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
Makukula, Ittmann (UKZN PMB) Math150 2012 2 / 22
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)
Makukula, Ittmann (UKZN PMB) Math150 2012 3 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 4 / 22
Binomial Theorem (cont.)
The binomial theorem can be written in Sigma notation as:
(a + b)n =n∑
r=0
(n
r
)an−rbr
Makukula, Ittmann (UKZN PMB) Math150 2012 5 / 22
Binomial Theorem (cont.)
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
Makukula, Ittmann (UKZN PMB) Math150 2012 6 / 22
Binomial Theorem (cont.)
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.
Makukula, Ittmann (UKZN PMB) Math150 2012 7 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 8 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 9 / 22
Pascal's Triangle
Pascal's Triangle (cont.)
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
Makukula, Ittmann (UKZN PMB) Math150 2012 10 / 22
Pascal's Triangle
Pascal's Triangle (cont.)
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
Makukula, Ittmann (UKZN PMB) Math150 2012 11 / 22
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
Pascal's Triangle (cont.)
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.
Makukula, Ittmann (UKZN PMB) Math150 2012 13 / 22
Pascal's Triangle
Pascal's Triangle (cont.)
This will be a very important point in Chapter 3
In particular,
(x + h)n ≈ xn + nxn−1h for small h (2)
Makukula, Ittmann (UKZN PMB) Math150 2012 14 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 15 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 16 / 22
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�
Makukula, Ittmann (UKZN PMB) Math150 2012 17 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 18 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 19 / 22
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)
Makukula, Ittmann (UKZN PMB) Math150 2012 20 / 22
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
Makukula, Ittmann (UKZN PMB) Math150 2012 21 / 22
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))
Makukula, Ittmann (UKZN PMB) Math150 2012 22 / 22