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Further Algebradeterminants : simultaneous equations : partial fractions: limits 

For example the determinant below is a second order determinant:
This is an example of a third order determinant:
Using the formula gives a value of 7:
Each second order determinant can then be evaluated individually and multiplied by the term outside because:
The evaluation is in two stages via three smaller determinants:
The evaluation of a determinant can be simplified. If a row or column has a common factor, that factor can be removed outside the determinant if all the rows / columns are then divided by that factor. This can lead to great simplifications of the arithmatic. In the determinant below, 6 is removed from column 1, then 2 is removed from row 1.
In the Introduction to Algebra, simultaneous equations were solved by manipulating them until one variable was left.
This is the method for solving simultaneous equations by using determinants.
Example 4: Solve the two simultaneous equations below:
Create and evaluate the three determinants.
Apply the formula and replace with the values of the determinants.
Equating the fraction containing x with the fraction containing only numbers gives:
Equating the fraction containing y with the fraction containing only numbers gives:
The solutions are therefore x = 5 and y = 2.
Three simultaneous equations in three unknowns (x, y, z) can be solved in simliar way. Four third order determinants are produced by using the coeficients and numbers as above to give D(n), D(x), D(y) and D(z). The formula for solving the simultaneous equations is:
Example 5: Solve the three simultaneous equations below:
First create the four determinants. D(n) contains the coeficients of x, y, and z. This has a value of 33.
D(x) contains all the coeficients and numbers except those for x. Its value is 33.
D(y) contains all the coeficients and numbers except those for y. Its value is 66.
D(z) contains all the coeficients and numbers except those for z. Its value is also 66.
Finally, apply the formula and replace with the values of the determinants.
By equating each fraction that contains an unknown with the fraction containing the numbers we can show that
x = 1, y = 2, z = 2.
Simultaneous equations of higher order with more unknowns can be solved just as easily.
In many problems it is useful to be able to express this in the form of partial fractions like so
where A, B and f, g are numbers. The expressions x + f and x + g are the two factors of the original function's denominator (the lower part).
The method for converting fractional functions to partial fractions involves factorising the denominator of the original fraction and then evaluating the numbers A and B. An example is the easiest way of showing the method.
Example 6: Convert the fractional function below into partial fractions.
Firstly we factorise the denominator as shown below. We are using the identity sign (three lines) rather than the equals sign (=) because we are changing the format without affecting any values.
We then separate the function into two new fractions each with one of the factors as a denominator. The numbers in the numerators (the top parts), A and B, are not known at this stage and will need to be calculated.
In the next stage we add the fractions using the normal rules of fractional algebra. The numerator of the first fraction (A) is multiplied by the denominator of the second fraction (x + 2). The numerator of the second fraction (B) is multiplied by the denominator of the first fraction (x  1). The combined numerator is then expanded and the various terms grouped together in terms of x and constants.
We are attempting to recreate the original fraction but with the A's and B's in the numerator:
We can now compare the original fractional function with the new version that we have created.
The left hand side is identical to the right hand side. This means that the coeficients of x can be equated on both sides and the numbers can also be equated. This gives a pair of simultaneous equations with two unknowns (A and B).
These equations can be easily solved to give A = 1 and B = 3, so
For example, consider the infinite series below:
The series has an infinite number of terms. The sum of the series (how much the total is) can be rewritten using the summation symbol (the Greek capital letter, sigma Σ) as follows
The symbol Σ means add up all terms like the following. In this case the terms are one over 2 raised to the power of n (1/2^{n}). The notation n= 0 below the Σ means begin when n is 0. The infinity symbol above the Σ means carry on until infinity. So the meaning of the notation above is:
Of course, we cannot do this sum because there are an infinite number of things to add. However, it can be shown that the more terms we take the closer the sum approaches to the value 2. We can use the idea of a limit to say that:
There is a concise way of writing this statement:
The following limit
means
A final example
This one means
Consider the function
This function has a value (y) for each value of x. To find a value for y, just square the value of x. When we find a derivative we are looking to see how this function varies at a given value of x. We can approximate this by adding a small amount to x and seeing how it affects y.
A small addition to x is given the symbol Δx. Similarly, a small addition to y is given the symbol Δy. The amounts have to be small otherwise we are not finding a variation at a particular point. When we do the addition we obtain:
We can expand the right hand side
Now subtract y from both sides
Remember that y = x^{2}
Divide both sides by Δx gives
The expression Δy / Δx is pronounced delta y by delta x. It is a measure of how y varies when x is changed by a small amount. The smaller we make Δx, the closer the expression approaches the rate of change at a given point, the instantaneous rate of change. We cannot just set Δx to 0 because we cannot divide by zero.
But we can use limits. We can allow Δx to approach 0 and see what the limit is. This can be expressed:
This limit is the derivative. It has a value 2x. The derivative of x^{2} is 2x. This can be expressed, using the dy/dx notation as
If the derivatives do not give a limit then, the process can be repeated. The use of this rule is best shown in an example.
Example 7: Find the limit of (1  cos x) / x^{2} as x approaches 0.
We are looking for the following limit.
It is not easy to see what the limit is because as x approaches zero, both top and bottom approach zero. The limit cannot be determined from this expression. However, according to L'Hôpital's Rule, we can find the derivatives of top and bottom to give a new expression; if that limit exists (for x approaching 0) it will equal the limit for the above expression.
The derivative of 1  cos(x) is sin(x). the derivative of x^{2} is 2x. Therefore we are looking for the following limit:
This is no better because the limit is 0 / 0 as x approaches zero.
We can continue and differentiate top and bottom again. The derivative of sin(x) is cos(x). The derivative of 2x is 2. Our third limit is therefore:
This does have a limit that is easy to see. As x approaches zero, cos(x) approaches 1. As x approaches zero, the denominator remains constant at 2.
Therefore the limit of this expression is 1/2. By L'Hôpital's Rule, 1/2 is also the limit of the original function when x approaches zero.
© 2004, 2009 KryssTal