ParthKohli
  • ParthKohli
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
So, the sequence is \(\{2,4,8,16\cdots \}\).
ParthKohli
  • ParthKohli
Am I correct?
ParthKohli
  • ParthKohli
@agentx5 I need a genius for this.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Powers of two, while n is greater than or equal to 1. Yep! You're good
anonymous
  • anonymous
Infinite sequence in fact, right? (i.e.: what's the limit as n goes to infinity for 2\(^n\) ? ) ;-)
ParthKohli
  • ParthKohli
Ah! Great stuff! \(\{ 1,2,3,4,5,6,7\} = \{n \} _{7 \ge n \ge 1}\)
ParthKohli
  • ParthKohli
It never reaches anywhere, so it's infinity. That was an oral question :P
anonymous
  • anonymous
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
ParthKohli
  • ParthKohli
|dw:1342971644614:dw|
ParthKohli
  • ParthKohli
But what if \(n\) is negative?
anonymous
  • anonymous
n isn't negative. or zero
anonymous
  • anonymous
It starts at 1.
ParthKohli
  • ParthKohli
Oh, okay.
ParthKohli
  • ParthKohli
What if we have negatives included?
anonymous
  • anonymous
\(\{n \} _{7 \ge n \ge 1}\) <--- this however has an upper boundary
anonymous
  • anonymous
2^7 = 128
anonymous
  • anonymous
so... {2,4,8,16,32,64,128}
ParthKohli
  • ParthKohli
The limit is 0 if we have \(\{ 2^n\}_{n \le 0}\) right?
ParthKohli
  • ParthKohli
I believe that the sequence \(\{2,4,8,16,32,64,128 \}\) doesn't have a limit.
anonymous
  • anonymous
Well for one thing you've got the wrong graph here, what happens when you raise to a negative power?
ParthKohli
  • ParthKohli
Exactly, we have it getting closer to 0.
anonymous
  • anonymous
|dw:1342971841230:dw|
ParthKohli
  • ParthKohli
Starts at 1 and keeps getting closer to 0.
ParthKohli
  • ParthKohli
Hmm.
ParthKohli
  • ParthKohli
Oops. I posted the graph for \(x^2\).
ParthKohli
  • ParthKohli
\[\lim _{x \to -\infty} 2^x = 0\] Good enough?
anonymous
  • anonymous
Yep sure does, as you approach negative infinity \[\huge \lim_{x \rightarrow -\infty} 2^x = 2^{-\infty} = (\frac{1}{2^{\infty}}) = \frac{1}{\infty} = 0\]
anonymous
  • anonymous
See how the Algebra keeps coming back?
ParthKohli
  • ParthKohli
It does =)
ParthKohli
  • ParthKohli
I believe that I'd owe you a lot when I get into MIT :)
anonymous
  • anonymous
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 ^_^
ParthKohli
  • ParthKohli
Heh. You're a nice guy!
ParthKohli
  • ParthKohli
And nice guys don't finish last ;)

Looking for something else?

Not the answer you are looking for? Search for more explanations.