# Can someone double check this, or tell me happened?

1. Apr 27, 2004

### JasonRox

I don't understand how they got rid of the negative.
l = absolute, but they aren't used anyways.

$$\mid \S_n - frac{a}{1 - R} \mid = \mid a( \frac{1 - r^n}{1-r} ) - \frac{a}{1-r}\mid$$

$$= \mid \frac{a r^n}{1-r} \mid$$

I'm getting...

$$= \mid \frac{-a r^n}{1-r} \mid$$

I always miss negatives, but I don't think I did this time.

Last edited: Apr 28, 2004
2. Apr 27, 2004

The absolute value's going to kill the negative either way.

3. Apr 28, 2004

### krab

Did you know that $|ab|=|a||b|$? Therefore $|-b|=|-1||b|=|b|$.

4. Apr 28, 2004

### JasonRox

The negative is gone after we use the absolutes.

We haven't used them, so where did it go?

5. Apr 28, 2004

### philosophking

I see what you did wrong. When you factor out the _a_ at the beginning, you get Sn -1, which is r^n -1, and that should solve your problem.

6. Apr 28, 2004

### JasonRox

Sn isn't even involved.

7. Apr 28, 2004

### JasonRox

I'll break it down even more, and I hope I find my mistake.

$$\mid S_n - \frac{a}{1 - R} \mid = \mid a( \frac{1 - r^n}{1-r} ) - \frac{a}{1-r}\mid$$

$$= \mid \frac{a - ar^n}{1 - r} - \frac{a}{1 - r} \mid$$

...same denominator...

$$= \mid \frac{a - ar^n - a}{1 - r} \mid$$

$$= \mid \frac{-ar^n}{1 - r} \mid$$

8. Apr 28, 2004

### HallsofIvy

Staff Emeritus
And you said before
What do you mean "we haven't used them"? You certainly are using the absolute value. That's why it's there and that's why
$$= \mid \frac{-ar^n}{1 - r} \mid = \frac{ar^n}{1-r}$$

If you mean that there is still an absolute value sign in the answer, that doesn't mean they haven't been used ||a||= |a| (absolute value of the absolute value of a is the same as the absolute value of a).

9. Apr 30, 2004

### JasonRox

Nevermind.

I shouldn't have put the absolutes in to begin with. It changes nothing!

It's like we right the number 4 instead of 4^1, which is the way it should be done.

There is no reason for the negative to "disappear" on its own.

Yes, I understand absolutes, but they only work if you use them.

10. Apr 30, 2004

I think the point is that |-a| is equivalent to |a|. If a can take both positive and negative values, then the absolute value sign is still present. But since the presence of the negative sign in the absolute value is useless (it's going to end up positive anyway), we drop it because it's less work to write it without.

On a similar note, you could have taken that negative sign and distributed it into the denominator if the r - 1 form was more convenient to work with. Or, if the negative weren't there, you could create it and then distribute it anyway because it's still equivalent.

All in all, the negative isn't a big deal.