1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of a sine

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Find this limit

    [tex] \lim_{n \rightarrow \infty} n\cdot sin(2 \pi n!e) [/tex]


    2. Relevant equations
    hint: use the euler power series


    3. The attempt at a solution
    Well I know that:

    [tex]e = \sum_{k=0}^{\infty} \frac{1}{k!} [/tex]

    But how can I use this information?
     
  2. jcsd
  3. Jul 8, 2008 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    That is the power series representation for e.

    What is the power series representation of the sine function?
     
  4. Jul 8, 2008 #3
    [tex] \sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots [/tex]

    Do I have to put one series into the other?
     
  5. Jul 8, 2008 #4
    Unfortunately, I have erred in solution
    Equal second incorrect
     
  6. Jul 8, 2008 #5
    What do you mean?
     
  7. Jul 8, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Look at the terms in the argument of the sine for a large value of n after you've used the power series expansion for e. Many are integer multiples of 2pi. Now look at the first one that isn't. Can you justify ignoring the terms after that? Think l'Hopital.
     
  8. Jul 8, 2008 #7

    G01

    User Avatar
    Homework Helper
    Gold Member

    Good call Dick. I don't know what I was trying to do when I posted before about the sine expansion. That's what I get for posting at work...
     
  9. Jul 9, 2008 #8
    i think answer is 0

    n!*e comes out to be nP0+nP1+nP2+nP3.................all integers
    sin(2pi*integer)==0
    infinity * 0=0
     
  10. Jul 9, 2008 #9
    You're wrong the answer isn't zero.
     
  11. Jul 9, 2008 #10
    would you please tell where i am going wrong....
     
  12. Jul 9, 2008 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Put in the infinite series for e. A lot of terms are multiples of two pi. They aren't all. The first one that isn't contributes 2*pi/(n+1) to the sine argument. (There are more but they can be ignored relative to this one for large n).
     
  13. Jul 9, 2008 #12
    how n+1??
    both the n(one of the n! and other of the e expansion) go upto infinity..

    so it would expand to:
    2pi + 2pi(n) + 2pi(n)(n-1)+ 2pi(n)(n-1)(n-2) +................................2pi(n!)
    2pi(n!) being the largest one.

    please, more explanation needed ..
     
  14. Jul 9, 2008 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Insert the power series first. Then let n go to infinity. Don't try and do both at the same time. The 2*pi*n! term comes from the k=0 term in the expansion, 2*pi comes from the k=n term, right? What about the k=n+1 term?
     
  15. Jul 9, 2008 #14
    i am not getting it.
    could you please send me the whole solution..
     
  16. Jul 9, 2008 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    We aren't in the business of providing 'whole solutions'. If you have another question ask it. I don't see what you aren't 'getting'. Put infinite series in sine argument - decide what terms are important. Take limit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Limit of a sine
Loading...