# Series expansion for (e^x-1)/x

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1. Dec 16, 2014

### Austin

1. The problem statement, all variables and given/known data
Give the first three terms and the general term for ((e^x)-1)/x

2. Relevant equations
None
3. The attempt at a solution

I wasn't sure if I would be able to take the e^x series and subtract 1 from each term and divide each term by x. I was also thinking of splitting it into e^x/x -1/x and adding both series, but I'm not sure how to calculate series for either of those. My textbook says the answer is 1+ x/2! + x^2/3! +...+ x^n/(n+1)! which I do not see how they got that answer. I'm pretty lost on this one, I'm new to series and trying to get a grasp on it so some help would be appreciated.

Some overall questions: what is series for 1/x? I would think you could say 1/(1-(1-x)) to use the series for 1/(1-x) but I don't see how you could get the answer in the book if this was true.

If I have a series for a function like f(x)/x can I just take a known series for f(x) and divide each term by x to get series for f(x)/x? I feel like you can but I'm new to this and not sure. Also same goes for f(x) -1, would you subtract 1 from each term?

Thanks

2. Dec 16, 2014

### Staff: Mentor

No, you don't subtract 1 from each term - just subtract 1 from the overall series. After doing that, then divide by x.

3. Dec 16, 2014

### Austin

Right, but subtracting from each term of a series is the same as subtracting from the overall series right?

4. Dec 16, 2014

### Austin

Wait I think I see what you are saying but I don't understand why that is

5. Dec 16, 2014

### Austin

Ok I actually understand what you are saying and I understand how to do the problem now. I was getting confused but I actually should have known how to do it I was just a little overwhelmed from getting used to this new stuff.

But I was wondering, wouldn't you get the same answer in theory by splitting it into (e^x)/x - (1/x) and subtracting each of those power series from each other? Because when I try to do that it looks nothing like the other way which you explained?

Additionally, this is a side question, how would you write series for 1/x? Would it be like how I asked earlier by writing 1/(1-(1-x)) and using the general form of 1/(1-x)?

Thanks

6. Dec 16, 2014

### Orodruin

Staff Emeritus
Yes, you should get the same. But to make any sense of it you need to do it correctly. Just as the -1 cancels the first term of the series of e^x, -1/x will cancel the first term of e^x/x.

If expanding around x = 0, you cannot write a Taylor expansion since it diverges there. However, you can do a Laurent series and 1/x happens to be its own Laurent series. It just so happens that the first term (the one proportional to 1/x) of the Laurent series of both e^x/x and 1/x are the same, leading to f(x) = (e^x-1)/x having a well defined Taylor series if defining f(0) = lim{x->0} f(x).

7. Dec 16, 2014

### Staff: Mentor

No it isn't. For example, if you have 3 + 8 + 2 and want to subtract 1 you don't subtract 1 from each term. That would be 3 - 1 + 8 - 1 + 2 - 1, which is 10. The correct value is 3 + 8 + 2 - 1 = 13 - 1 = 12.

8. Dec 16, 2014

### Ray Vickson

No, absolutely not! If you subtract 1 from each term of a convergent series you get a new series that is divergent to $-\infty$.

9. Dec 16, 2014

### Austin

Yes thank you I already said I made a mistake and understand that part