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Approximating Binomial Series I'm so close

  1. Apr 1, 2008 #1
    Use Binomial Series to approximate sqrt(35) with an accuracy of 10^(-7)

    Formulas for binomial series: (1-x)^r and sum{from 0 to n}(r k)(x)^k

    sqrt(35) = sqrt(35*36/36) = 6*sqrt(35/36) = 6*sqrt(1-(1/36))

    Now it looks more like the binomial series formula:
    let r = 1/2 because of the radical and x = 1/36

    6*[1-(1/36)]^(1/2) and use the second formula to expand:

    6* sum{from 0 to n}(1/2 k)(1/36)^k

    I did from k=0 to k=5:
    = 6[1-(1/2)*(1/36)+(1/8)*(1/36)^2-(1/16)*(1/36)^3+(1/32)*(1/36)^4-(1/64)*(1/36)^5+....]

    But I didn't get close enough with only 5 terms. First I used 4 terms beacause (1/32)*(1/36)^4 gave me 1.9*10^(-8) which I thought would be enough terms needed for an accuracy of 10^(-7) but wasn't.

    I eventually got 6[0.98620624] = 5.917237444
    and the actual sqrt(35) = 5.916079783

    +- 0.001 and I need +- 0.0000001
    My professor didn't teach us to use the alternating series estimation theorem which I though would help me find how many n terms I would need to use but I don't understand. Please help. I think I'm on the right track I'm just not quite there.

  2. jcsd
  3. Apr 1, 2008 #2
    I think you have used an incorrect formula
    The correct one is:
    [tex](1-x)^{\alpha}=\sum_{k=0}^{\infty}(-1)^k\[\left(\begin{array}{c}\alpha & k \end{array}\right)x^k\][/tex]

    [tex]\[\left(\begin{array}{c}\alpha & k \end{array}\right)=\frac{1}{r!}{\prod_{j=0}^{r-1}{(\alpha-j)}}\][/tex]
  4. Apr 1, 2008 #3


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    eys_physics is correct. The series isn't alternating. And starting at the (1/36)^4 term your binomial coefficients aren't correct either. I think you just extrapolated from the form of the earlier terms. Wrongly. What you've got going there is a geometric series. If that were correct you could sum it and show sqrt(35) is rational (!?!?!?!).
  5. Apr 2, 2008 #4
    (35) = sqrt(35*36/36) = 6*sqrt(35/36)
    6*sqrt(35/36) = 6[(1 - 1/36)^(1/2)] =

    from k=0 to k=4:
    = 6[(1 + (1/2)*(1/36) - (1/8)*(1/36)^2 + (1/16)*(1/36)^3 - (5/128)*(1/36)^4]

    = 6.082762528 which is different from sqrt(35) = 5.916079783

    I got a different series after calculating for the binomial. I calculated up to k=4 because the last term was 2.3*10^(-8) and it wanted an accuracy of 10^(-7).
  6. Apr 2, 2008 #5


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    Notice the (-1)^k in the formula eys_physics posted. Also notice the binomial coefficients generally alternate in sign. Most of the terms in your series should have the same sign.
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