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Math Help - [Binomial Series]Find 25^1/3 with precision of 3 decimals.

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    [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    Question. Find 25^1/3 with precision of 3 decimals.

    My book says:

    \sqrt[3]{25}=\sqrt[3]{27}\sqrt[3]{\frac{25}{27}}. Then we can say: \sqrt[3]{25}=\sqrt[3]{3(1-\frac{2}{27})}

    I didnt understand the logic. I mean, how did he go from the first affirmation to the second?

    Thanks in advance,
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  2. #2
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    Is that actually what is in your book? It is incorrect. What is true is that \sqrt[3]{27}= 3, of course. And \frac{25}{27}= \frac{27}{27}- \frac{2}{27}= 1- \frac{2}{27}. So \sqrt[3]{27}\sqrt[3]{\frac{25}{27}}= 3\sqrt[3]{1- \frac{2}{27}}. That is, the first "3" is not inside the cube root.
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    Quote Originally Posted by frank1 View Post
    Question. Find 25^1/3 with precision of 3 decimals.My book says:
    \sqrt[3]{25}=\sqrt[3]{27}\sqrt[3]{\frac{25}{27}}. Then we can say: \sqrt[3]{25}=\sqrt[3]{3(1-\frac{2}{27})}

    I absolutely agree with Halls in reply # 2 on this one.
    Moreover, I think that your title of this thread is totally misleading.

    Here is the standard approximation:
    f\left( {x_0  + \Delta x} \right) \simeq f'\left( {x_0 } \right)\Delta x + f\left( {x_0 } \right).

    In this problem f(x) = \sqrt[3]{x},\,x_0  = 25,\;\& \,\Delta x =  - 2\\ &\simeq \frac{1}{3}\left( {27} \right)^{\frac{{ - 2}}{3}} \left( { - 2} \right) + 3\\  &\simeq \frac{-2}{27}+3\\  &\simeq 3-\frac{2}{27}  .
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    You're right HallsofIvy,

    But why did you go from a cube root to 25/27?
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    Quote Originally Posted by frank1 View Post
    You're right HallsofIvy, But why did you go from a cube root to 25/27?
    Did you read reply # 3 ?
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    I don't agree with Plato's reply on two counts.

    First with the standard approximation stated, how do you know whether or not you will get three decimal accuracy ? You might get lucky (with the size of the correction term), but really, you need at least one more term in the Taylor expansion.

    Secondly I don't think that the title is misleading.

    We can write 25^{1/3}=(27-2)^{1/3}=3(1-2/27)^{1/3}, and now use the standard Binomial expansion for (1-x)^{n}.
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    Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

    Plato, I'm sorry but I didnt understand what you are doing in Reply 3.

    BobP, Thanks a lot! With your answer I understand the example of the book
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