# Homework Help: Ways of calculating e

1. Apr 5, 2009

### frestus

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. Apr 5, 2009

### Dick

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?

3. Apr 5, 2009

### frestus

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!

4. Apr 5, 2009

### Pengwuino

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

5. Apr 5, 2009

### Dick

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.

6. Apr 5, 2009

### frestus

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.

7. Apr 5, 2009

### Dick

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!

8. Apr 5, 2009

### frestus

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