Question

Expand x/(e^x-1) in powers of x.

looking for alternate methods,
i can find f(0), f'(0), f''(0)....and use taylor's series, but thats way too tedious and cumbersome!

Answer 1

can i use series for e^x directly...but then i'll be stuck at 1/(some series in x)

Answer 2

how about expanding e^x itself. exp x = 1 + x + x^2/2 + ...

Answer 3

yes, then i will have
1/(1+x/2+x^2/6+...)
how would i simplify that !?

Answer 4

but -1 will be canceled out. right?

Answer 5

did that, the x in the numerator was brought in the denominator

Answer 6

oh, got it. so what is the problem now?

Answer 7

how would i express that in powers of x

Answer 8

So first note, the x in the numerator does not matter much...right?

Answer 9

I have some thoughts, but nothing that's particularly compact and neat. Do you have a target final answer?

Answer 10

oh, so you're planning to expand 1/(e^x-1)
and then multiply x on both sides?

Answer 11

yes,
the answer should be
1-x/2+x^2/12-x^4/720+...

Answer 12

that actually will make the differentiation part quite easy!
and i can use taylor series now! :O

Answer 13

yes, that was the initial thought

Answer 14

awesome!
will go for that only
thanks :)

Answer 15

Ah okay, I wasn't going that route because you didn't want Taylor Series
but this works too

Answer 16

what route were you trying ?
always good to know alternate methods..

Answer 17

but f(0), f'(0) are coming out to be \(\infty \)

Answer 18

http://prntscr.com/4moyt6

Answer 19

continues ---> http://prntscr.com/4moz3a

Answer 20

I know it's gross looking...if you want to handle it abstractly
and then use the cauchy product. If we just do it concretely It will look simpler
I don't know... >.<

It really depends on you and what level you want to approach it from

Answer 21

can you first try taylor series method...
i am getting f(0), f'(0), f''(0) all as \(+\infty ~ or ~ -\infty \)

Answer 22

taylor/maclauring series wont work as f0) doesn't exist

Answer 23

oh right!
what will work ?

Answer 24

So what we'd be doing at this point is matching coefficients. the coefficient for the x term on the right hand side needs to be 1. And the coefficients for all the other terms need to be zero.
I think it's clearer if I write it like this:

http://prntscr.com/4moznh

Answer 25

mira's work looks good to me ^^

also binomial theorem is another similar approach

Answer 26

comparing the co-efficients!
yes, got it :)
how about binomial ?

Answer 27

Series always end up being that way in my opinion. So what was the main idea?

The main ideas was. (1) Assume we have some series form for the expression. sum i = 0 to infinity alpha_i* x^i
it has these coefficients alpha_i. we don't know what they are.
This is the series expansion of x/(e^x-1)

(2) Now use an algebraic trick and bring (e^x-1) to the other side of the equation and also use its series expansion

3) At this point, we have two series multiplying each other and that can get messy. If we just write out the series long form, like I did in red, we can get a feeling for how the multiplication goes.

By matching coefficients to the left side "x", we can solve for the alpha_i

Answer 28

If you want an abstract compact representation of the coefficients
Which is what I did in blue: http://prntscr.com/4moyt6

It's the same thing... just not written out in long form, and I didn't match the coefficients yet

Answer 29

right, need to be careful while doing multiplication of series

Answer 30

I made use of the cauchy product to simplify it

Answer 31

and correct. :)

Answer 32

It's very dangerous by hand.. haha

Answer 33

So on a scale of 1 to 10... 1 being totally unclear and 10 being easy enough to do in your sleep. How is this problem looking now to you? @hartnn

Answer 34

lol, just wanted to know the method, after that its just algebra, too easy, 10, piece of cake

Answer 35

using binomial theorem is more or less similar to above

Notice below first :
\(\large e^x - 1 = x(1 + x/2 + x^2/3! + ...)\)
\(\large = x(1 + A)\)

\[\large x(e^x-1)^{-1} = (1+A)^{-1} = \sum \limits_{k=0}^{\infty} \binom{-1}{k}A^k\]

Answer 36

ahh, but thats not so easy to click...also, after few steps, there will be comparing of co-efficients only...

Answer 37

hm, yes, so the cauchy product looks like the binomial theorem in a lot of ways ^

Answer 38

hartnn okay, but I would be sure that you can reproduce it by practicing with a similar problem. At least in my experience, I have thought I have known something when I read it because it all made sense, and then when I went and tried to solve another analogous problem myself without reference.... roadblock

Answer 39

noted, will look for similar problems and try to solve them :)