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Thread: Binomial Expansion

  1. #1
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    Binomial Expansion

    I am not a student, but I am trying to figure this out.

    The text book states:
    "We expand our previous forma(using a binomial expansion) and drop 'higher order' terms"

    The formula the text is referring to is
    \frac{F}{1-dF}

    And after binomial expansion reduces to:

    F^2 \time d

    I appreciate any help.




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  2. #2
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    Re: Binomial Expansion

    just do the division and truncate terms where the order of $F$ is greater than 1.

    I get $F + d F^2$ though

    are you sure the answer is just $d F^2$ ?
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  3. #3
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    Re: Binomial Expansion

    Yes, I just double checked.
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  4. #4
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    Re: Binomial Expansion

    well...

    a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

    and I confess I don't see the relevance of this for this problem.

    b)

    $\dfrac{1}{1 - d F} = \displaystyle{\sum_{k=0}^\infty}~(dF)^k = 1 + d F + d^2 F^2 + \dots,~\text{for }|d F|<1$

    so

    $\dfrac{F}{1- d F} = F + d F^2 + d^2 F^3 \dots$

    and we truncate this to

    $\dfrac{F}{1- d F} = F + d F^2$
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  5. #5
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    Re: Binomial Expansion

    Quote Originally Posted by romsek View Post
    well...

    a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

    and I confess I don't see the relevance of this for this problem.
    $\dfrac{F}{1 - dF} = $

    $F(1 - dF)^{-1} \approx $

    $F[(1)^{-1} - 1(1)^{-2}(-dF)] = $

    $F[1 + dF] = $

    $F + dF^2$
    Thanks from romsek
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  6. #6
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    Re: Binomial Expansion

    Quote Originally Posted by romsek View Post
    well...

    a) the binomial expansion is $(a + b)^n = \displaystyle{\sum_{k=0}^n}~\binom{n}{k} a^k b^{n-k}$

    and I confess I don't see the relevance of this for this problem.

    b)

    $\dfrac{1}{1 - d F} = \displaystyle{\sum_{k=0}^\infty}~(dF)^k = 1 + d F + d^2 F^2 + \dots,~\text{for }|d F|<1$

    so

    $\dfrac{F}{1- d F} = F + d F^2 + d^2 F^3 \dots$

    and we truncate this to

    $\dfrac{F}{1- d F} = F + d F^2$
    I am not sure what the relevance is. I just want to fully understand the text. Honestly the approximation is more work. I appreciate you help. Is the binomial theorem just for variables?

    Quote Originally Posted by greg1313 View Post
    $\dfrac{F}{1 - dF} = $

    $F(1 - dF)^{-1} \approx $

    $F[(1)^{-1} - 1(1)^{-2}(-dF)] = $

    $F[1 + dF] = $

    $F + dF^2$
    How did you go from
    $F[(1)^{-1} - 1(1)^{-2}(-dF)] = $

    to

    $F[1 + dF] = $

    This is what I came up with


    $\begin{aligned}\sum_{k=-1}^{n} 1^{n-k}dF^k\\
    F(1-dF)^{-1} &\approx \sum_{k=-1}^{2} 1^{n-(-1)}dF^{-1}\\
    &\approx 1^{2 -(-1)}dF^{-1} + 1^{2-0}dF^{0} + 1^{2-1}dF^{1} + 1^{2-2}dF^{2}\\
    &\approx dF^{-1} + dF + dF^2\\
    &\approx \frac{1}{dF} + dF + dF^2\\
    \end{aligned}$

    Thanks again
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  7. #7
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    Re: Binomial Expansion

    Quote Originally Posted by bmccardle View Post
    Is the binomial theorem just for variables? No. For example,

    it can be used on  \ \  (9 + 1)^{\frac{1}{2}}.See below.



    How did you go from

    $F[(1)^{-1} - 1(1)^{-2}(-dF)] = $

    to

    $F[1 + dF] = $
    $F[(1)^{-1} - 1(1)^{-2}(-dF)] = $

    $F[1 - 1(1)(-dF)] = $

    $F[1 + dF)] = $

    $F + dF^2$


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

    This is what I came up with


    $\begin{aligned}\sum_{k=-1}^{n} 1^{n-k}dF^k\\
    F(1-dF)^{-1} &\approx \sum_{k=-1}^{2} 1^{n-(-1)}dF^{-1}\\
    &\approx 1^{2 -(-1)}dF^{-1} + 1^{2-0}dF^{0} + 1^{2-1}dF^{1} + 1^{2-2}dF^{2}\\
    &\approx dF^{-1} + dF + dF^2\\
    &\approx \frac{1}{dF} + dF + dF^2\\
    \end{aligned}$
    Look at this instead:

    $\begin{aligned}F*\sum_{k=0}^{n} 1^{n-k}(dF)^k\\
    F(1-dF)^{-1} &\approx \sum_{k=0}^{2} 1^{n-(0)}(dF)^{-1}\\
    &\approx F[ 1^{2-0}(dF)^{0} + 1^{2-1}(dF)^{1} + 1^{2-2}(dF)^{2}]\\
    &\approx F(1 + dF + d^2F^2)\\
    &\approx F + dF^2 + d^2F^3 \\
    \end{aligned}$


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


     \ \  (9 + 1)^{\frac{1}{2}} \ \approx

    9^{\frac{1}{2}}  \ +  \ \bigg(\frac{1}{2}\bigg)9^{\frac{-1}{2}}(1)^1 \ =

    3 + \bigg(\dfrac{1}{2}\bigg)\bigg(\dfrac{1}{3}\bigg) =

    3 + \dfrac{1}{6} \approx

    3.167
    Last edited by greg1313; Mar 21st 2017 at 07:43 PM.
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  8. #8
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    Re: Binomial Expansion

    * * * This is an edit (correction) for the previous post that did not make it,
    because the time limit for edits ran out. * * *

    Look at this instead:

    $\begin{aligned}\sum_{k=0}^{n} 1^{n-k}(dF)^k\\
    F(1-dF)^{-1} &\approx F*\sum_{k=0}^{2} 1^{n-(0)}(dF)^{-1}\\
    &\approx F[ 1^{2-0}(dF)^{0} + 1^{2-1}(dF)^{1} + 1^{2-2}(dF)^{2}]\\
    &\approx F(1 + dF + d^2F^2)\\
    &\approx F + dF^2 + d^2F^3 \\
    \end{aligned}$


    Compare this end result to what was already posted in post # 4 by romsek.
    Last edited by greg1313; Mar 21st 2017 at 07:55 PM.
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  9. #9
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    Re: Binomial Expansion

    I think I understand now. Thank you greg and romsek!
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