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Finding a fraction using series.

  1. Oct 29, 2005 #1
    How do i find the fraction closest to 0.987654321 using the geometric series?

    I have tried taking the geometric series and its derivatives but none seem to fit it. Can anyone help?
     
  2. jcsd
  3. Oct 29, 2005 #2

    Tide

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    Is that supposed to be a repeating decimal?
     
  4. Oct 29, 2005 #3

    Tide

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    If it's not a repeating decimal then you want to factor out [itex]10^{-9}[/itex] then write the remaining factor as a sum of ascending powers of 10 and your earlier proposed method will give you the desired result.
     
  5. Oct 29, 2005 #4

    Curious3141

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    [tex]0.987654321 = \frac{987654321}{1000000000}[/tex]
    That's all. There's no sense in using geometric progression here, because it's a terminating decimal.
    OTOH,
    [tex]0.\bar{9}8765432\bar{1} = 987654321*\Sigma_{i=1}^{i=\infty}(10^{-9i})[/tex]
    and you can then work out the summation with the formula for the sum of a geometric series.
     
  6. Oct 31, 2005 #5

    CarlB

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    There is a method of finding rational approximations to real numbers called "continued fractions". You can do it on your calculator. I will illustrate:

    0.9876543210 invert and get
    1.0125000000 subtract 1 to get
    .01250000000 invert and get
    80.000000064 subtract 80 to get
    .000000064 invert to get
    15625000.000 subtract 15625000 to get
    0

    This sequence terminated, which won't happen for an irrational number like sqrt(2) or pi.

    The important sequence is the numbers are the things you subtracted, which were the integer parts of the numbers. The sequence is:

    1, 80, 15625000

    One can get a good fractional approximation by reversing the process. Since 15625000 is so big, I'll start with the "80". The reversed process (in rational numbers) is:

    80
    1/80
    1+1/80 = 81/80
    80/81.

    So a darned good fractional answer is 80/81.

    This is a very general algorithm. let me think a minute and I'll try and figure out how to do it with the geometric series and I'll edit that solution in.

    [tex]\frac{1-x^n}{1-x} = 1 +x + x^2 + ... + x^{n-1}[/tex]

    differentiate with respect to x get:

    [tex]\frac{1-x^n}{(1-x)^2} - \frac{nx^{n-1}}{1-x}
    = 1 + 2x + 3x^2 + ... + (n-1)x^{n-2}[/tex]

    Now put x = 0.1 and n-1=9. The right hand side will become 1.23456789 and you can get what you want with some artful subtraction and multiplication by constants. But I haven't worked it out to see if it gives the same answer as the continued fraction expansion (i.e. 80/81).

    Carl
     
    Last edited: Oct 31, 2005
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