breaking this expressions into areas..

**transgalactic** · December 25th 2009, 02:02 PM

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

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?

**transgalactic** · December 25th 2009, 11:19 PM

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

and i still cant see the logic in this separation

**CaptainBlack** · December 26th 2009, 12:05 AM

Originally Posted by transgalactic

i made a typing mistake
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

**transgalactic** · December 26th 2009, 12:08 AM

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?

**CaptainBlack** · December 26th 2009, 12:13 AM

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

**transgalactic** · December 26th 2009, 12:24 AM

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

**CaptainBlack** · December 26th 2009, 12:51 AM

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

**transgalactic** · December 26th 2009, 12:54 AM

but its absolute value
its minus for negative numbers

**HallsofIvy** · December 26th 2009, 03:35 AM

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

**transgalactic** · December 26th 2009, 03:42 AM

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

**HallsofIvy** · December 26th 2009, 03:54 AM

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

**transgalactic** · December 26th 2009, 04:05 AM

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

**transgalactic** · December 26th 2009, 06:11 AM

thanks i got the idea

**HallsofIvy** · December 26th 2009, 06:13 AM

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: $A= 1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}$ .

**transgalactic** · December 26th 2009, 06:20 AM

thanks i understand this thing now

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