Archive for the ‘calculus’ Category

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

The Chain Rule

November 21, 2009 Leave a comment

The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x2) or (5x3+2x+3)2. The rule is as follows
\frac{d}{dx}(f(g(x)) = \frac{dg}{dx}\frac{df}{dg}(g(x)) = g'(x)f'(g(x))
or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone).

Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.
Consider \frac{d}{dx}((ax + b)^n)
This is the composite of the functions ax+b and tn. So we differentiate them both to get a and ntn-1 and then apply the formula to get
\frac{d}{dx}((ax + b)^n) = an(ax+b)^{n-1}
Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function ntn-1 but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get
\frac{d}{dx}((f(x))^n) = f'(x)n(f(x))^{n-1}

Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).
Using the chain rule we get
\frac{d}{dx}(sin(f(x)) = f'(x)cos(f(x))
\frac{d}{dx}(cos(f(x)) = -f'(x)sin(f(x))

Fundamental Theorem of Calculus

October 31, 2009 Leave a comment

This theorem forms much of the basis of calculus and the uses of differentiation and integration. It basically states that differentiation and integration are opposites so if you differentiate and integral you’ll get the function you started with. This can be stated as follows:

if F(x) = \int_a(x)^b(x) \! f(t) \, dx then \frac{dF}{dx} = f(a(x))\frac{da}{dx} - f(b(x))\frac{db}{dx}

or in the more simple case

if F(x) = \int_0^x \! f(t) \, dx then \frac{dF}{dx} = f(x)- f(0)

It is this idea that allows us to know, for example,
\int \! \frac{1}{1+x^2} \, dx = tan^-1(x) + c
from the knowledge that
\frac{d(tan^-1(x))}{dx} = \frac{1}{1+x^2}

This makes much of integration easier as it is often much easier to work out the derivative a function than work out the integral of one so we can look for functions which when differentiated give us the function that we want to integrate and then know that the integral is that function plus a constant.

Integrating Fractions – using the natrual logarithm – Example tan(x)

October 27, 2009 Leave a comment

From result found be differentiating the natural logarithm,
\frac{d}{dx} (ln(f(x))) = \frac{f'(x)}{f(x)}
for some function f(x),

and the fundamental theorem of calculus we cay say that

\int \! \frac{f'(x)}{f(x)} \, dx = ln|f(x)| + c where c is the integration constant

Simple Example

The most basic example of this is the integration of 1/x,

\int \! \frac{1}{x} \, dx = ln|x| + c

More complex example: Integration of tan(x)

A slightly more complicated example of this is the integration of tan(x). To do this we must remember that tan(x) = \frac{sin(x)}{cos(x)} and notice that \frac{d}{dx}(cos(x)) = -sin(x). This means that -tan(x) is of the form \frac{f'(x)}{f(x)} as required. Using this we can get

\int \! tan(x) \, dx = \int \! \frac{sin(x)}{cos(x)} \, dx = lan|cos(x)| + c

Trick for using this identity

Sometimes we get integrals that are almost in this form but not exactly, eg) \int \! \frac{x}{5 + x^2} \, dx, however to solve these we can often factorise a constant so that it is in the required form. In this example we can take out a 2 so we get \frac{1}{2} \int \! \frac{2x}{ 5 + x^2} \, dx = ln|5 + x^2| + c

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!} + ...
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


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.

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.