Download MATH19730 Part 1 Section 1 Algebra and Graphs

MATH19730 Part 1
Algebra and Graphs
Section 1
In this section we will introduce (or revisit) some key mathematical ideas. It’s
important that you feel completely comfortable with this material as it will be
used later in the course.
1.1 Straight lines
When we plot points to see the relationship between two variables (e.g temperature against pressure) we often see that the data points lie close to a straight
line. Saying that a relationship is linear means that the graph which shows their
relation is a straight line and that, mathematically, is as simple as it can get.
Even if the true relationship is not linear, a linear “best fit” can be a useful
approximation.
Sometimes a change of variable (say replacing a variable by its square or
its log) can give us a linear relationship when the original one wasn’t linear.
Straight lines are much easier to deal with (easier even to draw) than curves.
A straight line relationship between two variables x and y means exactly
that the equation which relates them has the shape
y = mx + c
where m is a constant - the gradient (or slope) of the line and where c is a
constant - it’s the value of y when x = 0 i.e. where the line crosses the y-axis.
One common way to specify a line is to give two points that lie on it.
Suppose our line passes through the points with co-ordinates (a, b) and (k, l).
Let (x, y) be the co-ordinates of a general point on the line; then the equation
of the line has the form
y = mx + c.
The gradient of the line is given by
m=
b−l
.
a−k
To find c, substitute the co-ordinates of either of the points into the equation.
Another common way to specify a line is to give its slope and one point on
it.
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Example
Find the equation of the straight line through the points (1, 3) and (4, −2).
Another example
Find the equation of the straight line which has slope −3 and goes through the
point (2, 0).
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Now try these:
Examples
1. Find the straight line through the points (−1, 0) and (4, 6). What is the
gradient of the line? Find the point on the line where x = 7. Find the
point on the line where y = −3.
2. In a dry spell, water evaporates from the surface of a reservoir at a constant
rate of 5 × 105 L/day. The volume V of water in the reservoir at the
beginning of a hot spell is 4 × 104 m3 .
Draw a graph of V (in L) against time t (in days).
What is the volume remaining 20 days after the start of the dry spell?
How long will it take for all the water to evaporate?
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1.2 Powers
Let a be a positive real number. We define ap where p is some rational number
(=ratio of two integers, i.e. fraction) as follows.
Let n be a positive integer (whole number).
• an = a × a × · · · × a (n terms)
• a−n =
1
an
That dealt with both positive and negative integer powers. To deal with fractional powers, say m/n where m and n are positive integers, we make the further
definitions.
• a1/n is the n-th root of a
• am/n = (a1/n )m (this also equals (am )1/n )
(Note that there are “issues” if we let a be negative.)
Example
Calculate the following:
36 ,
41/4 ,
82/3 ,
57/2 ,
3−4 ,
3−3/4 .
Here are the rules for combining different powers (they’re consequences of
the definitions we just made and the usual arithmetic of real numbers):
• an × am = an+m
• an ÷ am = an−m
• (an )m = anm
• (ab)n = an .bn
• a1 = a, a0 = 1 if a 6= 0.
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However note that1 (a + b)n 6= an + bn ; for example2 a+b
6= a1 + 1b . Similarly,
√
√
√
for instance, a + b 6= a + b. Replacing 6= by = in either of these is a very
common mistake, especially when students are trying to “force” an answer!!
1 When
we say that such expressions are not equal we mean that, for general values of a
and b they are different; there might be a few special values of a and b which happen to make
the two expressions equal.
2 Which vlaue of n is this statement?
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Example
Simplify the following expressions.
(3am )3 a2/3
(a2 )5
2−7 .34
85 .9−3
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1.3 Exponentials
Again suppose a is a positive real number. We have seen how to define ar for
any integer or fraction r. What about ax for any
√ real number x? Not all real
numbers can be written as fractions e.g. neither 2 nor π is a rational number.
For example, π = 3.14159265... so, noting that, 3.14 is equal to 314/100, so
is a rational number, we could find an approximate value for 2π by calculating
2314/100 (which is approximately 8.82). By taking more and more decimal places
of π we get a better approximation to the actual value. To see this in action, let’s
compute successive values using the above decimal expansion. We get: 23 = 8,
231/10 = XX, 2314/100 = XX, 23141/1000 = XX, 231415/10000 = XX, . . . . It
looks as if the numbers are converging to some real number (they are), which
we define to be 2π . This works in general, so we make the following definition.
To make sense of ax when a is a positive real number and x is any real
number we approximate x by a sequence r1 , r2 , . . . of rational numbers, and
define ax to be the limit of the sequence ar1 , ar2 , . . . .
If a is a positive real number then we can plot the graph of the function
y = ax .
Try plotting the graph for various values of a. They all have the same basic
shape but some rise or fall more steeply than others. You might notice that all
the graphs pass through the point (0,1). Why?
Rather surprisingly, there is a special value of a. You might think at first
that a = 10 is a special value because that’s the basis of our decimal system but
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I mean value that is special for mathematical rather than evolutionary reasons
(assuming we use a decimal system because that’s how many fingers we have
to count on). The special value is usually written (and called) e; its value is
e = 2.71828182... (we can only give it approximately because it’s not a rational
number). And what’s so special about e? If we consider the graph of y = ex
then the gradient (= slope) at each point is equal to the (y-)value at that point.
That turns out to be a really useful property and the number e crops up all over
the place in mathematics, as you’ll see even just on this course.
We can use functions of the form y = Cex where C, k are constants3 with k
positive to model exponential growth. Similarly y = Cex can be used to model
exponential decay. This growth and decay is very rapid and outruns any power
function.
For example we calculate values of x10 .e−x for different values of x.
x
x10 e−x
5
10
15
20
25
30
Initially the power function x10 keeps the value high but, as x increases, the
exponential decay term e−x wins and the ratio rapidly approaches 0.
The notation exp(something) is sometimes used when the “something” is
better displayed in-line than as a superscript.
Examples
1. Find the values of the following to 3 d.p.
(i) 2−8
(ii) (5.1)4
(iii) (0.2)−3
(iv) 31/2
(v) e−π .
2. Plot y = x3 and y = ex on the same graph for 0 ≤ x ≤ 5. For which
values of x is ex > x3 ?
3 equivalently
y = ax where a is a constant
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1.4 Logarithms
Finding logarithms is essentially the opposite of taking powers. Let a and b be
positive real numbers. We define loga b, the logarithm to base a of b, to be the
power x that a has to be raised to to get b. We write that more mathematically:
loga b = x means that ax = b.
This can be rephrased as the following equations:
aloga b = b
loga (ax ) = x.
The rules for combining logarithms are just consequences of the rules for
combining powers.
• loga b + loga c = loga (bc)
• loga b − loga c = loga ( cb )
• loga (bc ) = c loga b
in particular loga
1
= − loga b
b
• loga a = 1
• loga 1 = 0
Just as with taking inverses and square roots, you have to play by the given
rules; and “loga (b+c) = loga b+loga c” isn’t one of them - it’s actually nonsense
(which you can see by putting in values for a, b and c).
Usually we only work with logarithms to base 10 and base e. Logs to base
e are called natural logarithms and we write
ln b = loge b.
We can switch between bases using the change of base formula:
logc b =
loga b
loga c
Examples
1. Express the following as a single logarithm.
log10 3 − 2 log10 5 + 0.5 log10 4
2. Solve for x.
log2 x2 − log2 (x + 1) = log2 (x + 2)
3. Solve the following equations, giving your answers to 3 decimal places.
(You may need to use the change of base formula to calculate the answers.)
(i)2x = 5
(ii) 53y+1 = 8
(iii) e5t−1 = 0.01
4. Suppose we know that y = Axn for some constants A and n. We measure
y for values of x as below:
x
2
3
4
5
y 20.93 57.68 118.40 206.84
Find the values of A and n.
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1.5 Binomial expansions
Here’s Pascal’s4 triangle:
1
1
1
1
1
2
3
1
3
1
................
Each number is the sum of the two numbers above it (one to the right, one to
the left) where we count blanks as 0.
The numbers in Pascal’s triangle are known as binomial coefficients. They
can be used to easily expand powers of (x + y); for instance:
(x + y)3 = (x + y)(x + y)2 = (x + y)(x2 + 2xy + y 2 ) = x3 + 3x2 y + 3xy 2 + y 3
Notice that the coefficients in the expansion are the same as the numbers
in the 3rd row of Pascal’s triangle. This is true in general. Let n be a positive
integer. Then
(x + y)n = a0 xn + a1 xn−1 y + a2 xn−2 y 2 + ... + an y n
where a0 , a1 , ..., an are the numbers from left to right in the appropriate row
(the one that starts “1 n”) of Pascal’s triangle.
Example
Expand (x + y)5 and (a + 2)6 using Pascal’s triangle.
.
4 Pascal was certainly not the first person to “discover” this, for instance in China it is
named after Yang Hui, but Pascal did develop a lot of mathematics around it.
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If n is large, using Pascal’s triangle can be tedious. So instead we use the
following formula:
(x + y)n = xn +n C1 xn−1 y +n C2 xn−2 y 2 + ... +n Cr xn−r y r + ... + y n
n!
.
r!(n − r)!
(If we define 0! = 1 then n C0 = 1 = n Cn so the first and last terms of the
expansion can also be written in the form n Ci xn−i y i .)
where n Cr =
This is known as the Binomial Theorem.
We will meet these binomial coefficients again later.
Examples
Use the Binomial Theorem to expand (x + y)9 .
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