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Binomial Series Help.

  1. Jan 30, 2005 #1
    I've got 1 question regarding the binomial series which I am currently stuck at.

    1. Expand [tex] (1-x)^-^3 [/tex] and express the coefficient of [tex] x^r [/tex] in terms of r.

    What i did was to first expand it, according to the binomial series, and I got,

    [tex] 1+3x+6x^2........\frac {(-3)(-3-1)....(-3-r+1)}{r!} (-x)^r [/tex]

    and the answer is [tex] \frac {(r+1)(r+2)}{2} [/tex]. How do i get from [tex] \frac {(-3)(-3-1)....(-3-r+1)}{r!} [/tex] to the answer? The dots confuse me.
     
  2. jcsd
  3. Jan 30, 2005 #2

    dextercioby

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    The general binomial formula is
    [tex] (1-a)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{k} a^{k} [/tex]

    So decide what is the coefficient of "a^{r}".


    Daniel.
     
  4. Jan 30, 2005 #3
    I know it seems like very absurd, daniel, but we haven't learnt about sequences and series as well as sigma notation but we started on binomial expansions first !

    But all I know for now, is that for the term [tex] x^r [/tex], in the expansion of [tex] (1+x)^n [/tex] is,

    [tex] \frac {n(n-1)(n-2)...(n-r+1)}{r!} x^r[/tex]

    what confuses me is how to get rid of those dots...
     
  5. Jan 30, 2005 #4

    Curious3141

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    Don't worry about the dots, all you have to remember is that you're multiplying a whole lot of terms to get the desired coefficient. Think simply.

    In this case, realise :

    a) the coefficients of [itex]x^r[/itex] are always positive.

    b) The numerator you're dealing with is very similar to a factorial - I'm assuming you've already studied up on those ? In this case, you're multiplying in increasing terms starting from 3,4, and going up to (r + 2). This is like taking [itex](r + 2) ![/itex] and dividing it by 2, do you see that ?

    Let's look at the general term (only the coefficient) :

    [tex]\frac{(3)(4)...(3 + r - 2)(3 + r - 1)}{r !} = \frac{(3)(4)...(r + 1)(r + 2)}{r !}[/tex]

    With me so far ? Then ...

    [tex]\frac{(3)(4)...(r + 1)(r + 2)}{r !} = \frac{(1)(2)(3)(4)...(r + 1)(r + 2)}{(1)(2)(r !)}
    [/tex]

    and here I'm multiplying both the numerator and the denominator by the same amount (2) to get the numerator to equal [itex](r + 2) ![/itex]

    Can you proceed from here ? :wink:
     
    Last edited: Jan 30, 2005
  6. Jan 30, 2005 #5
    Oh yes ! I've seen it....thanks alot man.

    And hey ! you're from Singapore too.....!
     
  7. Jan 30, 2005 #6

    Curious3141

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    Glad to help, and, yes, I'm from Sg. :smile:
     
  8. Jan 31, 2005 #7
    Some years ago, I came up with a formula for the binomial series expansion, which doesn't make use of factorials, but instead uses iterated products over a dummy variable inside the summation sign. From memory, the formula I got was:


    [tex] (1+U)^{\alpha}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{U(\alpha + 1 - k)}{k}[/tex]

    Eventually, I learned to solve many problems by assuming a general solution of the form:

    [tex] \sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} f(k)[/tex]

    where f(k) is an unknown function of k. Using that assumption, it is possible to solve for the spherical harmonics, for the hydrogen atom, without having to have memorized the legendre polynomials. But I can leave that for another time.

    As for your problem, [tex] U=-x [/tex], and [tex] \alpha = -3 [/tex]. Therefore, you want to expand the following combination sum/product:

    [tex] (1-x)^{-3}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{-x(-3 + 1 - k)}{k}[/tex]

    or equivalently

    [tex] (1-x)^{-3}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{x(2 + k)}{k}[/tex]

    the first term in the expansion (corresponding to n=0) is 1, the next term in the expansion corresponding to n=1, is given by

    [tex ] \prod_{k=1}^{k=1} \frac{x(2 + k)}{k} = \frac{x(2+1)}{1} = 3x [/tex]

    The third term in the expansion (corresponding to n=2) is given by:

    [tex] \prod_{k=1}^{k=2} \frac{x(2 + k)}{k} = (3x)(4x/2) = 6x^2 [/tex]

    Now you can see the pattern. The coefficient of x^r is given by:

    [tex] \prod_{k=1}^{k=r} \frac{x(2 + k)}{k} [/tex]


    Kind regards,

    Guru
     
  9. Feb 2, 2005 #8
    (Fixing latex error in previous post, I wanted to just edit it, but couldn't.)


    Some years ago, I came up with a formula for the binomial series expansion, which doesn't make use of factorials, but instead uses iterated products over a dummy variable inside the summation sign. From memory, the formula I got was:


    [tex] (1+U)^{\alpha}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{U(\alpha + 1 - k)}{k}[/tex]

    Eventually, I learned to solve many problems by assuming a general solution of the form:

    [tex] \sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} f(k)[/tex]

    where f(k) is an unknown function of k. Using that assumption, it is possible to solve for the spherical harmonics, for the hydrogen atom, without having to have memorized the legendre polynomials. But I can leave that for another time.

    As for your problem, [tex] U=-x [/tex], and [tex] \alpha = -3 [/tex]. Therefore, you want to expand the following combination sum/product:

    [tex] (1-x)^{-3}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{-x(-3 + 1 - k)}{k}[/tex]

    or equivalently

    [tex] (1-x)^{-3}=\sum_{n=0}^{n=\infty} \prod_{k=1}^{k=n} \frac{x(2 + k)}{k}[/tex]

    the first term in the expansion (corresponding to n=0) is 1, the next term in the expansion corresponding to n=1, is given by

    [tex] \prod_{k=1}^{k=1} \frac{x(2 + k)}{k} = \frac{x(2+1)}{1} = 3x [/tex]

    The third term in the expansion (corresponding to n=2) is given by:

    [tex] \prod_{k=1}^{k=2} \frac{x(2 + k)}{k} = (3x)(4x/2) = 6x^2 [/tex]

    Now you can see the pattern. The coefficient of x^r is given by:

    [tex] \prod_{k=1}^{k=r} \frac{(2 + k)}{k} [/tex]


    Kind regards,

    Guru
     
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