log_ab^c = c log_ab

For example,

log(9) = log(3²) = 2 log(3)

This can be used to either take a power outside of a log so it becomes a multiplication that is easier to work with or take a multiplication inside a log so you are only working with logs.

Proof when c is an integer

When c in the above equation is a positive integer it is simple. Since b^c means b multiplied by itself c times we can use the rules of log addition to get
log(b^c) = log(b….b) = log(b) + log(b) + … + log(b) = clog(b)

Similarly when c is negative we divide 1 by b c times so with using the rules of log subtraction we get
log(b^-c = log(1/b …. 1/b) = -log(b) – … -log(b) = -clog(b)

Proof when c is a real number

Noticing that b = a^log_ab
We can write
b^c = (a^log_ab)^c=a^clog_ab
So taking log a of both sides gives us
log(b^c) = c log_ab as required

Notice that this uses the fact that a log and a power cancel.
Ie) log_aa^b = b
and a^log_ab = b

Also see

• Binomial Theorem (Expansion) for positive integer powers (0)
• Differentiate Logs with Proof (0)
• Why anything to the power 0 is eqaul to 1 (2)
• Log Laws: Adding and Subtracting Logs (0)
• Matrix Multiplication (1)
• Matrix Calculator, C++ Program by David Woodford (5)

When multiply out this bracket we get all the combinations of powers for a and b such that the sum of their powers is n. So there is a term of aⁿwhere the power of b is 0 and one of bⁿ and then there are terms in between with both a and b raised to powers. Then all that is left is to find the co-efficents of each of these terms. For the case that n is a positive integer there are two ways of doing this, one is to use the Binomial Theorem or the other is to use Pascal’s triangle.

Binomial Theorem

The binomial Theorem is:

Example

If we know calculate the factorials this becomes

Which you can check since we know 3+1 = 4 and 4⁴ = 256.

However there is another simpler way of finding the coefficients for the power which is to use a Pascal’s triangle. This is an quick to calculate way for moderately large powers (the triangle can get quite large for big powers above say 10) by hand. A Pascal’s triangle with 5 rows is below.

Each item in the triangle is the sum of the two numbers above it and there is a 1 at the end of each row. When the binomial terms are written in order of powers of x (either way as it is symmetric) each number in the nth row is the coefficient of x in corresponding term (check it with the example above)

More complex sums

We can also use the binomial theorem on sums where the first term is something other than 1. In general when we want to find (a+b)ⁿ.
To do this we first factor out a within the sum to get

We then treat as x and use the binomial theorem to expand to get

we can then multiply through the sum by aⁿ to get

Example

Expand (3 + 2x)⁴.

Firstly factorise the 3 to get

Using Pascal’s triangle or our previous calculations we can find the coefficients and write out the sum

and now working out the arithmetic of each term we get our final answer

The binomial theorem can also be used to calculated sums raised to non integer or negative powers but that is for another post.

Also see

• Log Laws: Taking logs of powers (0)
• Why anything to the power 0 is eqaul to 1 (2)

1 + 2 + 4 + 8 + 16 + …       each term is twice the previous one
4 + 12 + 36 + 108 + …       each term is three times the previous one

In general the nth term of a geometric series is written as where a is the first term and r is the factor we multiply each term by. If a is 1 then the terms are just the powers of r.

We are often interested at find the value of a geometric sum, that is the sum of all the values of a geometric series to a certain point. We can also find the value of the sum of infinitely many terms if . To find both of these first let the sum of the first n terms be denoted S_n.

So

The formula for a finite sum that we will try to prove is

Proof of finite sum

Consider
from the definition of S_n


We can now factorise the left-hand side and tidy up the right multiply the r through the first sum and change the index down on the second,





We can now break up the sums into parts in order that we cancel bits of them in the next step





So cancelling out the two sums and factorising the a noting that anything the the power of 0 is 1 gives





and then dividing through by (r-1) gives us our required result


.

Infinite Sums

As noted earlier is we can also find the value of an infinite sum. It may seem like a strange idea to be able to find the value of the sum of an infinite number of things but if the size of each term keeps getting smaller so the amount the sum increases by each time gets smaller and smaller.

A simple way of visualising how this can work is to imagine a piece of rope two meters long. We are going to cut this rope up into an infinite number of pieces with the length of each piece representing a term in the sum. Firstly cut the rope in two. Put one piece to the side as this is our first term so a=1 as this piece is 1 meter long. r=1/2 because each piece of rope will be half the length of the previous one. Now cut the remaining piece of the rope in half and again put one piece to the side as the second term and cut the remainder in half. Theoretically we will always be able to repeat this process because you can always halve something. This will mean that we will now have an infinite number of pieces of rope with lengths 1m, 1/2m, 1/4m, 1/8m, … so the lengths of the pieces of rope form a geometric progression and we know that the sum of their lengths must be less than two as we only started out with two meters of rope. In fact it is exactly two which we can prove.

In general,


if


To show this simply note that rⁿ tends to 0 as n tends to infinity. Then using the algebra of limits and multiplying the top and bottom y -1 we find the above expression to be true.

Also see

• Arithmetic Series and Progression (2)
• Proof there are infinitley many prime numbers (0)

Example 1: simplifying single fractions

Consider

We want to find things that divide both the numerator (top) and denominator (bottom). In this case 2x does. We can then divide both the top and bottom by 2x (known as cancelling out the 2x) to gain the fraction

.

Note that you can only cancel things that divide the whole of the top and bottom not part of the top and bottom or things that are added to the top and bottom. For example

Sometimes the common factor that we are cancelling isn’t immediately obvious, for example if we have quadratic expressions. In these cases you should try to factorise the quadratic to see if there is anything you should cancel.

Note on quadratics: If a fraction has quadratics that can be factorised you should always leave the denominator factorised whilst it is common to multiply out the numerator (unless there is a need to leave it factorised or it is very complex)

Example 2: adding fractions

Consider the algebraic expression

Before we consider adding them we should make each of the fractions in the sum into its simplest form.

Firstly we want to have one denominator. To do this we need to find the simplest common denominator and then make the denominator of both these fractions this. This is done in a similar way in which we find the lowest common denominator when we are adding normal fractions. The easiest way to do this is to multiply numerator and the denominator of both factions by the denominator of the other so the expression becomes:

Now, as when adding normal fractions, because the denominator of both fractions is the same we can add their numerators together and put them over a single denominator to get

Example 3: Solving Equations

To solve an equation involving an algebraic fraction you should first multiply out the fractions as explained above.

Consider the equation

which is already simplified. To solve this we need to cross multiply, this involves multiplying both sides of the equation by the both the denominators and then simplifying both sides. However, because after the multiplication the denominator of each fraction will be multiplied by the numerator they will cancel out with the result that we have multiplied each side of the equation by the denominator of the other side and have removed the denominators from the fractions. This is most easily show by example

ie)

becomes

with intermediate step

We can then solve the equation using normal methods (note that if the result is a quadratic you may have to use the quadratic formula)

Also see

• First Order Differential Equations (6)
• Finding the Inverse of a Function (0)
• Find equation of tangent to a curve (2)
• Inequalities (0)
• Why the proof 2=1 is wrong (0)
• Simultaneous Equations (0)

These can be defined as follows,
if for all x₁ < x₂

• f(x₁) ≤ f(x₂) then f is increasing
• f(x₁ ) < f(x₁ ) then f is strictly increasing
• f(x₁ ) ≥ f(x₂) the f is decreasing
• f(x₁) < f(x₂) then f is strictly decreasing

Monotonicity and Derivatives

If a function f(x) is increasing then what we mean is that the slope is always positive, so if f is continuous we can relate the the properties of increasing and decreasing to the derivative as shown in the table below.

Its important to note that these rules only work if the function is continuous, for example consider f(x) =1/x, which is discontinuous at 0.
We can differentiate it to get f’(x) = -1/x² which we know is always negative (because the squared term is always positive) so we would expect it to be a decreasing function. However if we consider two point either side o, 1 and -1 say  we find that f is not a decreasing function because whilst -1 < 1, -1/x < 1/x contrary to our definition of decreasing

Also see

• Stationary Points (Maximum and Minimums) and Differentiation (1)
• Integrating Fractions – using the natrual logarithm – Example tan(x) (0)
• Hyperbolic Functions (0)

So for example f(x) = x+2 (from reals to reals) has an inverse as every element y is mapped to only by y-2 but f(x) = x² (from reals to reals) doesnt because both -1 and 1 map to 1 so how would we decide which of these would be the unique result of the inverse applied to 1 (we could of course define the domain to be the positive reals in which case it would have an inverse).

When finding the inverse of a function you are really looking to see what maps to each element of the range or codomain, to find the inverse of f you are looking for the element x in the domain for a given y in the range such that f(x) = y. This basically means you are reversing the process of the function.

Finding the Inverse when the function is a formula

When a function is given by a formula what you need to try and do is apply the operations of that formula backwards. This easiest way of doing this is to let the function f(x) = y. Now you know the formula to get from x to y so substitue this in for f(x). All that now needs to be done is to rearrange this equation so that x is the subject and the resulting rexpression on the otherside is only in terms of y. This function is the inverse of f, to show this lets denote it g so we g(y) = x.

Then g(f(x)) = g(y) = x since we started by letting f(x) = y and created g such that g(y) =x.
and f(g(y)) = f(x) = y

so g is indeed the inverse of f.

Example

This is most easily demostrated through an example.
Let
Then to find the inverse of f we first write

and rearrange as follows

So the inverse of f is given by the formula so we can write

Note that we have replaced the y’s with x’s, this doesnt matter as we can put any variable we like into the function but it important to make sure that you use the same variable as the parameter of the function and in the formula that defines it.

Also see

• Introduction to functions and maps (0)
• Simplifying and Solving equations with Algebraic Fractions (0)
• CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi (0)
• Exponential Functions (1)
• Why the proof 2=1 is wrong (0)
• Proof by Mathematical Induction (4)

1. A domain
2. A co-domain or range
3. A “rule” for assigning each element of the domain to a unique element of the co-domain.

Important Points

There are some important things about functions, in particular part 3, which you need to remember.

Firstly the function must be able to be applied to every element of the domain so f(x) = 1/x isn’t a function from the real numbers to the real numbers since 0 cant be assigned a value, as 1/0 isn’t a real number. There are two ways round this problem, we can define the co-domain not to include 0 (ie ) or can give 1/0 a different rule, eg f(x) = 1/x for all and f(0) = 7.

However not every element in the co-domain needs to be assigned to something so is a valid function even though nothing goes to -1

Every element of the domain is assigned a unique element of the c0-domain. This uniqueness is often a source of confusion – here what we mean is that the function only assigns a single element of the co-domain to each element of the domain, but more than one element of the domain can be assigned the same element of the co-domain.

Simple examples

Here are some simple examples of functions from the real numbers to the real numbers.

– This simply applys this operation to any number and give you another number.

for $ x \not 0 $ and – This takes the reciprocal of a number unless that number is zero where the reciprocal is undefined so gives 7.

-Note that  f(a) = f(-a) and nothing gives f(x) = -1 however this is still a function as f(x) is defined for all x and f(x) gives only one answer.

Composing Functions

You can compose two or more functions to form a new function. Consider two functions f and g both from the reals to the reals, f composed with g, written, fg or is given by

so if f = x+1 and g=2x fg(x) = f(2x) = 2x+1

Note that in order to do this the domain of g has to be the same as the co-domain of f.

Inverses and the Identity map

The identity map or function is the function which does nothing. If I is the identity on the reals the I(x) = x.

The inverse of a function , written , is the function which when composed with f gives the identity. There are two types of inverse, a right sided inverse and a left sided inverse, depending upon which side you compose them with f.

g is the left sided inverse of f is

gf(x) = I(x) = x

and h is the right sided inverse of f if

fh(x) = I(x) = x

If there is a function h such that

hf = fh = I

then h is said to be a two sided inverse of f and is often simply refereed to as the inverse of f.

Note that not all functions have inverses, for example x² has no inverse, but when a function does have one it is unique.

Also see

• Finding the Inverse of a Function (0)
• CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi (0)
• Exponential Functions (1)
• Differentiate Inverse Sine (0)

Finding the Stationary Points

We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.

For example lets consider the graph . We cab differentiate this to find

We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.

So in our example we form the equation

by equating our expression for , , to 0
Solving this equation we find that stationary points occur exactly when

Note that there can be more than solution to this equation, each of which is a valid stationary point.

Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example
So the only stationary point is at

Nature of Stationary Points

The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, (found by differentiating the derivative). There are three types of stationary point:

1. Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
   Here
2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

   Here

3. Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.

   At a point of inflection but isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point

Checking the nature of a Stationary Point when
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection

Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible

Also see

• Points of Inflection (0)
• Product Rule for Differentiation (0)
• Monotonic (Increasing and Decreasing) Functions (0)
• Fundamental Theorem of Calculus (0)
• Integrating Fractions – using the natrual logarithm – Example tan(x) (0)
• Integration by Parts (0)

Example y=x²
Find the equation of the tangent to the curve when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.

so at the point x=t the gradient is 2t.

From this we can get a general equation for the tangent using the equation for the gradient of a straight line
grad =
to get the general equation for the tangent at the point x=t by substituting x₁=t, y₁=t² and m=2t

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get

which is our final answer.

Also see

• Simplifying and Solving equations with Algebraic Fractions (0)
• Tan Graph – y=tan(x) (0)
• First Order Differential Equations (6)
• Auxiliary Angle Method for Solving Trigonometry Equations (1)
• Sine and Cos Graphs Differentiating sin and cos (7)
• Quadratic Formula (9)

Please note: I am not responsible for the content of external websites and cannot guarantee that these features will be working, available or accurate.

Shodor Graph Sketcherr

This enables you to plot most functions in Cartesian form. It requires you to enter the graph as a function of x and then clicking “graph” will plot it for you. Using the tools in the top right hand corner you can zoom in/out and more around the graph, you can also adjust the axis.

A screen shot of the graph y=x^2 using the Shodor Graph Sketcher

Polar Co-ordinates Graph Sketcher

This applet will allow you to sketch graphs using polar co-ordinates. You can enter a range and your graph as a function of t (the angle). It also allows you to zoom in and out of the graph and includes a list of the functions it supports and how to enter them.

Screenshot of the graph r=sin(t) as drawn by the polar co-ordinates graph sketcher

Also see

• Equations using Polar Co-ordinates (0)
• Exponential Functions (1)
• Using Polar Co-ordinates and Converting to and from Cartesian (0)
• Tan Graph – y=tan(x) (0)