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Math Help - Binomial expansion question

  1. #1
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    Binomial expansion question

    Prove that , if x is so small that terms in x^3 and higher powers may be neglected, then Binomial expansion question-msp520319eed9935g5h93hf000055b8ic10787h09a9.gif.

    By substituting a suitable value of x in your result, show that (11)^1/2 is approximately equal to 663/200.



    I can solve the first part, but i have difficulty in the second part (show that (11)^1/2 is approximately equal to 663/200). Can anyone help me? Thanks.
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  2. #2
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    If \displaystyle \sqrt{11} = \sqrt{\frac{1 + x}{1 - x}}, then what is \displaystyle x?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    If \displaystyle \sqrt{11} = \sqrt{\frac{1 + x}{1 - x}}, then what is \displaystyle x?
    By solving that, i get x= 5/6. Substituting x=5/6 into , i get (11)^1/2=157/72 which is incorrect...
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  4. #4
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    Indeed, I was getting this too. I think there is something wrong with the question. The issue is surely due to the fact that 5/6 is not small enough for x^3 and higher powers to be neglected.

    Either that, or I'm missing something.
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  5. #5
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    "Approximately equal" is a loose term. Yes, using x= 5/6 gives 1+ 5/6+ 25/72= 1.18 while \sqrt{11}= .32 approximately. That's not very close but then, as Unknown008 said, 5/6 is not all that "small"! \left(\frac{5}{6}\right)^3 is larger than 1/2- hardly "negligible".
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  6. #6
    MHF Contributor Unknown008's Avatar
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    \sqrt{11} = 3.316...

    1 + \dfrac56 + \dfrac12 \left(\dfrac{25}{36}\right) = 2.18...



    It must eirler be very late, or very early where you are HallsofIvy
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    Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.
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  8. #8
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    Quote Originally Posted by awkward View Post
    Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.
    But how do i get 'square root of 11/9'? And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..
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  9. #9
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    Quote Originally Posted by MichaelLight View Post
    But how do i get 'square root of 11/9'?
    The same way you found the square root of 11 previously.

    And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..
    Simplify \sqrt{\frac{11}{9}}.
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