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!

Ways of calculating e

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that all these equations are equal to calculate e:

    e1= lim (1+1/n)^n - for n to infinity
    e2= E 1/k! - for k from 0 to m
    e3= 1 / E[(-1)^k/k!] - for k from 0 to m
    e4= E(k^2)/2k! - for k from 1 to m
    e5= E(k^3)/5k! - for k from 1 to m
    e6= E(k^4)15k! - for k from 1 to m

    I used E as the symbol for summing a series of numbers. Guess it´s the most similar letter to that symbol. Hope you understand the equations...if not I have them in a word document.

    2. Relevant equations

    MacLaurin series, binomial expansion...etc

    3. The attempt at a solution

    e1 is done since it´s the most common definition for e.
    e2 is done thanks to the binomial expansion

    e3 have no idea
    e4 to e6 I think it has something to do with Bell´s number, but don´t know how to prove these are equal. I tried expanding them over several pages and finding some common factors but to no use.

    Can someone help me? I´ve browsed the web for days but I can only find people saying that they are ways of calculating e and not the reason why they are equal.
    Many thanks!
     
    Last edited: Apr 5, 2009
  2. jcsd
  3. Apr 5, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You should have seen the third one. You know e^x=sum x^k/k! for k=0 to infinity, right? If you look at the denominator of e3, it's e^(-1). To get e4 look at the derivative of e^(x) at x=1 by differentiating the power series. To get the others look at higher derivatives of e^(x) at x=1. They are all equal to e, right?
     
  4. Apr 5, 2009 #3
    Thanks Dick for the answer. But I don´t understand ¨look at the derivative of e^(x) at x=1 by differentiating the power series¨. I do calculus in spanish so couldn´t get your idea.
    How does Bell´s Number play into the last three formulas?
    Why does k start from 1 and not from 0 in those last 3 formulas?

    e3 is like you said. Can´t see why I didn´t get that one. Must be too much coffee blocks my brain!!!

    Cheers and thanks again!
     
  5. Apr 5, 2009 #4

    Pengwuino

    User Avatar
    Gold Member

    The n=0 terms in the last 3 do not contribute anything as opposed to the first 3 where they do (0! = 1).
     
  6. Apr 5, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I mean that e^(x)=sum x^k/k!. The derivative of e^(x) is e^(x). The derivative of x^k/k! is k*x^(k-1)/k!, right? So e^(x) is also equal to sum k*x^(k-1)/k!. Try evaluating at x=1. Now take higher derivatives. I don't think you need Bell's numbers.
     
  7. Apr 5, 2009 #6
    ok now I get for the first derivative e^1 = k.(1)^k-1 / k!, which is the same as writing 1/(k-1)! That´s not the same as e4, but lets see the next derivative:
    d2 e^x/d2 x = e^x = k . (k-1) . x^(k-2) / k!

    which evaluated in x= 1 gives something like [ k^(2)-k ] / k! which isn´t the same as e4= k^2/ 2k!

    That 2 underneath has me confused. The same as the 5 and 15 in e5 and e6 respectively.
     
  8. Apr 5, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are getting there. You've got sum k/k!=e. Now you also have sum k^2/k!-k/k!=e. So k^2/k!=2e, right? Solve for e!
     
  9. Apr 5, 2009 #8
    Done...the last two are just adding and subtracting blocks that are equal to e that we already know.

    Many thanks to both of you!

    Have a great week.
    Nico
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook