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ParthKohli

  • 2 years ago

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.

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  1. ParthKohli
    • 2 years ago
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    So, the sequence is \(\{2,4,8,16\cdots \}\).

  2. ParthKohli
    • 2 years ago
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    Am I correct?

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

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

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

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

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

  8. agentx5
    • 2 years ago
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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

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

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

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

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

  13. ParthKohli
    • 2 years ago
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    Oh, okay.

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

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

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

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

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

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

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

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

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

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

  24. ParthKohli
    • 2 years ago
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    Hmm.

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

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

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

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

  29. ParthKohli
    • 2 years ago
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    It does =)

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

  31. agentx5
    • 2 years ago
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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 ^_^

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

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

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