Homework Help: Geometric Series

1. Apr 4, 2005

Briggs

I am having a little trouble with some questions on geometric series'
For example, Find the values of x for which the following geometric series converge
I have done the first one easy enough
$$2+4x+8x^2+16x^3.....$$
$$r=2x$$
$$|r|<1$$
$$|x|<\frac{1}_{2}$$
$$\frac{-1}{2} < x < \frac{1}_{2}$$

But then it gets more difficult and adds more to the question for example
$$5+25(3x+4)+125(3x+4)^2$$
I get $$r=5(3x+4)$$
so $$|5(3x+4)|<1$$

But then I am stuck as to what to do next any hints on how to go about these questions would be appreciated.

2. Apr 4, 2005

dextercioby

It's the same kind of story to explicitate the modulus...

$$\left|a\right|<b\Leftrightarrow -b<a<b$$

Thankfully,the ratio is linear in "x"...

Daniel.

EDITalls,this is the kind of terminology i had been used to in high school.Always the teacher said about "modulus explicitation" (sic!)...

Last edited: Apr 4, 2005
3. Apr 4, 2005

HallsofIvy

"explicate the modulus"?? Wow! Them big words is too much for me!

Briggs, what dextercioby means is: |5(3x+4)|< 1 is exactly the same as

-1< 5(3x+4)< 1. Now solve for x.

4. Apr 4, 2005

Data

It's explicitate! :tongue:

5. Apr 4, 2005

Briggs

Thanks for the help, it seems quite simple now that you have shown me the next step, I get

$$-1<15x+20<1$$

$$\frac{-1-20}{15}<x<\frac{1-20}{15}$$

$$\frac{-7}{5}<x<\frac{-19}{15}$$

Which is the correct answer so thanks a lot for the help

6. Apr 4, 2005

Gokul43201

Staff Emeritus
To help vizualize this geometrically, imagine a point P on the number line. Then, all points X that are strictly inside a distance d from the fixed point P, are given by the equation : |X-P| < d

In other words, this means that X is the set of points in (P-d, P+d). This follows directly from the definition of the modulus :

Definition : |x| = x if x > 0 and |x| = -x if x <= 0

Derivation of Geometric Statement : |X-P| = X-P if X > P and |X-P| = P - X, otherwise (from def'n.)

In the first case (when X > P), |X-P| < d means that X-P < d or X < P+d. So : P < X < P+d

In the second case (when X <= P) |X-P| < d means that P-X < d or X > P-d. So : P-d < X <= P

Combining the above two cases, you see that |X-P| < d gives P-d < X < P+d.

Example : |ax + b| < d is the same as writing |ax - (-b)| < d, which translates as "the distance of points ax from the point b is less than d. So, b-d < ax < b+d, or (b-d)/a < x < (b+d)/a