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    f '(x) = (1-x)^-2 - (1-x/3)^-2

    show that when f '(x) is expanded in powers of x the coefficients of all the powers of x are positive.

    now the mark scheme says:
    [(-2)(-3)...(r/1)]/r! [(-1)^r - (-1/3)^r ]
    = (r+1) [1-(1/3)^r] > 0

    what on earth??

    also how do you work out what the limits for an binomial expansion are?

    Thank you
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    (Original post by Asphyxiating)
    f '(x) = (1-x)^-2 - (1-x/3)^-2

    show that when f '(x) is expanded in powers of x the coefficients of all the powers of x are positive.

    now the mark scheme says:
    [(-2)(-3)...(r/1)]/r! [(-1)^r - (-1/3)^r ]
    = (r+1) [1-(1/3)^r] > 0

    what on earth??

    also how do you work out what the limits for an binomial expansion are?

    Thank you
    Do you know how to expand it? cos the coeffiecint of x is negative, and the power isnegative , so yeah all the powers of x are positive.
    As for range i think it's -1/3<x<1/3 cos you take the smallest range out of the two
    anyone correct me on this??
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    (1 - x)^-2
    = 1 + [(-2)/1!](-x) + [(-2)(-3)/2!](-x)^2 + [(-2)(-3)(-4)/3!](-x)^3 + ...
    = 1 + 2x + 3x^2 + 4x^3 + ...

    (1 - x/3)^-2
    = 1 + (2/3)x + (3/3^2)x^2 + (4/3^3)x^3 + ...

    (1 - x)^-2 - (1 - x/3)^-2
    = [1 + 2x + 3x^2 + 4x^3 + ...]
    - [1 + (2/3)x + (3/3^2)x^2 + (4/3^3)x^3 + ...]
    = 2[1 - 1/3]x + 3[1 - (1/3)^2]x^2 + 4[1 - (1/3)^3]x^3 + ...

    For all r >= 1 the coefficient of x^r in the last expansion is (as the mark scheme says)

    (r + 1) * [1 - (1/3)^r] . . . (*).

    We have to prove that (*) is positive. That's easy because (*) is the product of two positive terms, (r + 1) and [1 - (1/3)^r].

    Now on to limits of validity. The binomial expansion of (1 - ax)^k is valid when |ax| < 1. In your problem, the expansion of (1 - x)^-2 is valid when |x| < 1 and the expansion of (1 - x/3)^-2 is valid when |x| < 3. The expansion of (1 - x)^-2 - (1 - x/3)^-2 is valid when both of the sub-expansions are: ie, when |x| < 1 and |x| < 3; ie, when |x| < 1.
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    (Original post by Jonny W)
    (1 - x)^-2
    = 1 + [(-2)/1!](-x) + [(-2)(-3)/2!](-x)^2 + [(-2)(-3)(-4)/3!](-x)^3 + ...
    = 1 + 2x + 3x^2 + 4x^3 + ...

    (1 - x/3)^-2
    = 1 + (2/3)x + (3/3^2)x^2 + (4/3^3)x^3 + ...

    (1 - x)^-2 - (1 - x/3)^-2
    = [1 + 2x + 3x^2 + 4x^3 + ...]
    - [1 + (2/3)x + (3/3^2)x^2 + (4/3^3)x^3 + ...]
    = 2[1 - 1/3]x + 3[1 - (1/3)^2]x^2 + 4[1 - (1/3)^3]x^3 + ...

    For all r >= 1 the coefficient of x^r in the last expansion is (as the mark scheme says)

    (r + 1) * [1 - (1/3)^r] . . . (*).

    We have to prove that (*) is positive. That's easy because (*) is the product of two positive terms, (r + 1) and [1 - (1/3)^r].

    Now on to limits of validity. The binomial expansion of (1 - ax)^k is valid when |ax| < 1. In your problem, the expansion of (1 - x)^-2 is valid when |x| < 1 and the expansion of (1 - x/3)^-2 is valid when |x| < 3. The expansion of (1 - x)^-2 - (1 - x/3)^-2 is valid when both of the sub-expansions are: ie, when |x| < 1 and |x| < 3; ie, when |x| < 1.
    Wow thank you! That is so much clearer now Just one thing, why does an expansion have to be valid when |ax| < 1 ? Is that just a rule? Thank you so much
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    (Original post by Asphyxiating)
    Wow thank you! That is so much clearer now Just one thing, why does an expansion have to be valid when |ax| < 1 ? Is that just a rule? Thank you so much
    Its just a rule.
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    (Original post by Asphyxiating)
    Wow thank you! That is so much clearer now Just one thing, why does an expansion have to be valid when |ax| < 1 ? Is that just a rule? Thank you so much
    When using the binomial expansion for (1+ax)^n, you have to note that the series becomes infinite when the power is not a positive integer. So you have to have a rule that |ax| < 1 for the series to converge.
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    (Original post by Fermat)
    When using the binomial expansion for (1+ax)^n, you have to note that the series becomes infinite when the power is not a positive integer. So you have to have a rule that |ax| < 1 for the series to converge.

    How come you can't have infinite series that diverge? Um how does |ax| < 1 make it converge though?
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    (Original post by Asphyxiating)
    why does an expansion have to be valid when |ax| < 1 ? Is that just a rule?
    For example,

    (1 - ax)^-2
    = 1 + [(-2)/1!](-ax) + [(-2)(-3)/2!](-ax)^2 + [(-2)(-3)(-4)/3!](-ax)^3 + ...
    = 1 + 2ax + 3(ax)^2 + 4(ax)^3 + ... [1].

    If |ax| >= 1 then the terms in [1] get bigger (in absolute value) as you move to the right. So you can't sum them.

    If |ax| < 1 then the terms in [1] get close to zero rapidly as you move to the right. So you can sum them.

    If ax = 1.1 the first hundred terms are

    1.0, 2.2, 3.63, 5.324, 7.3205, 9.66306, 12.40093, 15.58974, 19.2923, 23.57948, 28.53117, 34.2374, 40.79957, 48.3318, 56.96248, 66.83597, 78.11454, 90.98047, 105.63843, 122.31818, 141.2775, 162.8055, 187.22632, 214.90326, 246.24332, 281.70235, 321.79077, 367.07984, 418.20881, 475.89279, 540.93147, 614.21896, 696.75463, 789.65525, 894.16845, 1011.68773, 1143.76918, 1292.15005, 1458.76939, 1645.79111, 1855.62948, 2090.97761, 2354.83907, 2650.56304, 2981.88342, 3352.96225, 3768.43801, 4233.47929, 4753.84446, 5335.94786, 5986.9335, 6714.75678, 7528.2754, 8437.35016, 9452.95712, 10587.31198, 11854.00823, 13268.17062, 14846.6254, 16608.08943, 18573.38001, 20765.64782, 23210.63538, 25936.96398, 28976.45195, 32364.46787, 36140.32245, 40347.70328, 45035.15704, 50256.62452, 56072.03393, 62547.95898, 69758.34869, 77785.33676, 86720.13896, 96664.04823, 107729.53796, 120041.48515, 133738.52641, 148974.56106, 165920.41738, 184765.69936, 205720.83355, 229019.33759, 254920.3341, 283711.33654, 315711.33613, 351274.22227, 390792.57228, 434701.85006, 483485.05768, 537677.88832, 597874.43451, 664733.51106, 738985.65857, 821440.90047, 912997.33417, 1014650.64561, 1127504.64599, 1252782.93998

    If ax = 0.9 the first hundred terms are

    1.0, 1.8, 2.43, 2.916, 3.2805, 3.54294, 3.72009, 3.82638, 3.8742, 3.8742, 3.83546, 3.76573, 3.67158, 3.55861, 3.43152, 3.29426, 3.15013, 3.00189, 2.8518, 2.7017, 2.55311, 2.40722, 2.26497, 2.12711, 1.99416, 1.86653, 1.74449, 1.62819, 1.51771, 1.41304, 1.31413, 1.22087, 1.13312, 1.05071, 0.97345, 0.90114, 0.83355, 0.77047, 0.71167, 0.65693, 0.60602, 0.55872, 0.51482, 0.47411, 0.4364, 0.40149, 0.36919, 0.33934, 0.31177, 0.28632, 0.26284, 0.2412, 0.22125, 0.20288, 0.18598, 0.17042, 0.15612, 0.14297, 0.13089, 0.1198, 0.10962, 0.10027, 0.0917, 0.08384, 0.07664, 0.07003, 0.06399, 0.05845, 0.05338, 0.04873, 0.04449, 0.0406, 0.03705, 0.0338, 0.03083, 0.02812, 0.02564, 0.02338, 0.02131, 0.01942, 0.0177, 0.01612, 0.01469, 0.01338, 0.01218, 0.01109, 0.0101, 0.0092, 0.00837, 0.00762, 0.00693, 0.00631, 0.00574, 0.00522, 0.00475, 0.00432, 0.00393, 0.00357, 0.00325, 0.00295
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    (Original post by Jonny W)

    Now on to limits of validity. The binomial expansion of (1 - ax)^k is valid when |ax| < 1. In your problem, the expansion of (1 - x)^-2 is valid when |x| < 1 and the expansion of (1 - x/3)^-2 is valid when |x| < 3. The expansion of (1 - x)^-2 - (1 - x/3)^-2 is valid when both of the sub-expansions are: ie, when |x| < 1 and |x| < 3; ie, when |x| < 1.
    Say you have (1 + x/3)^-2 ..is it still x < 3 ? thank yous
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    (Original post by Asphyxiating)
    Say you have (1 + x/3)^-2 ..is it still x < 3 ? thank yous
    What you should have is,

    |x/3| < 1
    -1 < x/3 < 1
    -3 < x < 3
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    (Original post by Fermat)
    What you should have is,

    |x/3| < 1
    -1 < x/3 < 1
    -3 < x < 3
    Alright thank you
 
 
 
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