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Math Help - Binomial series expansion

  1. #1
    Newbie SuumEorum's Avatar
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    Binomial series expansion

    Hey everyone, I'm stuck on the last part of this question, but it might require something from the previous parts, so I'm going to start from the beginning, including my working out which should hopefully be correct. Also, this is my first time typing in codes, so don't laugh if I make a mistake .

    I have been asked to expand the following equation up to and including the term y^3, for |y|<1:

     (1-y)^{-2}

    This expands to:
     1+2y+3y^3+4y^3

    Hence, or otherwise, I had to show that:

     1+\frac{2x}{1+x}+\frac{3x^2}{(1+x)^2}+...+\frac{rx  ^{r-1}}{(1+x)^{r-1}}

    I did that by substituting y as  \frac{x}{1+x}

    I then had to write it in the form (a+x)^n where a and n are integers, which should be  (1+x)^2 .

    This last part asks me to 'find the set of values of x for which the series (above) is convergent", and I haven't a clue what to do next. I'm really stuck and any help is appreciated .
    Last edited by SuumEorum; June 16th 2008 at 04:08 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by SuumEorum View Post
    Hey everyone, I'm stuck on the last part of this question, but it might require something from the previous parts, so I'm going to start from the beginning, including my working out which should hopefully be correct. Also, this is my first time typing in codes, so don't laugh if I make a mistake .

    I have been asked to expand the following equation up to and including the term y^3, for |y|<1:

     (1-y)^{-2}

    This expands to:
     1+2y+3y^3+4y^3

    Hence, or otherwise, I had to show that:

     1+\frac{2x}{1+x}+\frac{3x^2}{(1+x)^2}+...+\frac{rx  ^{r-1}}{(1+x)^{r-1}}

    I did that by substituting y as  \frac{x}{1+x}

    I then had to write it in the form (a+x)^n where a and n are integers, which should be  (1+x)^2 .

    This last part asks me to 'find the set of values of x for which the series (above) is convergent", and I haven't a clue what to do next. I'm really stuck and any help is appreciated .
    If you are talking about \sum_{n=1}^{\infty}\frac{nx^{n-1}}{(x+1)^{n-1}}

    THen applying the root test we would get

    \bigg|\frac{x}{x+1}\bigg|<1

    Which implies that

    \frac{x}{1+x}<1

    and \frac{x}{1+x}>-1

    I think you can go from there
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  3. #3
    Newbie SuumEorum's Avatar
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    Oooh thanks for your help, I've never encountered the root test before but I've read all about it on Wiki now . But I'm still slightly confused, how come, after you've applied the root test you get \bigg|\frac{x}{x+1}\bigg|<1 but not \bigg|\frac{nx}{x+1}\bigg|<1?
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by SuumEorum View Post
    Oooh thanks for your help, I've never encountered the root test before but I've read all about it on Wiki now . But I'm still slightly confused, how come, after you've applied the root test you get \bigg|\frac{x}{x+1}\bigg|<1 but not \bigg|\frac{nx}{x+1}\bigg|<1?
    Note that \lim_{n\rightarrow{+\infty}}\sqrt[n]{n}=1 while \lim_{n\rightarrow{+\infty}}\sqrt[n]{\frac{x^{n-1}}{(1+x)^{n-1}}}=\frac{x}{1+x} ( x\geq{0})
    Last edited by PaulRS; June 16th 2008 at 05:15 PM.
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  5. #5
    Newbie SuumEorum's Avatar
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    Aaah I see it now, thanks , I can finally go to sleep .
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