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Thread: Substitute suitable value of y to find 5^{1/3}

  1. #1
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    Substitute suitable value of y to find 5^{1/3}

    Given the equation $\displaystyle (1+3y)^{\frac{1}{3}} \approx 1+y-y^2$ is valid for $\displaystyle |y|<\frac{1}{3}$, by substituting a suitable value for $\displaystyle y$, find an approximate value of $\displaystyle 5^{\frac{1}{3}} $.

    I found the answer by substituting $\displaystyle y=\frac{1}{5}$ and the answer is $\displaystyle 1\frac{21}{29}$

    But I dont understand why, when I used $\displaystyle y=-\frac{4}{15}$, i dont get the answer.

    Here are my workings.

    When $\displaystyle y=-\frac{4}{15},$

    $\displaystyle (\frac{1}{5})^{\frac{1}{3}}\approx 1+\frac{-4}{15}-\frac{16}{225}=\frac{149}{225}$

    Therefore, $\displaystyle 5^{\frac{1}{3}}=\frac{225}{149} $ which is wrong
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  2. #2
    Newbie AgentSmith's Avatar
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    Re: Substitute suitable value of y to find 5^{1/3}

    $\displaystyle \sqrt[3]{5} \approx 1.709$ and $\displaystyle \frac{149}{225} \approx 1.510$.

    They are almost equal.
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  3. #3
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    Re: Substitute suitable value of y to find 5^{1/3}

    Quote Originally Posted by AgentSmith View Post
    $\displaystyle \sqrt[3]{5} \approx 1.709$ and $\displaystyle \frac{149}{225} \approx 1.510$.

    They are almost equal.
    In my opinion, they differ very much because if $\displaystyle y=\frac{1}{5}$ is used instead, the answer is $\displaystyle 1.72$ which is very much closer to the exact answer. The problem over here is that I don't understand why such a huge difference was generated when it seemed all reasonable to use $\displaystyle y=-\frac{4}{15}$
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Substitute suitable value of y to find 5^{1/3}

    Quote Originally Posted by Punch View Post
    I found the answer by substituting $\displaystyle y=\frac{1}{5}$ and the answer is $\displaystyle 1\frac{21}{29}$ But I dont understand why, when I used $\displaystyle y=-\frac{4}{15}$, i dont get the answer.
    Because $\displaystyle 1/5$ is closer to $\displaystyle 0$ than $\displaystyle -4/15$ .
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