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Binomial series - Finding square root of number problem

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Expand ##(1+x)^(1/3)## in ascending powers of x as far as the term ##x^3##, simplifying the terms as much as possible. By substituting 0.08 for x in your result, obtain an approximate value of the cube root of 5, giving your answer to four places of decimals.

    2. Relevant equations

    Binomial series.

    3. The attempt at a solution
    Expansion as per binomial series :
    ##1+(1/3)x - (1/9)x^2 + (5/81)x^3##.

    No idea how to find cube root of 5 by using x = 0.08.
    Help please!
    Sorry if I am asking too many questions - I don't have a tutor for this, I'm on my own

    Thanks! :)
  2. jcsd
  3. Apr 27, 2014 #2


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    Are you sure you have read the problem correctly? Putting x= 0.08 into that formula will give you [itex](0.08+ 1)^{1/3}= 1.08^{1/3}[/itex], not [itex]\sqrt[3]{5}[/itex]. 5= 4+ 1 so you should set x= 4 to use that formula to find [itex]\sqrt[3]{5}[/itex].
  4. Apr 27, 2014 #3

    Ray Vickson

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    That would approximate ##\sqrt[3]{5}## by a few terms of a divergent series, which does not look like a good idea. However, using ##2^3 = 8## we can write
    [tex] 5 = 8 - 3 = 8 \left(1 - \frac{3}{8}\right) \Longrightarrow 5^{1/3} = 2 (1 - 0.375)^{1/3}[/tex]

    BTW: please try not to confuse everybody by using a bad thread title. Don't say "square root" when you mean "cube root".
    Last edited: Apr 27, 2014
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