Steps for solving standard deviation problems

A standard deviation example would typically have a set of observations x_1, x_2, x_3,  …… x_n. The number of observations being n. In general it can be written like this: x_i, where i = 1, 2, 3, … n. To calculate the standard deviation of such a problem the following steps are followed:

Step 1: List all the observations in the first column of a table. The title of column would be x_i and the observations would be x_1, x_2, x_3,  …… x_n.

Step 2: Find the mean of the observations. If we denote the mean by x ¯, then the formula for calculating the sample mean would be:  x ¯ = (x_1+ x_2 + x_3+  ……+ x_n)/n = [∑x_i]/n

Step 3: We now come to the second column. In this column we find the deviation of each of the observation from the mean. For that we calculate the value of x_i - x ¯ for each of the observations.
Note that in this step it is possible that we get negative values. We shall take care of the negatives in the next step.

Step 4: Now we construct the third column. In this column we square the results of column (3) for each of the observations. That means we find the value of (x_i - x ¯)^2 for each of the n observations. Note here that since we squared the difference, the negatives are got rid of so we need not worry about the negative numbers canceling out the positive numbers and affecting our end result.

Step 5: Now we find the standard deviation for each of the observation. For that we compute the following value: (x_i - x ¯)^2 / n. That would be placed in our fourth column.

Step 6: Adding standard deviations for each of the observations we get the value of the expression : ∑((x_i - x ¯)^2 / n). That will give us our sample standard deviation formula. The symbol for standard deviation is the Greek small alphabet ‘s’ pronounced as ‘sigma’. Therefore we say that:
s = sqrt[∑((x_i - x ¯)^2 / n)] or s = sqrt [∑(x_i - x ¯)^2/n]

Inverse Functions and their derivatives in a nutshell

In Math, Inverse Function is a function whose relation to a given function is such that their composite is the identity function. Inverse Function can also be defined as the function obtained by expressing the dependent variable of one function as the independent variable of another; for instance f and g are inverse-functions if f(x)=y and f(y)=x. It is denoted by (-1) in the power, inverse of f(x) is f-1(x).

One important point we need to remember is that the inverse of a function may not always be a function. While finding Inverse of Function of a function y in terms of x, we just switch the x and y and then solve for y. The new function [y-1] we get is the inverse-function of the given original function. For instance, finding Inverse of a Function f(x)= (x/3) -1 involves a list of steps. First let us take f(x)=y=(x/3) -1. In the next step we switch the x and y, that gives us, x = (y/3)-1. The last step would be to solve for y, finally we get y= y-1= 3x+3 which is the inverse-function of f(x) = (x/3) -1

There are four main steps involved in Solving Inverse Functions. Given a function f(x) in terms of x,
1. First step, we write the given function f(x) equal to y
2. Second step involves interchanging the x and y
3. In the third step, we solve for y
4. Fourth step, we write y as f-1(x) which is the inverse-function of f(x)

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

A square root function is the inverse-function of a square function f(x) = x2. So, the derivative of an Inverse Function of square function f(x) = x2 can be solved using the power rule

An Introduction to Quartiles in Statistics

Quartile is a term derived from the word ‘quarter’ which is one fourth of something. So, we can say  Quartiles Statistics is a certain fourth of a data set. When a given data is arranged in the ascending order that is from the lowest value to the highest value and this data is divided into groups of four, we get what we call Statistics Quartiles.

In Quartiles Statistics there are three quartiles which are, the first quartile also called as lower quartile is denoted as Q1 which lies in the twenty five percent of the bottom data. The second quartile is the median of the data that divides the data in the middle and has fifty percent of the data below it and the other fifty percent of the data above it. It is denoted as Q2.The third quartile also called the upper quartile denoted as Q3 has seventy five percent of the data below it and twenty five percent of the data above it.

We can state the Quartiles Definition as, Quartiles are the three values that divide an ordered data set into four approximately equal parts. Definition of Quartiles can also be given as, A statistical term describing a division of a data set into four defined intervals based upon the values of the data and how they compare the entire data set.

Now that we are a bit familiar with Quartiles Statistics, let us learn how to calculate Quartiles of a given data set. For a better understanding of the steps involved let us consider a data set. Calculate the three
 quartiles of the data set of scores,  27 18 20 20 23 29 24.

The steps involved are:
• Count the total number of scores in the given data; n= 7
• Arrange the data set of scores in the ascending order, that is, from the lowest score to the highest score, we get, 18 20, 20, 23, 24, 27, 29
• Find the median (Q2) which is the middle value of the data set
Median = ½(n+1) = ½(7+1)= 4
So, the median (Q2)=23 [the fourth term of the ordered data set]
• Next we shall calculate the lower quartile (Q1) which is the median of the lower half of the data set.    Q1 = ¼(n+1) = ¼(7+1) = 2
So, the lower quartile (Q1) = 20 [the second term of the ordered data set]
• Finally we shall calculate the upper quartile (Q3) which is the median of the upper half of the data set. Q3=3/4(n+1) = ¾(7+1) = 6
So, the upper quartile (Q3)= 27 [the sixth term of the ordered data set]

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Derivatives of Cot functions for some special values

We can find the derivative of the cot function using the quotient rule of differentiation. The quotient rule can be states as follows: For a function f such that f(x) = u/v, where u and v are both function of x, the derivative of f(x) is given by the formula:
f’(x) = (v*u’ – u*v’)/v^2, where u’ is derivative of function u and v’ is derivative of function v.
We know that, cot x = cos x / sin x,
Therefore, (d/dx) cot x = (d/dx) (cos x / sin x),
= [sin x (d/dx) (cos x) – cos x (d/dx) (sin x)]/sin^2 (x)  … using the quotient rule stated above.
=[ sin x (-sin x) – cos x * cos x]/sin^2(x)
= [-sin^2(x) – cos^2(x)]/sin^2(x)
= -[sin^2(x) + cos^2(x)]/sin^2(x)
= -1/sin^2(x)
= -csc^2(x)
Therefore the derivative of cot x is –csc^2(x)

The above derivative can also be proved using the definition of derivative which is like this:
F’(x) = lim(t->x) [(f(t) – f(x)]/(t-x), for the function y = f(x) = cot (x) the definition would be like this:
(d/dx) cot x = lim(t->x) [cot t – cot x]/(t-x)
= lim(t->x) [1/tan t – 1/tan x]/(t-x)
  = lim(t->x) [tan x – tan t]/[(tan t * tan x)(t-x)]
= lim(t->x) {[tan x – tan t]/(t-x)} * 1/[tan t * tan x]
= lim(t->x) {-[tan t – tan x]/(t-x)} * lim(t->x) 1/[tan t * tan x]
= lim(t->x) {-[ tan(t-x) (1+tan t tan x)]/(t-x)} * 1/tan^2(x)
We know that lim(t->x) of tan(t-x)/(t-x) = 1. Therefore,
= -{1+tan^2(x)} *1/tan^2(x)
= - sec^2(x)/tan^2(x)
= [-1/cos^2(x)]/[sin^2(x)/cos^2(x)] … here the cos^2(x) cancel each other out.
= -1/sin^2(x)
= -csc^2(x)

Derivative of cot-x:
For finding the derivative of cot(-x) we use the chain rule.
Let –x = u, then du/dx = -1.
Cot (-x) = cot u
(d/du) cot u = - csc^2(u) …  from what we proved above.
Using the chain rule we have:
(d/dx) cot (-x) = (d/dx) cot u * du/dx = -csc^2(u) * (-1)
= -csc^2(-x) * (-1)
= csc^2(-x)
Thus we see that the derivative of cot(-x) is csc^2(-x)

Derivative of cot^-x (k):
Cot^(-x) (k) where k is some constant can be written like this also
= (cot(k))^(-x)
= 1/(cot k)^x
= (1/cot k)^x
Since k is a constant, cot k is also a constant and in turn, 1/cot k is also constant. So let 1/cot k = some constant say p. Therefore,
= p^x is the function.
(d/dx) p^x = p^x ln p
= (1/cot k)^x ln(1/cot k)
That is the derivative of the above function.

Triple integrals spherical co-ordinates

The concept of definite integration can be extended to two or three or multi dimensional system. For a bounded function f(x,y,z) defined on a rectangular box B (x0 = x = x1, y0 = y = y1, z0 = z = z1), the triple integral of f over B,
Integral (B) [f(x,y,z)] dV or  Integral  (B) [f(x,y,z)] dx dy dz,
can be defined as a suitable limit of Riemann sums corresponding to partitions of B into sub boxes by planes parallel to each of the co-ordinate planes. Triple integrals over more general domains are defined by extending the functions to be zero outside the domain and integrating over a rectangular box containing the domain.
All properties of double integrals hold good for triple integrals as well. In particular, a continuous function is integrable over a closed, bounded domain. If f(x,y,z) = 1 on the domain D, then the triple integral gives the volume of D.
Volume of D =  Integral (B)  dV
Triple integrals spherical coordinates: The spherical co-ordinates related to Cartesian co-ordinates x, y and z are given by the equations:

The triple integrals in spherical coordinates relationships are illustrated in the picture below

For the co-ordinate surfaces in the spherical co-ordinates, see picture below:

The volume in spherical co ordinates would be like this:

Spherical co ordinates are suited to problems involving  spherical symmetry and, in particular, to regions bounded by spheres centered at origin, circular cones with axes along the z axis, and vertical plains containing the z axis.

Infinite and removable discontinuity

Introduction To Discontinuity : Let us first define discontinuity  , a function f(x) is said to be continuous in an interval if it is continuous at every point in the domain of f(x). If the interval is closed, say [a, b], the function f(x) is continuous at every point of the interval including the end points a and b. If the interval is open, say ]a, b[, we exclude the end points of the interval. If the function is not continuous at a point, say c, we say that the function is discontinuous at c and c is called a point of discontinuity of f(x) .The graph of the function y = f(x)= 5 for x > 0 and 1 for x = 0, when drawn, we see that the function is discontinuous as there is a jump in the graph at x = 0. If we carefully observe that the function is continuous to the left of O and also to the right of O.

In other words, we say that left hand limit is existing and also the right hand limit is existing but they are not equal. ).Now let us more understand what is discontinuity? The function f is said to be discontinuous at a point a of its domain D if it is not continuous thereat. The point a is then called the point of discontinuity of the function. The discontinuity may arise due to any of the following situations: (i) lim x -> a+ f(x) or lim x -> a- f(x) of both may not exist. (ii) lim x -> a+ f(x) as well as lim x -> a- f(x) may exist, but are unequal. (iii) lim x -> a+ f(x) as well as lim x -> a- f(x) both may exist, but either of the two or both may not be equal to f(a).

We classify the points of discontinuity according to the various situations discussed above:
1. Removable discontinuity: Now let us understand What is a Removable Discontinuity ? A function f is said to have removable discontinuity at x = a if lim x -> a- f(x) = lim x -> a+ f(x) but their common value is not equal to f(a). Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.

2. Discontinuity of the first kind: What is infinite discontinuity ? A function f is said to have a discontinuity of the first kind at x = a if lim x -> a- f(x) and lim x -> a+ f(x) both exist but are not equal. f is said to have a discontinuity of the first kind from the left at x = a if lim x -> a- f(x) exists but not equal to f(a). Discontinuity of the first kind from the right is similarly defined.

3. Discontinuity of the second kind: A function f is said to have a discontinuity of the second kind at x = a if neither lim x -> a- f(x) nor lim x-> a+ f(x) exists. F is said to have discontinuity of the second kind from the left at x = a if lim x -> a- f(x) does not exist. Similarly, if lim x -> a+ f(x) does not exist, then f is said to have discontinuity of the second kind from the right at x = a.

Know more about Free calculus help.
Let I be an open interval let c belongs to I, and let f be a function defined on I except possibly at the point c. The function f has a jump discontinuity at c if the one sided limits of f exists at c but are not equal. The function f is said to have a removable discontinuity at c if lim x -> c f(x) exists, but f(x) is neither not defined or has a value different from the limit. As a specific example, the function g defined by g(x) = x/|x| has a jump discontinuity at 0 since the left hand limit is -1 and the right hand limit is 1. A removable discontinuity can be removed by giving the function the value of the limit at the given point. The function h defined by h(x) = x sin (1/x) has a removable discontinuity at 0; assigning h(0) = 0 makes h a continuous function at 0.

Instantly calculate vertical asymptotes

An asymptote is a line that runs very close to a curve but the curve never meets the line. At infinity the curve coincides with the asymptote.  In other words, an asymptote of a curve is a line such that the distance between the curve and the line tends to zero. The asymptote gets very very close to the curve but never touches it.

Vertical asymptote:
If the above defined asymptote is vertical, that means, that if the asymptote is parallel to the y axis it is called the vertical asymptote. The general equation of the vertical asymptote is y = a, where a is any real number.

How to find vertical asymptotes:
The simplest way to find vertical asymptotes of functions is to graph them.  Consider for example that we need to find the vertical asymptote of the function: f(x) = 1/x. Let us graph this function and see what we get. To be able to graph the function we first need the table of values. So let us make a table of value of x and the corresponding values of f(x).

Now let us plot those values on a graph.

Vertical asymptotes can also be found algebraically. In case of rational functions, there is a denominator. If we equate the denominator to zero and then solve for x, the values of x that we thus get would be the vertical asymptote. Let us try to understand that with an example:

Find the vertical asymptotes of the function f(x) = (2x^2-4x+5)/(x^2-2x+1)
Solution: In such a function, we shall work only on the denominator. Here the denominator is x^2 - 2x + 1. Therefore we equate that to zero. So we have:
x^2 - 2x + 1 = 0 now we factor the left hand side
(x-1)(x-1) = 0 now use the zero product rule
(x-1) = 0, (x-1) = 0 now solve for x
X = 1. That is the equation of the vertical asymptote