# Thread: breaking this expressions into areas..

1. ## breaking this expressions into areas..

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

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?

2. i made a typing mistake
there is no absolute value in the expression

and i still cant see the logic in this separation

3. Originally Posted by transgalactic
there is no absolute value in the expression

and i still cant see the logic in this separation
Still does not look right.

C B

4. ok how would you break the exression
the left most is for n=1.. infinity the central is for n=0
the right most for n=-1 ..-infinity
how it should look like?

5. Originally Posted by transgalactic
ok how would you break the exression
the left most is for n=1.. infinity the central is for n=0
the right most for n=-1 ..-infinity
how it should look like?
$\sum_{n=-\infty}^{\infty}r^ne^{inx}=\sum_{n=-\infty}^{-1} r^ne^{inx}+r^0e^0 +\sum_{n=1}^{\infty}r^ne^{inx}$

CB

6. but the power of should be -n on the minus case
correct?

7. Originally Posted by transgalactic
but the power of should be -n on the minus case
correct?
No, I can change it to minus, but then I will have to change a few more things.

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

or we could have (if you want the absolute value in the exponent of $r$ on the left):

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

Now if you could clarify what the sum on the left is supposed to be and what you are trying to do, that would help. At the moment I feel I am wasting my time as you have not ma
de it clear what you are trying to do or why.

CB

8. but its absolute value
its minus for negative numbers

9. Yes, it is- but notice that Captain Black has switched indices also.

The sum in question is $\sum_{n=-\infty}^1 r^{|n|}e^{inx}$. Since n is negative, that is the same as $\sum_{n= -\infty}^1 r^{-n}e^{inx}$. Now let k= -n so that n= -k, when n= -1, k= 1 and when $n= -\infty$, $k= \infty$. In terms of index k the sum is $\sum_{k=1}^\infty r^{k}e^{-ikx}$.

Finally, since this is a "dummy index" anyway, it doesn't matter whether you call it "k" or "n". That is exactly the same as $\sum_{n=1}^\infty r^n e^{-inx}$

10. if you say that n=-k
than you cant say that k=n because that gives us
n=-n wich is true only for n=0
not for n=-1 to - infinty

11. Originally Posted by transgalactic
if you say that n=-k
than you cant say that k=n because that gives us
I didn't say that k=n! I said that k= -n.

n=-n wich is true only for n=0
not for n=-1 to - infinty

12. you said -n=k
$
\sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}
$

and then you said k is just a dummy and we can call k as n
which gives us
$
\sum_{n=1}^\infty r^n e^{-inx}
$

but saying that contradicts the logic of
n=-n wich is true only for n=0
not for n=-1 to - infinty

13. thanks i got the idea

14. Originally Posted by transgalactic
you said -n=k
$
\sum_{-k= -\infty}^{-1} r^{k}e^{i(-k)x}= \sum_{k=1}^\infty r^k e^{-ikx}
$

and then you said k is just a dummy and we can call k as n
which gives us
$
\sum_{n=1}^\infty r^n e^{-inx}
$

but saying that contradicts the logic of
n=-n wich is true only for n=0
not for n=-1 to - infinty
The two "n"s have nothing to do with each other.

If n+ 3= 6, I can replace n by "-k" and get -k+ 3= 6.

I could then replace k by n and get -n+ 3= 6. The two equations "n+3= 6" and "-n+ 3= 6" [b]would[b] contradict each other if they were the same n.

But in sums we have, as I said "dummy" indices, if $A= \sum_{n=1}^4 \FRAC{1}{n}$, I can replace n by -k and have $A= \sum_{k= -1}^{-4}\frac{1}{-k}$ and then replace k by n to get $
A= \sum_{n= -1}\frac{1}{-n}$
.

Those all say the same thing: $A= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}$.

15. thanks i understand this thing now