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Leaving Cert Maths: Preparing for the Mocks

January 18, 2011 in Grinds, Leaving Cert | 1 comment

This note is primarily about LC Project Maths but also contains details on the Old Syllabus

The Leaving Cert mocks are coming up very fast but don’t fret. There is a lot of time in between February and June to learn maths – but there will be no more chances to test your skills as an examinee. This short note will focus on five areas which you must address to get the maximum benefit from your mocks.

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Leaving Cert: Integration

November 22, 2010 in General, Grinds, Leaving Cert, MA 1008 | Leave a comment

A short note covering integration for Leaving Cert maths.

 

(Please note that the proof of the Fundamental Theorem of Calculus inside isn’t quite correct. We need the Mean Value Theorem to prove it but the one in here is just for illustrative purposes.)

West Kerry Live Problem 1

November 22, 2010 in General, Leaving Cert | 2 comments

The Binomial Theorem is easier and more naturally proven in a combinatorics context but can be proven by induction.

Problem: Prove the Binomial Theorem by Induction.

Solution: Let P(n) be the proposition that for x,y\in\mathbb{R}, n\in\mathbb{N}

(x+y)^n=\sum_{i=0}^n{n\choose i}x^{n-i}y^i

(Binomial Theorem)

P(1):

\sum_{i=0}^1{1\choose i}x^{1-i}y^i={1\choose 0}x^{1-0}y^0=x+y=(x+y)^1

(P(1) is true)

Now assume P(k) is true; that is:

(x+y)^k=\sum_{i=0}^k{k\choose i}x^{k-i}y^i

Now

(x+y)^{k+1}=(x+y)^k(x+y)

=\left(\sum_{i=0}^k{k\choose i}x^{k-i}y^i\right)(x+y)

=\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k+1-i}y^i\right)}_{=S_1}+\underbrace{\left(\sum_{i=0}^k{k\choose i}x^{k-i}y^{i+1}\right)}_{=S_2}

Now all terms are of the form c(i)x^{k+1-i}y^i as i runs from 0\rightarrow k+1. Let j\in\{0,1,\dots,k+1\}. Now the x^{k+1-j}y^j term has constant from S_1 and S_2:

x^{k+1-j}y^j\left[{k\choose j}+{k\choose j-1}\right]

\Rightarrow (x+y)^{k+1}=\sum_{i=0}^{k+1}\left[{k\choose j}+{k\choose j-1}\right]x^{k+1-j}y^j

It is a straightforward exercise to show:

{n+1\choose k}={n\choose k}+{n\choose k-1}

Hence

(x+y)^{k+1}=S_1+S_2=\sum_{i=0}^{k+1}{k+1\choose i}x^{k+1-i}y^ii

(P(k+1) is true)

P(1) is true. P(k)\Rightarrow P(k+1). Hence P(n) is true for all n\in\mathbb{N}; i.e. the Binomial Theorem is true \bullet

Leaving Cert Project Maths

November 15, 2010 in Grinds, Leaving Cert | Leave a comment

For those doing Project Maths at Leaving Cert, an initial stumbling block will be on how to revise. The best way to revise mathematics is by doing exercises and a ready supply of these is traditionally found in past exam papers:

http://examinations.ie/index.php?l=en&mc=en&sc=ep&formAction=agree

However you might be under the impression that these past papers are useless with the new syllabus: not necessarily. For those doing Project Maths in 2011, the syllabus is to be found at:

http://ncca.ie/en/Curriculum_and_Assessment/Post-Primary_Education/Review_of_Mathematics/Project_Maths/LCM_str1-4_sep09_ex11.pdf

Strand 5, “Functions”, doesn’t become examinable until 2012. However this topic includes differentiation and integration, which will certainly be examinable as per the old syllabus.

Essentially the past papers may be done partially. Paper 1, Q. 6-8 is definitely fair game. Figuring out which of the other questions on past papers are still relevant involves cross-checking the new syllabus with the old.

The Quotient Rule for Differentiation

November 3, 2010 in Grinds, Leaving Cert, MA 1008, MS 2001 | 5 comments

Here we present the proof of the following theorem:

Let f,g:\mathbb{R}\rightarrow\mathbb{R} be functions that are differentiable at some a\in\mathbb{R}.  If g(a)\neq 0, then f/g is differentiable at a with

\left(\frac{f}{g}\right)'(a)=\frac{f'(a)g(a)-f(a)g'(a)}{[g(a)]^2}

Quotient Rule

Remark: In the Leibniz notation,

\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}

Proof: Let q=f/g:

q(a+h)-q(a)=\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}

=\frac{f(a+h)g(a)-f(a)g(a+h)}{g(a+h)g(a)}

=\frac{f(a+h)g(a)\overbrace{-f(a)g(a)+f(a)g(a)}^{=0}-f(a)g(a+h)}{g(a+h)g(a)}

=\frac{g(a)[f(a+h)-f(a)]-f(a)[g(a+h)-g(a)]}{g(a+h)g(a)}

\Rightarrow \frac{q(a+h)-q(a)}{h}=\frac{g(a)\left[\frac{f(a+h)-f(a)}{h}\right]-f(a)\left[\frac{g(a+h)-g(a)}{h}\right]}{g(a+h)g(a)}

Letting h\rightarrow 0 on both sides:

q'(a)=\left(\frac{f}{g}\right)'(a)=\frac{g(a)f'(a)-f(a)g'(a)}{[g(a)]^2} \bullet

Making sense of Algebra

October 26, 2010 in Grinds, Leaving Cert, MA 1008, MS 2001 | Leave a comment

Hopefully. The following note (in progress) might help you understand the power and proper functioning of basic real algebra Short_note_on_algebra

The Chain Rule

October 26, 2010 in General, Grinds, Leaving Cert, MA 1008, MATH6037, MS 2001 | 1 comment

MATH6037 please skip to the end of this entry.

The sum, product and quotient rules show us how to differentiate a great many different functions from the reals to the reals. However some functions, such as f(x)=\sin 2x are a composition of functions, and these rules don’t tell us what the derivative of \sin 2x is. There is, however, a theorem called the chain rule that tells us how to differentiate these functions. Here we present the proof. In class we won’t prove this assertion but we will make one attempt to explain why it takes the form it does. In general only practise can make you proficient in the use of the chain rule. See http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf or any other textbook (such as a LC text book) with exercises.

Proposition 4.1.8 (Chain Rule)

Let f, g:\mathbb{R}\rightarrow\mathbb{R} be functions, and let F denote the composition F=g\circ f (that is F(x)=g(f(x)) for each x\in\mathbb{R}). If a\in\mathbb{R} such that f is differentiable at a and g is differentiable at f(a), then F is differentiable at a with

F'(a)=g'(f(a))f'(a)

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Calculus of Limits: Product and Quotient Rules

October 6, 2010 in General, Grinds, Leaving Cert, MA 1008, MS 2001 | 4 comments

Here we present the proof of assertions 3. and 4. of the following proposition. The proofs of 1. and 2. will be presented in class and here they are assumed. The proofs presented here will not be presented in class.

Proposition 3.1.4 (Calculus of Limits)
Suppose that f and g are two functions \mathbb{R}\rightarrow \mathbb{R}, and that for some a\in\mathbb{R} we have

\lim_{x\rightarrow a}f(x)=p , and  \lim_{x\rightarrow a}g(x)=q.

for some p,q\in\mathbb{R}. Then

  1. \lim_{x\rightarrow a} (f(x)+g(x))=p+q.
  2. If k\in\mathbb{R}, \lim_{x\rightarrow a} kf(x)=kp.
  3. \lim_{x\rightarrow a} (f(x)g(x))=pq.
  4. If q\neq0, \lim_{x\rightarrow a} (f(x)/g(x))=p/q.
  5. If n\in\mathbb{N}, and p>0 then \lim_{x\rightarrow a} \sqrt[n]{f(x)}=\sqrt[n]{p}.
  6. Read the rest of this entry »

An inductive proof of the Factor Theorem

September 8, 2010 in Leaving Cert, MS 2001 | 2 comments

At Leaving Cert you are only required to prove the Factor Theorem for cubics. This is a more general proof.

The Factor Theorem is an important theorem in the factorisation of polynomials. When (x-k) is a factor of a polynomial p(x) then p(x)=(x-k)q(x) for some polynomial q(x) and clearly k is a root. In fact the converse is also true. Most proofs rely on the division algorithm and the remainder theorem; here a proof using strong induction on the degree of the polynomial is used. See Hungerford, T.W., (1997), Abstract Algebra: An Introduction. Brooks-Cole: U.S.A for the standard proof (this reference also describes strong induction).

Factor Theorem
k\in\mathbb{C} is a root of a polynomial p(x) if and only if (x-k) is a factor: p(x)=(x-k)q(x) where q(x) is a polynomial.

Proof:
Suppose deg p(x)=1. Then p(x)=ax+b and p(k)=ak+b=0\Rightarrow k=-b/a. Clearly p(x)=(x-(-b/a))a and q(x)=a is a polynomial (of degree 0).

Suppose for all polynomials of degree less than or equal to n-1, that k a root implies (x-k) a factor.  Let p(x)=\sum_ia_ix^i be a polynomial of degree n and assume k\in\mathbb{C} a root. Now

p(x)-p(k)=\sum_{i=0}^na_ix^i-\sum_{i=0}^na_ik^i

\underset{p(k)=0}{\Rightarrow} p(x)=\sum_{i=1}^n a_i(x^i-k^i)

Now each x^i-k^i for i=1,\dots,n-1 is a polynomial of degree less than or equal to n-1, with a root x=k, hence (x-k) is a factor of each. Thence x^i-k^i=(x-k)q_i(x) where q_i(x) is a polynomial:

p(x)=\sum_{i=1}^{n-1}a_i(x-k)q_i(x)+a_n(x^n-k^n)

Now x^n-k^n=(x-k)x^{n-1}+kx^{n-1}-k^n but kx^{n-1}-k^n is a polynomial of degree n-1 with root k and hence (x-k) is a factor:

x^n-k^n=(x-k)x^{n-1}+(x-k)h(x)

Hence

p(x)=(x-k)\lbrack x^{n-1}+h(x)+\sum_{i=1}^{n-1}a_iq_i(x)\rbrack

 

Hence (x-k) a factor \bullet