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

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mathaddict4471Best ResponseYou've already chosen the best response.0
\[\lim_{n \rightarrow \infty} n \sin (2pien!)\]
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
sry compute equation above using e^1
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
Is the expression n.sin(2.pi.e.n!) ?
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
...which can't be right. What exactly are we taking the limit of?
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
the first thing you said
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
we're taking the limit of n sin i think
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
i'll use wolfram
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
http://www.math.utah.edu/ugrad/calc_challenge/calcchallenge2006.pdf it's number 6
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
I see. Ok, let me think.
 one year ago

HoaBest 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
 one year ago

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

JamesJBest ResponseYou've already chosen the best response.2
Ok. 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
 one year ago

HoaBest ResponseYou've already chosen the best response.0
@JamesJ : are you Math Professor?
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
In 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
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
Oh, 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.
 one year ago

HoaBest ResponseYou've already chosen the best response.0
I 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)
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
This is very counterintuitive result. Adding this to the the archives.
 one year ago

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

JamesJBest ResponseYou've already chosen the best response.2
Better to post the question and then mail me with the link.
 one year ago

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

JamesJBest ResponseYou've already chosen the best response.2
Just copy it from your browser and paste it into a message
 one year ago

HoaBest ResponseYou've already chosen the best response.0
do 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
 one year ago

JamesJBest ResponseYou've already chosen the best response.2
Hm, possibly. I'm not sure.
 one year ago

mathaddict4471Best ResponseYou've already chosen the best response.0
thank you soo much for your greatly intuitive answer James, you are the master. best of luck to both of you JamesJ and Hoa
 one year ago
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