Help with C1!

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    (Original post by ihatePE)


    What does this notation mean?

    i'm on binomial expansion and came across this, i understand the substitution part, the factorial equation substitution but what does that notation mean? 'n' on top of 'r' . does it mean n take away r? cos thats the pattern i see in the equation
    It means \displaystyle \binom{n}{r} = \frac{n!}{(n-r)! r!}, so for example \displaystyle \binom{4}{2} = \frac{4!}{2!2!} = \frac{24}{4} = 6.
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    (Original post by 34908seikj)
    It means n "choose" r if that makes sense.
    Whatever is within that matrix substitute accordingly into it.
    It's read as n choose r, it doesn't mean that's what it is.

    It's not a matrix.
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    (Original post by ihatePE)
    i still dont get it :rofl:
    I edited my post to clarify.
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    (Original post by ihatePE)
    So to reduce making threads like every night, im going to concentrate every C1 related issues i have on this thread

    Can anyone check my answer for this inequalities question?




    find the set value for
    a) 1 - 2x < 4
    b) x^2 - 6x + 8 ≥ 0


    for a) my answer is x > 3/2
    for b) my answer is 2 ≥ x ≥ 4
    You look so cute in your new profile picture ; you have great facial structure and your glasses are very trendy . Your old pic was so bad but this one is a massive improvement. You look like such a nice person
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    (Original post by Zacken)
    It means \displaystyle \binom{n}{r} = \frac{n!}{(n-r)! r!}, so for example \displaystyle \binom{4}{2} = \frac{4!}{2!2!} = \frac{24}{4} = 6.
    so in that case (n r) = 6?

    the (n r) is short for that equation with the factorial, in simple terms?
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    (Original post by fefssdf)
    You look so cute in your new profile picture ; you have great facial structure and your glasses are very trendy . Your old pic was so bad but this one is a massive improvement. You look like such a nice person
    im trying to find the right expression to share so that it gives an accurate impression of me. but I'm a bit more sarcastic online so this smiley face doesnt sometimes match with what im typing thanks anyways :hoppy:
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    (Original post by ihatePE)
    so in that case (n r) = 6?
    No. (4 2) = 6. (n r) = n!/(r!(n-r)!)

    the (n r) is short for that equation with the factorial, in simple terms?
    Yes.
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    (Original post by Zacken)
    No. (4 2) = 6. (n r) = n!/(r!(n-r)!)



    Yes.
    thankssssss another problem off my list :hoppy:
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    (Original post by ihatePE)
    so in that case (n r) = 6?

    the (n r) is short for that equation with the factorial, in simple terms?
    Yes, that's the binomial function, (n c) is a short way of writing it just like if I said f(x) rather than 2x+1.
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    (Original post by ihatePE)
    i still dont get it :rofl:
    You know on your calculator, there is a button called nCr and you use that when dealing with this.
    It ha to do with factorials and the top number is 'n' and the bottom number is 'r'.
    It's used to expand brackets in C2 (basically something like (2x+5)^10


    Just an example if you want to expand (x+1)^2 you do
    0C2(1^2)(x^0) + 1C2(1^1)(x^1) + 2C2(1^0)(x^2) =
    = 1(1) + 2(1)(x) + 1(x^2) =
    = 1+ 2x + x^2

    Obviously in C2 you won't get asked this easy question and they will want the first 3/4 terms only.

    Also may I suggest that you look at examsolutions videos on C1/C2, they are really helpful
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    (Original post by ihatePE)
    im trying to find the right expression to share so that it gives an accurate impression of me. but I'm a bit more sarcastic online so this smiley face doesnt sometimes match with what im typing thanks anyways :hoppy:
    Awww haha ; I like your emojis as well with the dancing bear but could you tell us what type of animal it actually is ?
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    (Original post by ihatePE)
    thankssssss another problem off my list :hoppy:
    I think it would be useful for you to also note that this binomial function also follows Pascal's Triangle:


    As you may notice, you begin with the 3 1's at the tip, and to get a number the row below, you simply add the two above that number; eg: 2=1+1, 4=1+3, 10=4+6 as these are above them. The top row is the 0th row, and the left-most number is the 0th one.

    Here's the example:


    When you say something like \displaystyle \binom{4}{2} You are looking at the 4th row's 2nd number from the left. As you can see, the binomial function has symmetry which is useful.

    I think this can prove useful to some people who may wish to memorise the first few rows of the binomial function.

    Edit: This also shows you that within \displaystyle \binom{n}{r}, r takes any positive integer values from 0 up to n and nothing above or below that restriction. This can be notated by 0\leq r \leq n for r,n\in\mathbb{N}.
    You can refer to the actual function to see what if r&gt;n, then you would be having a factorial of a negative number on the denominator which can't be done as they are undefined due to division by 0; which you should notice if you know the pattern amongst the factorials.
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    (Original post by RDKGames)
    I think it would be useful for you to also note that this binomial function also follows Pascal's Triangle:


    As you may notice, you begin with the 3 1's at the tip, and to get a number the row below, you simply add the two above that number; eg: 2=1+1, 4=1+3, 10=4+6 as these are above them. The top row is the 0th row, and the left-most number is the 0th one.

    Here's the example:


    When you say something like \displaystyle \binom{4}{2} You are looking at the 4th row's 2nd number from the left. As you can see, the binomial function has symmetry which is useful.

    I think this can prove useful to some people who may wish to memorise the first few rows of the binomial function.

    Edit: This also shows you that within \displaystyle \binom{n}{r}, r takes any positive integer values from 0 up to n and nothing above or below that restriction. This can be notated by 0\leq r \leq n for r,n\in\mathbb{N}.
    You can refer to the actual function to see what if r&gt;n, then you would be having a factorial of a negative number on the denominator which can't be done as they are undefined due to division by 0; which you should notice if you know the pattern amongst the factorials.
    thank you, you've helped me answer the question ''why though'' :rofl:
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    (Original post by ihatePE)
    thank you, you've helped me answer the question ''why though'' :rofl:
    If you want "why" then you can interpret \binom{n}{r} as the number of ways of choosing r elements (disregarding order) from a set of n elements.
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    I'm currently doing differentiation from first principles and the formula f(x+h) - f(x) / h is confusing me

    say we have a curve 5x^2 - x - 1

    why would the f(x+h) become 5(x+h)^2 - (x+h) ?

    i dont get the substitution in this part. i get that f(x) = 5x^2 - x - 1
    but how come having a ''h'' in f(x+h) make it 5(x+h)^2 - (x+h)?

    from my common sense it would be 5x^2 - x - 1 + h but obviously my common sense is wrong
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    (Original post by ihatePE)
    I'm currently doing differentiation from first principles and the formula f(x+h) - f(x) / h is confusing me

    say we have a curve 5x^2 - x - 1

    why would the f(x+h) become 5(x+h)^2 - (x+h) ?

    i dont get the substitution in this part. i get that f(x) = 5x^2 - x - 1
    but how come having a ''h'' in f(x+h) make it 5(x+h)^2 - (x+h)?

    from my common sense it would be 5x^2 - x - 1 + h but obviously my common sense is wrong
    If  f(x)=5x^2-x-1 then for  f(x+h) you replace all x's with (x+h). So  f(x+h)= 5(x+h)^2-(x+h)-1 .
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    (Original post by B_9710)
    If  f(x)=5x^2-x-1 then for  f(x+h) you replace all x's with (x+h). So  f(x+h)= 5(x+h)^2-(x+h)-1 .
    Ok thats a bit clearer now in my head but i dont think they had the -1 at the end in the answer video im watching...or is my mind tricking me at the moment ?
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    This motivated me to do some maths. Tight now I'm on complex numbers


    Posted from TSR Mobile
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    If you consider substituting "x" with say, 5 for f(5), then the equation would be 5(5)2 - 5 - 1. If you substitute "x" with "x + h" as the new function is f(x + h), then it becomes 5(x + h)2 - (x + h) - 1. The expression in the brackets substitutes into the variable in your original function.
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    (Original post by ihatePE)
    I'm currently doing differentiation from first principles and the formula f(x+h) - f(x) / h is confusing me

    say we have a curve 5x^2 - x - 1

    why would the f(x+h) become 5(x+h)^2 - (x+h) ?

    i dont get the substitution in this part. i get that f(x) = 5x^2 - x - 1
    but how come having a ''h'' in f(x+h) make it 5(x+h)^2 - (x+h)?

    from my common sense it would be 5x^2 - x - 1 + h but obviously my common sense is wrong
    Isn't differentiation from first principles from FP1? Or did I miss somewhere that you'd doing FM? I don't think it's in C1.
 
 
 
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