Gaussian Elimination Tutor

Introduction 

A system of equations in which all the given quantities are present only in the first degree is called a linear system of equations. The system of equations can be solved by various methods , one of the methods called gaussian elimination method.Here tutor will help you on gaussian elimination

Let us consider the following set of m linear equations with n unknowns
b11 x1 + b12 x2 + ....................+ b1n xn = c1
b21 x1 + b22 x2 + ....................+ b2n xn = c2
............................................................................
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bm1 x1 +bm2 x2 + ....................+ bmn xn = cm
Here all b's and c's are constants.
If c1, c2 upto cm.... are all zero, then the system is called homogeneous. The set of values x1, x2, upto ..xn which satisfy all the equations simultaneously is called a solution to the system of equations. Let us discuss how to find this solution online by gaussian elimination method.

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Gaussian Elimination Tutor : Steps

Step 1: Place the leftmost column in the matrix not consisting totally of zeros.

Step 2: If there is zero entry present , then substitute the top row by any another row to get non-zero entry to the top of this column.
Step 3: If this non-zero way used in a multiply the first row through 1/a to bring in a top 1.
Step 4: Add the suitable possible multiples of the top rows to each other row in arrange to contain only zero entries in the column below this important step1.
Step 5: Now remove the top row of the matrix and create once more with step 1 applied to the secondary matrix that remains. Carry on pending the whole matrix is in row-echelon structure.
Step 6: Then create with the last non-zero row and effective upwards, add suitable multiples of every row to the rows over in order to carry in zeroes over the leading 1s.Consider a system of three independent equations with 3 unknowns.

b11 x1 + b12 x2 + b13 x3 = c1
b21 x1 + b22 x2 + b23 x3 = c2
b31 x1 + b32 x2 + b33 x3 = c3
The system of equations is equivalent to the matrix equation BX  =  C
where B = `[[b11,b12,b13],[b21,b22,b23],[b31,b32,b33]]`    ,        X = `[[x1],[x2],[x3]]`,        C = `[[c1],[c2],[c3]]`
Now we have to reduce the coefficient of matrix B to a possible upper triangular matrix
[ A : B ] = `[[b11,b12,b13,:,c1],[b21,b22,b23,:,c2],[b31,b32,b33,:,c3]]`

Gaussian Elimination Tutor : Examples

Answer the given set of  linear system using Gaussian elimination Method:

x - 3y + z = 4,
2x + 8y - 8z = -2,
-6x + 3y + 15z = 9.
Solution:
The increased matrix for the method is
1        -3      1        |        4
2           8     -8      |        -2
-6        3       15      |        9
Add -2*row 1 to row 2 to get a zero directly below the leading 1. Also add 6*row 1 to row 3 to get a second zero.
1        -3        1     |          4
0        -2       6      |        -10
0        -15   -9      |        33
Consequently multiply row 2 by 1/2 to get a leading 1 and multiply row 3 by 1/3.
1        -3           1  |              4
0        -1          3   |             -5
0        -5         -3  |             11
Add -5*row 2 to row 3 to get a zero under this leading 1.
1        -3           1      |        4
0          -1         3      |       -5
0           0           -18 |      36
We at the present twist around the change procedure and turn the increased matrix into a system of equations we have
x - 3y + z = 4,
0x - y + 3z = -5,
0x + 0y - 18z = 36.
- 18 * z = 36.
z = - 2. (Answer)
- y + 3(-2) = -5
- y - 6 = -5
y = - 1
x - 3y + z = 4,
x - 3(-1) - 2 = 4,
x + 3 – 2 = 4
x + 1 = 4
x = 3
Then the solution is (x, y, z) = (3, -1, -2)

Partial Differentiation Versus Implicit Differentiation

Differentiation Definition
Differentiation is the rate of change of a function with respect to one of the variables. It is something like finding the slope of the tangent line of a function at a point. We can define differentiation of a function f(x) with respect to x as f’(x) = lim(h->0)[f(x+h) – f(x)]/h

Partial Differentiation 
A function in two variables given by f(x,y) when differentiated by keeping one of the variables constant, like differentiating y with respect to x keeping y constant we get what is called Partial differentiation of the function, it is denoted as doe(f)/doe(x) or fx when it is with respect to x and when the function in two variables is differentiated with respect to y keeping x constant it is denoted as doe(f)/doe(y) or fy.
For instance, let us find the partial differentiation of the function, f(x,y) = 2x^2y+ 3x +y
Solution: Keeping y constant let us differentiate first with respect to x, we get
fx = doe(f)/doe(x) = doe/doe(x) [2x^2y+3x+y]
   = doe/doe(x)[2x^2y] + doe/doe(x) [3x] + doe/doe(x) [y]
   = 4xy +3 +0 = 4xy + 3
Now keeping x constant let us differentiate the function with respect to y, we get
fy = doe(f)/doe(y) = d/dy [2x^2y+3x+y]
    = doe/doe(y)[2x^2y] = d/dy[3x] + d/dy [y]
    = 2x^2 + 0 + 1 = 2x^2 + 1
So, the partial differentiation of the funcntion f(x,y) = 2x^2y+3x + y is fx= 4xy =3 and fy =2x^2+1

Implicit Differentiation Solver
In Implicit Differentiation the differentiation is done with respect to x when the given function involves two variables, x and y. But here we do not assume the variable y to be constant, y treated as it is.  Usually x is assumed to be the independent variable and the variable y is assumed to be the function of x, the independent variable. For differentiation, chain rule is applied with respect to x followed by some algebraic solving steps to get dy/dx. The formula for Implicit Differentiation solver is, given a function F(x,y) equals zero, which defines a differential relationship of the two variables x and y, then,

dy/dx = - [doe(F)/doe(x)]/[doe(F)/doe(y)] here, doe(F)/doe(x) is the partial derivative of F with respect to x and doe(F)/doe(y) is the partial derivative of F with respect to y.

Let us solve the Implicit differentiation of the function, x^3y^2- xy = 5
Solution: Implicit differentiation, dy/dx [x^3y^2 – xy] = dy/dx [5]
Applying chain rule, we get
dy/dx [x^3y^2] – dy/dx [xy] =0
3x^2y^2 + x^3. 2y. dy/dx – 1.y – x. 1. dy/dx = 0
(2x^3y- x) dy/dx = y – 3x^2y^2
dy/dx = [y-3x^2y^2]/[2x^3y – x]

Know more information on Differentiation and Differential Equations.

Standard Deviation Examples

What is Standard Deviation?
Standard deviation is used in probability theory and statistics.  It shows how much variation exists in the data set from the average. The average can also be called as mean or expected value. If the variance of a data set is known, then the standard deviation can be obtained by finding the square root of it. The standard deviation can be calculated for a data set, random variable, probability distribution or statistical population.  An important property of standard deviation that makes it the preferred method of statistical measure is that the standard deviation computed for a data set has the same unit as that of the data set.

Standard Deviation in Statistics
Statistics Standard Deviation is used to measure the dispersion of the set of data given.   Standard Deviation Statistics shows how some data has high standard deviation and some have low standard deviation. Standard Deviation examples such as volatile stock and blue chip stock demonstrate this difference. The high standard deviation in volatile stock and low standard deviation in stable blue chip stock shows how standard deviation is used for statistical analysis of the given data sets. If the standard deviation is low, it indicates that the data values in the data set are very closely spread around the mean. If the standard deviation is high, it shows the wide dispersal of the data values over a larger range.

Steps to compute Standard Deviation
The first step in computing standard deviation is to find the mean of the data values of the given data set. The next step is to find out how far each data value is deviated from the mean. To compute this deviation, the mean is subtracted from each data value of the given data set and then the values are squared. Then find out the mean for the squared values. Then find the square root of this computed mean. Thus, by finding the square root of the mean of the squared values, you get the standard deviation.

Example of Standard Deviation 
Let us consider an example for standard deviation: The students of a school have participated in Mathematics Olympiad exam. To know their level of knowledge in Mathematics, marks of few students is taken as a sample from the whole population of students who attended the exam and the standard deviation is determined for this sample. This is demonstrated below:

Here is the sample: the marks of 6 students are 91, 100, 100, 98, 96, and 100
Step 1: The mean of these marks will be (92+100+100+99+97+100)/6 which will be 98
Step 2: The next step is to find the squared differences from the mean and sum it: (92-98)^2 +(100-98)^2 +(100-98)^2 +(99-98)^2 +(97-98)^2 +(100-98)^2  = 50
Step 3: Find the square root of the above sum divided by one less than the number of students. Thus you will end up in Square root (50/5), which is equivalent to 3.1623. This resultant value 3.1623 is the standard deviation.

As this standard deviation is low, we can understand that the actual marks scored by the students are very near to their expected marks i.e. 100. As this is a sample data set, it implies that all the students who have attended the Mathematics Olympiad exam have scored good marks.

An overview of the fundamental theorem of calculus

The second fundamental problem addressed by calculus is the area of a region of the plane bounded by various curves. Many practical problems in various disciplines require the evaluation of areas for their solution, and the solution to the problem of areas necessarily involves the notion of limits. On the surface the problem of areas appears unrelated to the problem of tangents. However, the two problems are very closely related. One is the inverse of the other. Finding area is equivalent to finding the anti derivative or finding the integral. The relationship between derivatives and integrals is called the fundamental calculus theorem.

We shall now demonstrate the relationship between definite integral and indefinite integral. A consequence of this relationship is that we will be able to calculate definite integrals of functions whose anti derivatives we can find.
1^st fundamental theorem of calculus:
Suppose that a function f is continuous on an interval I that contains the point a. Let the function F be defined on I  by:

Then F is differentiable on I and F’(x) = f(x) there. Thus, F is an anti derivative of f on I:

2nd fundamental theorem of calculus:
If G(x) is any anti derivative of f(x) on l, so that G’(x) is equal to f(x) on l, then for any b in l we have,

Fundamental theorem of calculus proof for part I:

Using the definition of derivatives we can calculate

Second fundamental theorem of calculus proof:
Continuing from the proof above, if G’(x) = f(x), then F(x) = G(x) + C on l for some constant C. Hence,

Let x = a and obtain 0 = G(a) + C, so C = -G(a). Now let x= b, then we get,

For fundamental theorem of calculus example, anywhere else, we can the t with x or any other variable as the variable of integration on the left hand side.

Solve Trigonometry Half angle formula

Trigonometry Half angle formula
In trigonometry, we may come across half angle formula examples, where in we are required to find a particular trigonometric ratio of half of the given angle ratio. For example, we know that sin 45 degrees = 1/v2. Can we use that to find the value of sin (45/2) = sin (22.5 degrees)? That is what half angle formula is all about.

Sine half angle formula:
The sine half angle formula is like this:
Sin (b/2) = v[(1 – Cos b)/2]
Proof: This formula can be proved using the double angle formula for cosine. The double angle formula for cosine is like this:
Cos (2a) = 1 – 2 sin^2 (a)
Here, let 2a = b, then a = b/2. So now the formula would become,
Cos (b) = 1 – 2 sin^2 (b/2)
=> 2 sin^2 (b/2) = 1 – Cos b
=> Sin^2 (b/2) = (1 – Cos b)/2
=> Sin (b/2) = v[(1 – Cos b)/2].
So that is how we prove the formula of half angle of sine.

Cos half angle formula:
The half angle formula for cosine is as follows:
Cos (b/2) = v[(1 + Cos b)/2]
Proof: This formula can be proved using the double angle formula for cosine. The double angle formula that we will use here is:
Cos (2a) = 2 cos^2 (a) – 1
Here, let 2a = b, then a = b/2. So the above formula now would become,
Cos (b) = 2 cos^2 (b/2) – 1
=> Cos b + 1 = 2 cos^2 (b/2)
=> 2 cos^2 (b/2) = Cos b + 1
=> Cos^2 (b/2) = (Cos b + 1)/2
=> Cos (b/2) = v[(1 + Cos b)/2]
So we see how we can prove the half angle formula for cosine as well.

Half angle formula for tangent function:
This one can be stated like this:
Tan (b/2) = v[(1 – cos b)/(1 + cos b)]
Proof: As proved above we know that
Sin (b/2) = v[(1 – cos b)/2] and Cos (b/2) = v[(1 + Cos b)/2]
We also know that tan (b/2) = sin (b/2)/Cos (b/2)
Substituting values of sin and Cos into the above formula, we have,
tan (b/2) = {v[(1 – Cos b)/2}/{v[(1 + Cos b)/2]}
=> tan (b/2) = v{((1 – Cos b)/2) / ((1 + Cos b)/2)}
=> tan (b/2) = v[(1 – Cos b)/(1 + Cos b)]
Hence proved.

Derivative of Arcsin X and Arcsin X 2

Derivative of Arcsin X
If there is an arcsin function X, then the derivative of arcsin is given by dividing one by the square root of the value obtained from subtracting the square of x from one. This is shown as a formula below:

Derivative of ArcSin = 1/square root of (1-square of X)

Here X belongs to real number in such a way that its value lies between -1 and +1.

Let us prove the formula for derivative of arc-sin X.  If Y = arcsin X where X is a real number lying between -1 and +1, then we can say that X = sin Y. By using the derivative of sine function, we can rewrite in the derivative form as dX/dY = cos Y. By applying the derivative of inverse function, it becomes dY/dX = 1/cos Y. We know that cos square Y + sin square Y is equal to 1. We can rewrite it as cos Y equal to plus or minus of the square root of one minus sin square Y. As cos Y is greater than or equal to Zero on the arcsin X range that ranges from –pi/2 to +pi/2, we can say that cos Y always takes a positive value. So we get the value of dy/dx as one divided by the square root of the value obtained from subtracting sin square y from one.  In other words, the differentiation or the derivative of arc-sin X is equal to one divided by the square root of the value obtained from subtracting the square of X from one, as we know X = sin Y. Thus, Derivative of Arc-Sin = 1/square root of (1-square of X)

Derivative of Arcsin X 2
As per the chain rule, the derivative of a square of any function is given by twice the function itself. By applying this rule, we can find out the derivative of arc-sin X2 from the formula of       arcsin X.

We have seen that g(X) = arcsin X. We know the differentiation of arcsin X i.e. d/dx arcsin(X) is obtained from the formula 1/square root of (1-square of X). By applying the chain rule, we get 2arcsin(X) multiplied with (1/square root of (1-square of X)). Thus, Derivative of arc-sin X2 = 2arcsin(X)/square root (1-square of X).

Derivative of arcsin(X)+arcos(X)
The differentiation of the summation of arcsin(y) and the arcos(y) is given by pi divided by 2, i.e. arcsin(X) +arccos(X) = pi/2.

Applications of Arcsin
An important application of the arcsin transformation can be seen in the analysis of binary data.

Measures of variation

Measures of variation definition that is the difference between highest value and the lowest value. That is a particular range.  In another way we also simplify that Measures of variation means the quantities that show the measure of variation in terms of random variable. For better understand the measure of variation we should have knowledge of random variables.

Random variable is determined by the outcome of the experiment which we have done. We may assign probabilities to the possible values of the random variable. We can also differ random variable in two ways. First one is discrete random variable in which a variable that can take one value from a discrete set of values. And the second one is continuous random variable, which means a variable that can take one value from a continuous range of values.

Measures of variation in statistics, statistics is a branch of mathematics which gives us the tools to deal with large quantities of data and derive meaningful conclusions about the data. To do this, statistics uses some numbers or a measure which describes the general feature contained in the data. In other word using statistics we can summaries large quantities of data by a few descriptive measures.  In this section we describe how any value is carried out from a given set of data. Measures of the variations sometimes are also known as measure of dispersion or measure of spread.

In statistics there are three measures of the variation…

1. The range
2. The standard deviation
3. The variance

The standard deviation is a measures of the variation or measure of dispersion amongst data. Instead of taking absolute deviation from the arithmetic mean we may square each deviation and obtain the arithmetic mean of squared deviations. This gives us the variance of the values. The positive square root of the variance is called the standard deviation of the given values. The standard deviation may be for raw data and grouped data.

The variance means the square of standard deviation. Alternatively square root of variance is equal to standard deviation. The larger the standard deviation, larger will be the variance. The standard deviation is an absolute measure of variation and hence cannot used for comparing variability of two data sets with different means. Therefore such comparisons are done by using a relative measure of variation known as coefficient of variance.

At last we must take measure of variation examples. Suppose, we have a set in which five different numbers so to find measure of variation we have to calculate only range means difference between highest and lowest value in given set of value.