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Advanced Higher Maths 2012-2013 : Discussion and Help Thread

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    (Original post by ukdragon37)
    .
    The thing is I'm getting everything right when it comes to the binomial theorem stuff but I don't understand by looking at the formulae what it means?

    Also, when I say ''how does 2! (2-n)! equal 2'' I am referring to when you have to work out n on your own. The question is, for example:

    (n)
    (2) = 15

    So I got:

    n! / 2!(n-2)! = 15

    which goes to

    n(n-1) / 2 = 15

    which you can then solve through to get n=6 for a positive integer. However I don't understand how the bottom lone goes to 2 in this case from 2!(n-2)!
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    Just started today!

    I love it so much more interesting than higher scared that I get a C for higher as a boy last year missed out by one mark and had to resit higher

    Doing this with adv physics and crashing adv graphic comm



    This was posted from The Student Room's iPhone/iPad App
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    Everyone's starting yet I've helped with the P7 visit and am again tomorrow... yaaaaay :boring:
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    That strange feeling of jealousy and loss when you realise that ukd will never be helping you with maths ever again... :moon:

    (Original post by JaggySnake95)
    The thing is I'm getting everything right when it comes to the binomial theorem stuff but I don't understand by looking at the formulae what it means?

    Also, when I say ''how does 2! (2-n)! equal 2'' I am referring to when you have to work out n on your own. The question is, for example:

    (n)
    (2) = 15

    So I got:

    n! / 2!(n-2)! = 15

    which goes to

    n(n-1) / 2 = 15

    which you can then solve through to get n=6 for a positive integer. However I don't understand how the bottom lone goes to 2 in this case from 2!(n-2)!
    What do you mean by looking at the formula? You mean this?

    \displaystyle\sum_{r=0}^n \displaystyle \binom{n}{r} a^{n-r}b^r
    As for algebraic long division, it is a pretty imperative skill at Advanced Higher. So get good at it, you'll need it for complex numbers and some calculus (usually integration) questions, if there's no other method you can find, that is.

    I'd definitely suggest learning LaTeX if you're going to be posting in here, it makes it easier for everyone involved so we don't have to decipher what you mean. It's actually much easier than you'd think!

    \displaystyle \binom{n}{2}= 15 \\ \frac{n!}{2!(n-2)!} = 15

    As for how that becomes what you said, it's really simple. 2! = 2, as to how the bottom seems to "switch" to the top is from an understanding of factorials.

    n! = 1 \times 2 \times 3 \times \ldots \times (n-2) \times (n -1) \times n
    and (n-2)! is what you've got on the bottom which is obviously...

    (n-2)! = 1 \times 2 \times 3 \times \ldots \times (n - 3) \times (n -2)

    It's then as simple as terms cancelling out. If you set these series on top of each other in a fraction

    \frac{n!}{(n - 2)!} (I'm ignoring the 2! for now, as it doesn't really effect anything in terms of cancelling out, if you want a more mathematical way of putting it, I'm taking 1/2 out as a factor of the fraction)

    Then if we expand this above fraction we get:

    \dfrac{1 \times 2 \times 3 \ldots (n-2) \times (n -1) \times n}{1 \times 2 \times 3 \ldots (n - 3) \times (n -2)}

    And as you can see, if we cancel out terms we end up left with:

    \frac{(n - 1) \times n}{1} which can be rewritten as n(n - 1)

    So once again for emphasis:

    \frac{n!}{2!(n - 2)!} \\ = \frac{n(n - 1)}{2!} \\ = \frac{n(n - 1)}{2} and since 2! = 2...

    Set that equal to 15 and solve for n. (solution in spoiler)

    Spoiler:
    Show
    \frac{n(n - 1)}{2} = 15 \\ n(n - 1) = 30 \\ n^2 - n = 30 \\ n^2 - n - 30 = 0 \\ (n - 6)(n + 5) = 0 \\ n = -5, 6


    Edit: (I need to stop adding exclamation points for emphasis when I'm talking about factorials. So much confusion could be caused. )
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    Some members might be interested in the draft (i.e. this won't apply to you!) new course specifications for maths, mechanics and statistics. At this stage, there's not a huge amount of information - the wording suggests there are minor changes afoot in the 'pure' maths course, but more significantly that they're scrapping the 'maths for applied maths' unit and replacing it with mechanics- and statistics-specific units instead.

    Spoiler:
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    (Original post by ukdragon37)
    ...
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    (Original post by JaggySnake95)
    The thing is I'm getting everything right when it comes to the binomial theorem stuff but I don't understand by looking at the formulae what it means?

    Also, when I say ''how does 2! (2-n)! equal 2'' I am referring to when you have to work out n on your own. The question is, for example:

    (n)
    (2) = 15

    So I got:

    n! / 2!(n-2)! = 15

    which goes to

    n(n-1) / 2 = 15

    which you can then solve through to get n=6 for a positive integer. However I don't understand how the bottom lone goes to 2 in this case from 2!(n-2)!
    See Hype en Ecosse's post
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    (Original post by ukdragon37)
    See Hype en Ecosse's post
    And here I thought we were on first name basis, ukd. :cry2:
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    (Original post by Hype en Ecosse)
    And here I thought we were on first name basis, ukd. :cry2:
    :console: I wasn't sure which word to use as your first name
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    (Original post by ukdragon37)
    :console: I wasn't sure which word to use as your first name
    Hype. <3
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    Good luck guys! Advanced Maths is so much more interesting than Higher. It's also not as difficult as everyone says. There is just soooooo much in the course. I would guess about double what there is in higher (based on my notes jitters from both levels). As long as you keep on top of things and don't let yoursel trail behind you should be ok.
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    (Original post by ukdragon37)
    This only works if n = 1 or n = 2. If n = 1 then 2!(2 - 1)! = 2! * 1! = 2 * 1 = 2. If n = 2 then you have (2 - 2)! = 0! = 1 and it works similarly.



    That is not correct. The correct formula is \displaystyle\binom{n}{k - 1} + \displaystyle\binom{n}{k} = \displaystyle\binom{n + 1}{k}, i.e. it only works when the bottom number differs by 1.

    Proof

    \displaystyle\binom{n}{k - 1} + \displaystyle\binom{n}{k} \\\\

= \dfrac{n!}{\left(k - 1\right)!\left[n - \left(k - 1\right)\right]!} + \dfrac{n!}{k!\left(n - k\right)!} \\\\

= \dfrac{n!k}{k\left(k - 1\right)!\left[\left(n + 1\right) - k\right]!} + \dfrac{n!\left(n + 1 - k\right)}{k!\left(n - k\right)!\left(n + 1 - k\right)}\\\\

= \dfrac{n!k}{k!\left[\left(n + 1\right) - k\right]!} + \dfrac{n!\left(n + 1\right) - n!k}{k!\left[\left(n + 1\right) - k\right]!}\\\\

= \dfrac{n!k + \left(n + 1\right)! - n!k}{k!\left[\left(n + 1\right) - k\right]!}\\\\

= \dfrac{\left(n + 1\right)!}{k!\left[\left(n + 1\right) - k\right]!} = \displaystyle\binom{n+1}{k}

    N.B. this proof is the more general form of the answer to 2010 Paper Q5.
    In the proof bit I'm nor sure how n! equals n!k and how n! equals n!(n+1-k)

    So far whenever we have done questions involved n! we just end up with n x n-1 x n-2 etc.
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    (Original post by JaggySnake95)
    In the proof bit I'm nor sure how n! equals n!k and how n! equals n!(n+1-k)
    It doesn't. He multiplies the top and bottom of each fraction with k and (n+1-k), respectively.
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    Everyones started and is actually working, yet my school just started back and, due to an assembly, my class hasn't even started yet

    Feeling behind already


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    (Original post by TheUnbeliever)
    It doesn't. He multiplies the top and bottom of each fraction with k and (n+1-k), respectively.
    Oh. Why does he do this? I was looking at the 2 binomial theorem proofs in my notes and it does something similar. Why do proofs require this step?

    We haven't actually some these proofs yet I'm just curious.
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    (Original post by JaggySnake95)
    Oh. Why does he do this? I was looking at the 2 binomial theorem proofs in my notes and it does something similar. Why do proofs require this step?
    To enable some manipulation. For example, to allow us to turn the (k-1)! on the bottom left into k!.
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    (Original post by JaggySnake95)
    Oh. Why does he do this? I was looking at the 2 binomial theorem proofs in my notes and it does something similar. Why do proofs require this step?

    We haven't actually some these proofs yet I'm just curious.
    To combine the two fractions together you need to get them to have the same denominator. In the two fractions, one part of the denominator is k! and (k-1)! respectively. To make them the same you multiply top and bottom of the (k-1)! fraction by k since k! = (k-1)!k. The other part of the denominator is (n - k)! and (n - k + 1)!. To make them the same you multiply top and bottom of the (n - k)! fraction by (n - k + 1), since (n - k)!(n - k + 1) = (n - k + 1)!.
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    (Original post by ukdragon37)
    This only works if n = 1 or n = 2. If n = 1 then 2!(2 - 1)! = 2! * 1! = 2 * 1 = 2. If n = 2 then you have (2 - 2)! = 0! = 1 and it works similarly.



    That is not correct. The correct formula is \displaystyle\binom{n}{k - 1} + \displaystyle\binom{n}{k} = \displaystyle\binom{n + 1}{k}, i.e. it only works when the bottom number differs by 1.

    Proof

    \displaystyle\binom{n}{k - 1} + \displaystyle\binom{n}{k} \\\\

= \dfrac{n!}{\left(k - 1\right)!\left[n - \left(k - 1\right)\right]!} + \dfrac{n!}{k!\left(n - k\right)!} \\\\

= \dfrac{n!k}{k\left(k - 1\right)!\left[\left(n + 1\right) - k\right]!} + \dfrac{n!\left(n + 1 - k\right)}{k!\left(n - k\right)!\left(n + 1 - k\right)}\\\\

= \dfrac{n!k}{k!\left[\left(n + 1\right) - k\right]!} + \dfrac{n!\left(n + 1\right) - n!k}{k!\left[\left(n + 1\right) - k\right]!}\\\\

= \dfrac{n!k + \left(n + 1\right)! - n!k}{k!\left[\left(n + 1\right) - k\right]!}\\\\

= \dfrac{\left(n + 1\right)!}{k!\left[\left(n + 1\right) - k\right]!} = \displaystyle\binom{n+1}{k}

    N.B. this proof is the more general form of the answer to 2010 Paper Q5.
    So if the bottom numbers differ by one then it take 1 away from the bottom number of the lefthand side?

    And what if they don't differ by 1?
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    (Original post by JaggySnake95)
    So if the bottom numbers differ by one then it take 1 away from the bottom number of the lefthand side?

    And what if they don't differ by 1?
    No by k-1 and k it just shows that the bottom number on the left differ by 1, rather than you having to do anything with them. Say if you have 5C3 + 5C4, then n = 5 and k = 4 in this case, so you get 5C3 + 5C4 = (n+1)Ck = 6C4.

    If they don't differ by 1 then this rule doesn't apply and you need to use other rules depending on the situation.
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    (Original post by JaggySnake95)
    So if the bottom numbers differ by one then it take 1 away from the bottom number of the lefthand side?

    And what if they don't differ by 1?
    Like ukd said, you may be asked to provide a proof like that in your exam. So it's not enough to just know the rule. You really have to understand it, and it looks like you're thinking about the maths the wrong way. e.g. you're looking at the fraction and wondering why n! = n!k (it doesn't!), instead of looking at the entire fraction to see that ukd is only doing a bit of manipulation. He's multiplied the fraction by 1. \frac{k}{k}. Expressed the fraction in a different, but equivalent, way.
    Sometimes if you want to turn one expression into another, like above with nC(k-1) +nCk = (n + 1)Ck it's helpful to look at what you want it to be in the end, and work your way backwards, that makes the manipulation easier. I did this all the time with my proofs. Stops you going off in the wrong direction.

    You should never be singling out a single part of expressions like in the proof above, you have to look at the entire equation to see what's been done to it. Follow the logic line-by-line, rather than term-by-term. This is the fun part of maths, I think.
    Memorising rules was all fine and dandy at Higher, but you really have to understand everything now to get by at AH.

    I struggled with that question when I was starting off AH, too. Don't worry about it, bud. I asked some bloody obvious questions in last year's thread!
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    Is anyone else starting the course at differentiation? Don't know why my school are doing that :/


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