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Homework Help: Geometric Series

  1. Apr 4, 2005 #1
    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
    [tex]2+4x+8x^2+16x^3.....[/tex]
    [tex]r=2x[/tex]
    [tex]|r|<1[/tex]
    [tex]|x|<\frac{1}_{2}[/tex]
    [tex]\frac{-1}{2} < x < \frac{1}_{2}[/tex]

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

    But then I am stuck as to what to do next any hints on how to go about these questions would be appreciated.
     
  2. jcsd
  3. Apr 4, 2005 #2

    dextercioby

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

    [tex] \left|a\right|<b\Leftrightarrow -b<a<b [/tex]

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

    Daniel.

    EDIT:Halls,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
  4. Apr 4, 2005 #3

    HallsofIvy

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    "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.
     
  5. Apr 4, 2005 #4
    It's explicitate! :tongue:
     
  6. Apr 4, 2005 #5
    Thanks for the help, it seems quite simple now that you have shown me the next step, I get

    [tex]-1<15x+20<1[/tex]

    [tex]\frac{-1-20}{15}<x<\frac{1-20}{15}[/tex]

    [tex]\frac{-7}{5}<x<\frac{-19}{15}[/tex]

    Which is the correct answer so thanks a lot for the help :smile:
     
  7. Apr 4, 2005 #6

    Gokul43201

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    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
     
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