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Math Help - Cube root of 5

  1. #1
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    Cube root of 5

    Hi Guys,

    Need a little hint on this one.

    Expand (1 + x)^{\frac{1}{3}} in ascending powers of x as far as the term in 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, upto four decimal places.



    I expanded the problem as a binomial series,

    1 + \dfrac{x}{3} - \dfrac{x^2}{9} + \dfrac{5x^3}{81}

    The rest of the question is confusing me. How do I break up 5 to include 0.08?

    The only approach I could think of was,

    <br />
\sqrt[3]{5} = \sqrt[3]{8\left(1 - \frac{3}{8}\right)} = 2\sqrt[3]{1 - \frac{3}{8}}<br />

    Then using x = -3/8, to put in the above equation.

    This gives me 1.7122 which is incorrect, and also does not use the suggested 0.08. Any ideas where I am going wrong?

    Thanks for your help.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    (1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{25}

    Can you proceed?
    Last edited by alexmahone; May 6th 2011 at 03:51 AM. Reason: Typo
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  3. #3
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    Surely if you're evaluating the cube root of 5, you need to let x = 4...
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  4. #4
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    Quote Originally Posted by alexmahone View Post
    (1+0.08)^\frac{1}{3}=(\frac{27}{25})^\frac{1}{3}=3/\sqrt[3]{5}

    Can you proceed?
    Thanks for the quick reply!

    I am missing something here, Wouldn't that be 3/\sqrt[3]{25}? How does that lead to \sqrt[3]{5}?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mathguy80 View Post
    Thanks for the quick reply!

    I am missing something here, Wouldn't that be 3/\sqrt[3]{25}? How does that lead to \sqrt[3]{5}?
    Of course. I'm sorry, that was a typo.

    Anyway, 3/\sqrt[3]{25}=3/(\sqrt[3]{5})^2
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    Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

    One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?
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    @Prove It, I am not sure I follow your suggestion.
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    Quote Originally Posted by mathguy80 View Post
    Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

    One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?
    \sqrt[3]{1.08}=\sqrt[3]{\frac{108}{100}}=\sqrt[3]{\frac{(4)27}{(4)25}}

    \Rightarrow\sqrt[3]{5}=\sqrt[3]{\frac{500}{100}}=\sqrt[3]{\left(\frac{108}{108}\right)\frac{500}{100}}

    =\sqrt[3]{\frac{500}{108}\left(\frac{108}{100}\right)}}

    =\sqrt[3]{\frac{500}{108}}\sqrt[3]{1+0.08}=\sqrt[3]{\frac{(4)125}{(4)27}}\sqrt[3]{1+0.08}

    =\sqrt[3]{\frac{125}{27}}\sqrt[3]{1+0.08}
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  9. #9
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    Quote Originally Posted by mathguy80 View Post
    @Prove It, I am not sure I follow your suggestion.
    Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...
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  10. #10
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Prove It View Post
    Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...
    But the binomial series (when the index is a fraction) converges only if |x|<1.
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  11. #11
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    Quote Originally Posted by alexmahone View Post
    But the binomial series (when the index is a fraction) converges only if |x|<1.
    Point taken. I'm sure, however, that a series can be found which is centred somewhere near x = 4 so that you can evaluate it at x = 4.
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  12. #12
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    Nice! This is why I love this site! Thanks @Archie Meade.
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  13. #13
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    Interesting point @Prove it. The |x| < 1 for the convergent binomial series has tripped me up as well. Don't know enough Math to know if there is such a series.. Thanks for all the help today, guys, much appreciated.
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