# Maclaurin Series for e^x

1. May 6, 2012

### forestmine

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

I'm just trying to understand a few things about the Maclaurin series for e^x...

So, in one case, if you have a series from 1 to infinity of [(-1)^n * 3^n ]/n!, how is it that it is equal to e^-3 - 1? I understand the e^-3 part, as -3 is simply our x value from the series. Is the negative one simply because the series starts at 1, or does it have something to do with the alternating portion of the series? I'm a bit confused there...

And then say we have xe^x. If I wanted to find the Maclaurin series for this one, well, I know that the value that e is raised to will take the place of x in the series. As for the x hanging out in the front, do I simply add that to the series? So the series would be from 0 to infinity, x^n+1 / n!

I guess I'm just looking for some general rules when dealing with the Maclaurin series for e^x...

Any help in the right direction would be great.

Thanks!

2. Relevant equations

e^x = series (0 to infinity) x^n/n!

3. The attempt at a solution

2. May 6, 2012

### cjc0117

the sums start at different values of n.

$e^{x}=\sum^{∞}_{n=0}\frac{x^{n}}{n!}$

$e^{-3}=\sum^{∞}_{n=0}\frac{(-3)^{n}}{n!}=\sum^{∞}_{n=0}\frac{(-1)^{n}3^{n}}{n!}=1+\sum^{∞}_{n=1}\frac{(-1)^{n}3^{n}}{n!}$

Therefore...

$\sum^{∞}_{n=1}\frac{(-1)^{n}3^{n}}{n!}=e^{-3}-1$

And yes, you are right about the second part with xex

3. May 6, 2012

### forestmine

I'm sorry, I don't really follow. If the series starts at 1, we'd like to account for that since we know the expansion which starts from 0, right?

I guess I don't really understand what exactly is going on in the last two terms you wrote in the second l understand right up until you have the 1 + ... in front of the series. I see you've changed the index to 1 at that point, and yet you're adding 1 out front?

I'm sorry, I'm just not seeing it!

4. May 6, 2012

### cjc0117

Yes, because if you plug n=0 into (-1)n3n/n! you get 1. So the first term of the series is 1. The index becomes 1 because we just wrote the first term ((-1)030/0!=1) outside of the sum.

5. May 6, 2012

### forestmine

Ahh ok, I see. So in your last line, you're compensating for the + 1 by subtracting it from e^-3. Ok, thanks a lot. That makes sense. Thanks a lot.

So, in general, if I'm given a series of e^x starting at say 2, then I can write the series as being the sum of the first two terms of the series of e^x, 0 and 1, plus my series from n=2 to infinity of what's given?

6. May 6, 2012

### cjc0117

Yes, exactly. Only the first term of the series ex is 1 and the second term is x. So...

$\sum^{∞}_{n=2}\frac{x^{n}}{n!}=-1-x+\sum^{∞}_{n=0}\frac{x^{n}}{n!}$

Last edited: May 6, 2012
7. May 6, 2012

### forestmine

Ok, got it!

And then if I'm starting in the reverse direction, looking for the Maclaurin series of some form of e^x, what's a good way to think of that? I know that for e^x, it's just the standard series we've been talking about. And then if e^x is divided, multiplied, or added to anything, do I do the same exact operation to the series?

8. May 6, 2012

### forestmine

Hm, how come it's -1 and -x?

Ah nevermind, it's because the series it's being added to starts from 0, right?

Can that also be written as,

1 + x + series (2 to infinity) x^n/n!

9. May 6, 2012

### cjc0117

Okay, sorry. Let me be more clear:

$e^x=\sum^{∞}_{n=0}\frac{x^{n}}{n!}=1+x+\sum^{∞}_{n=2}\frac{x^{n}}{n!}$

Therefore, if we subtract 1 and x from both sides...

$\sum^{∞}_{n=2}\frac{x^{n}}{n!}=-1-x+e^{x}=-1-x+\sum^{∞}_{n=0}\frac{x^n}{n!}$

Better?

10. May 6, 2012

### forestmine

Yes, perfect. Thanks a lot for all the help.

One last question...

If I'm starting in the reverse direction, looking for the Maclaurin series of some form of e^x, what's a good way to think of that? I know that for e^x, it's just the standard series we've been talking about. And then if e^x is divided, multiplied, or added to anything, do I do the same exact operation to the series?

11. May 6, 2012

### cjc0117

If you add two series, make sure the indices start at the same value of n:

$\sum^{∞}_{n=0}x+\sum^{∞}_{n=0}x^{2}=\sum^{∞}_{n=0}(x+x^{2})$

If you are trying to find the series of a series multiplied or divided by x, you just add or subtract to the exponent:

$\frac{1}{x}\sum^{∞}_{n=0}x^{3}=\sum^{∞}_{n=0}x^{2}$

But if you are multiplying or dividing two completely different Maclaurin series, you'd have to use multiplication or polynomial long division. For instance....

$\frac{sinx}{cosx}=\frac{\sum^{∞}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}}{\sum^{∞}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}}≠\sum^{∞}_{n=0}\frac{(-1)^{n}x^{2n+1}(2n)!}{(2n+1)!(-1)^{n}x^{2n}}$

Last edited: May 6, 2012
12. May 6, 2012

### forestmine

Ok, excellent. Thank you so much!