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Math Help - Newton's Binomial Theorem

  1. #1
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    Newton's Binomial Theorem

    (1 + x)^(1/3) = 1 + (1/3)x - (1/9)x^2 + (5/81)x^3 - (10/243)x^4 + (22/729)x^5

    Use this to get a quick numerical estimate of 70^(1/3).

    I started by observing that 70 = 10*(1+6)

    so... 70^(1/3) = (10*(1+6))^(1/3) = 10^(1/3) * (1+6)^(1/3)

    I then used the binomial expansion of (1 + x)^(1/3), plugging in x = 6

    I am getting an answer no where close to the true value of 70^(1/3)
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  2. #2
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    Hello, thamathkid1729!

    (1 + x)^{\frac{1}{3}} \;=\; 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{729}x^5 + \hdots

    \text{Use this to get a quick numerical estimate of }70^{\frac{1}{3}}

    First, we must get 70^{\frac{1}{3}} in the form (1 + x)^{\frac{1}{3}}
    . . This requires some Olympic-level gymnastics.

    70^{\frac{1}{3}} \;=\;(64 + 6)^{\frac{1}{3}}\;=\;\bigg[64\left(1 + \frac{6}{64}\right)\bigg]^{\frac{1}{3}}

    . . . . =\;64^{\frac{1}{3}}\left(1 + \frac{3}{32}\right)^{\frac{1}{3}} \;=\;4\left(1 + \frac{3}{32}\right)^{\frac{1}{3}}


    70^{\frac{1}{3}} \;=\;4\left(1 + \frac{3}{32}\right)^{\frac{1}{3}}

    . . . . =\;4\,\bigg[1 + \frac{1}{3}\left(\frac{3}{32}\right) - \frac{1}{9}\left(\frac{3}{32}\right)^2 + \frac{5}{81}\left(\frac{3}{32}\right)^3 - \frac{10}{243}\left(\frac{3}{32}\right)^4 + \hdots \bigg]

    . . . . =\;4(1.030321121)

    70^{\frac{1}{3}} \;=\;\boxed{4.121284485}


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


    Check:

    . . (4.121284485)^3 \;=\;69.9999567 . . . close enough!

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  3. #3
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    Thank you! Do you know why it was not working for me when I used 70^(1/3) = (10*(1+6))^(1/3)
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  4. #4
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    Quote Originally Posted by thamathkid1729 View Post
    Thank you! Do you know why it was not working for me when I used 70^(1/3) = (10*(1+6))^(1/3)
    When the Binomial series is infinite, we require that |x|<1 for convergence.
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