# Homework Help: Bernoulli numbers problem

1. Mar 12, 2007

### sara_87

1. The problem statement, all variables and given/known data

Let

x/((e^x)-1)= 1 + B1x+B2((x^2)/2!) + B3((x^3)/3!)+…

where Bn are called the Bernoulli numbers. Determine B1 , B2 and B3.

2. Relevant equations

I think this one: (1+x)^n = 1 + nx + ((n)(n-1)/2!)(x^2)+...

3. The attempt at a solution

i wrote it in this form -x(1-e^x)^(-1) to perhaps make it easier to use the maclaurin series but i still don't get what the question asks us to get

Last edited: Mar 12, 2007
2. Mar 12, 2007

### matt grime

Expand the LHS in power series. Read off the coefficients.

3. Mar 12, 2007

### sara_87

i understand what you are saying but i don't know what to do...when i expand the first tem isn't 1 it's -x

4. Mar 12, 2007

### matt grime

Well google for Bernoulli numbers to see what the correct definition is.

5. Mar 12, 2007

### Dick

You are trying to use a formula for the expansion of 1/(1-s) and apply it to the case where s=1. Don't do it. As matt said, the B's are just the derivatives of x/(e^x-1) taking the limit as x->0.

6. Mar 12, 2007

### HallsofIvy

?? Then you expanded wrong! The MacLaurin series for x/(ex- 1) has first term the limit as x goes to 0. That certainly is 1!

7. Mar 13, 2007

### sara_87

thanx all for your help and time; i googled bernoulli numbers and i got the answer, but i still don't understand i think i should ask a real person and they might be able to talk me through it

8. Mar 13, 2007

### matt grime

You can't just use the taylor series of 1/(1-s) and then substitute e^x for s. That series does not converge in the correct place when s=e^x (e^x >1 for x>0). So you just have to find the Taylor series of x/(e^x - 1) directly from the definition of a Taylor series. I.e. differentiate and let x tend to 0.

So what's B_0? It is the limit as x tends to zero of x/(e^x-1)

We can't take that limit directly, but by l'Hopital it is the same as the limit f 1/e^x as x tends to 0, and that is just 1.

What about B_1? The derivative is (e^x-1 - xe^x)/(e^x - 1)^2. We're going to have to apply l'hopital again.

Doing it once, we need the limit of (-xe^x)/(2e^2x-2e^x), but we see we need to do it again.

So we want to find the limit of (-e^x -xe^x)/(4e^2x - 2e^x) as x tends to zero, and that is -1/2. Which Wolfram tells me is what I should have been expecting.

Last edited: Mar 13, 2007
9. Mar 13, 2007

### sara_87

'You can't just use the taylor series of 1/(1-s) and then substitute e^x for s. That series does not converge in the correct place when s=e^x (e^x >1 for x>0). So you just have to find the Taylor series of x/(e^x - 1) directly from the definition of a Taylor series. I.e. differentiate and let x tend to 0.'

yeah that i managed to pick up! thank you though; i'm at a maths center and i have half a dozen questions for the maths teacher so i'll just add this one in there so that i fully understand...i did actually understand everything you said and what the other tips were so you didn't waste your time :)

10. Mar 13, 2007

### matt grime

What else is there to understand? You know you can't just use an extant Taylor series, and thus have to find your own. I gave the derviation of B_0 and B_1 in the above edited post. What more do you need to know?

11. Mar 13, 2007

### matt grime

If you want to do it directly, then consider putting in the series expansion of e^x first.

e^x -1 = x+x^2/2! + x^3/3! +...

Now divide by x.

(e^x-1)/x = 1+x/2! +x^2/3! +... x^n/(n+1)! +...

so x/(e^x-1) = 1/(1+x/2! +...)

now use the expansion of 1/(1-s) and assume it all works, and I think you get the same answer as before. The first two terms are certainly correct, 1 and -1/2.

12. Mar 13, 2007

### sara_87

i get most of it lol
just need somenone to talk me through it to make sure it sinks in

(i need someone to tell me 'yes sara! you're right! that was a difficult question!')

13. Mar 13, 2007

### Dick

We ARE 'real people'! Aren't we??

Last edited: Mar 13, 2007
14. Mar 13, 2007

### matt grime

Personally, I don't think it was a difficult question, but I do think you tried to make it difficult. I.e. instead of actually working out the Taylor series by hand you wanted to make a short cut and substitute into a divergent series.

Please don't think that is belittling your efforts - many's the time I've made a simple thing far too complicated as well.

It was certainly a tedious question - I was fed up just working out B_0 and B_1, never mind getting on to do B_2 and B_3.

If someone asked you to find the Taylor series about 0 of x/(e^x-1) you'd be able to do it.

Last edited: Mar 13, 2007