About the Newton Raphson method
Consider the following equation:
Y = x^2 + 2x – 3
If you are asked to find the roots of that equation, it is a pretty simple process of factoring the equation and then using the zero product rule to solve for possible values of x.
Illustrated as follows:
Y = x^2 + 2x – 3
= (x+3)(x-1)
For finding zeros, y = 0
So, (x+3)(x-1) = 0
Using the zero product rule,
(x+3) = 0 ; (x-1) = 0
Thus, x = -3 or x = 1. Therefore these are the two zeros of the given equation.
Now consider the following equation:
Y = 2x^2 + x+ 5
Since in this case the left hand side polynomial is prime, we can simply use the quadratic formula to find the real roots of this equation (if any).
But now suppose if the function is like this:
Y = 3x^5 + x^3 – 2x^2 - x + 7
There are no simple methods or formulae for finding the roots (or zeros) of such a function. This is where we use some method of approximation. One such method is the Newton – Raphson method named after the mathematicians Issac Newton and Joseph Raphson who invented the method. The idea here is to start with some approximate zero of the function and then through iterative process come to closer and closer approximations of the zeros of the polynomial function. For a function of one variable the Newton -
Raphson algorithm can be stated as follows:
1. For a function f, we guess the zero = x0 to begin our iterations.
2. For the next best approximation of the zero of f we use the following formula:
X1 = x0 – f(x0)/f’(x0). Thus f(x1) is closer to 0 than f(x0).
3. The next best approximation would be x2 given by the formula:
X2 = x1 – f(x1)/f’(x1). Thus f(x2) is closer to 0 than f(x1)
4. The above iterative process can be continued for as many number of times as we like. Each time the zero we get would be closer and closer to the actual zero of the polynomial.
5. The general formula for the (n+1)th approximation is like this:
X(x+1) = x(n) – f(x(n))/f’(x(n))
For any newton – raphson example, we begin by guessing a zero. For functions with smaller coefficients, we can conveniently assume 0 as an approximate zero. For other functions, we may graph the function using few test values to come to some approximate first guess.
Graphing and solving logarithmic functions
A logarithmic function is the inverse of an exponential function. The general form of an exponential function would be y = b^x, where b is the base of the exponent and x is the exponent or the index. b belongs to positive real numbers and x is any real number. Therefore we see that the domain of an exponential function is all real numbers; whereas the range of this exponential function would be all positive real numbers.
The general form of a logarithmic function would be y = log_b?x, where b is the base of the logarithm. It is the same base that we used above in the exponential function. The x and y have switched places. Therefore, if we were to write the exponential function y = b^x in logarithmic form it would be log_b?y = x.
Graphs of logarithmic functions: To graph logarithmic functions online could be easy, but to graph them manually is not difficult either. Since we know that the logarithmic function is the inverse of an exponential function, we can make a table of values and plot the points, and then join the points with a curve. Let us try to see that with an example.
Example 1: Graph the function y = log_2?x.
Tabulating the results we have:
X (1/4) (1/2) 1 2 4 8
y -2 -1 0 1 2 3
Now we plot those ordered pairs on a graph sheet and run a curve through it. See picture below:
The blue curve above is the graph of the logarithmic function y = log_2?x. The red curve is the graph of the inverse exponential function y = 2^x.
Solving logarithmic functions: There are various methods to solve logarithmic functions. One way is to convert the logarithmic function to the corresponding exponential function and then solve.
Example 2: Solve log_5?125 = y.
Solution: Converting to exponential form we have, 5^y = 125
5^y = 5^3
Since in the above equation the bases are equal, the exponents would also be equal.
Therefore, y = 3 is the answer.
The general form of a logarithmic function would be y = log_b?x, where b is the base of the logarithm. It is the same base that we used above in the exponential function. The x and y have switched places. Therefore, if we were to write the exponential function y = b^x in logarithmic form it would be log_b?y = x.
Graphs of logarithmic functions: To graph logarithmic functions online could be easy, but to graph them manually is not difficult either. Since we know that the logarithmic function is the inverse of an exponential function, we can make a table of values and plot the points, and then join the points with a curve. Let us try to see that with an example.
Example 1: Graph the function y = log_2?x.
Tabulating the results we have:
X (1/4) (1/2) 1 2 4 8
y -2 -1 0 1 2 3
Now we plot those ordered pairs on a graph sheet and run a curve through it. See picture below:
The blue curve above is the graph of the logarithmic function y = log_2?x. The red curve is the graph of the inverse exponential function y = 2^x.
Solving logarithmic functions: There are various methods to solve logarithmic functions. One way is to convert the logarithmic function to the corresponding exponential function and then solve.
Example 2: Solve log_5?125 = y.
Solution: Converting to exponential form we have, 5^y = 125
5^y = 5^3
Since in the above equation the bases are equal, the exponents would also be equal.
Therefore, y = 3 is the answer.
Linear approximation Formulae
Introduction: Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions are called linearization. They are based on tangents. We introduce new variables dx and dy and define them in a way that gives new meaning to the Leibniz notation dy/dx. We will use dy to estimate error in measurement and sensitivity to change.
Linear approximation: If we see the graph of y = x^2 and y = 2x – 1, the tangent to a curve y = f(x) lies close to the curve near the point of tangency. For a brief interval to either side, the y-values along the tangent line give a good approximation to the y-values on the curve.
The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent. In the graph of the function y = f(x), the tangent line passes through the point (a, f(a)), so its point slope equation is y = f(a) + f ’(a) (x – a). Thus, the tangent is the graph of the function L(x) = f(a) + f ’ (a) (x – a). For as long as the line remains close to the graph of f, L(x) gives a good approximation to f(x).
According to the definition, Linear approximation formula can be given as below: If f is differentiable at x = a, then the approximating function L (x) = f(a) + f ’ (a) (x – a)is the linearization of f at a. The approximation f (x) ˜ L (x) of f by L is the standard linear approximate of ‘f’ at a” and is also known as Local Linear Approximation .The point x = a is the centre of the approximation. Let us take some Linear approximation examples
Example: Find the linearization of f(x) = sqrt(1 + x) at x = 0.
Solution: With f ’ (x) = ½ (1 + x)^-1/2,
We have f(0) = 1, f ’ (0) = ½, and
L (x) = f (a) + f ’ (a) (x – a) = 1 + ½ (x – 0) = 1 + x/2.
The approximation sqrt(1 + x) ˜ 1 + x/2 gives
Sqrt(1.2) ˜ 1 + 0.2/2 = 1.10
Sqrt(1.05) ˜ 1 + 0.05/2 = 1.025
Sqrt(1.005) ˜ 1 + 0.005/2 = 1.00250
A linear approximate normally loses accuracy away from its centre. The approximation sqrt(1 + x) ˜ 1 + x/2 will probably be too crude to be useful near x = 3. There, we need linearization at x = 3.
Derivatives of Power Functions of e Explained
What is the Derivative of Ex
Ex here is the natural exponential function which is written as e^x, where x is any real number. To find the derivative of e^x first let us understand what a derivative is. Derivative can be defined as the instantaneous rate of change of a function with respect to one of its variables which in simple words is the slope of the tangent line to the function at a point. The derivative of a function f(x) with respect to ‘x’ is the function f’(x) given by , f’(x) = lim(h?0) [f(x+h) – f(x)]/h where f’(x) is read as ‘f prime of x’. Let us now find what is the derivative of Ex. We need to find the derivative of e^x to get the derivative of Ex. Let us take y= e^x, which is the inverse of y= ln x, for which we can obtain the derivative. y= e^x implies ln y = ln e^x = x. Taking derivatives on both sides of lny = x, we get d[ln y]/dx = d[x]/dx and applying the chain rule to ln y, we get, 1/y. y’ = 1 which gives, y’ = y . So, in this case we can see that the derivative of y is y which means derivative of e^x is itself. We can conclude that the Derivative of e^x is e^x
Let us now find the derivative of e with a functional exponent. Let y = e^u(x), applying the chain rule, we get, d[e^u(x)]/dx= d[e^u(x)]/dx. du(x)/dx which gives us e^u(x). du(x)/dx. Using the derivative of e with a functional exponent, let us find the derivative of E 4x. Here u=4x, d[e^4x]/dx would be, e^u. d[u]/dx. du/dx = d(4x)/dx = 4. So, substituting u=4x and du/dx = 4, we get, derivative of E 4x as 4e^4x.
Derivative of E Ax would be the derivative of e^Ax. Taking g(x)=Ax where A is some constant, let us apply the chain rule to find the derivative of E Ax. As per the chain rule, f(g(x)’ = f’(g(x)).g’(x). Here f(x) = e^Ax and g(x)=Ax. Let us find the derivatives of each functions, f’[g(x)] and g’(x); f’g(x) = e^Ax and g’(x)=A. Substituting these values we get Derivative of E Ax as A.e^Ax
Derivative of E -1 is written as derivative of e^-1. e^-1 is something like e^x where x = -1. We know that the derivative of e^x is e^x and hence the derivative of e^-1 is e^-1. So, Derivative of E -1 is e^-1
Standard deviation of the mean
Standard deviation of the mean: If the values of x or f are large, the calculation of AM by the direct method is quite tedious and time consuming, because calculations involved are lengthy. In such a case to minimize the time involved in calculation, we take deviations from an arbitrary point as discussed below. Let x1, x2, x3, …., xn be values of a variable x with corresponding frequencies f1, f2, f3,…, fn respectively. Taking deviations about an arbitrary point ‘A’, we have
Taking the deviation d of the mid-point of class interval from the mean, squaring it to get d2, multiplying this by the frequency of the class, i.e., fd2, adding all the items, i.e., Sfd2, taking the average, Sfd2/n, and then taking the square root, we get, Standard deviations from the mean is given by S.D. = s = sqrt(Sfd2/n)
Standard deviation from the mean: To understand the concept of standard deviation from mean, we should know The population standard deviation, or s, is simply the square root of the population variance. Because the variance is the average of the squared distances of the observations from the mean, the standard deviation is the square root of the average of the squared distances of the observations from the mean. While the variance is expressed in the square of the units used in the data, the standard deviation is in the same units as those used in the data.
The standard deviation of the mean formula is
s = sqrt(s2) = sqrt(S(x – µ)2/N) = sqrt((Sx2/n) – µ2)
where x = observation, µ = population mean, N = total number of elements in the population, S = sum of all values (x – µ)2, or all the values x2, s = population standard deviation,s2 = population variance.The square root of a positive number may be either positive or negative because a2 = (-a)2. When taking the square root of the variance to calculate the standard deviation, however, statisticians consider only the positive square root.
Standard deviation of the mean example
Example: Find out the standard deviation of the following items: 8, 10, 12, 14, 16, 18, 20, 22, 24, 26.
Solution: Calculation of standard deviation.
Size of items (x) Deviation from mean = 17(d) (d2)
8 -9 81
10 -7 49
12 -5 25
14 -3 9
16 -1 1
18 +1 1
20 +3 9
22 +5 25
24 +7 49
26 +9 81
Sx = 170 Sd2 = 330
Arithmetic average or = Sx/n = 170/10 = 17
Standard deviation or s = sqrt(Sd2/n) = sqrt(330/10) = sqrt(33) = 5.74
Taking the deviation d of the mid-point of class interval from the mean, squaring it to get d2, multiplying this by the frequency of the class, i.e., fd2, adding all the items, i.e., Sfd2, taking the average, Sfd2/n, and then taking the square root, we get, Standard deviations from the mean is given by S.D. = s = sqrt(Sfd2/n)
Standard deviation from the mean: To understand the concept of standard deviation from mean, we should know The population standard deviation, or s, is simply the square root of the population variance. Because the variance is the average of the squared distances of the observations from the mean, the standard deviation is the square root of the average of the squared distances of the observations from the mean. While the variance is expressed in the square of the units used in the data, the standard deviation is in the same units as those used in the data.
The standard deviation of the mean formula is
s = sqrt(s2) = sqrt(S(x – µ)2/N) = sqrt((Sx2/n) – µ2)
where x = observation, µ = population mean, N = total number of elements in the population, S = sum of all values (x – µ)2, or all the values x2, s = population standard deviation,s2 = population variance.The square root of a positive number may be either positive or negative because a2 = (-a)2. When taking the square root of the variance to calculate the standard deviation, however, statisticians consider only the positive square root.
Standard deviation of the mean example
Example: Find out the standard deviation of the following items: 8, 10, 12, 14, 16, 18, 20, 22, 24, 26.
Solution: Calculation of standard deviation.
Size of items (x) Deviation from mean = 17(d) (d2)
8 -9 81
10 -7 49
12 -5 25
14 -3 9
16 -1 1
18 +1 1
20 +3 9
22 +5 25
24 +7 49
26 +9 81
Sx = 170 Sd2 = 330
Arithmetic average or = Sx/n = 170/10 = 17
Standard deviation or s = sqrt(Sd2/n) = sqrt(330/10) = sqrt(33) = 5.74
Geometry of circle
A circle is a conic section. When a cone is cut by a plane that is exactly perpendicular to the axis of the cone, the cross section we get is a circle. All points on the circle are equidistant from a fixed point in the circle. This fixed point is called the centre.
Radius of a circle:
The distance between the centre of a circle and any point on the circle is called the radius of that circle. The radius is half the diameter of the circle. So if we denote the radius by r and the diameter by d then, r = d/2
Circle formula:
We can measure the circumference of a circular object by winding a piece of fine string around the curved surface of the object exactly once and then measuring the length of the string with a meter scale. On measuring the circumference of a number of objects we find that the value of the ratio Circumference/diameter in each case is almost the same. This would always be some number between 3.1 and 3.2. This constant ratio is named by the Greek letter pi (pronounced as pi). Therefore we can write the formula for circumference of a circle as : C = pid, where d = diameter of the circle.
Area of a circle is given by the formula: A = pi*r^2 = pi * (d/2)^2 = (pi/4)d^2
Area of a semicircle:
We know that a semicircle is formed when a diameter divides a circle into two equal halves. So obviously the area of a semicircle is exactly half the area of the circle of the same diameter. Mathematically it is written like this,A.S. = (pi/2)*r^2, where A.S. = area of semi circle, r = radius of the semi circle.
Area of a quadrant of a circle = (1/4)* pi*r^2
Circle Geometry:
Consider a unit circle centered at the origin(O) of a co-ordinate axis. A point Q on the circle is such that the line segment OQ makes an angle of h with the positive x axis. Then the co-ordinates of the point Q would be (cos(h), sin(h)). If the radius of the circle is r, then the co-ordinates of the point Q would be (r*cos(h), r*sin(h)).
From the above figure we see that triangle OQS is a right triangle. The radius of the circle is 1, so OQ = 1. Therefore adjacent side to angle h = cos(h) and opposite side to angle h = sin(h).
When and How to Use Law of Sines and Cosines
What type of triangles use Law of Cosines and Sines
In a right triangle, we can find the unknown sides or angles using the Pythagorean Theorem. But if the given triangle is not right triangle and is an oblique triangle then how do we go about? In such triangles we use the Law of Cosines and Sines to solve triangles. Sine and Cosine Laws are given as follows: Law of Sines: a/Sin(A)=b/Sin(B)=c/Sin(C)
Law of Cosines: a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C
In a given oblique triangle, when do we use Law of Sines and Law of Cosines
The Law of Sines are used when we know two sides and one opposite side or when we know two angles and one opposite side of an oblique triangle. The Law of Cosines are used when we know two sides and the included angle or given the three sides of an oblique triangle.
Law of Sines and Cosines
The Law of Sines help to establish a relationship between the side lengths and angles of a triangle ABC. There are three Sines and hence the relationship explains the plural ‘s’ of Law of Sines.
The Law of Sines are, a/Sin[A] = b/Sin[B] =c/Sin[C] or we can even write them as Sin[A]/a=Sin[B]/b=Sin[C]/c ; where a,b,c are the side lengths and A, B and C are the opposite angles of the respective sides a,b,c in the oblique triangle ABC
The Law of Cosines is used most widely than the Law of Sines. When we know two sides of a triangle and their included angle, then Law of Cosines enables us to find the third side. The plural‘s’ in the law of Cosines is used as there are three cosines and hence by rotation similar formulas are valid for other angles. Law of Cosines are, a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C. The Law of Sines and Cosines, are also known as the Sine Rule and the Cosine Rule.
Solving Law of Sines and Cosines Word Problems
Let us solve some Law of Sines and Cosines Problems with the given side lengths and angles
Example: Peter wants to measure the height of a tree. He walks 100ft from the base of the tree and looks up. The angle of elevation found is 33 degrees. This particular tree grows at an angle of 83 degrees with respect to the ground rather than vertically. Calculate the height of the tree.
Solution:
Angle B=83 degrees , Angle A= 33 degrees, c=100ft [two angles and included side]
Angle C= 180-[angle A+angleB] = 180-116= 64 degrees
Using Sine Rule, a/SinA = c/SinC
a = c . SinA/SinC = 100. Sin(33)/Sin(64) = 100. (0.606) = 60.6
So, the height of the tree calculated by Peter is 60.6 ft
Know more about the Math Homework Help,online Math help. This article gives basic information about Standard deviation. Next article will cover more statistics concept and its advantages,problems and many more. Please share your comments.
In a right triangle, we can find the unknown sides or angles using the Pythagorean Theorem. But if the given triangle is not right triangle and is an oblique triangle then how do we go about? In such triangles we use the Law of Cosines and Sines to solve triangles. Sine and Cosine Laws are given as follows: Law of Sines: a/Sin(A)=b/Sin(B)=c/Sin(C)
Law of Cosines: a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C
In a given oblique triangle, when do we use Law of Sines and Law of Cosines
The Law of Sines are used when we know two sides and one opposite side or when we know two angles and one opposite side of an oblique triangle. The Law of Cosines are used when we know two sides and the included angle or given the three sides of an oblique triangle.
Law of Sines and Cosines
The Law of Sines help to establish a relationship between the side lengths and angles of a triangle ABC. There are three Sines and hence the relationship explains the plural ‘s’ of Law of Sines.
The Law of Sines are, a/Sin[A] = b/Sin[B] =c/Sin[C] or we can even write them as Sin[A]/a=Sin[B]/b=Sin[C]/c ; where a,b,c are the side lengths and A, B and C are the opposite angles of the respective sides a,b,c in the oblique triangle ABC
The Law of Cosines is used most widely than the Law of Sines. When we know two sides of a triangle and their included angle, then Law of Cosines enables us to find the third side. The plural‘s’ in the law of Cosines is used as there are three cosines and hence by rotation similar formulas are valid for other angles. Law of Cosines are, a^2=b^2+c^2-2bc Cos A; b^2=c^2+a^2-2ca Cos B ; c^2=a^2+b^2-2ab Cos C. The Law of Sines and Cosines, are also known as the Sine Rule and the Cosine Rule.
Solving Law of Sines and Cosines Word Problems
Let us solve some Law of Sines and Cosines Problems with the given side lengths and angles
Example: Peter wants to measure the height of a tree. He walks 100ft from the base of the tree and looks up. The angle of elevation found is 33 degrees. This particular tree grows at an angle of 83 degrees with respect to the ground rather than vertically. Calculate the height of the tree.
Solution:
Angle B=83 degrees , Angle A= 33 degrees, c=100ft [two angles and included side]
Angle C= 180-[angle A+angleB] = 180-116= 64 degrees
Using Sine Rule, a/SinA = c/SinC
a = c . SinA/SinC = 100. Sin(33)/Sin(64) = 100. (0.606) = 60.6
So, the height of the tree calculated by Peter is 60.6 ft
Know more about the Math Homework Help,online Math help. This article gives basic information about Standard deviation. Next article will cover more statistics concept and its advantages,problems and many more. Please share your comments.
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