Question

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

Answer 1

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

Answer 2

sry compute equation above using e^1

Answer 3

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

Answer 4

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

Answer 5

the first thing you said

Answer 6

we're taking the limit of n sin i think

Answer 7

i'll use wolfram

Answer 8

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

Answer 9

I see. Ok, let me think.

Answer 10

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

Answer 11

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

Answer 12

got it

Answer 13

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

Answer 14

@JamesJ : are you Math Professor?

Answer 15

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

Answer 16

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.

Answer 17

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)

Answer 18

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

Answer 19

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 .

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

ok, let me try once

Answer 24

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

Answer 25

Hm, possibly. I'm not sure.

Answer 26

thank you soo much for your greatly intuitive answer James, you are the master. best of luck to both of you JamesJ and Hoa