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Stationary Points (Maximum and Minimums) and Differentiation

December 5, 2009 Leave a comment

On a graph a stationary point is any point where the gradient is 0 so where the graph is flat. For example the graph y=x2 has one stationary point at the origin.

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 y = 3x^2 + 2x - 7. We cab differentiate this to find
\frac{dy}{dx} = 6x + 2

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
6x + 2 = 0
by equating our expression for \frac{dy}{dx}, 6x + 2, to 0
Solving this equation we find that stationary points occur exactly when
x = \frac{2}{6} = \frac{1}{3}
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 y= 3 \cdot \frac{1}{3}^2 + 2 \cdot \frac{1}{3} - 7 = -6
So the only stationary point is at (\frac{1}{3},-6)

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, \frac{d^{2}y}{ dx^2} (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 {d^{2}y}{dx^2} < 0
  2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)

    Here {d^{2}y}{dx^2} > 0
  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 {d^{2}y}{dx^2} = 0 but {d^{2}y}{dx^2} = 0 isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point
  4. Checking the nature of a Stationary Point when {d^{2}y}{dx^2} = 0
    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

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Find equation of tangent to a curve

November 19, 2009 Leave a comment

The tangent to a curve is a line which touches the curve at a point without intersecting it at that point so the gradient of the curve at that point and the gradient of the tangent are the same. So we can work out the point the tangent passes though and the gradient of the tangent from the equation of the curve, which will give us enough information to find the equation of the tangent.

Example y=x2
Find the equation of the tangent to the curve y=x^2 when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.
\frac{d}{dx}(x^2) = 2x
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 = $latex \frac{y – y_1}{x – x_1}
to get the general equation for the tangent at the point x=t by substituting x1=t, y1=t^sup>2 and m=2t
2t = \frac{y - t^2}{x - t} \rightarrow 2xt - 2t^2 + t^2 = y \rightarrow y=2xt-t^2

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get
y=8x-16
which is our final answer.

Inequalities

November 1, 2009 Leave a comment

An inequality(or inequation) is similar to an equation accept for instead of saying both side of the inequality are equal we say one side is greater than (or equal to depending upon the type of inequality) the other, this is done using the greater than (> ), greater than or equal to (\geq ), less than (< ) and less than or equal to (\leq ).

Examples

Some simple examples which contain only one variable are:

5x > 3
2x-7 < x+5
x^2 - 4 \leq x

Solving and Manipulating Inequalities

Inequalities can be solved by rearranging them and isolating the variable you want to find in a similar way to normal equation (see the post quadratic inequalities to see how to solve quadratics). However, rather than getting an exact value such as x=3 we get a range (open or closed) of values such as x<2 or -3<-1.

Much of the manipulation is the same though there are slight variations when dividing or multiplying by negative numbers or taking the reciprocal. The important thing to remember is that like normal equations we must do the same to both sides.

Addition and Subtraction

Addition and subtraction are exactly the same to equalities. We can add or subtract whatever we like as long as we do the same to both the sides. This enables us to take expressions “to the other side” by reversing their sign. For example all the following manipulations are valid.

x + 3 < 4  \Leftrightarrow x < 4-3 = 1
x - 3 < 4  \Leftrightarrow x < 4+3 = 7
x < 4  \Leftrightarrow x + 3 < 4+3 = 7

Multiplication and Division

Again we can perform multiplication and division in a similar way to the way we perform it with equalities by doing the same to both sides. However, if we are multiplying or dividing by a negative number we must reverse the direction of the inequality since
-x < y \Leftrightarrow x>-y
This means we must be careful when diving by an unknown since by definition we don’t know whether or not it is positive or negative. If this has to be done you should consider both the cases it is positive and negative separately and if it is only positive or negative then the other inequality should lead to a contradiction which can easily be spotted such as x<0 and x>3.

Examples of valid manipulation are below:
2x < 6 \Leftrightarrow x<3
-2x < 6 \Leftrightarrow x>3
\frac{4}{x} < 3 \Leftrightarrow \frac{4}{3} < x for x \geq 0 and/or \frac{4}{3} > x for x<0

Reciprocals

When taking the reciprocal or “one over” of an expression you must reverse the inequality so
x < y \Leftrightarrow \frac{1}{x} > \frac{1}{y}

Exponential Functions

October 24, 2009 Leave a comment

Exponential functions are any function of the form
y = a^{bx} latex for some constants a and b.

If a and b are both positive then the graph will be an upward curve which tends to infinity as x tends to infinity and tends to 0 as x tends to negative infinity and looks something like the below. Note that all exponential graphs cut the y axis at 1.

The graph of y = 2^x (y equals 2 to the power x)

The graph of y = 2^x (y equals 2 to the power x)

If a is positive and b is negative the graph is simply a reflection of this about the y axis to give the following graph:

The graph of y=2^-x (y equals 2 to the power of minus x)

The graph of y=2^-x (y equals 2 to the power of minus x)

The most import exponential graph is y=e^x because the gradient of this graph is always equal to the value of e^x at that point.

Categories: algebra, calculus Tags: , ,

Taylor Series with example cos(x)

October 13, 2009 Leave a comment

The Taylor series is the general case of the Maclaurin Series for calculating the value of a function. It enables you calculate the value of a function at any point if you can find the value of the function and and all its derivatives at any point. This is done as a power series. The series is as follows:

f(x) = \sum^{\infty}_{n=0}  \frac{f^{(n)}(x-a)^n}{n!} = f(a) + \frac{f^{(1)}(a)(x-a)}{1} + \frac{f^{(2)}(a)(x-a)^2}{2!} + \frac{f^{(3)}(a)(x-a)^3}{3!} + ...

The series is said to be taken about a meaning we calculate the derivatives of the function at the point a and then from these we find the value of the function. Because the series is infinite we can never find the value of the function exactly but we can give it to any required degree of accuracy by taking the first i terms of the series. The Maclaurin Series is just the Taylor series about 0.

Example cos(x)

An example of the Taylor series is to find a power series for cos(x). We can choose to do this about any value of a so in this example we will use a = \frac{\pi}{2} . In this example we will look at the first 2 no zero terms.

The first we need to find the value of the derivatives and the function at pi/2.
The first one is:
cos(\frac{\pi}{2}) = 0

We now need to differentiate the function to get:
\frac{d}{dx} cos(x) = -sin(x)
and then take its value at \frac{\pi}{2} which is:
-sin({\pi}{2}) = -1

Similarly for the second term we differentiate again to find the second derivative is:
\frac{d}{dx} (-sin(x)) = -cos(x)
so at pi/2 this is
-cos(\frac{\pi}{2}) = 0

Since the last term was zero we need to find the next one which is:
\frac{d}{dx} (-cos(x)) = sin(x)
so sin(\frac{\pi}{2}) = 1

Now we have found all the values of the function and its derivative we need for the level of accuracy required we can simply put these values into the series to get

cos(x) = 0 + (-1)(x-\frac{\pi}{2}) + \frac{0(x-\frac{\pi}{2})^2}{2!} + \frac{1(x-\frac{\pi}{2})^3}{3!} + ...
cos(x) = -(x-\frac{\pi}{2})  + \frac{(x-\frac{\pi}{2})^3}{3!} - ...

To test this we can try x = \frac{\pi}{3}
cos(\frac{\pi}{3}) = -(\frac{\pi}{3}-\frac{\pi}{2})  + \frac{(\frac{\pi}{3}-\frac{\pi}{2})^3}{3!} - ...
cos(\frac{\pi}{3}) = 0.49967
which is close to the 1/2 it really is.

Maclaurin Series with example sin(x)

September 23, 2009 Leave a comment

The Maclaurin series is a way of approximating a function f(x) using a power series of x. It only works on functions you can differentiate but you can gain any level of accuracy by stopping the series at different points. It is equal to the taylor series for a function about 0. One example of a use of the Maclaurin series is to calculate a value for sin(x).

The Maclaurin series for a function f(x) is as follows:

f(x) = f(0) + xf'(0) + \frac{x^2 f''(x)}{2!} + \frac{x^3 f'''(x)}{3!} + ...
or
f(x) = \sum_{i=0}^{\infty} \frac{x^i f^i (0)}{i!}

Example sin(x)

An expression for sin(x) can be calculated using this method. Firstly we must decide what level of accuracy to go to, for the example we will go to the 3rdterm. The next step is to differentiate sinx. Firstly let f(x) = sinx

then

f(0) = 0

f'(x) = cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'(0) = 1

f''(x) = -sin(x) \quad \quad \quad \Rightarrow \quad \quad \quad f''(0) = 0

f'''(x) = -cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'''(0) = -1

f''''(x) = sin(x) \quad \quad \quad \Rightarrow \quad \quad \quad f''''(0) = 0

f'''''(x) = cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'''''(0) = 1

We can now combine these into the series to get

f(x) = sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...

f(x) = sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ...

which can be used to calculate the value of sin(x)  — though only for the radian measure of angle.

eg)
sin(\frac{\pi}{3}) = \frac{\pi}{3} - \frac{\pi ^3}{6 \cdot 3^3} + \frac{\pi ^5}{120 \cdot 3^5} + ... \simeq   0.8663

Which is approximately the value you would get if you type sin(pi/3) into a calculator.

Complex roots of unity

September 22, 2009 Leave a comment

Without complex numbers taking the square root of any positive integer, such as 1, will give you two answers, in this case +/- 1, but taking the cube roots will only five you one answer, 1. However when we consider the complex roots you will find that the nth root of any number will give you n roots.

This can be shown using De Moivres theorem. Firstly consider the complex number z = cosΘ + isinΘ and let

zn = 1
(cosΘ+isinΘ)n = 1

which using De Moivres theomrm gives

cos nΘ + isin nΘ = 1

We can now compare the real and imaginary parts to find the values of n

cos nΘ = 1 and sin nΘ = 0

Therefore nΘ = 2pπ for integral values of p
So \theta = \frac{2p\pi}{n}

which gives

\theta = \frac{2\pi}{n} , \frac{4\pi}{n} , \frac{6\pi}{n} , \frac{8\pi}{n} , ... , 2\pi

so the roots of unity are
z_1 = cos \frac{2\pi}{n} + i sin \frac{2\pi}{n}
z_2 = cos \frac{4\pi}{n} + i sin \frac{4\pi}{n}
z_3 = cos \frac{6\pi}{n} + i sin \frac{6\pi}{n}
etc

However what is interesting about these is when you plot them on an argand diagram they are evenly spaced around a circle of radius 1 and centred at the origin. An example for n=8 is shown below

The roots of unity are space evenly around a circle of radius 1 and center at the origin

The roots of unity are space evenly around a circle of radius 1 and centre at the origin

This happens becuase the increase in the angle for each successive root is equal since we divided 2pi by n.