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Thread: Binominal series

  1. #1
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    Binominal series

    hi im new here and not very good with math terms of english words

    i have this to make and i just dont know how i tried but it just wouldnt happen

    Write (find) the first 3 terms in binomianl series to calculate approximate value for n-th root ; n=3.


    28^(1/n) ?

    thx in advanced
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  2. #2
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    Hello, Paps!

    Welcome aboard!


    Find the first 3 terms in binomianl series to calculate approximate value of: $\displaystyle 28^{\frac{1}{3}}$

    The binomial series looks like this:

    $\displaystyle (a + b)^n\;=\;a^n + n\!\cdot\!a^{n-1}b + \frac{n(n-1)}{2!}a^{n-2}b^2 \,+$ $\displaystyle \frac{n(n-1)(n-2)}{3!}a^{n-3}b^3 + \hdots + b^n
    $


    We have: .$\displaystyle 28^{\frac{1}{3}}\;=\;(27 + 1)^{\frac{1}{3}}$ . . . where: $\displaystyle a = 27,\;\;b = 1,\;\;n = \frac{1}{3}$

    The first three terms are: .$\displaystyle (27 + 1)^{\frac{1}{3}}\;=\;27^{\frac{1}{3}} + \frac{1}{3}(27^{-\frac{2}{3}})(1) + \frac{\frac{1}{3}(\frac{1}{3} - 1)}{2!}(27^{-\frac{5}{3}})(1^2) $

    . . $\displaystyle = \;\sqrt[3]{27} + \frac{1}{3(\sqrt[3]{27})^2} + \frac{\frac{1}{3}(-\frac{2}{3})}{2(\sqrt[3]{27})^5} \;=$ $\displaystyle 3 + \frac{1}{3\cdot9} + \frac{-\frac{2}{9}}{2\cdot243} \;= \;3 + \frac{1}{27} - \frac{1}{2187}$


    Therefore: .$\displaystyle \boxed{\sqrt[3]{28} \;\approx \;3.03657979}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This is a very good approximation!

    The actual value is: .$\displaystyle \sqrt[3]{28}\;=\;3.036588972$

    Our answer gives us: .$\displaystyle 3.03657979^3\;=\;27.9997...$


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  3. #3
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    OH thank you very very much

    i couldnt find that equasion, actually i didnt really understand what this wants from me

    again really appreciate it
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