# Breaking this expressions into areas.

• nhrock3
In summary: However, in the second equation, you wrote \sum_{n=1}^{\infty}r^{n}e^{inx} for the rightmost term, which should be \sum_{n=-\infty}^{\infty}r^{n}e^{-inx}. So the final correct equation is:\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}\ =\ \sum_{n=-\infty}^{-1}r^{n}e^{-inx} + r^{0}e^{0} + \sum_{n=1}^{\infty}r^{n}e^{inx}
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:
nhrock3 said:
$$\sum_{n=-\infty}^{\infty}r^{|n|}e^{inx}=$$$$\sum_{n=1}^{\infty}e^{inx}$$+ $$r^{0}e^{0}$$ +$$\sum_{n=-\infty}^{\infty}r^{n}e^{-inx}$$

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?

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}$$
?​

why we always have $$r^n$$ why in the minus case we don't have $$r^{-n}$$

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

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 )

i can't 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 don't need to change it to minus

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]. nhrock3 said: i can't see how the minus indexes changes the original expression Try changing m to -n in the right-hand term in my equation … what do you get? no you can't use is as dummy if n=-k and k=n then n=-n which works only for n=0 HallsofIvy said: Do you see that both sums are for positive n? Your original sum is $$\sum_{n=-\infty}^\infty r^{|n|}e^{inx}$$ 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, [itex]r^0e^{i(0)x}= 1$, and, for n> 0, |n|= n, so the entire sum becomes
$\sum_{n= 1}^\infty r^n e^{-inx}+ 1+ \sum_{n=1}^\infty r^n e^{inx}[/tex]. no you can't use it as dummy if n=-k and k=n then n=-n which works only for n=0 and for n= -1 to -infinity nhrock3 said: no you can't use it as dummy if n=-k and k=n then n=-n which works only for n=0 and for n= -1 to -infinity What that tells me is that you don't know what a "dummy" index is. If I had an equation that said k+ n= 3, then I could not replace k by either n or -n- they mean different things. But if my equation is [itex]\sum_{k=1}^5 \frac{1}{k}+ \sum{n=1}^5 \frac{(-1)^n}{n}$ 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!

aahhh thanks :)

## 1. What does it mean to "break an expression into areas"?

Breaking an expression into areas refers to breaking down a mathematical or scientific expression into smaller, more manageable parts in order to better understand or solve the overall problem.

## 2. Why is it important to break an expression into areas?

Breaking an expression into areas can make complex problems more approachable and easier to solve. It allows for a more organized and systematic approach to problem-solving, and can also help identify patterns and relationships within the expression.

## 3. How do you break an expression into areas?

The specific method for breaking an expression into areas may vary depending on the expression and the problem at hand. However, some common techniques include factoring, substitution, and using geometric or algebraic models to visualize the expression.

## 4. Can breaking an expression into areas be used for all types of expressions?

Yes, breaking expressions into areas can be used for various types of expressions, including mathematical equations, chemical reactions, and physical phenomena. It is a useful problem-solving strategy in many fields of science.

## 5. Are there any drawbacks to breaking an expression into areas?

While breaking an expression into areas can be helpful, it may not always lead to a solution or may not be the most efficient approach for every problem. It is important to consider other problem-solving strategies as well and choose the most appropriate one for the specific situation.

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