502-3 confidentiality, hipaa & case records - procedure

Math150Mathematics for Natural Sciences

Z. Makukula P. Ittmann

UKZN, Pietermaritzburg

Semester 1, 2012

Makukula, Ittmann (UKZN PMB) Math150 2012 1 / 22

Tutorial Questions

The questions for the �rst tutorial are:

Questions 1 to 31 on Pages 20 - 23 of the Textbook

The questions for the second tutorial (held next week) are:

Questions 32 to 56 on Pages 23 - 25, and

Questions 1 to 8 on Pages 56 - 57 of the Textbook


Binomial Theorem

The binomial theorem is used to expand expressions of the form

(a + b)n

Theorem

Binomial Theorem If n ∈ N, then

(a+b)n = an+

(n

1

)an−1b1+

(n

2

)an−2b2+· · ·+

(n

r

)an−rbr+· · ·+bn (1)


Binomial Theorem (cont.)

To understand why this works let us look at the term containing

an−2b2

This comes from multiplying out

(a + b)(a + b) · · · (a + b)︸︷︷︸n times

How do we get a term in an−2b2?

Well we must choose 2 b from the n brackets, which can be done(n

2

)ways

This is why the coe�cient of an−2b2 in the Binomial Theorem is

exactly(n

2

)The same sort of consideration applies to the other terms



The binomial theorem can be written in Sigma notation as:

(a + b)n =n∑

r=0

(n

r

)an−rbr



Example

Expand (x − 2y)4 in powers of x and y

Now,

(x − 2y)4 = x4 +

(4

1

)x3(−2y)1 +

(4

2

)x2(−2y)2 +

(4

3

)x1(−2y)3 + (−2y)4

= x4 − 8x3y1 + 24x2y2 − 32x1y3 + 16y4

Note: The sum of the powers in each term is the same



Example

In the expansion of(

x

2− y)11

�nd the term involving x6y5

The term involving x6y5 is(11

5

)(x2

)6(−y)5 =

(11

5

)x6

26(−1)5y5 = −231

32x6y5.


Pascal's Triangle

Pascal's Triangle

For small values of n, the binomial coe�cients(n

r

)in the binomial

theorem are most readily obtained from Pascal's Triangle

We show the �rst 6 rows of the triangle below:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1


Pascal's Triangle

Pascal's Triangle (cont.)

This is constructed as follows:

Write down 1 in the �rst row and then two 1s in the second row as

shown

Each new row is started o� and ended with 1s

The intervening numbers are obtained by adding the two numbers

diagonally above as shown by the arrows

In the third row we �nd the numbers 1 2 1

Note that

(a + b)2 = 1.a2 + 2.ab + 1.b2


Pascal's Triangle


In the fourth row we have 1 3 3 1 and

(a + b)3 = 1.a3 + 3.a2b + 3.ab2 + 1.b3

In the same way, using the next two rows, we have

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

and

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5


Pascal's Triangle


Please note that

(a + b)n 6= an + bn (unless n = 1)

Thus,

(a + b)7 6= a7 + b7 and (a + b)1/2 6= a1/2 + b1/2


Pascal's Triangle

Pascal's Triangle (cont.)Example

Write out the binomial expansion of (1+ x)5, and hence evaluate

1.025 to two decimal places

Using the 5th row of Pascal's triangle

(1+ x)5 = 1+ 5x + 10x2 + 10x3 + 5x4 + x5

Thus setting x = 0.02,

1.025 = (1+ 0.02)5 = 1+ 5(0.02) + 10(0.02)2 + 10(0.02)3 + 5(0.02)4 + (0.02)5

≈ 1+ 5(0.02)

= 1.10

We need only use �rst two terms, since the others are all too small to

a�ect the �rst 2 decimal placesMakukula, Ittmann (UKZN PMB) Math150 2012 12 / 22

Pascal's Triangle


The last example illustrates a very important point

Consider the expansion of (x + h)n

We have

(x + h)n = xn +

(n

1

)xn−1h +

(n

2

)xn−2h2 + · · ·+

(n

n

)hn

If h is small, then h2 and all higher powers are very small

So,

(x + h)n ≈ xn +

(n

1

)xn−1h = 1+ nxn−1h.


Pascal's Triangle


This will be a very important point in Chapter 3

In particular,

(x + h)n ≈ xn + nxn−1h for small h (2)


Pascal's Triangle

Chapter 1 Summary

We have covered the following in Chapter 1:

Number systems

Equations, Functions and

Graphs

Proportionality

Units

Powers, Exponents and

Scienti�c Notation

Sigma Notation

Binomial Expansions

Pascal's Triangle


Pascal's Triangle

Functions

If two variables x and y are related in such a way that when any value

is given to x there is one and only one corresponding value of y , then

y is a function of x

We write y = f (x)

Example

The height of a tree is a function of its age

The area of a circle is a function of its radius

At constant pressure, the volume of a �xed mass of gas is a function

of its temperature

The cosines of angles are functions of the angles


Pascal's Triangle

Functions (cont.)

One can regard a function f as a machine, with input x and output

f (x)

The function is f , whereas f (x) is the value the f produces when

given input x

Example

If f (x) = 2x , then f is the rule �multiply by 2�

If g(x) = x2, then g is the rule �square the input�


Pascal's Triangle

Functions (cont.)

We take a number x in a set A

The function f then produces the corresponding number in the set B

If y = f (x), then the set A of all allowable x values is the domain

The set B of all resulting y -values is the range

We write f : A→ B


Pascal's Triangle

Functions (cont.)

Example

Let f (x) = x2

The domain is all of R

The range is the set of non-negative real numbers, since y = x2 ≥ 0


Pascal's Triangle

Functions (cont.)

Example

Let g(x) = 1/x

The domain is everything except zero (since you can't divide by zero)

The range is also everything except zero (since y = 1/x cannot be

zero for any x)


Pascal's Triangle

Functions (cont.)

Example

Let f (x) = x2 − 5x + 7

Find (a) f (3) (b) f (f (2)) (c) All points x such that f (x) = 2

(a)

f (3) = 32 − 5.3+ 7 = 1

(b)

f (f (2)) = f (22 − 5.2+ 7)

= f (1) = 3

(c) We want all x such that

x2 − 5x + 7 = 2, that is

x2 − 5x + 5 = 0. Using the

quadratic formula

x =5±√52 − 4.5

2=

5±√5

2


Pascal's Triangle

Functions (cont.)

Let f and g be two functions where the domain of g is the range of f

Consider a point x in the domain of f

Calculating f (x) we get a point f (x) which is allowed by g

Use this as the input for g to get g(f (x))

This gives a new function, called the composition of g and f , written

g ◦ f(g ◦ f )(x) = g(f (x))


502-3 confidentiality, hipaa & case records - procedure

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