty\) or "Diverges" as they say

    • 2 years ago
  9. ParthKohli Group Title
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    |dw:1342971644614:dw|

    • 2 years ago
  10. ParthKohli Group Title
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    But what if \(n\) is negative?

    • 2 years ago
  11. agentx5 Group Title
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    n isn't negative. or zero

    • 2 years ago
  12. agentx5 Group Title
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    It starts at 1.

    • 2 years ago
  13. ParthKohli Group Title
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    Oh, okay.

    • 2 years ago
  14. ParthKohli Group Title
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    What if we have negatives included?

    • 2 years ago
  15. agentx5 Group Title
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    \(\{n \} _{7 \ge n \ge 1}\) <--- this however has an upper boundary

    • 2 years ago
  16. agentx5 Group Title
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    2^7 = 128

    • 2 years ago
  17. agentx5 Group Title
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    so... {2,4,8,16,32,64,128}

    • 2 years ago
  18. ParthKohli Group Title
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    The limit is 0 if we have \(\{ 2^n\}_{n \le 0}\) right?

    • 2 years ago
  19. ParthKohli Group Title
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    I believe that the sequence \(\{2,4,8,16,32,64,128 \}\) doesn't have a limit.

    • 2 years ago
  20. agentx5 Group Title
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    Well for one thing you've got the wrong graph here, what happens when you raise to a negative power?

    • 2 years ago
  21. ParthKohli Group Title
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    Exactly, we have it getting closer to 0.

    • 2 years ago
  22. agentx5 Group Title
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    |dw:1342971841230:dw|

    • 2 years ago
  23. ParthKohli Group Title
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    Starts at 1 and keeps getting closer to 0.

    • 2 years ago
  24. ParthKohli Group Title
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    Hmm.

    • 2 years ago
  25. ParthKohli Group Title
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    Oops. I posted the graph for \(x^2\).

    • 2 years ago
  26. ParthKohli Group Title
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    \[\lim _{x \to -\infty} 2^x = 0\] Good enough?

    • 2 years ago
  27. agentx5 Group Title
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    Yep sure does, as you approach negative infinity \[\huge \lim_{x \rightarrow -\infty} 2^x = 2^{-\infty} = (\frac{1}{2^{\infty}}) = \frac{1}{\infty} = 0\]

    • 2 years ago
  28. agentx5 Group Title
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    See how the Algebra keeps coming back?

    • 2 years ago
  29. ParthKohli Group Title
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    It does =)

    • 2 years ago
  30. ParthKohli Group Title
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    I believe that I'd owe you a lot when I get into MIT :)

    • 2 years ago
  31. agentx5 Group Title
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    I'm used to tutoring students with learning disabilities :-) That's my part-time job on campus. Maybe you can help me find a job at some point in the future, ya never know ^_^

    • 2 years ago
  32. ParthKohli Group Title
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    Heh. You're a nice guy!

    • 2 years ago
  33. ParthKohli Group Title
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    And nice guys don't finish last ;)

    • 2 years ago
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