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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

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)

Read the rest of this entry »

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 »

How to define ‘Greater Than’

September 20, 2010 in General, MS 2001 | Leave a comment

How can a statement like “5 is greater than 4” be quantified? Or is it just obvious? If we know anything about mathematics we know that there is no way we can assume something as obvious, there must be an axiomatical contruct that puts a rigorous meaning on “5 is greater than 4“.

The first attempt would be to say that 5=4+1 so 5 is “1 more” than 4 so must be bigger. This translates to 5-4=1: “5 is greater than 4 because 5-4 is positive“. Careful! 4=5+(-1) so 4-5=-1: “4-5 is negative“. But what does positive and negative mean? Easy? Positive is greater than zero… At this point a stronger construct is needed:

Definition: Call a set P\subset \mathbb{R} positive if for all a,b\in P

  1. a+b\in P
  2. ab\in P
  3. Given x\in \mathbb{R} either x\in P, -x\in P or x=0

If we think carefully, this definition concurs exactly with that of the naive notion of positive. So we can say that “5 is greater than 4 because 5-4 is positive.”

Definition: Given a,b\in\mathbb{R},  a is said to be greater than b, a>b, if a-b is positive.