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    part a) find binomial expansion of (1+2x)^(-1/2)
    my answer: 1-x+3x/4
    part b) using part a show with x=-0.1 sqr(5)=2.23
    by substituting in -0.1 and multiplying by 2 i only get 2.215

    What have i done wrong?
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    (Original post by TheGreatPumpkin)
    part a) find binomial expansion of (1+2x)^(1/2)
    my answer: 1-x+3x/4
    part b) using part a show with x=-0.1 sqr(5)=2.23
    by substituting in -0.1 and multiplying by 2 i only get 2.215

    What have i done wrong?
    Your expansion.

    It should be 1 + x - x^2 / 2.

    Check it again.
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    (Original post by Zacken)
    Your expansion.

    It should be 1 + x - x^2 / 2.

    Check it again.
    Sorry I made a mistake typing it was (1+2x)^-(1/2) I've changed it now
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    (Original post by TheGreatPumpkin)
    Sorry I made a mistake typing it was (1+2x)^-(1/2) I've changed it now
    Your expansion is still wrong. It should be 1-x+\frac{3}{2}x^2 + \cdots
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    (Original post by Zacken)
    Your expansion is still wrong. It should be 1-x+\frac{3}{2}x^2 + \cdots
    I don't understand, to work out the co-efficient of x^2 don't you (-0.5)(-1.5)(2)/2 which gives 3/4,
    like the formula n(n-1)/2
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    (Original post by TheGreatPumpkin)
    I don't understand, to work out the co-efficient of x^2 don't you (-0.5)(-1.5)(2)/2 which gives 3/4,
    like the formula n(n-1)/2
    You might want to find the expansion of (1+y)^{-1/2} and then replace y by 2x...
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    (Original post by DFranklin)
    You might want to find the expansion of (1+y)^{-1/2} and then replace y by 2x...
    That's just 1 - x/2 + 3x^2 /8
    1/2^2= 1/4
    1/4 +1/2 is 3/4
    (3/4)/2=3/8
    can I see your working out please?
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    the coefficient of x is also squared in the x^2 term
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    (Original post by TheGreatPumpkin)
    I don't understand, to work out the co-efficient of x^2 don't you (-0.5)(-1.5)(2)/2 which gives 3/4,
    like the formula n(n-1)/2
    As the above posts suggest, it'll be (-0.5)(-1.5)(2)/2 * (2x)^2
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    (Original post by Andrew Dainty)
    the coefficient of x is also squared in the x^2 term
    Oh thank you
 
 
 
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