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Advanced Higher Maths 2012-2013 : Discussion and Help Thread

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Original post by Shengis14
Is anyone else starting the course at differentiation? Don't know why my school are doing that :/


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Perhaps your teacher thinks it leads on more naturally from Higher than the Binomial Theorem? For this reason the AH teacher at my school used to start with Vectors (unit 3).
Reply 41
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Direct15
2
Original post by Shengis14
Is anyone else starting the course at differentiation? Don't know why my school are doing that :/


This was posted from The Student Room's iPhone/iPad App


Yes one of our teachers are doing this the other was very upset as they have to teach binomial therom ( both love calculus)


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Reply 42
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hazzm2
Original post by Shengis14
Is anyone else starting the course at differentiation? Don't know why my school are doing that :/


This was posted from The Student Room's iPhone/iPad App


we are. have you done diffirentiation from first principles? it seems so pointless that a simple diffirentiation of f(x)=x^2 takes a whole page haha.
Original post by hazzm2
we are. have you done diffirentiation from first principles? it seems so pointless that a simple diffirentiation of f(x)=x^2 takes a whole page haha.



We've just done product rule, quotient rule and higher derivatives.
The actual advanced higher part is easy, but, for some reason, the simplification is murder


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Original post by hazzm2
it seems so pointless that a simple diffirentiation of f(x)=x^2 takes a whole page haha.


Pointless? :p: It's to provide a formal basis for differentiation, without which you've got little basis to justify the operation.
Original post by hazzm2
we are. have you done diffirentiation from first principles? it seems so pointless that a simple diffirentiation of f(x)=x^2 takes a whole page haha.


Thou shall not besmirch the work of Sir Isaac Newton.

(down with Leibniz.)
I'm having trouble with the proof type questions for (n r) = (n n-2) and (n r-1) = (n r)

Can anyone help? :s-smilie:
Original post by JaggySnake95
I'm having trouble with the proof type questions for (n r) = (n n-2) and (n r-1) = (n r)

Can anyone help? :s-smilie:


Is there more context you could give us? Those two statements are in general false.
Original post by ukdragon37
Is there more context you could give us? Those two statements are in general false.


Sorry I have no idea where I got those from. :lol: We were given a proof the other day and asked to make sure we understand the proofs:

(n) = (n)
(r)---(n-r)

and

(n) + (n) = (n+1)
(r-1) + (r) --(r)

What I'm having trouble is a) How do I know what to multiply one of the fractions by so I can get a common denominator (for the 2nd proof). Do I just multiply by whatever so that the bottom half of both factions are equal? and b) when I get to the end of both proofs I don't understand how they equal what you are asked to prove. For example:

In 2010 Q5 I don't understand how n! / 2! (n-2)! =

(n)
(2)
Original post by JaggySnake95
Sorry I have no idea where I got those from. :lol: We were given a proof the other day and asked to make sure we understand the proofs:

(n) = (n)
(r)---(n-r)

and

(n) + (n) = (n+1)
(r-1) + (r) --(r)

What I'm having trouble is a) How do I know what to multiply one of the fractions by so I can get a common denominator (for the 2nd proof). Do I just multiply by whatever so that the bottom half of both factions are equal? and b) when I get to the end of both proofs I don't understand how they equal what you are asked to prove. For example:

In 2010 Q5 I don't understand how n! / 2! (n-2)! =

(n)
(2)


Those two questions are linked together. First you need know the key identity for this topic:

(nk)=n!k!(n−k)!\displaystyle\binom{n}{k} = \dfrac{n!}{k!\left(n-k\right)!}

and from this you can see that nC2 = n! / 2! (n-2)! by letting k = 2.

This identity is also used in converting a binomial coefficient from a fraction to nCr form. Since you know that you want to arrive at (n+1)Cr when you are doing nC(r-1) + nCr, you know the fraction you want to arrive at must have r!(n+1-r)! as the denominator. Hence you would want to multiply top and bottom of the component fractions such that it becomes that denominator.
Original post by ukdragon37
Those two questions are linked together. First you need know the key identity for this topic:

(nk)=n!k!(n−k)!\displaystyle\binom{n}{k} = \dfrac{n!}{k!\left(n-k\right)!}

and from this you can see that nC2 = n! / 2! (n-2)! by letting k = 2.

This identity is also used in converting a binomial coefficient from a fraction to nCr form. Since you know that you want to arrive at (n+1)Cr when you are doing nC(r-1) + nCr, you know the fraction you want to arrive at must have r!(n+1-r)! as the denominator. Hence you would want to multiply top and bottom of the component fractions such that it becomes that denominator.


Okay I understand how to get the denominator but sometimes at the end of the proof question like this,

(nk)=n!k!(n−k)!\displaystyle\binom{n}{k} = \dfrac{n!}{k!\left(n-k\right)!}

if I try and simplify it, it doesn't reach what I want...
(edited 11 years ago)
Original post by JaggySnake95
Okay I understand how to get the denominator but sometimes at the end of the proof question like this,

(nk)=n!k!(n−k)!\displaystyle\binom{n}{k} = \dfrac{n!}{k!\left(n-k\right)!}

if I try and simplify it, it doesn't reach what I want...


Can you give me an example?
Original post by ukdragon37
Can you give me an example?


It's okay, I've got it now thanks. :smile:

Just one other thing: We were doing questions today like (1-0.1)^5 by Binomial Theorem which involves squaring, cubing etc. of decimals. Do we need to know how to do this because I'm really bad at knowing how many decimal places they become...
Original post by JaggySnake95
It's okay, I've got it now thanks. :smile:

Just one other thing: We were doing questions today like (1-0.1)^5 by Binomial Theorem which involves squaring, cubing etc. of decimals. Do we need to know how to do this because I'm really bad at knowing how many decimal places they become...


You are allowed a calculator in the exam :smile:
Original post by JaggySnake95
It's okay, I've got it now thanks. :smile:

Just one other thing: We were doing questions today like (1-0.1)^5 by Binomial Theorem which involves squaring, cubing etc. of decimals. Do we need to know how to do this because I'm really bad at knowing how many decimal places they become...


If you weren't allowed to use a calculator, though. You can always just express 0.1 as a fraction and THEN do all your multiplication stuffs. Makes it easier than in decimal form for me.
Reply 55
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Reece:)
Yeaayyy! finally a reason to post on here again!
But honestly, i have been to 1 period of AH maths so far... mostly because I'm lazy, i have some major catching up to do!


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When you split up 3x+1/(x+2)^2 into partial fractions how come you would have:

(x+1)^2 as one denominator and (x+1) as the other because multiply those together and you get (x+1)^3? Do the denominators have to be different to allow you to find out what A and B equal?
Ok so I'm starting work through the course now, at Differentiation. I was finding it ok until I got to the Product rule, and while I get the rule fine (I think) I am really having problems simplifying the questions I've been asked. I went for help and I still never understood it. An example is:

y= x{root} (x-6) {close root}

The answer given at the back is:

3(x-4)/2{root}(x-6) {close root}

I really don't know how to get to this :'(
Original post by JaggySnake95
When you split up 3x+1/(x+2)^2 into partial fractions how come you would have:

(x+1)^2 as one denominator and (x+1) as the other because multiply those together and you get (x+1)^3? Do the denominators have to be different to allow you to find out what A and B equal?


You don't multiply them together to get the same denominator in both. You just need to multiply top and bottom of the one with (x+1) as the denominator by (x+1) to get (x+1)^2 in the bottom.

Essentially what you are doing is saying that:

Unparseable latex formula:

\dfrac{3x+1}{\left(x+1\right)^2} = \dfrac{A\left(x+1) + B}{\left(x+1\right)^2} = \dfrac{A}{x+1} + \dfrac{B}{\left(x+1\right)^2}



From which you can see quite clearly from equating coefficients that A = 3 and A + B = 1, so B = -2.

Original post by TheFOMaster
Ok so I'm starting work through the course now, at Differentiation. I was finding it ok until I got to the Product rule, and while I get the rule fine (I think) I am really having problems simplifying the questions I've been asked. I went for help and I still never understood it. An example is:

y= x{root} (x-6) {close root}

The answer given at the back is:

3(x-4)/2{root}(x-6) {close root}

I really don't know how to get to this :'(


d(xx−6)dx=dxdxx−6+xdx−6dx=x−6+x2x−6 \dfrac{d\left(x\sqrt{x-6}\right)}{dx} = \dfrac{dx}{dx}\sqrt{x-6} + x\dfrac{d\sqrt{x-6}}{dx} = \sqrt{x-6} + \dfrac{x}{2\sqrt{x-6}}

=2x−6x−62x−6+x2x−6=2(x−6)+x2x−6=3(x−4)2x−6 = \dfrac{2\sqrt{x-6}\sqrt{x-6}}{2\sqrt{x-6}} + \dfrac{x}{2\sqrt{x-6}} = \dfrac{2\left(x-6\right)+x}{2\sqrt{x-6}} = \dfrac{3\left(x-4\right)}{2\sqrt{x-6}}
Original post by ukdragon37
You don't multiply them together to get the same denominator in both. You just need to multiply top and bottom of the one with (x+1) as the denominator by (x+1) to get (x+1)^2 in the bottom.

Essentially what you are doing is saying that:

Unparseable latex formula:

\dfrac{3x+1}{\left(x+1\right)^2} = \dfrac{A\left(x+1) + B}{\left(x+1\right)^2} = \dfrac{A}{x+1} + \dfrac{B}{\left(x+1\right)^2}



From which you can see quite clearly from equating coefficients that A = 3 and A + B = 1, so B = -2.



d(xx−6)dx=dxdxx−6+xdx−6dx=x−6+x2x−6 \dfrac{d\left(x\sqrt{x-6}\right)}{dx} = \dfrac{dx}{dx}\sqrt{x-6} + x\dfrac{d\sqrt{x-6}}{dx} = \sqrt{x-6} + \dfrac{x}{2\sqrt{x-6}}

=2x−6x−62x−6+x2x−6=2(x−6)+x2x−6=3(x−4)2x−6 = \dfrac{2\sqrt{x-6}\sqrt{x-6}}{2\sqrt{x-6}} + \dfrac{x}{2\sqrt{x-6}} = \dfrac{2\left(x-6\right)+x}{2\sqrt{x-6}} = \dfrac{3\left(x-4\right)}{2\sqrt{x-6}}


Thank you :biggrin:

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