# Homework Help: Breaking this expressions into areas.

1. Dec 26, 2009

### nhrock3

$$\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}=$$$$\sum_{n=1}^{\infty}e^{inx}$$+ $$r^{0}e^{0}$$ +$$\sum_{n=-\infty}^{\infty}r^{n}e^{-inx}$$

if we look at the left side
i have been told that it was broken by intervals
the left most is for n=1.. infinity the central is for n=0
the right most for n=-1 ..-infinity

i know the definition of the |x| function
for some values its x for other its -x

but here the power of e changes too
why?

Last edited: Dec 26, 2009
2. Dec 26, 2009

### tiny-tim

Hi nhrock3!

I suspect you've copied it wrong, is it …

$$\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}\ =\$$ $$\sum_{n=1}^{\infty}r^{n}e^{inx}$$+ $$r^{0}e^{0}$$ +$$\sum_{n=1}^{\infty}}r^{n}e^{-inx}$$
?​

3. Dec 26, 2009

### nhrock3

why we always have $$r^n$$ why in the minus case we dont have $$r^{-n}$$

and why the power of the exponent changes to minus in the negative case
?

4. Dec 26, 2009

### tiny-tim

Because it depends what the limits are.

Perhaps it's clearer if I write it …

$$\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}\ =\$$ $$\sum_{n=1}^{\infty}r^{n}e^{inx}$$+ $$r^{0}e^{0}$$ +$$\sum_{m=1}^{\infty}}r^{m}e^{-imx}$$

where m = -n, so |n| = m, and inx = -imx.

(and m and n are "dummy" indices, so you can then change m to n )

5. Dec 26, 2009

### nhrock3

i cant see how the minus indexes changes the original expression

?

why the power of the exponent changes to minus in the negative case

they should be negative
and we dont need to change it to minus

6. Dec 26, 2009

### HallsofIvy

Do you see that both sums are for positive n?
$$\sum_{n=-\infty}^\infty r^{|n|}e^{inx}$$

You can break that into three parts, as you say:
$$\sum_{n=-\infty}^{-1} r^{|n|}e^{inx}+ r^0e^{i0x}+ \sum_{n=1}^\infty r^{|x|}e^{inx}$$

Now look at each of that first sum separately. For n< 0, |n|= -n so
$$\sum_{n=-\infty}^{-1} r^{|n|}e^{inx}= \sum_{n=-\infty}^{-1} r^{-n}e^{inx}$$
Change the index: let k= -n so n= -k:
$$= \sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}$$

Since the index here is "dummy" we can use any letter we like, say "n", in place of k:
$$= \sum_{n= 1}^\infty r^n e^{-inx}$$

Of course, $r^0e^{i(0)x}= 1$, and, for n> 0, |n|= n, so the entire sum becomes
$\sum_{n= 1}^\infty r^n e^{-inx}+ 1+ \sum_{n=1}^\infty r^n e^{inx}[/tex]. 7. Dec 26, 2009 ### tiny-tim Try changing m to -n in the right-hand term in my equation … what do you get? 8. Dec 26, 2009 ### nhrock3 no you cant use is as dummy if n=-k and k=n then n=-n which works only for n=0 9. Dec 26, 2009 ### nhrock3 no you cant use it as dummy if n=-k and k=n then n=-n which works only for n=0 and for n= -1 to -infinity 10. Dec 26, 2009 ### HallsofIvy What that tells me is that you don't know what a "dummy" index is. If I had an equation that said k+ n= 3, then I could not replace k by either n or -n- they mean different things. But if my equation is [itex]\sum_{k=1}^5 \frac{1}{k}+ \sum{n=1}^5 \frac{(-1)^n}{n}$ I could replace the "k" in the first sum by "n" because $\sum_{k=1}^5 \frac{1}{k}= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}$ and $\sum_{n=1}^5 \frac{1}{n}= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+ \frac{1}{5}$ are both numbers and do not depend upon either "n" or "k". Having changed from "n" to "k", I could now make the change from k to -n as well and have $\sum_{n=-1}^{-5}\frac{1}{-n}= \frac{1}{-(-1)}+ \frac{1}{-(-2)}+ \frac{1}{-(-3)}+ \frac{1}{-(4)}+ \frac{1}{-(-5)}$ which is exactly the same thing. This "n" has nothing to do with the original n. That does NOT imply "k= n" and "k= -n" for the same n!

11. Dec 26, 2009

### nhrock3

aahhh thanks :)