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

- anonymous

- katieb

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- anonymous

\[\lim_{n \rightarrow \infty} n \sin (2pien!)\]

- anonymous

sry compute equation above using e^1

- JamesJ

Is the expression
n.sin(2.pi.e.n!) ?

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## More answers

- JamesJ

...which can't be right. What exactly are we taking the limit of?

- anonymous

the first thing you said

- anonymous

we're taking the limit of n sin i think

- anonymous

i'll use wolfram

- anonymous

http://www.math.utah.edu/ugrad/calc_challenge/calc-challenge-2006.pdf it's number 6

- JamesJ

I see. Ok, let me think.

- anonymous

@mathaddict: the equation must be e^x , right? to apply Maclaurin series, how can you apply to e^1

- JamesJ

The problem actually says: Using the Maclaurin series for \( e^1 \), calculate
\[ \lim_{n \rightarrow \infty} n.\sin(2\pi.e.n!) \]

- anonymous

got it

- 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

@JamesJ : are you Math Professor?

- 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

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

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

This is very counter-intuitive result. Adding this to the the archives.

- 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

Better to post the question and then mail me with the link.

- anonymous

how to post the link. once , other asked me to do that but I didn't know how to send the link

- JamesJ

Just copy it from your browser and paste it into a message

- anonymous

ok, let me try once

- 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

Hm, possibly. I'm not sure.

- 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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