anonymous
  • anonymous
challenge calculus question without a calculator use the Maclaurin series to compute e^1 medal and fan rewarded
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} n \sin (2pien!)\]
anonymous
  • anonymous
sry compute equation above using e^1
JamesJ
  • JamesJ
Is the expression n.sin(2.pi.e.n!) ?

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JamesJ
  • JamesJ
...which can't be right. What exactly are we taking the limit of?
anonymous
  • anonymous
the first thing you said
anonymous
  • anonymous
we're taking the limit of n sin i think
anonymous
  • anonymous
i'll use wolfram
anonymous
  • anonymous
http://www.math.utah.edu/ugrad/calc_challenge/calc-challenge-2006.pdf it's number 6
JamesJ
  • JamesJ
I see. Ok, let me think.
anonymous
  • anonymous
@mathaddict: the equation must be e^x , right? to apply Maclaurin series, how can you apply to e^1
JamesJ
  • JamesJ
The problem actually says: Using the Maclaurin series for \( e^1 \), calculate \[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]
anonymous
  • anonymous
got it
JamesJ
  • JamesJ
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
anonymous
  • anonymous
@JamesJ : are you Math Professor?
JamesJ
  • JamesJ
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
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
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)
JamesJ
  • JamesJ
This is very counter-intuitive result. Adding this to the the archives.
anonymous
  • anonymous
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 .
JamesJ
  • JamesJ
Better to post the question and then mail me with the link.
anonymous
  • anonymous
how to post the link. once , other asked me to do that but I didn't know how to send the link
JamesJ
  • JamesJ
Just copy it from your browser and paste it into a message
anonymous
  • anonymous
ok, let me try once
anonymous
  • anonymous
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
JamesJ
  • JamesJ
Hm, possibly. I'm not sure.
anonymous
  • anonymous
thank 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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