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mathaddict4471
 3 years ago
challenge calculus question without a calculator use the Maclaurin series to compute e^1 medal and fan rewarded
mathaddict4471
 3 years ago
challenge calculus question without a calculator use the Maclaurin series to compute e^1 medal and fan rewarded

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mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} n \sin (2pien!)\]

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0sry compute equation above using e^1

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Is the expression n.sin(2.pi.e.n!) ?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2...which can't be right. What exactly are we taking the limit of?

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0the first thing you said

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0we're taking the limit of n sin i think

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0i'll use wolfram

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.math.utah.edu/ugrad/calc_challenge/calcchallenge2006.pdf it's number 6

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2I see. Ok, let me think.

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0@mathaddict: the equation must be e^x , right? to apply Maclaurin series, how can you apply to e^1

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2The problem actually says: Using the Maclaurin series for \( e^1 \), calculate \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Ok. This is what I've got so far. When n = 2, n!.e = 2!(1 + 1 + 1/2 + 1/3! + ...) = 2 + 2 + 1 + 2/3! + ... Now sin(2.pi.e.n!) = sin(2.pi ( 2 + 2 + 1 + 2/3! + ... ) ) = sin(2.pi ( 5 + stuff) ) Recall that sin(a + b) = sin a . cos b + cos a . sin b hence sin(2.pi.e.n!) = sin(2.pi.5).cos(stuff) + cos(2.pi.5).sin(stuff) = 0 .cos(stuff) + 1 .sin(stuff) = sin(stuff) This argument generalizes to \[ \sin(2\pi e n!) = \sin \left[ 2\pi \left( \sum_{i=0}^n n!/i! + \sum_{i=n+1}^{\infty} n!/i! \right) \right] = \sin \left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Thus the limit is equal to this limit \[ \lim_{n \rightarrow \infty} n . \sin\left[ 2\pi\sum_{i=n+1}^{\infty} n!/i! \right] \] Now we want to figure out how to bound the quantity in square brackets and show it goes to zero sufficiently fast to cancel out the n outside altogether or end up with some finite number

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0@JamesJ : are you Math Professor?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2In fact that sum is 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... < 1/(n+1) + 1/(n+1)^2 + 1/(n+1)^3 + ... and this infinite geometric series is equal to 1/n Hence the sum is bounded above by the limit as n > infinity of n.sin(2pi/n) = sin(2pi/n)/(1/n) > 2pi So I imagine the limit is in fact 2pi. Now a little more work is required to show that it is exactly 2pi

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Oh, and that's easy because 1/(n+1) + 1/(n+1)(n+2) + 1/(n+1)(n+2)(n+3) + .... > 1/(n+1) and hence the limit is bounded below by the limit of n.sin(2pi/(n+1) which has the same limit. **** Nice question! Yes, I was once.

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0I love your neatness. It means you used to be a Math Professor, right? (sorry for Capitalize the P in professor, because to me, those people deserve that C)

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2This is very counterintuitive result. Adding this to the the archives.

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0Can I make question when i have some via message? I don't want to post and wait wait wait..... until someone have free time .

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Better to post the question and then mail me with the link.

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0how to post the link. once , other asked me to do that but I didn't know how to send the link

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Just copy it from your browser and paste it into a message

Hoa
 3 years ago
Best ResponseYou've already chosen the best response.0do I have to be your fan to send the message? I try reply yours and send message to your profile. both are fail. I cannot submit the mail

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2Hm, possibly. I'm not sure.

mathaddict4471
 3 years ago
Best ResponseYou've already chosen the best response.0thank you soo much for your greatly intuitive answer James, you are the master. best of luck to both of you JamesJ and Hoa
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