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Binomial expansion for fractional power

  1. Mar 23, 2016 #1
    1. The problem statement, all variables and given/known data
    So, I'm solving a dipole thing and I have these vectors:
    |r + d - r'| = (r² + d² - r'²)(1/2)
    2. Relevant equations
    I want to expand this but I have no idea how! I know I may have an infinite power series, but I may expand at the square terms tops...
    Before I needed to do the same with something like (a + b)(1/2) and I got it, I found how to do it. But in this case I can't find a conection...
    Can anyone please explain to me how to? Or at least indicate somewhere where I can find it? Because I looked over and couldn't find...

    Thanks...
     
  2. jcsd
  3. Mar 23, 2016 #2

    Ray Vickson

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    What do you want to expand with respect to? Expand in powers of ##r^2##? Expand in powers of ##d^2##? Expand in powers of ##r'^2##? Expand in something else?

    Just expand ##(a+x)^{1/2}## in powers of ##x##, then identify ##a## and ##x## for your problem.
     
  4. Mar 23, 2016 #3
    For example,if I have ##(a+b)^n## , where n can be a fraction...I know I solve like:

    ##(a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...##

    If I had ##(a-b)^n## I should alternate the signal + and -...

    But in the case ##(a+b+c)^n## or more precisely ##(a+b-c)^n## I don't know how to do it...
    I just want to expand like the one above...
     
  5. Mar 23, 2016 #4

    WWGD

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    (a+b+c)=((a+b)+c) is the original form.
     
  6. Mar 23, 2016 #5

    Ray Vickson

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    You have still have not answered my original question: expand with respect to what?

    You could expand in power of ##a## as
    [tex] (a+b+c)^n = (b+c)^n \sum_{k=0}^{\infty} {n \choose k} \left( \frac{a}{b+c} \right)^k, [/tex]
    where
    [tex] {n \choose k}= \frac{n(n-1) \cdots (n-k+1)}{k!} [/tex]
    for any real ##n## and non-negative integer ##k##.

    Or, you could try to expand the whole thing in powers like ##a^k b^j c^l##, perhaps by re-writing the above as
    [tex] (a+b+c)^n = \sum_{k=0}^{\infty} {n \choose k} a^k (b+c)^{n-k}, [/tex]
    then expanding each of the ##(c+b)^{n-k}## in powers of ##b## and ##c##, and then collecting terms. It would be messy.

    Or, you could use the trinomial expansion, which says that
    [tex] (a+b+c)^n = \sum_{k_1,k_2,k_3} c(k_1,k_2,k_3) a^{k_1} b^{k_2} c^{k_3}, [/tex]
    where
    [tex] c(k_1,k_2,k_3) = {n \choose k_1} {n-k_1 \choose k_2} {n-k_1-k_2 \choose k_3} [/tex]
     
    Last edited: Mar 23, 2016
  7. Mar 23, 2016 #6

    vela

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    What vectors? How are you managing to equate the lefthand side with the righthand side?
     
  8. Mar 23, 2016 #7
    Sorry, I know I have not been very clear, but I'm trying to understand (and failing) that's why...
    Here what I have (in the figure):
    I need to expand the first term, conserve only the linear term in "d". I need to get the last equation...but I have no idea about what I'm doing :( 1.jpg
     
  9. Mar 23, 2016 #8

    Ray Vickson

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    Much better: write ##\vec{p} = \frac{1}{2} \vec{d}## so you have
    [tex] \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{|\vec{r} - \vec{p}|} - \frac{1}{|\vec{r} + \vec{p}|} \right] [/tex]
    Now
    [tex] \frac{1}{|\vec{r} - \vec{p}|}=\frac{1}{r} (1-x_1)^{-1/2}, \;\; x_1 = \frac{2 \vec{r}\cdot \vec{p}}{r^2} - \frac{p^2}{r^2} [/tex]
    with as similar expression for ##1/|\vec{r} + \vec{p}|##. Now use the binomial expansion for ##(1-x_1)^{-1/2}## in powers of ##x_1##, exactly as I had suggested in post #2.

    In your case, stopping at the first power (##x_1^1##) will be good enough.
     
  10. Mar 24, 2016 #9

    vela

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    You have a sign error in your diagram, but I think your expression for the potential is okay for the way I think you're defining ##\vec{d}##. Still, you probably should start over and go with Ray's suggestion since ##\theta## is defined in terms of ##\vec{r}## and the dipole axis.

    Remember that ##\lvert \vec{x} \rvert## is defined as ##\sqrt{\vec{x}\cdot\vec{x}}##. Use that fact and then factor the ##r^2## out of the square root to get the expression Ray wrote above on the righthand side of the equation.
     
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