Solving Geometric Series: Find x Values |r|<1

[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.

 

[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!)...

 

[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.

 

[Data]

"explicate the modulus"??

It's explicitate!

 

[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

 

[Gokul43201]

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