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    • Thread Starter
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    So I was doing STEP I 2003 Question 5 which had Binomial Expansions in it.

    As part of the Advanced Higher course we've only come across Binomial Expansions involving natural values for n. So I've never come across the expansion of say:

    \displaystyle(x+y)^\frac{a}{b}

    Or:

    \displaystyle(x+y)^{-n}

    I'm assuming the standard expansion formula doesnt hold? If anyone has any tips or would fancy explaining anything I would be really greatful!

    Cheers

    Malcolm
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    You would use maclaurin expansion, after all the binomial expansion is a special case of maclaurin.
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    Get it into the form (1+p)^n... like this, using your example:

    (x+y)^{a/b} = x^{a/b}(1 + (y/x))^{a/b}

    and you can expand it like this:

    (1+p)^n = 1 + np + \frac{n(n-1)}{2!} p^2 + \frac{n(n-1)(n-2)}{3!} p^3 + \frac{n(n-1)(n-2)(n-3)}{4!} p^4 + \dots
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    Why can't you use the 1 + nx + 1/2*n(n-1)x^2... formula?

    (x+y)^(a/b)
    (x(1+y/x))^(a/b)
    x^(a/b)(1+y/x)^(a/b)

    Then expand out (1+y/x)^(a/b) using the 1 + nx... formula.
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    Ahh, the penny drops.

    Much thanks to Pencil King and the General.
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    Newton discovered this during the plague when he was sent from Cambridge and he said that these years were 'the most fruitful' of his life.

    If you write out the binomial series going down from 0 you get

    1
    11
    121
    1331
    ...

    What newton did was go back up instead

    1 -2 3 -4 5 -6 7
    1 -1 1 -1 1 -1 1
    1 0 0 0 0 0 0
    1 1 0 0 0 0 0
    1 2 1 0 0 0 0
    1 3 3 1 0 0 0
    ...
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    I think he wanted to work out how to expand any rational n, not just naturals (or just integers).
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    (Original post by DeathAwaitsU)
    I think he wanted to work out how to expand any rational n, not just naturals (or just integers).
    Interesting, though.
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    (Original post by thomas795135)
    Newton discovered this during the plague when he was sent from Cambridge and he said that these years were 'the most fruitful' of his life.


    What newton did was go back up instead

    1 -2 3 -4 5 -6 7
    1 -1 1 -1 1 -1 1
    1 0 0 0 0 0 0
    1 1 0 0 0 0 0
    1 2 1 0 0 0 0
    1 3 3 1 0 0 0
    ...
    i'm confused - how does this relate to the first post?
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    When you expand out (a+b)^5, the expansion goes sort of like this:

    (a^5)(b^0) + (a^4)(b^1) + (a^3)(b^2) + (a^2)(b^3) + (a^1)(b^4) + (a^0)(b^5)

    The pattern is pretty clear there.

    Except that every term is multiplied by a constant. The constant can be found from Pascal's triangle:

    n=0 : 1
    n=1 : 1 1
    n=2 : 1 2 1
    n=3 : 1 3 3 1
    n=4 : 1 4 6 4 1
    n=5 : 1 5 10 10 5 1

    The line n in the triangle can be formed using the line n-1. So to work out what n = 4 is, you take n = 3 and, for each term, add together two terms in n-1. So:

    n=4: 1 + 0, 3+1, 3+3, 3+1, 1+0
    n=4: 1, 4, 6, 4, 1
    n=5: 0+1, 1+4, 4+6, 6+4, 4+1, 1+0
    n=5: 1, 5, 10, 10, 5, 1

    Although, to add to history, Pascal wasn't the first to discover those numbers. Chinese monks got to it first aggeesss ago.

    He was just saying that Newton, instead of increasing n, decreased it. This allowed him to do the expansion for any integer n, rather than any natural n.
 
 
 
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