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Math Help - Binomial series (range of validity)

  1. #1
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    Binomial series (range of validity)

    Expand (1-x)^(1/2) * (1+2x)^(1/2)

    (1-x)^(1/2) * (1+2x)^(1/2)
    = 1 + (1/2) x - (9/8)x^2 + (9/16)^3 - (117/128)x^4 + (279/256)x^5 + ...

    the range of validity for the expansion of (1-x)^(1/2) is |x|<1
    the range of validity for the expansion of (1+2x)^(1/2) is |x|< 1/2
    taking the intersection, the expansion of (1-x)^(1/2) * (1+2x)^(1/2) is valid for |x|< 1/2.

    Here come my questions,

    (1-x)^(1/2) * (1+2x)^(1/2)
    =(1+x-2x^2)^(1/2)

    if I expand using Binomial theorem, the result is exactly the same.
    However, when I solve |x-2x^2|<1 for the range of validity,
    I got -0.414<x<1 or 1<x<2.414 (rejected as 1-x should be positive)
    So -0.414<x<1!!! which is different from the previous |x|<1/2

    May I ask why is the range of validity different? I thought since I am expanding the same expression so the range of validity should be unique, right?

    Pls enlighten me
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  2. #2
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    Thumbs up Re: Binomial series (range of validity)

    Quote Originally Posted by acc100jt View Post
    Expand (1-x)^(1/2) * (1+2x)^(1/2)

    (1-x)^(1/2) * (1+2x)^(1/2)
    = 1 + (1/2) x - (9/8)x^2 + (9/16)^3 - (117/128)x^4 + (279/256)x^5 + ...

    the range of validity for the expansion of (1-x)^(1/2) is |x|<1
    the range of validity for the expansion of (1+2x)^(1/2) is |x|< 1/2
    taking the intersection, the expansion of (1-x)^(1/2) * (1+2x)^(1/2) is valid for |x|< 1/2.

    Here come my questions,

    (1-x)^(1/2) * (1+2x)^(1/2)
    =(1+x-2x^2)^(1/2)

    if I expand using Binomial theorem, the result is exactly the same.
    However, when I solve |x-2x^2|<1 for the range of validity,
    I got -0.414<x<1 or 1<x<2.414 (rejected as 1-x should be positive)
    So -0.414<x<1!!! which is different from the previous |x|<1/2

    You have solved |x-2x^2|<1 incorrectly.

    May I ask why is the range of validity different? I thought since I am expanding the same expression so the range of validity should be unique, right?

    Pls enlighten me
    |x-2x^{2}|=\left\{\begin{array}{ccc}x-2x^{2}&\mbox{if}& x-2x^{2}\geq 0\\ 2x^2-x & \mbox{if}& x-2x^2<0\end{array}\right.

    Therefore, |x-2x^{2}|=\left\{\begin{array}{ccc}x-2x^{2}&\mbox{if}& 0\leq x\leq \frac{1}{2}\\ 2x^2-x & \mbox{if}& x<0\mbox{ or }x>\frac{1}{2} \end{array}\right.

    Case 1:

    x-2x^{2}<1 \mbox{ and }0\leq x\leq \frac{1}{2}\Rightarrow 0\leq x\leq \frac{1}{2}------(1)

    Case 2:

     2x^2-x <1\mbox{ and }\left(x<0\mbox{ or }x>\frac{1}{2}\right)\Rightarrow -\frac{1}{2}<x<0\mbox{ or }\frac{1}{2}<x<1-------(2)

    By (1) and (2);

    -\frac{1}{2}<x\leq 1

    So the two methods you have done does tally with each other.
    Last edited by Sudharaka; June 29th 2011 at 01:49 AM. Reason: Corrected the mistake pointed in post #5
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  3. #3
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    Re: Binomial series (range of validity)

    Ooops! Sorry for the mistake.

    I solved |x-2x^2|<1 using graphic calculator, but the answer is -0.5<x<1.

    So same question, why are the two range of validity different? Which one should be the correct range of validity?
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    Re: Binomial series (range of validity)

    Quote Originally Posted by acc100jt View Post
    Ooops! Sorry for the mistake.

    I solved |x-2x^2|<1 using graphic calculator, but the answer is -0.5<x<1.

    So same question, why are the two range of validity different? Which one should be the correct range of validity?
    Your calculator is giving you a wrong answer. I suggest you try to do the problem yourself without relying on the calculator.
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    Re: Binomial series (range of validity)

    really? let's clarify first the solution to |x-2x^2|<1.

    I got -0.5 < x <1. It should be correct right?

    For example, |0.9 - 2*0.9^2| = |-0.72| = 0.72 is smaller than 1.

    But 0.9 is not captured in your solution.
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  6. #6
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    Re: Binomial series (range of validity)

    Quote Originally Posted by acc100jt View Post
    really? let's clarify first the solution to |x-2x^2|<1.

    I got -0.5 < x <1. It should be correct right?

    For example, |0.9 - 2*0.9^2| = |-0.72| = 0.72 is smaller than 1.

    But 0.9 is not captured in your solution.
    Yep. I am sorry, there's a mistake in my previous post (in the line numbered as 2). As you can see I have corrected it now. So the answer the calculator gave is correct. It agrees with the value obtained in the first method. As you can see in the first method you obtained that the expansion is valid in the interval |x|<1/2. In the second method we obtained that the expansion is valid in the interval -1/2<x<1 which is a larger set than |x|<1/2. So its just that in the second method we get additional information. Do you understand?
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