Statistical Mean a stepwise approach

Mean Statistics
In Statistics a branch of Mathematics, the expression for the mathematical mean of a statistical distribution is the mathematical average of all the terms in the data. Here we add up the values of the terms given and divide the sum by the number of terms in the data. This expression is also called the Arithmetic Mean. For example, let us find the Statistical Mean of the following data 6, 8, 5, 10, 10, 12, 8, 6. Here first we need to find the summation of the data values which would be 6 + 8 + 5 +10 + 10 +12+8 +6 = 65. The Arithmetic Mean is got by dividing this sum with the total number of data values. The total number of data values is 8 and so Mean = 65/8= 8.125

Mean Math of the given data is the average of the total number of given data values. It is very simple to calculate we just add up the data values and divide by the count of data values. When the give data values are positive, we just add them and divide by the count to get the mean. The mean of 5, 8, 10, 12, 15 would be, (5+8+10+12+15)/5 = 50/5= 10. When the given data has negative data values, the method would be the same except that we need to combine the like terms. For example, The mean of 5, -3, 7,       12, - 2 would be (5 +(-3) +7 + 12- 2). Combining the like terms, [5+7+12 +(-3-2)] = [24 – 5] = 19. The Mean would be, sum of the data values/total number of data values = 19/5 = 3.8

Short cut Method
A short cut method of calculating the arithmetic mean is based on the property of arithmetic average. In this method the deviations (D) of the items from an assumed mean are first calculated and then multiplied with their respective frequencies (f). Then, the total of these products [summation(fD)] is divided by the total frequencies [summation(f)]and added to the assumed mean(A). The figure we get is the actual arithmetic average or the Arithmetic Mean.

Formula used in the short-cut method of calculating the arithmetic mean:
X(bar) = A + summation(fD)/summation(f)
Given the following data, calculate the Mean using the short cut Method
Weight(Kg) 68 70 72 74 76
Number of students 3 4 2 1 2
Summation of f(i) = 3+4+2+1+2 = 12

Let us assumed mean A= 72, let us tabulate the deviation
X(i) – A f(i).x(i)
(68 – 72) = -4 -4x 3 = -12
(70-72) = -2 -2x4  = -8
(72 – 72) = 0 0 x 2 = 0
(74 – 72) = 2 2 x 1 = 2
(76 – 72) = 4 4 x 2 = 8

Summation(f(i)x(i) = -10
So, we have, summation f(i) = 12, Summation(f(i)x(i) = -10 and A = 72
X(bar) = 72 + (1/12)(-10) = 72 – 0.833 =  71.17 kg

Derivatives of Inverse Functions made simple

Derivatives of Inverse Functions can be understood easily when we first learn about inverse functions. Consider an example, f(x) = 2x and g(x)=x/2  here, the first function doubles the input values and the second function halves the input values, they are going in opposite direction. These functions f and g are called inverse functions. The inverse is showed by using ‘-1’ in the power. It does not mean x-1=1/x as in exponential functions, it just denotes that the function is the inverse of the original given function.

An Inverse Function Solver helps us to find the inverse function of the given function. We can find the inverse function y in terms of x by first solving for x and then interchanging the x and y.  The function we get by doing so will be the inverse function of the original given function. We can also find inverse function solver online.  For better understanding let us consider Inverse Function Examples as given:

Example: Find the inverse of y= 2x-3
Solution: f(x) = y = 2x-3 [original function]
Solving for x, we get
2x = y+3
x = (y+3)/2
to  get the inverse function we interchange the x, y in the above step
here, y = y-1 =(x+3)/2  is the required inverse function of 2x-3

While Graphing Inverse Functions we use the same method as we use in graphing functions. The only difference being, first we find the inverse of the given function which will be taken as y and then various x values are plugged in, to get the coordinates(x,y). These coordinates are plotted on the graph which when joined give the required graph of inverse function.
Now that we have learnt about inverse functions, let us learn about Derivative of Inverse Function. If f(x) and g(x) are inverse functions then, the derivative of inverse function is given by the formula:
g’(x) = 1/f’[g(x)]

To find the derivative of inverse function of f(x) = 2x-1, first we need to find the inverse function of f(x), which is, f-1(x)=(x+1)/2. In the next step we find the derivative of this inverse function. [f-1(x)]’= d/dx[x/2 +1/2]=1/2 [applying the derivative rule] . So, the derivative of inverse function of f(x)=2x-1 is 1/2

Let us now go through Inverse Sine Function which is given as, y= sin-1(x) which implies sin(y)=x where y lies between – pi/2 and pi/2. Evaluating an inverse sine function is same as asking what angle we need to plug into the sine function to get the value x. Let us evaluate sin-1[1/sqrt(2)]. Here we are asked for what angle ‘ y’ we arrive to the value 1/sqrt(2), written as sin(y) = 1/sqrt(2). We have already learnt that sin(45) gives us the value 1/sqrt(2) and hence we get, y = pi/4.
Learn more about how to solve Calculus Problems.

Trigonometric function integrals tables

Trigonometric functions Integrals: If u is a differentiable function of x , then sin u is a differentiable  of x . The chain rule gives the derivative of sin u as d/dx sin u = cos u du/dx . From another  point of view , however , this same equation says that sin u is one of the  antiderivatives  of the product  cos u .(du/dx) .
Therefore , integration (cos u du/dx ) dx = sin u + C .

A formal cancellation of the dx’s in the integral on the left leads to the following rule . If u is a differentiable function , then  integration  cos u du = Sin u + c .

Let us take an example of Trigonometric Integral  top understand the concept . suppose vwe have given to integrate cos ( 7 (theta) +5) d(theta) so , to solve this let us take u = 7 (theta) + 5 , du = 7 d(theta) => 1/7 du = d(theta)  so plugging in the problem we have  integration  cos u . 1/7 du = 1/7 integration cos u du = 1/7 sin u + c  .

The chain rule formulae for the derivatives of the tangent , cotangent , secant , and cosecant of a differentiable function  will give us the following Trigonometric Integrals Table:

Table of Trigonometric Integrals
(i) Integration of sin  x dx = cos x + c  ,  Extension sin( ax +b) dx = -cos (ax+b) / a+ c
(ii) Integration of cos x dx = sin x + c  , extension integration cos (ax + b) = sin (ax +b) / a+  c
(iii) Integration tan x dx = -log mod (cos x ) + c , extension integration tan ( ax+ b) = -log mod ( cos (ax +b)/ a+ c
(iv) Integration  cot x dx = - cosec ^2 x  + c , extension  integration cot (ax +b) = -cosec^2(ax +b) / a+ c
(v) Integration cosec x dx = log mod (cosec x  - cot x ) + c , extension integration cosec (ax +b) dx = - log mod( cosec (ax +b) – cot (ax +b))/ a+ c
(vi) Integration sec x dx = log mod (sec x + tanx ) + c  , extension sec  (ax + b) dx = log mod sec (ax +b) + tan (ax +b)/a + c
(vii) Integration sec x . tan x = sec x + c , extension  integration sec (ax + b).tan (ax+b) dx = sec (ax +b)/ a+ c
(viii) Integration cosec x . cot x dx –cosec x + c , extension cosec (ax + b) . cot (ax + b) x = -cosec (ax +b) / a+ c

Integration formula

The following is a non comprehensive list of integration formulas:

Understanding logarithmic functions

The function f(x) = a^x is a one to one function provided that a > 0 and a =! 1. Therefore f would have an inverse which we will call the logarithmic function.

What are logarithmic functions?
If a > 0 and a =! 1, then the function  log_a(x), called the logarithm of x to the base a, is the inverse of the one to one function a^x:
Y = log_a(x)  x = a^y, (a > 0, a =! 1)
Since a^x has the domain (-inf, inf), log_a(x) has the range (-inf, inf). Since a^x has the range (0,inf), log_a(x) has the domain (0, inf). Since a^x and log_a(x) are inverse functions, the following cancellation identities hold:
log_a (a^x) = x for all real x and a^(log_a x) = x for all x > 0.
The graphs of some examples of logarithmic functions are shown in the picture (a) below. They all pass through the point (0,1). Each graph is the reflection in the line y = x of the corresponding exponential graph in the picture (b).

From the laws of exponents we can derive the following laws of logarithmic functions.
Laws of logarithms: If x > 0, y > 0, a > 0, b > 0, a =! 1, b =! 1, then
(i) log_a(1) = 0
Proof: We know that a^0 = 1 for any real number a. Therefore by the definition of log we can say that log_a(1) = 0. Hence proved.
(ii) log_a(xy) = log_a (x) + log_a(y)
Proof: Let u = log_a(x) and v = log_a(y). By defining property of inverse of log function we know that x = a^u and y = a^v. Thus xy = a^u * a^v = a^(u+v). Inverting again we have, log_a(xy) = u + v = log_a(x) + log_a(y). Hence proved!
(iii) log_a(1/x) = - log_a(x)
(iv) log_a(x/y) = log_a(x) - log_a(y)
Proof: Let u = log_a(x) and v = log_a(y). Then again by defining property of inverse of log function we know that x = a^u and y = a^y. Thus a^u/a^v = a^(u-v). Inverting again we have, log_a(x/y) = u - v = log_a(x) - log_a(y). Hence proved. The rule number (iii) can be proved in the same manner.
(v) log_a(x^y) = y log_a(x)
(vi) log_a(x) = (log_b x)/(  log_b a)
The logarithm law (vi) presented above shows that if you know the logarithms to a particular base b, you can calculate the logarithms to any other base a.

Some standard derivatives: Derivative of csc function

CSC is the acronym for cosecant function. Csc is the reciprocal of sine function. Symbolically it can be written as : csc (x) = 1/sin(x).

Derivative of csc function:
The derivative of csc function is the slope of tangent to the curve of the equation y = csc (x) at any point x. It can also be called the gradient of the csc function. To find the derivative of the function csc (x) we use the quotient rule which is as follows:
If a function f is such that it is a quotient of two functions g and h, symbolically,
f(x) = h(x)/g(x), then the derivative of f, represented by f’(x) is given by the formula,
f’(x) = [g(x)*h’(x) – h(x)*g’(x)]/(g(x))^2
Using the above rule we can find the derivative of csc function as follows:
(d/dx) csc (x) = (d/dx) (1/sin x) = [sin x * (d/dx)(1) – 1* (d/dx) (sin x)]/sin ^2 (x)
= -cos x/sin^2(x)
= -csc x cot x
Therefore derivative of csc is –csc x cot x.

Derivative of csc^-1 (x):
Inverse of the cosecant function is written as csc^(-1) (x). The derivative of the inverse of csc function can be found as follows:
Let y = csc^-1 (x), |x| >1
Thus, x = csc y, y belongs to (0,pi) – {pi/2}
So, dx/dy = - csc y cot y ? 0 because csc y ? 0 and since y belongs to (0,pi) – {pi/2}, cot y ? 0
Therefore, dy/dx = -1/(csc y cot y)
Since y belongs to (0,pi) – {pi/2} that means, either y belongs to (0,pi/2) or y belongs to (pi/2 , pi)
In both cases the following holds.
Then x = csc y > 0 and so |x| = x
Also cot y > 0 and so cot y = sqrt(csc^2 (x) – 1) = sqrt(x^2 - 1)
Therefore dy/dx = -1/x*sqrt(x^2 – 1)
Thus derivative of csc^-1 (x) = -1/x*sqrt(x^2 – 1)

Derivative of csc^squared (x):
Derivative of csc^2 (x) can be found using the chain rule.
Let y = csc^2 (x) and let csc x = u
Then y = u^2
Therefore dy/du = 2u
And since u  = csc x
So, du/dx = -csc x  cot x
Thus dy/dx = dy/du * du/dx
= 2u * (-csc x cot x)
= 2*csc x * (-csc x cot x)
= -2 csc^2 (x) cot x
Thus we see that derivative of csc^2 (x) = -2 csc^2 (x) cot x

Vertical Asymptotes: The Vertical Lines that never intersect the function

The vertical lines that correspond to the zeroes of the denominator of a rational function are called the Vertical Asymptotes. They occur only for those values of ‘x’ that produce zero in the denominator but not in the numerator. If  0/0 occurs, then we simply say we have a ‘hole’ in the graph. The next thing that comes to the mind is, How to find Vertical Asymptote of a given function. It involves a simple method where we just need to follow few steps to find all the values of x for which the denominator equals zero.

Following are the steps involved in finding the Vertical Asymptotes:
1. Vertical asymptote is in the form of an equation in x, x=a where f(x) is a function and f(a) does not exist.  A vertical asymptote is a vertical line which never intersects the function f(x).
2. We know that a fraction is undefined if its denominator is equal to zero. So, to find the vertical asymptote of a rational function, we need to  find the value of x such that the denominator is equal to zero.
3. In the next step, we need to equate the denominator to zero.
4. Then, we need to solve the equation in x to get the value of a.
5. Finally we need to plug in the value of a in the equation x=a, which is the vertical asymptote of the given function.

Note:  We can get more than one vertical asymptote depending on the given function

Finding Vertical Asymptote of a rational function, f(x) =(x^2+2x+3)/(x^2-5x+6). Let us first find all the x values by setting the denominator (x^2-5x+6) equal to zero and solving for ‘x’. Factorizing (x^2-5x+6) we get (x-6)(x+1)=0 and hence x = 6 and x=-1 which will make the denominator zero. Hence the vertical asymptotes of the given function are, x = 6 and x=-1.

Let us now Find Vertical and Horizontal Asymptotes of f(x)= (2x^2-5x+3)/x^2-1. The vertical asymptote is got by equating the denominator to zero, x^2-1=0. On factorization, we get (x+1)(x-1)=0. This gives us x=1 and x=-1 as the vertical asymptotes. To find the horizontal asymptote we need to first find the degree of the numerator and the denominator. Here the degree of the numerator and the denominator is the same, so, the horizontal asymptote is given by y= the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator. That gives us y=2/1=2, which is the horizontal asymptote.

Know more about the Asymptote Calculator. This article gives basic information about Asymptotes.