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

[member 508213]

Homework Statement

Give the first three terms and the general term for ((e^x)-1)/x

Homework Equations

None

The Attempt at a Solution

[/B]
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

 

[Mark44]

Homework Statement

Give the first three terms and the general term for ((e^x)-1)/x

Homework Equations

None

The Attempt at a Solution

[/B]
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.

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

 

[member 508213]

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

 

[member 508213]

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

 

[member 508213]

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

 

[Orodruin]

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?

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.

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)?

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).

 

[Mark44]

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

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.

 

[Ray Vickson]

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

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

 

[member 508213]

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

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