Measure of Central tendency

Statistics is the study of data. Data is collected resources that is translated into a meaningful information. Data is a measured values and it can be classified into four different perspectives. तथ्याङक शाश्त्र भनेको डाटाको अध्ययन गर्ने गणितको एउटा खण्ड हो। डाटा भन्नाले संकलन गरिएको कच्चा संसाधन हो जसलाई अर्थपूर्ण सुचनाको रुपमा मा प्रशोधन गर्नु पर्ने हुन्छ । समग्रमा, डाटा भन्नाले मापन गरिएको मान हो र जसलाई चार फरक दृष्टिकोणका आधारमा वर्गीकृत गर्न सकिन्छ।

  1. Based on collection (Primary and Secondary) संग्रहमा आधारित (प्राथमिक र माध्यमिक)
  2. Based on Scale/ Measurement (Nominal, Ordinal, Interval, Ratio) मापनमा आधारित (नाम बुझाउने, क्रम बुझाउने, अन्तराल बुझाउने, अनुपात बुझाउने)
  3. Based on nature (Qualitative and Quantitative) प्रकृतिका आधारित (गुणात्मक र मात्रात्मक)
  4. Based on distribution (Individual, Discrete, Continuous) बर्गिकरण \ वितरणका आधारमा (व्यक्तिगत, खण्डित, निरन्तर)
Based on these data, there are two common types of statistics.
  1. Descriptive statistics
  2. Inferential statistics
Descriptive statistics
A statistics that collects, organize and summarize the information is called Descriptive statistics. For example bar graph and mean.
Inferential statistics
A statistics that utilize current data and predicts it for future reference, is called inferential statistics. For example hypothesis test or regression analysis.


Measure of Central tendency

Measure of central tendency लाई केन्द्रीय प्रवृत्तिको मापन भनिन्छ । यसले तथ्याङकको केन्द्रमा हुने प्रवृत्तिको एकल मान (डाटा सेटको प्रतिनिधि मान) लाई जनाउदछ जसले डाटाको सम्पूर्ण मात्रात्मक सेटको प्रतिनिधित्व गर्दछ। यस केन्द्रीय प्रवृत्तिको मापनलाई स्थान वा स्थितिको मापन पनि भनिन्छ, यसैलाई औसत मापन पनि भनिन्छ।
The Measure of central tendency is a statistic that summarizes the entire quantitative set of data in a single value (a representative value of the data set) having a tendency to concentrate somewhere in the center of the data. Therefore, the tendency of the observations to cluster in the central part of the data is called the central tendency. It measures the central location (or position) of data set. It is also known as average.

NOTE
  1. केन्द्रीय प्रवृत्तिको मापन जहिले पनि डाटा सेटको दायरा भित्र पर्दछ। The Measure of central tendency lies somewhere within the range of the data set
  2. डाटा लाई फरक अर्डरमा पुनर्व्यवस्थित गर्दा केन्द्रीय प्रवृत्तिको मापनमा परिवर्तन हुदैन । The Measure of central tendency remain unchanged by a rearrangement of the data set
The most common types of such central tendencies are:
  1. मध्ययक (Mean)
  2. मध्यिका (Median), Quartile, Decile, Percentile
  3. रीत (Mode)}



Mean

Mean is measure of central tendency that utilize each and every data to give a single best value. The arithmetic mean or simply mean is also knows as average, which is obtained by dividing the sum of all the observations by total number of observations (summed).
It is denoted by \bar{X} and define as follows.

Individual data Discrete data Continuous data
Arithmetic Mean \bar{X}=\frac{\sum{x}}{n}\bar{X}=\frac{\sum{fx}}{n}\bar{X}=\frac{\sum{fm}}{n}
Geometric Mean \bar{X}= \left (\prod x \right)^{\frac{1}{n}}\bar{X}=\left (\prod f x \right )^{\frac{1}{n}}\bar{X}=\left( \prod f m \right )^{\frac{1}{n}}
Harmonic Mean \bar{X}= \frac{n}{\sum \left( \frac{1}{x}\right )}\bar{X}=\frac{n}{\sum \left( \frac{f}{x}\right )}\bar{X}=\frac{n}{\sum \left( \frac{f}{m}\right )}
Weighted Mean \bar{X}= \frac{\sum (w.x)}{\sum w}\bar{X}= \frac{\sum (w.x)}{\sum w}\bar{X}= \frac{\sum (w.x)}{\sum w}
The common type of mean are
  1. अंकगणितिय मध्यक (AM) Arithmetic Mean
    The arithmetic mean answers the question, “if all the quantities have same value, what is the value to achieve the same total?” The answer is AM. For example, let Ram has Rs 100 and Shyam has Rs 120, then the avarage amount is AM, which is answered by
    AM =\frac{a+b}{2} =\frac{100+120}{2} =Rs 110
    In the figure below, a+b is same as AM+AM.
  2. ज्यामितीय मध्यक (GM) Geometric Mean
    The geometric mean answers the question, “if all the quantities have same value, what is the value to achieve the same product?”. The geometric mean is a useful when we expect changes in data in percentages as rate of change or ratios. It is utilised in the field of finance for the purpose of determining average growth rates, which are also known as the compounded annual growth rate. For example, let Ram deposited Rs 100 in a bank, on which 80% growth in first year and 25% growth in second year, then the average profit is GM, which is answered by
    GM =\sqrt{1.80 \times 1.25}=1.50, the average growth is 50%
    Please note that, the situation can NOT be explained by \frac{80+25}{2} =52.5\%
    In the figure below, a*b is same as GM*GM.
  3. हार्मोनिक मध्यक (HM)Harmonic Mean
    Harmonic Mean is used to calculate average speeds of various distances covered.For example, Let Ram traveled 100km with fuel efficiency 25KM per liter and next 100km with fuel efficiency 16KM per liter, then the average fuel efficiency is HM, which is answered by
    HM =\frac{2*25*16}{25+16}=19.51
    Please note that, the situation can NOT be explained by AM or GM
    Because
    The full efficiency for first 100 km is \frac{100}{25}=4 liter
    The full efficiency for second 100 km is \frac{100}{16}=6.25 liter
    The total fuel efficiency is
    \frac{200}{4+6.25}=19.51
  4. भारित मध्यक (WM)Weighted Mean}
    A weighted mean is a kind of average where some data points contribute more “weight” than others. If all the weights are equal, then the weighted mean equals the arithmetic mean.



Application of Mean
The mean is calculated from all data value, so it is affected by each and every value of data set. It is applicable if the data distribution represents
  1. Quantitative data
  2. Closed ended
  3. Normally distributed data



Relation between AM, GM and HM

Let a and b are two non-negative numbers, then

  1. GM^2=AM \times HM
  2. AM \ge GM \ge HM Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,
AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}
The proof are as follows:

  1. Now, we have
    GM^2=ab
    or GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b}
    or GM^2=AM \times HM
  2. Now, we have
    AM-GM=\frac{a+b}{2}-\sqrt{ab}
    or AM-GM=\frac{a+b-2\sqrt{ab}}{2}
    or AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}
    or AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}
    or AM\ge GM (1)
    Similarly,
    GM-HM=\sqrt{ab}-\frac{2ab}{a+b}
    or GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}
    or GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}
    or GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} )
    or GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}
    or GM\ge HM(2)
    Combining (1) and (2), we get
    AM\ge GM\ge HM
Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

  1. Visualization of AM
    By the property of radius and diameter, we get that
    AM =\frac{a+b}{2}

  2. Visualization of GM
    By the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
    GM =\sqrt{ab}

  3. Visualization of HM
    By using proportionality, we get
    Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
    \frac{GM}{a}=\frac{b}{GM}
    or GM= \sqrt{ab}
    Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
    HM =\frac{2ab}{a+b}
    By using proportionality, we get
    Triangle DRQ and ODQ are similar with QR=GM,QD=\sqrt{ab}, OD=\frac{a-b}{2}, so we have
    \frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}}
    or HM= \frac{2ab}{a+b}




Example 1

Find the mean of the numbers 3, −7, 5, 13, −2
The sum of the numbers is
\sum X= 3 − 7 + 5 + 13 − 2 = 12
There are 5 numbers, so n=5.
Hence, the mean of the numbers is
\bar{X}=\frac{\sum X}{n}=\frac{12}{5}=2.4




Example 2
Find the mean of the wages from the following data
Wages507090110130150
Number of Workers245621
Based on the data given above, the frequency table is prepared as below.
Wages (X) Number of workers (f) f.x
50 2 100
70 4 280
90 5 450
110 6 660
130 2 260
150 1 150
\sum f=n=20 \sum f x=1900
Based on the formula, the mean wages is
\bar{X}=\frac{\sum fx}{n}=\frac{1900}{20}=95




Example 3
Find the average marks from the following data
Marks of the Students0-2020-4040-6060-8080-100
Number of Students2050554015
Based on the data given above, the frequency table is prepared as below.
Marks of students (X) Mid value of marks m Number of students (f) f.m
0-20 10 20 200
20-40 30 50 1500
40-60 50 55 2750
60-80 70 40 2800
80-100 90 15 1350
\sum f=n=180 \sum fm=8600
Based on the formula, the average marks is
\bar{X}=\frac{\sum fx}{n}=\frac{8600}{180}=47.8




Median

Median is a measure of central tendency that utilize middle portion of the data to give a single best value. The median is the middle value of the data series when the values are placed in order of magnitude (in ascending or descending order). Therefore, Median is not affected by extreme values. It is denoted by Md and define as follows.

Individual Discrete Continuous
Median M_d=\frac{n+1}{2} \text{th item} M_d=\frac{n+1}{2} \text{th item} M_{d-class}=\frac{n+1}{2} \text{th item}
with M_d=L+\frac{\frac{N}{2}-cf}{f} \times i
Calculating the median is also very simple. Here are the steps:
  1. Sort the data in an ascending order.
  2. Find the middle number of the sorted data.
  3. If there’s an odd number of data, get the value exactly in the middle.
  4. If there’s an even number of data, get the mean of the two middle values.
Application of Median: The median doesn’t know how far the data is. It only help to split data in two parts. It is applicable if the distribution represents
  1. Qualitative data
  2. Open ended or Skewed data



Example 1

Find the median of the following wages(in hundreds): 40,30,35,42,32,45,48
Given wages (in hundreds) are
40,30,35,42,32,45,48
Arranging the wages in ascending order, we get

30323540424548
1st item2nd item3rd item4th item5th item6th item7th item

Here, the number of data are n=7, thus, based on the formula, the Median is
M_d= \left (\frac{n+1}{2} \right )^{th} item
or M_d= \left (\frac{7+1}{2} \right )^{th} item
or M_d= 4^th item
or M_d= 40 hundreds

Example 2
Find the median of the wages from the following data
Wages507090110130150
Number of Workers245621
Based on the data given above, the frequency table is prepared as below.
Wages X Number of Workers f Cumulative frequency cf
50 2 2
70 4 6
90 5 11
110 6 17
130 2 19
150 1 20
Here, the number of data are n=20 , thus, based on the formula, the Median is
M_d= \left (\frac{n+1}{2} \right )^{th} item
or M_d= \left (\frac{20+1}{2} \right )^{th} item
or M_d= 10.5^th item
or M_d= 90
Example 3
Find the median marks from the following data
Marks of the Students0-2020-4040-6060-8080-100
Number of Students2050554015
Based on the data given above, the frequency table is prepared as below.
Marks of the Students X Number of Students f Cumulative frequency cf
0-20 20 20
20-40 50 70
40-60 55 125
60-80 40 165
80-100 15 180

Here, the number of data are n=180 , thus, based on the formula, the Median class is
Md Class= \left (\frac{n}{2} \right )^{th} item
or Md Class= \left (\frac{180}{2} \right )^{th} item
or Md Class= 90^th item
Here, 90^th item lies in the cf of 125, thus
L=40,f=55, cf=70,i=20
Hence, the Median is
M_d=L+\frac{\frac{N}{2}-cf}{f} \times i
or M_d=40+\frac{\frac{180}{2}-70}{55} \times 20=47.27

Example 4
Find the median marks from the following data
Marks of the Students0-2020-4040-6060-8080-100
Number of Students23546
Based on the data given above, the frequency table is prepared as below. \
Marks of the Students X Number of Students f Cumulative frequency cf
0-20 2 2
20-40 3 5
40-60 5 10
60-80 4 14
80-100 6 20

Here, the number of data are n=20, thus, based on the formula, the Median class is
Md Class= \left (\frac{n}{2} \right )^{th} item
or Md Class = \left (\frac{20}{2} \right )^{th} item
or Md Class = 10^th item
Here, 10^th item lies in the cf of 10, thus
L=40,f=5, cf=5,i=20
Hence, the Median is
M_d=L+\frac{\frac{N}{2}-cf}{f} \times i
or M_d=40+\frac{\frac{20}{2}-5}{5} \times 20=60
NOTE
In the example above, student may ask that the median 60 does not lie in the class 40-60 as instructed for inclusive data groupings, teaches need to encourage the usual rules for computing.




Quartile, Decile and Percentile

The formula for Quartile, Decile and Percentile are similar as of Median.
Individual Discrete Continuous
Quartile
k=1,2,3
Q_k=\frac{k(n+1)}{4} \text{th item} Q_k=\frac{k(n+1)}{4} \text{th item} Q_{k-class}=\frac{k(n)}{4} \text{th item}
with Q_k=L+\frac{\frac{kn}{4}-cf}{f} \times i
Decile
k=1,2,\cdots 9
D_k=\frac{k(n+1)}{4} \text{th item} D_k=\frac{k(n+1)}{4} \text{th item} D_{k-class}=\frac{k(n)}{4} \text{th item}
with D_k=L+\frac{\frac{kn}{4}-cf}{f} \times i
Percentile
k=1,2,\cdots 99
P_k=\frac{k(n+1)}{4} \text{th item} P_k=\frac{k(n+1)}{4} \text{th item} P_{k-class}=\frac{k(n)}{4} \text{th item}
with P_k=L+\frac{\frac{kn}{4}-cf}{f} \times i



Mode

The concept of mode, as a measure of central tendency, is preferable when it is desired to know the most typical value, e.g., the most common size of shoes, the most common size of a ready-made garment, the most common size of income, the most common size of pocket expenditure of a college student, the most common size of a family in a locality, the most common duration of cure of viral-fever, the most popular candidate in an election, etc.
Thus, Mode is a measure of central tendency that utilize fashionable (most repeated data) information to give a single best value. So, Mode is an average value which occurs most frequently in a set of data i.e. it indicates the most frequent (common) results. It is not affected by every values. It is denoted by Mo and define as follows.

Individual Discrete Continuous
Mode Repeated data Repeated data/ Table analysis M_0=L+\frac{f_1-f_0}{2f_1-f_0-f_2} \times i
M_0=L+\frac{f_2}{f_0+f_2} \times i
Application of Mode: The mode doesn’t know anything about any number in the collection but the one which appears most frequently. It is best applicable when concerning about
  1. Frequency related data
  2. Fashionable data
Example 1
Find the mode value of the following data: 3, 7, 5, 13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29
Given data set are
3, 7, 5, 13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29
In frequency table, the data set becomes
X3 5 7 12 13 14 20 23 29 39 40 56
f1 1 1 1 1 1 1 4 1 1 1 1
Being highest frequency 4, the mode value is 23.
Example 2
Find the Mode of the wages from the following data\par
Wages507090110130150
Number of Workers245621
Being highest frequency 6, the mode value is 110.
Example 3
Find the Mode of from the following data
Wages0-1010-2020-3030-4040-5050-6060-70
Number of Workers4121518321413

Being highest frequency 4, the model class is 40-50. Thus,
L=40,f_0=18,f_1=32,f_2=14,i=10
Hence, using formula, the Mode is
M_0=L+\frac{f_1-f_0}{2f_1-f_0-f_2} \times i
or M_0=40+\frac{32-18}{2 \times 32-18-14} \times 10=44.47

Analytical method to find the Mode

If the frequency distribution is regular, then mode is determined by the value corresponding to maximum frequency. There may be a situation where frequency distribution is NOT regular, means the concentration of observations around a value having maximum frequency is less than the concentration of observations around some other value. In such a situation, mode cannot be determined by the use of maximum frequency criterion. Further, there may be concentration of observations around more than one value of the variable and, accordingly, the distribution is said to be bi-modal or multi-modal depending upon whether it is around two or more than two values. In such cases, we use analytical method (also called tabular or grouping or empirical method) to find the Mode.
दिएको श्रेणिमा Mode अस्पष्ट भएमा वा तलका निम्न अवस्थामा यो बिधीको प्रयोग गरिन्छ ।

  1. highest frequency सख्या एक भन्दा बढी भएमा
  2. highest frequency तथ्याङकको सुरु वा अन्यतिर भएमाा
  3. highest frequency को वरिपरि ठुला frequency भएमाा
  4. frequency को अनियमित घटबढ भएमाा
यस अवस्थामा Mode पत्ता लगाउन Empirical Method (Mode=3 Median -2 Mean) वा analytical method प्रयोग गर्न सकिन्छ । तर यि दुबै बिधीमध्ये analytical method लाई बढी बिश्वासनिय मानिन्छ।
Example 4
Find the Mode of from the following data
Wages102030405060708090
Number of Workers1517222120934

Here, the maximum frequency is 22, however three are big frequencies around 22, thus we use analytical method to find the Mode.
Hence, based on the rule, the analytic table is given as below.

Wages f1st + 2nd2nd+ 3rd1st+2nd+3rd 2nd+3rd+4th3rd+4th+5th
10 1
6
20 5 23
22
30 17 44
39
40 22 60
43
50 21 63
41
60 20 50
29
70 9 32
12
80 3 16
7
90 4
  1. Prepare a table consisting of 7 column, 1st column for X, 2nd column for frequencies of X.
  2. In third column, add the frequencies, starting from the top and grouped in twos.
  3. In forth column, add the frequencies, starting from the second and grouped in twos.
  4. In fifth column, add the frequencies, starting from the top and grouped in threes .
  5. In sixth column, add frequencies, starting from the top second and grouped in threes.
  6. In seventh column, add the frequencies, starting from the top third and grouped in threes.
  7. Finally, prepare frequency chart based on the analytic table
Based on the analytic table, the frequency chart is prepared as below.
Column102030405060708090
11
211
311
4111
5111
6111
Total24531
Here, the highest frequency is aligned with 50, therefore, Mode=50.


Relation between Mean Median and Mode

A distribution in which the values of mean, median and mode coincide (i.e. mean = median = mode) is known as a symmetrical distribution.
Conversely, when values of mean, median and mode are not equal the distribution is known as asymmetrical or skewed distribution. In moderately skewed or asymmetrical distribution, a very important relationship exists among these three measures of central tendency. In such distributions
Mode = 3 Median – 2 Mean

Symmetrical Distribution

Sequence and Series

In mathematics, the word, “sequence” we usually mean a ordered collection with an identified first member, second member, third member and so on. is also called progression (AP).

Introduction

Sequence is a pattern of ordered numbers. Since the numbers follow a pattern, we can relate each number to its numerical position with a rule. Such a rule is called sequence. Each number in a sequence is a term of the sequence.

Presumably, a sequence continues by following the pattern that first few “terms” suggest. To understand the pattern more explicitely, it is also useful to think of a sequence as a function.
Thus,
So, a real-valued function defined on N = {1, 2, 3, . . .} is a sequence.
The range of the function is real numbers, the term of sequences
Mathematically, it is written as
f : N \to R .

A sequence can be defined with two different ways

  1. Recursive definition (Syntactic definition), Implicit definition
  2. Formal definition (Semantic definition), Explicit definition
Recursive definition

A sequence can be described by comparing each term to the one that comes before it, i.e., by defining the later term in relationto previous terms. Such rule of describing sequence is called recursive definition. The recursive definition contains two parts. twwo parts, the initial condition and definition. For example, in a sequence
133,130,127,124,…,
each term after the first term is equal to three less than the previous term.
Therefore, a recursive definition for this sequence is as follows

  1. an initial condition (the value of the first term): a1=133
  2. a recursive definition (relates each term after the first term to the one before it): an=an-1-3 for n>1
Example 1
Writing a recursive definition for a sequence

The number of blocks in two dimensional pyramid is a sequence that follows a recursive formula. What is the recursive definition of the sequence?

The solution is as follows.

  • Let us count the number of blocks in each pyramid: it is 1,3,6,10,….
  • Now, subtract consecutive terms to find out what happens from one term to the next
    a2-a1=3-1=2
    a 3-a2=6-3=3
    a 4-a3=10-6=4
  • Now, use n to express the relationship between successive terms
    a n-an-1=n
  • To write the recursive definition, state the initial condition and the recursive formula
    a 1=1 and an=an-1+n
Example 2

What is the 100th term of the pyramid sequence in the example given below?

Solution
To find the explicit formula, expand the first few terms of the pyramid sequence. which is as below.

a1 a2 a3 a4 an
1 3 6 10 an
1 1+2 1+2+3 1+2+3+4 1+=2+3+4+…+n

Therefore,
a n=1+2+3+4+—+(n-2)+(n-1)+n (1)
Writing in reverse order, we get
a n=n+(n-1)+(n-2)+…+4+3+2+1 (2)
Adding (1) and (2), we get

an =1 +2 3 +—+ (n-2) +(n-1) +n
an =n +(n-1) +(n-2) +—+ +3 +2 +1
2an =(n+1) +(n+1) +(n+1) +—+ +(n+1) +(n+1) +(n+1)
2an =n(n+1)
an =\frac{n(n+1)}{2}

Therefore, the explicit formula, for this sequence is
a n=\frac{n(n+1)}{2}
Now, we substitute n by 100 to find the 100th term. Which is
a n=\frac{n(n+1)}{2}
or a 100=\frac{100(100+1)}{2}
or a 100=5050
Now, we define sequence mathematically.

Semantic Definition

A sequence is a function defined on the set \mathbb{N}. For example, f(an)=3an. There are different types of such sequences. Among them we discuss three basic types of sequence in the following section.

  1. Arithmetic sequence
  2. Geometric sequence
  3. Harmonic sequence



Arithmetic Sequence

An arithmetic sequence is a list of numbers with a definite pattern based on addition. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence.

In the arithemetic sequence
The constant difference in all pairs of consecutive or successive numbers is called the common difference. It is denoted by the letter d.
Please note that
Difference here means the second minus the first. We add the common difference to go from one term to another.

Therefore,
An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
For example,
the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with common difference of 2.

General term of arithmetic sequence

An arithmetic sequence with a starting value a and common difference d is a sequence of the form
a,a+d,a+2d,a+3d,\cdots
A recursive definition for this sequence has two parts
t_1=a: initial condition
t_n=t_{n-1}+d for n>1:recursive formula
Therefore, an semantic definition for this sequence is a single formula as general terms given following calulations
t_n=a+(n-1)d for n > 1

Proof.

t_1=a
t_2=a+(d)
or t_2=a+(2-1)d)
t_3=a+(2-1)d+d
or t_3=a+(3-1)d)
t_4=a+(3-1)d+d
or t_4=a+(4-1)d)
Similarly,
t_n=a+(n-1)d for n > 1

Arithmetic mean

Let a, b, and c are three terms in an arithmetic sequence, then b is called Arithemetic mean of a and c.
In this case
, first two terms a and b will have the difference which will be equal to the next two terms b and c.
So we can Thus,
b-a = c-b.
Rearranging the terms, we get
or 2b = a + c
or b =  \frac{a+c}{2}

Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence, denoted by S_n
To find the explicit formula for the Sn: Sum of arithmetic sequence up to n th term, we write

S_n=a+(a+d)+(a+2d)+… +[a+(n-2)d]+[a+(n-1)d]
S_n=[a+(n-1)d]+[a+(n-2)d]+… +(a+2d)+(a+d)+a
2S_n=[2a+(n-1)d]+[2a+(n-1)d]+…+… +[2a+(n-1)d]+[2a+(n-1)d]
Thus

2S_n=n[2a+(n-1)d])
or S_n=\frac{n}{2}[2a+(n-1)d]

Example 3

If the first term of an AP is 67 and the common difference is -13, find the sum of the first 20 terms.
Here, a = 67 and d= -13, thus
Sn = \frac{n}{2}[2a+(n-1)d]
or Sn = \frac{20}{2}[2 \times 67+(20-1)(-13)]=-1130
So, the sum of first 20 terms is -1130.

Example 4

[Modeling]:Look at the figure below. The length of the side of each cube is 1cm. Copy the figure in isometric dot paper Based on the pattern

  1. Draw Figure no. 4 of this pattern on the dot paper
  2. Find the volume of the four Figures
  3. What would be the volume of Figure no. 12 of this pattern
  4. Write an equation to represent the volume of Figure n
Example 5

Show that sum of first n odd natural number is Sn=n^2
Here, a=1,d=2, thus the sum is
Sn=\frac{n}{2} [2a+(n-1)d]
or Sn=\frac{n}{2} [2.1+(n-1)2]
or Sn=\frac{n}{2} [2+2n-2]
or Sn=n^2

Example 6

Show that sum of first n even natural number is Sn=n(n+1)
Here, a=2,d=2, thus the sum is
Sn=\frac{n}{2} [2a+(n-1)d]
or Sn=\frac{n}{2} [2.2+(n-1)2]
or Sn=\frac{n}{2} [4+2n-2]
or Sn=n(n+1)




Geometric sequence

Euclid’s book The Elements (300 BC, Book VIII) introduces a “geometric progression” as a progression in which the ratio of any element to the previous element is a constant.
The Greeks, over two thousand years ago, considered sequences such as
\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},....,\frac{1}{2^k},...
and their sums, such as
\displaystyle \sum_{k=1}^{k=\infty} \frac{1}{2^k}
The sum, above, at any finite step, is always less than the number 1.
Since the sum is less than 1 at any finite step , we can conclude that the series converges in the limit to 1

Does a series have a sum?
\sum_{n=1}^\infty \frac{1}{2^n}
Answer
\sum_{n=1}^\infty \frac{1}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...=1
The sum of the series is 1.
The visualization is as follows
Imagine that we paint a blank canvas in steps. At each step, we paint half of the unpainted area. The total area painted after “n” steps is therefore the “n”th partial sum
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+ \frac{1}{2^n}
The total area remaining unpainted is
\frac{1}{2^n}
After an infinite number of steps we will have painted all of the canvas, of which the area is 1.

Geometric sequence

An geometric sequence is a list of numbers with a definite pattern based on multiplication. If you take any number in the sequence then divide it by the previous one, and the result is always the same or constant then it is an geometric sequence.

In the geometric sequence,
The constant ratio in all pairs of consecutive or successive numbers in a sequence is called the common ratio. It is denoted by a letter r.
Please note that
Ratio here means the second divide the first. We use the common ratio to go from one term to another.

Therefore, a geometric sequence is a sequence of numbers in which the ratio between the consecutive terms is always constant. For instance,
the sequence 5, 15, 45,135, . . . is an geometric progression with common ratio of 3.

General term of geometric sequence

A geometric sequence with a starting value a and common ratio r is a sequence of the form
a,ar,ar^2,ar^3,...
A recursive definition for this sequence has two parts
t_1=a: initial condition
t_n=t_{n-1}r for n>1 recursive formula
Therefore, an explicit definition for this sequence is a single formula as general terms given by
t_n=ar^{n-1}for n > 1

Example 1

When a ball bounces as shown below, the height of consecutive bounces become a geometric sequence. If a_1=100,a_3=49, thn what is the height of 5th bounce?

The height of the first and third bounces are given in the figure above.

Solution
a_1=100,a_3=49
Use the explicit formula to relate a1 to a3 and to find r.
or an=a_1 r^{n-1}
or a_3=a_1 r^{3-1}
or 49=100 r^2
or r=\frac{7}{10}
Find the fifth using explicit formula, thus we have
a_5=a_1 r^4
or a_5=100 \times \left ( \frac{7}{10} \right ) ^4 \approx 24

Geometric mean

In the geometric sequence, if a, b and c are three consecutive terms, then b is called geometric mean of a and b.
In this sequence, the first two terms a and b will have the ratio which will be equal to the next two terms b and c.
So we can say,
\frac{b}{a}=\frac{c}{b}
Rearranging the terms, we get
b^2 = a \times c
or b = \sqrt{ac}

Geometric Series

A geometric series is the sum of the terms of an geometric sequence, denoted by S_n
Let a geometric sequence is given by
a,ar,ar^2,ar^3,...
If we are adding up the terms of the geometric sequence given above, then we have a geometric sum or geometric series given by
\displaystyle \sum_{k=0}^{k=\infty} ar^k

Method 1

The explicit formula for the sum of finite terms is given as below
Sum of geometric sequence up to n th term: a,ar,ar^2,ar^3,...ar^{n-1} is written as
Sn=a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]
Taking a common, we get
Sn=a[1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]
Multiplying by \frac{1-r}{1-r} , we get
Sn=a \frac{1-r}{1-r} [1+r+r^2+r^3+---+r^{n-2}+r^{n-1}]
orS_n=\frac{a(r^n-1)}{r-1}

Method 2

Sum of geometric sequence up to n th term: a,ar,ar^2,ar^3,...ar^{n-1} is written as
Sn=a+(a r)+(a r^2)+(a r^3)+---+[a r^{n-2}]+[a r^{n-1}]
Multiplying by r, we get
rSn=(a r)+(a r^2)+(a r^3)+---+[a r^{n-1}]+[a r^n]
Substracting , we get
rSn-Sn=ar^n-a
orS_n=\frac{a(r^n-1)}{r-1}

Infinite Sum

We know that, sum of the first n terms of a geometric series with first term a and common ratio r is
S_n=\frac{a(r^n-1)}{r-1} ; r \ne 1
In the case when r has magnitude less than 1, the term r^n approaches 0 as n becomes very large. So, in this case, the sequence of partial sums S1,S2,S3,... has a limit:
So,
S_\infty= \displaystyle \lim_{n \to \infty} \frac{a(r^n-1)}{r-1} = \frac{a}{1-r}
The limiting sum is usually referred to as the sum to infinity of the series and denoted by S∞. Thus, for a geometric series with common ratio r such that |r | < 1, we have
S_\infty= \frac{a}{1-r}

Example 2

A person saves NRs100 in a bank account at the beginning of each month. The bank offers a return of 12% compounded monthly.
(a) Determine the total amount saved after 12 months.
Solution
Let
Sn: Sum of geometric sequence up to n th term.
It is written as
Sn=\frac{a(1-r^n)}{1-r}
Based on the formula, the solution of above problem with
a = 100, r=1.12, n = 12
The solution is
S_{12}=\frac{a(r^n-1)}{r-1}= 1268.25

Example 3

What is the sum of the finite geometric series 3+6+12+24+...+3072
Given that, the first term is 3, the common ratio is 2, and the nth term is 3072, therefore, we use explicit formula to find the n

an=a r^{n-1}
or 3072=3 \times 2^{n-1}
or 1024=2^{n-1}
or n=11
Therefore, the sum up to 11th term is
S_n= \frac{a(1-r^n)}{1-r}
or S_{11}= \frac{3(1-2^{11})}{1-2}
or S_{11}= 6141 


Harmonic sequence

In mathematics, a harmonic sequence formed by taking the reciprocals of an arithmetic progression.
The sequence 1,2,3,4,5,6,... is an arithmetic progression, so its reciprocals
\frac{1}{1}, \frac{1}{2},             \frac{1}{3}, \frac{1}{4}, ...
is a harmonic sequence.

If each term of an harmonic sequence is multiplied or divide by a constant, the sequence of the resulting number are also harmonic sequence. For example,

if a,b,c,d, \cdots is harmonic sequence, then
  1. then ka,kb,kc,kd, \cdots is harmonic sequence where k is constant and k \ne 0
  2. \frac{a}{k},\frac{b}{k},\frac{c}{k},\frac{d}{k},\cdots is a harmonic sequence where k \ne 0
General term of Harmonic sequence

Let a,a+d,a+2d,a+3d, \cdots are the terms of arithemetic sequence
Then
terms of Harmonic sequence are given by
\frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d}+\frac{1}{a+3d},\cdots
Now,
The nth term of the sequence is given by
a_n=\frac{1}{a+(n-1)d}

Harmonic Mean

Harmonic mean is finding taking the reciprocal of the arithmetic mean of the reciprocals. The formula to calculate the harmonic mean is given by:
Harmonic Mean =\frac{2}{(\frac{1}{a})+(\frac{1}{b})}=\frac{2ab}{a+b}
Where a,b are the values of arithmetic sequence.

Sum of Harmonic Series

If \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d} is given harmonic progression, the formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]
where,
“a” is the first term of A.P
“d” is the common difference of A.P
“ln” is the natural logarithm

Proof
Given \frac{1}{a},\frac{1}{a+d},\frac{1}{a+2d},…,\frac{1}{a+(n-1)d} are in harmonic progression,
We set the values as follows,
a=1,d=1
Then, \frac{1}{1},\frac{1}{1+1},\frac{1}{1+2.1},…,\frac{1}{1+(n-1)1} are the terms
1,\frac{1}{2},\frac{1}{3}, ...,are the terms
Now, the Riemann sum of the function f(x)=\frac{1}{x} approximates the sum of the harmonic series given above
Therefore,
We set the values as follows,
S_n= \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx
For any common difference,the formula becoms
S_n=\frac{1}{d} \int_{a+(0-\frac{1}{2})d}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx
or S_n= \frac{1}{d} \int_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d} \frac{1}{x} dx
or S_n= \frac{1}{d} \left [ \ln x \right]_{a-\frac{d}{2}}^{a+(n-\frac{1}{2})d}
or S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]
Therefore The formula to find the sum of n terms in the harmonic progression is given by the formula:
Sum of nth terms, S_n= \frac{1}{d} \ln \left [ \frac{2a+(2n-1)d}{2a-d} \right ]




A.M, G.M, H.M and their relations

A finite sequence consisting more than two terms has one or more terms in between the first and last terms. Theses between terms are called means of the sequence. Precisely

  1. if a,b,c,d are in arithmetic sequence then b,c are arithmetic means (AM)
    So, given n terms, the AM is
    AM=\frac{a+b+ \cdots}{n}
  2. item if a,b,c,d are in geometric sequence then b,c are geometric means (GM)
    So, given n terms, the GM is
    GM=\sqrt[n]{a \times \times \cdots }
  3. if a,b,c,d are in harmonic sequence then b,c are harmonic means (HM)
    So, given n terms, the HM is
    HM=\frac{n}{(\frac{1}{a})+(\frac{1}{b})+ \cdots}
Theorem 1

Given any two numbers a and b the AM, GM, and HM are as follows.

  1. AM=\frac{a+b}{2}
  2. GM=\sqrt{ab}
  3. HM=\frac{2ab}{a+b}

The proof are as follows:

  1. Let AM is the single mean between a and b, then
    AM-a=b-AM
    or 2AM=a+b
    or AM=\frac{a+b}{2}
  2. Let GM is the single mean between a and b, then
    \frac{GM}{a}=\frac{b}{GM}
    or GM^2=ab
    or GM=\sqrt{ab}
  3. Let HM is the single mean between a and b, then
    \frac{1}{HM}-\frac{1}{a}=\frac{1}{b}-\frac{1}{HM}
    or \frac{2}{HM}=\frac{1}{a}+\frac{1}{b}
    or HM=\frac{2ab}{a+b}
Theorem 2

Let a and b are two non-negative numbers, then

  1. GM^2=AM \times HM
  2. AM \ge GM \ge HM Arithmetic mean is greater than geometric mean and harmonic mean, and geometric mean is greater than harmonic mean.

Let a and b are two non-negative numbers then,
AM=\frac{a+b}{2}, GM=\sqrt{ab}, HM=\frac{2ab}{a+b}
The proof are as follows:

  1. Now, we have
    GM^2=ab
    or GM^2=\frac{a+b}{2} \times \frac{2ab}{a+b}
    or GM^2=AM \times HM
  2. Now, we have
    AM-GM=\frac{a+b}{2}-\sqrt{ab}
    or AM-GM=\frac{a+b-2\sqrt{ab}}{2}
    or AM-GM=\frac{{{\sqrt{a}}^{2}}+{{\sqrt{b}}^{2}}-2\sqrt{a}\sqrt{b}}{2}
    or AM-GM=\frac{{{( \sqrt{a}-\sqrt{b} )}^{2}}}{2}
    or AM\ge GM (1)
    Similarly,
    GM-HM=\sqrt{ab}-\frac{2ab}{a+b}
    or GM-HM=\frac{\sqrt{ab}( a+b )-2ab}{a+b}
    or GM-HM=\frac{\sqrt{ab}( a+b )-2\sqrt{ab}\sqrt{ab}}{a+b}
    or GM-HM=\frac{\sqrt{ab}}{a+b}( a+b-2\sqrt{ab} )
    or GM-HM=\frac{\sqrt{ab}}{a+b}{{( \sqrt{a}-\sqrt{b} )}^{2}}
    or GM\ge HM(2)
    Combining (1) and (2), we get
    AM\ge GM\ge HM
Visualization of the proof

Let us suppose that a and b are two given numbers. Now, draw a semi circle with diameter a+b.

  1. Visualization of AM
    By the property of radius and diameter, we get that
    AM =\frac{a+b}{2}

  2. Visualization of GM
    By the mean proportionality property (squaring a rectangle), we can obtain by using the property of similarity that, DQ is the geometric mean given by
    GM =\sqrt{ab}

  3. Visualization of HM
    By using proportionality, we get
    Triangle ADQ and QDB are similar with AD=a, DB=b, so we have
    \frac{GM}{a}=\frac{b}{GM}
    or GM= \sqrt{ab}
    Again, by using the property of similarity on OCDE, we get that, QR is the harmonic mean given by
    HM =\frac{2ab}{a+b}
    By using proportionality, we get
    Triangle DRQ and ODQ are similar with QR=GM,QD=\sqrt{ab}, OD=\frac{a-b}{2}, so we have
    \frac{HM}{\sqrt{ab}}=\frac{\sqrt{ab}}{\frac{a+b}{2}}
    or HM= \frac{2ab}{a+b}




Solved Example

Compare the sum of n terms of the series: 1+2a+3a^2+4a^3+........ . and a+2a+3a+ 4a ... up to n term
Solution
Given that
S_n=1+2a+3a^2+4a^3+\cdots+na^{n-1}
T_n=a+2a+3a+ 4a +\cdots+na
Now, substracting, we get
S_n-T_n= (1+2a+3a^2+4a^3+\cdots+na^{n-1})-(a+2a+3a+ 4a +\cdots+na)
or S_n-T_n= (1+2a+3a^2-a-2a-3a)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)
or S_n-T_n= (1+3a^2-a-3a)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)
or S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)

Case 1: If a=1, then
S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)
S_n-T_n= (0)+(0)+(0)+\cdots+(0)
or S_n-T_n= 0
or S_n =T_n (1)
Case 1: If a >1, then
S_n-T_n= (a-1)(3a-1)+4a(a^2-1)+5a(a^3-1)+\cdots + na(a^{n-2}-1)
S_n-T_n= (+ve)+(+ve)+(+ve)+\cdots+(+ve)
or S_n-T_n= Positive
or S_n -T_n>0
or S_n >T_n= 0 (2)

Thus, the comparision is
S_n = T_nif a =1
S_n > T_nif a > 1




Exercise

  1. Integration to Investment and Growth: Determine the monthly repayments needed to repay a 100000 loan which is paid back over 25 years when the interest rate is 8% compounded annually[Ans: 780.66]
  2. Show that if three quantities form any two of the three sequences AS,GS,and HS then they also form the remaining third sequence
  3. The sum of three numbers in AP is 36. When the numbers are increased by 1,4,43 respectively, the resulting numbers are in GP, find the numbers.
  4. Divide 69 in three parts in AP and the product of first two parts is 483
  5. If a be the AM between b and c, b be the GM between c and a, then prove that c will be HM between a and b.
  6. Show that b^2 is greater than, equal to, or less than according as a,b,c are in AP, GP or HP
  7. If a and b are two positive numbers then prove that AM,GM,HM form GP.
  8. If the 4^{th} and 15^{th} terms of an arithmetic series are 11 and 44 respectively then, find the sum of its first 20 terms. (Ans:610)
  9. Puja gets an employment of Rs.25000 in a month with an annual increment of Rs. 1500. How much does she earn in six years? Find her salary at 12^{th} year. (Ans: Rs.190500, Rs.41500)
  10. In a geometric series, if the sixth term is 16 times the second term and the sum of first seven terms is \frac{127}{4} , then find the sum of the first 15 term. (Ans: \frac{32767}{4})
  11. There are 8 bags full of books. The numbers of books in each bag form a GP. If the 4^{th} and 6^{th} bags contain 24 and 96 books respectively, find the number of the books in the 1^{th} and last bags. ( Ans: 3, 384)
  12. One side fo a square is 10 cm . The mid point of its sides are joined to form another square, whose mid point are again joined t form one more square. The process is continued indefinitely. Find the sum of the areas of all the squares so formed. (Ans:200 square cm)



Exercise: Miscellaneous Exercise [NCERT, pg 147]

  1. If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and \displaystyle \sum_{x=1}^{n} f(x)=120,find the value of n.

    Solution 👉 Click Here

  2. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
  3. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
  4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

    Solution 👉 Click Here

  5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

    Solution 👉 Click Here

  6. If \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}(x≠ 0), then show that a, b, c and d are in G.P.

    Solution 👉 Click Here

  7. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P^2R^n = S^n.

    Solution 👉 Click Here

  8. If a, b, c, d are in GP, prove that (a^n + b^n), (b^n + c^n), (c^n + d^n) are in GP

    Solution 👉 Click Here

  9. If a and b are the roots of x^2 - 3x + p = 0 and c, d are roots of x^2 - 12x + q = 0, where a, b, c, d form a GP. Prove that (q + p) : (q – p) = 17:15.

    Solution 👉 Click Here

  10. The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a:b=\left ( m+\sqrt{m^2-n^2} \right ) : \left ( m-\sqrt{m^2-n^2} \right )

    Solution 👉 Click Here

  11. Find the sum of the following series up to n terms:
    (i) 5 + 55 +555 + …
    (ii) 0.6 +0.66 +0.666+…
  12. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ... + n terms.

    Solution 👉 Click Here

  13. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
  14. Bidhan buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

    Solution 👉 Click Here

  15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

    Solution 👉 Click Here

  16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
  17. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
  18. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

    Solution 👉 Click Here




EXERCISE 8.2 [NCERT, page 145]

  1. Find the 20th and nth terms of the G.P. \frac{5}{2},\frac{5}{4},\frac{5}{8},...

    Solution 👉 Click Here

  2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

    Solution 👉 Click Here

  3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q^2 = ps.

    Solution 👉 Click Here

  4. The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.

    Solution 👉 Click Here

  5. Which term of the following sequences:
    1. 2,2\sqrt{2},4,... is 128 ?

      Solution 👉 Click Here

    2. \sqrt{3},3,3\sqrt{3},... is 729 ?

      Solution 👉 Click Here

    3. \frac{1}{3},\frac{1}{9},\frac{1}{27},... is \frac{1}{19683}

      Solution 👉 Click Here

  6. For what values of x, the numbers \frac{2}{7},x,\frac{7}{2} are in G.P.?

    Solution 👉 Click Here


  7. Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

  8. 0.15, 0.015, 0.0015, ... 20 terms

    Solution 👉 Click Here

  9. \sqrt{7},\sqrt{21},3\sqrt{7},... n terms

    Solution 👉 Click Here

  10. 1,-a,a^2,-a^3,... n terms (if a≠-1)

    Solution 👉 Click Here

  11. x^3,x^5,x^7,... n terms (if n \ne \pm 1)

    Solution 👉 Click Here

  12. Evaluate \displaystyle \sum_{k=1}^{11} (2+3^k)

    Solution 👉 Click Here

  13. The sum of first three terms of a G.P. is \frac{39}{10} and their product is 1. Find the common ratio and the terms.
  14. How many terms of G.P. 3, 3^2, 3^3, … are needed to give the sum 120?
  15. The sum of first three terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
  16. Given a G.P. with a = 729 and 7th term 64, determine S_7.
  17. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
  18. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
  19. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
  20. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2,\frac{1}{2}
  21. Show that the products of the corresponding terms of the sequences a, ar, ar^2,…ar^{n - 1} and A, AR, AR^2, … AR^{n - 1} form a G.P, and find the common ratio.
  22. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
  23. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that a^{q - r} b^{r - p}c^{p - q} = 1.

    Solution 👉 Click Here

  24. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that p^2 = (ab)^n.

    Solution 👉 Click Here

  25. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)^{th} to (2n)^{th} term is \frac{1}{r^n}
  26. If a, b, c and d are in G.P. show that (a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2 .
  27. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
  28. Find the value of n so that \frac{a^{n+1}+b^{n+1}}{a^n+b^n} may be the geometric mean between a and b.
  29. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \left ( 3+2\sqrt{2} \right ): \left ( 3-2\sqrt{2}\right )
  30. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A \pm \sqrt{(A+G)(A-G)}
  31. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour ?
  32. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
  33. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.



Linear Programming




LPP is a technique of mathematical modeling to solve real life problems whose requirements are given in linear relations. It helps to find “best” solution (minimum and maximum) under given conditions.
It is a type of optimization problem that determines the feasible region (a region that contains all the possible solutions of an LPP) and optimizes the solution to get the maximum or minimum function value.
In simple terms, a linear programming problem is a set of linear equations that are used to find the value of variables to optimize the objective functions. LPP can be solved in two different ways.

LPP (linear programming problem) वास्तविक जीवनका समस्याहरू समाधान गर्ने एक गणितीय मोडलिङ हो जसको शर्तहरु रेखिय सम्बन्धमा दिएको हुन्छ । यसले दिइएको अवस्थाहरूमा “उत्तम” समाधान (न्यूनतम वा अधिकतम) को बारेमा थाहा पाउन मद्दत गर्दछ।
यो एक प्रकारको अप्टिमाइजेसन समस्या हो जसले सम्भाव्य क्षेत्र (सबै सम्भावित समाधानहरू समावेश गर्ने क्षेत्र) निर्धारण गर्छ र अधिकतम वा न्यूनतम मान प्राप्त गर्न मद्दत गर्छ ।
समान्य भाडषामा, LPP (linear programming problem) भनेको रेखिय समीकरणहरूको समुह हो जसबाट “उत्तम” समाधान (न्यूनतम वा अधिकतम) को खोज गरिन्छ।




Linear Programming Applications_Example 1

Let us take a real-life problem to understand linear programming.
A company produces two types of pots X and Y by using copper and steel. Type X pot requires 300 g of copper and 100 g of steel while Y requires 100 gram of copper and 200 grams of steel. The X pot brings profit of Rs. 400 and the Y pot brings profit of Rs. 500. Find the number of units of each type of pots that the company should produce with 5kg copper and 12kg of steel to achieve maximum profit.
Given the situation, let us take up different scenarios to analyse how the profit can be maximized.

  1. The company can decide to produces X type pots, in this case, he can create 5000/100= 50 copper slot, 12000/200=60 steel slot. This would give him to produce 50 pots with a profit of Rs 500×50 = Rs 12500.
  2. The company can decide to produces Y type pots, in this case, he can create 5000/300= 16 copper slot, 12000/100=120 steel slot. This would give him to produce 16 pots with a profit of Rs 400×14 = Rs 6400.
  3. Similarly, there can be many strategies which the company can devise to maximize his profit by allocating the available row materials to the two types of pots. We do a mathematical formulation of the discussed LPP to find out the strategy which would lead to maximum profit.



Linear Programming Applications_Example 2

A home decorator company received an order to manufacture tables. The first consignment requires up to 50 tables. There are two types of tables, The first type requires 15 hours of the labour force (per piece) to be constructed and gives a profit of Rs 5000 per piece to the company. Whereas, the second type requires 9 hours of the labour force and makes a profit of Rs 3000 per piece. However, the company has only 540 hours of workforce available for the manufacture of the tables. With this information given, you are required to find a deal which gives the maximum profit to the company.

Given the situation, let us take up different scenarios to analyse how the profit can be maximized.
  1. He decides to construct all the tablesof the first type. In this case, he can create 540/15 = 36 tables. This would give him a profit of Rs 5000 × 36 = Rs 180,000.
  2. He decides to construct all the tables of the second type. In this case, he can create 540/9 = 60 tables. But the first consignment requires only up to 50 cabinets. Hence, he can make profit of Rs 3000 × 50 = Rs 150,000.
  3. He decides to make 15 tables of type 1 and 35 of type 2. In this case, his profit is (5000 × 15 + 3000 × 35) Rs 180,000.

Similarly, there can be many strategies which he can devise to maximize his profit by allocating the different amount of labour force to the two types of tables. We do a mathematical formulation of the discussed LPP to find out the strategy which would lead to maximum profit.




Basic terminologies of LPP

  1. Decision Variable: These are quantities to be determined.
  2. Objective Function: The function that need to optimized.
  3. Constraints: Linear equations that represents decision variables on how to use the available resource (for example, amount of time, number of people, etc.)
  4. Feasible Solution: Set all the possible solutions that satisfy the given constants.
  5. Optimal Solution: It is the best possible solution among all the possible feasible solutions.
  6. Slack Variables : Slack variable represents an unused quaintly of resources ; it is added to less than or equal (<) to type constraints in order to get an equality constraint.
  7. Surplus Variables : A surplus variable represents the amount by which solution values exceed a resource. These variables are also called ‘Negative Slack Variables’. Surplus variables carry a zero coefficient in the objective function. it is added to greater than or equal to (>) type constraints in order to get an equality constraint.



Characteristics of LPP

  1. Decision variables will decide the output.
  2. The objective function should be specified quantitatively.
  3. Constraints (limitations) should be expressed in mathematical form.
  4. Relationships between two or more variables should be linear.
  5. The values of the variables should always be non-negative or zero.

Methods of Solving LPP

LPP बिभिन्न तरिकाबाट हल गर्न सकिन्छ।
  1. Graphical method
  2. Simplex method
  3. North West Corner Method
  4. Least Square Methods



Graphical Methods of Solving LPP

LPP models can be solved by graphical method if two variable are presented in the model.
The general process while solving LPP by graphical methods, are given below
  1. graph the inequalities (constraints)
  2. form walled-off region (feasible region)
  3. find the corner points in feasible region
  4. test corner points in the object function
  5. find the solution (maximum or minimum)



Simplex Methods of Solving LPP

LPP models can be solved by simplex method if two or more variable are presented in the model.
The general process while solving LPP by simplex methods, are given below
  1. Convert subjective function with ≤ inequality (standard form)
  2. Introduce slack variables to convert inequality into equation
  3. Prepare simplex table
  4. Find pivot element [column (negative, lowest), Find pivot row (positive, lowest)]
  5. Convert pivot element to 1 and else to zero
  6. If optimal solution is NOT reached, iterate the process from 4 onwards






A company produces two types of pots X and Y by using copper and steel. Type X pot requires 300 g of copper and 100 g of steel while Y requires 100 gram of copper and 200 grams of steel. The X pot brings profit of Rs. 400 and the Y pot brings profit of Rs. 500. Find the number of units of each type of pots that the company should produce with 5kg copper and 12kg of steel to achieve maximum profit.

Solution

  1. The decision variables: Since the question has asked for an optimum number of pots, that’s what our decision variables is. Let us say
    x = number of X pots produced
    y = number of Y pots produced
  2. The constraints: Since the company can’t produce a negative number of pots, a natural constraint is:
    x ≥ 0
    y ≥ 0
  3. The subjective function
    pots copper steel Profit
    X 300 100 400
    Y 100 200 500
    Availability 5000 12000 ?
    Given
    Each unit of X requires 300 unit of copper and 100 units of steel
    Each unit of Y requires 100 unit of copper and 200 units of steel
    The company has a total of 5 kg (5000 g) of copper and 12 kg (12000 g) of steel.
    Thus, the subjective functions are
    300x+100y≤5000
    100x+200y≤12000
  4. The objective function
    We need to optimize the Profit. On each sale, the company makes a profit of
    Rs 400 per unit X sold.
    Rs 500 per unit Y sold.
    The Profit Function is:
    Z =400x + 500y
  5. LPP modeling
    Therefore, the LPP model is (in 100) is
    Maximize Z = 4x + 5y, subject to:
    3x+1y≤50
    1x+2y≤120
    x ≥ 0
    y ≥ 0

Solution

  1. Step 1: graph the inequalities (constraints)
    We plot the following inequalities in a graph.
    1. 2x + y \le 100
      Let us take a table for plot points
      x050
      y1000
      The graph is given below.
    2. x + y \le 80
      Let us take a table for plot points
      x080
      y800
      The graph is shown with a label.
    3. x \ge 0,x \ge 0
      The graph is shown with a label.
  2. Step 2: form a walled-off region (feasible region)
    In this LPP, the feasible region is a polygon OABC labeled in the graph.
  3. Step 3: find the corner points in the feasible region
    In this LPP, the corner points of polygon OABC are O = (0,0), A = (16.5,0), D= (0,50)
    NOTE: The corner point are obtained by solving two equations, which do intersect
  4. Step 4: test corner points in the object function
    Objective function Point Value Remarks
    Z=4x+5y (0,0) Z=4\times 0+5\times 0=0 Minimum
    Z=4x+5y (16.5,0) Z=4\times 16.5+5\times 0=66
    Z=4x+5y (0,50) Z=4\times 0+5\times 50=250 Maximum
  5. Step 5: find the solution (maximum or minimum)
    The maximum value of the function Z = 4x + 5y is 250 at the point (0,50)
    The minimum value of the function Z = 4x + 5y is 0 at the point (0,0).
    The graphical solution of this modeling can be shown below.



In the LPP, the graphical method is very useful in lower dimensions. If we have more than two variables then we’re looking at some higher dimensional shape, then we can’t interpret anything easily in three dimensions visually. Therefore, the simplex method gives an algorithm for solving these problems with any number of variables. Graphical method can be applied in 2D. In 3D and higher dimensions+, identifying an optimal solution using the graphical method is no longer feasible.So, simplex method can be applied to 1D, 2D, 3D and 3D+ linear programs. In other words, simplex method can be used for theoretically unlimited amounts of optimization variables

The general process for solving the maximization LPP model by the simplex method is given below

  1. Convert subjective function with \le inequality (standard form)
  2. Introduce slack variables to convert inequality into an equation
  3. Prepare simplex table
  4. Find the pivot column (negative, lowest)
  5. Find pivot row (positive, lowest)
  6. Find pivot element
  7. Convert pivot element to 1 and else to zero
  8. Iterate the process from 4 onwards

Test your understandings

  1. Why do we make an objective function with a negative coefficient?
    The initial simplex table represents an augmented matrix, where slack variables form an identity matrix.

    In an LPP model
    Maximize z = 400x + 500y, subject to:
    300x+100y \le 5000
    100X+200y \le 12000

    x y u v z
    300 100 1 0 0 5000
    100 200 0 1 0 12000
    -400 -500 0 0 1 0

    the objective function is z=400x+500y, which is in the linear form written as z-400x-500y=0 in order to produce a unit matrix as shown in the table above

  2. Why do we choose the most negative entry in the bottom row?

    We know that the simplex method begins at a corner point where all the main variables are zero. It then moves from a corner point to the adjacent corner point always increasing the value of the objective function.
    when we choose the most negative entry in the bottom row, we are trying to increase the value of the objective function ,
    for example, the coefficient of y (which is 500) , this entry will increase the value of the objective function the quickest

  3. Why do we find quotients, and why does the smallest quotient identify a row?
    When we choose the most negative entry in the bottom row, we are trying to increase the value of the objective function by bringing in the variable, say y [in the example below]. But we cannot choose any value for y. Therefore, using the lowest quotients guarantees that we do not violate the constraints and that the effect is high to increase the value of the objective function.
    x y u v z ratio
    300 100 1 0 0 5000 5000/100=50
    100 200 0 1 0 12000 12000/200=60
    -400 -500 0 0 1 0
    For example,

    in a model, Maximize z = 400x + 500y, subject to:
    300𝑥+100𝑦 \le 5000
    100x+200𝑦 \le 12000
    the smallest ratio is 50, which ensures the validity of the constraints

  4. Why do we identify the pivot element?
    The simplex method begins with a corner point and then moves to the next corner point to improve the value of the objective function. Therefore, the value of the objective function is improved by changing the number of units of the variables. We may add the number of units of one variable while throwing away the units of another. Pivoting allows us to do just that and help us to utilize available resources
    x y u v z ratio
    300 100 1 0 0 5000 5000/100=50
    100 200 0 1 0 12000 12000/200=60
    -400 -500 0 0 1 0

    For example,

    in a model, Maximize z = 400x + 500y, subject to:
    300𝑥+100𝑦 \le 5000
    100x+200𝑦 \le 12000

    the value of the objective function is improved by changing the number 100[a12] to 1 and else to zero
  5. Why are we finished when there are no negative entries in the bottom row?
    Since all variables are non-negative, the highest value that can be achieved is guaranteed.

Example 1

Find the maximum value of the following LPP problem.
Maximize: Z = 50x + 18y
Subject to: 2x + y \le 100, x + y \le 80, x \ge 0 , y\ge 0

Example 2

Find the maximum value of the following LPP problem.
Maximize: Z = 25x + 40y
Subject to: 2x + y \le 10, x + 2y \le 6, x \ge 0 , y\ge 0

Solution

  1. Step 1: Since all the subjective function are with \le inequality, Prepare a simplex Tableaux
    x y u v
    2 1 1 0 10
    1 2 0 1 6
    -25 -40 0 0 0
  2. Step 2: Find the pivot column, pivot row, and pivot element.
    (a) Largest negative entry is -40, thus C_2 is the pivot column.
    (b) Smallest positive ratio is 3, thus R_2 is the pivot row.
    (c) Based on (a) and (b), 2 is the pivot element.
    x y u v ratio
    2 1 1 0 10 \frac{10}{1}=10
    1 2 0 1 6 \frac{6}{2}=3
    -25 -40 0 0 0
  3. Step 3: Operating Row equivalent operation, convert pivot element 2 to 1, and else to 0.
    x y u v Operation Stage
    3/2 0 1 -1/2 7 R_1 \approx R_1-R_2 2
    1/2 1 0 1/2 3 R_2 \approx \frac{1}{2} R_2 1
    -5 0 0 20 120 R_3 \approx R_3+40 R_2 3
  4. Step 4: Since a negative entry is found in row 3, again iterate the process to find the pivot column, pivot row, and pivot element.
    (a) Largest negative entry is -5, thus C_1 is the pivot column.
    (b) Smallest positive ratio is 14/3, thus R_1 is the pivot row.
    (c) Based on (a) and (b), 3/2 is the pivot element.
    x y u v ratio
    3/2 0 1 -1/2 7 \frac{7}{3/2}=14/3
    1/2 1 0 1/2 3 \frac{3}{1/2}=6
    -5 0 0 20 120
  5. Step 5: Operating Row equivalent operation, convert pivot element 3/2 to 1, and else to 0.
    x y u v Operation Stage
    1 0 2/3 -1/3 14/3 R_1 \approx \frac{2}{3}R_1 1
    0 1 -1/3 2/3 2/3 R_2 \approx R_2-\frac{1}{2} R_1 2
    0 0 10/3 55/3 430 R_3 \approx R_3+5 R_1 3
  6. Step 6: Since all entries in row 3 are non-negative, the solution is achieved. Here, the maximum value is
    Z=\frac{430}{3} at x=14/3,y=2/3

Question

Find the maximum value of the following LPP problem.
Maximize: P=x+2y+z
Subject to: x-y+2z\le 10, 2x+y+3z \le 12, x \ge 0 , y\ge 0,z\ge 0

Complex Number

Introduction

A complex number is an extended version of real number in the form
x + iy; 𝑥∈ℝ
Euler (1707 – 1783) introduced the imaginary unit ‘i’ (read as iota) for √-1 with property
i^2+ 1 = 0
Therefore, imaginary unit i is the solution of an equation
x^2+ 1 = 0

Acomplex is written in the STANDARD form as z = x+iy where x and y are real numbers

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Note

  1. The cartesian form of complex number can be written z=x+iy [standard form] or z=(x,y) [order pair form]
  2. In z = (x,y,), x is real part denoted as Re (z), and y is imaginary part denoted as Im (z)
    In z = (x,y), Re (z) = x and Im (z) =y
  3. A complex number z = x+iy, is purely real if y = 0 i.e. Re (z) = 0, and purely imaginary if x = 0 i.e. Im (z) = 0.
  4. A complex number z = x+iy, is zero if x = y = 0 i.e. Re (z) = Im (z) = 0

The introduction of complex numbers gives rise to the fundamental theorem of algebra. In the 16th century, Italian mathematician Gerolamo Cardano used complex numbers to find solutions to cubic equations.

Then Italian mathematician Rafael Bombelli developed the rules for addition, subtraction, multiplication, and division of complex numbers. A more abstract formalism for complex numbers was developed by the Irish mathematician William Rowan Hamilton.




Meaning of i

In the complex number system, i is called imaginary unit. Tthe value of i is (0, 1). Thus, we can write i=(0,1).

If we expand a complex number z=(x,y) as z=(1,0)x+(0,1)y then (1,0) is unit of real part denoted by 1 and (0,1) is unit of imaginary part and denoted by i. Here i is imaginary unit with
i^2=-1.
A complex number is visually represented in Argand diagram. Here i is operator giving anticlockwise quarter turn such that i^2=-1.

NOTE

  1. We should NOT mean i as non-existence number nor a number that exist only in imagination
  2. i is a number that denotes imaginary unit (0,1)
  3. i is an anticlockwise quarter turn operator for (x,y), thus
    i^2=(0,1) (0,1) = i (0,1) =(-1,0) =-1 (1,0)=-1

Positive powers of i

In general, for any integer k,
i^{4k} = 1, i^{4k+1} = i, i^{4k+2} = -1, i^{4k+3} = -i.
For example:
i^{23} = i^{4.5+3} = (i^4)5. i ^3= 1^5. (-i) = -i.

The verification on the positive powers of i are as follows.
i^2 = -1
i^3 = i2. I = (-1).i = - i
i^4 = (i^2)^2 = (-1)^2 = 1
i^5 = i^{4 + 1}= i^4. i = 1. i = i
i^6 = i^{4 + 2} = i^4. i^2= 1. 1 = 1 and so on




Absolute Value

In a complex number
z= x + iy
The absolute value is Modulus, a non-negative real number denoted by
| z | and defined by
|𝑧|=\sqrt{x^2+y^2}
Geometrically, 
|𝑧| is distance of z from origin

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Note
Due to order property, z1 <z2 is meaningless unless z1 and z2 both are real.
However,| z1 | <| z2 | means zis closer than z2.
Distance between z_1 =x_1 +iy_1 and z_2 =x_2 +iy_2 is denoted by
| z_1 -z_2 |
and defined by
| z_1 -z_2 |=\sqrt{( x_1 -x_2 )^2 +( y_1 -y_2 )^2 }

NOTE:
( x+iy )^2 =( x+iy )( x-iy )




Conjugate

The conjugate of a complex number z=x+iy is denoted \bar{z} and defined by
\bar{z}=x-iy
Geometrically, \bar{z} is reflection of z on real axis

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Properties of Conjugate numbers

  1. If z_1 =x_1 +iy_1 and z_2 =x_2 +iy_2 then \overline{z_1 +z_2 }=\bar{z_1} +\bar{z_2}
    Proof
    \overline{z_1 +z_2 }=\overline{( x_1 +iy_1 )+( x_2 +iy_2 )}
    or \overline{z_1 +z_2 }=\overline{( x_1 +x_2 )+i( y_1 +y_2 )}
    or \overline{z_1 +z_2 }=( x_1 +x_2 )-i( y_1 +y_2 )
    or \overline{z_1 +z_2 }=( x_1 -iy_1 )+( x_2 -iy_2 )
    or \overline{z_1 +z_2 }=\bar{z}_1 +\bar{z}_2
  2. The sum of a complex number z and its conjugate \bar{z} is twice of its real part
    Proof
    If z=x+iy be a complex number then \bar{z}=x-iy , where
    z+\bar{z}=2x
    or x=\frac{z+\bar{z}}{2}
    or Rez=\frac{z+\bar{z}}{2}
  3. The difference of z and its conjugate \bar{z} is twice of its imaginary part
    Proof
    If z=x+iy be a complex number then \bar{z}=x-iy , where
    z-\bar{z}=2iy
    or y=\frac{z-\bar{z}}{2i}
    or Imz=\frac{z-\bar{z}}{2i}
  4. The real and imaginary parts of a complex number can be extracted using its conjugate

Theorem: An important property

Let z=x+iy be a complex number then z.\bar{z}=|z|^2

Proof
Given z=x+iy be a complex number, then \bar{z}=xiy .
Now,
z.\bar{z}=( x+iy )( x-iy )
or z.\bar{z}=( x )^2 -( iy )^2
or z.\bar{z}=x^2 +y^2 (i)
Also
| z |=\sqrt{x^2 +y^2}
or |z|^2 =x^2 +y^2  (ii)
From (i) and (ii), we get
z.\bar{z}=|z|^2

Question

Theorem: Triangle inequality: | z_1 +z_2 |\le | z_1 |+| z_2 |

Proof>br> We know that
| z_1 +z_2 |^2 =( z_1 +z_2 )( \overline{z_1 +z_2 } )
or | z_1 +z_2 |^2 =( z_1 +z_2 )( \bar{z}_1 +\bar{z}_2 )
or | z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +z_2 \bar{z}_1 +z_2 \bar{z}_2 )
or | z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\bar{\bar{z}}_2 \bar{z}_1 +z_2 \bar{z}_2 )
or | z_1 +z_2 |^2 =( z_1 \bar{z}_1 +z_1 \bar{z}_2 +\overline{z_1 \bar{z}_2 }+z_2 \bar{z}_2 )
or | z_1 +z_2 |^2 =z_1 \bar{z}_1 +2Re( z_1 \bar{z}_2 )+z_2 \bar{z}_2
or | z_1 +z_2 |^2 \le z_1 \bar{z}_1 +2| z_1 \bar{z}_2 |+z_2 \bar{z}_2
or | z_1 +z_2 |^2 =|z_1|^2 +2| z_1 || \bar{z}_2 |+|z_2|^2
or | z_1 +z_2 |^2 =( | z_1 |+| z_2 | )^2
or | z_1 +z_2 |\le | z_1 |+| z_2 |
This completes the proof.

Corollary

  1. | z_1 +z_2 |\ge | z_1 |-| z_2 |


    Proof
    or| z_1 |=| z_1 +z_2 +( -z_2 ) |
    or \le | z_1 +z_2 |+| -z_2 |
    or =| z_1 +z_2 |+| z_2 |
    Thus, | z_1 +z_2 |\ge | z_1 |-| z_2 |
  2. | z_1 -z_2 |\le | z_1 |+| z_2 |


    Proof
    | z_1 +z_2 | \le | z_1 |+| z_2 |
    Now, replacing z_2 by -z_2 we get
    | z_1 -z_2 |\le | z_1 |+| -z_2 |
    Thus, | z_1 -z_2 |\le | z_1 |+| z_2 |
  3. | z_1 -z_2 |\ge | z_1 |-| z_2 |


    Proof
    | z_1 +z_2 | \ge | z_1 |-| z_2 |
    Now, replacing z_2 by -z_2 we get
    | z_1 -z_2 |\ge | z_1 |-| -z_2 |
    Thus, | z_1 -z_2 |\ge | z_1 |-| z_2 |



Algebra of complex number

Complex plane looks like an ordinary two-dimensional plane of z=( x,y ), but z=( x,y ) is a single number losing order axioms. Fundamental operations on complex number are defined as below.

  1. Equality
    Two complex numbers z_1 =x_1 +iy_1 and z_2 =x_2 +iy_2 are equal if
    x_1 =x_2 \text{ and } y_1 =y_2
  2. Addition
    Sum of two complex numbersz_1 =x_1 +iy_1 and z_2 =x_2 +iy_2 is defined as
    z_1 +z_2 =( x_1 +iy_1 )+( x_2 +iy_2 )=( x_1 +x_2 )+i( y_1 +y_2 )
    According to definition,z_1 +z_2 corresponds to resultant vector addition.
  3. Multiplication
    Product/multiplication of two complex numbers z_1 =x_1 +iy_1 and z_2 =x_2 +iy_2 is defined as
    z_1 .z_2 =( x_1 +iy_1 ).( x_1 +iy_2 )
    or z_1 .z_2 =x_1 x_2 +ix_1 y_2 +ix_2 y_1 +{{i}^{2}}y_1 y_2
    or z_1 .z_2 =( x_1 x_2 -y_1 y_2 )+i( x_1 y_2 +x_2 y_1 )
    The product is neither scalar nor the vector product of ordinary vector analysis. This departure is due to i^2=-1

Also, complex number C is a field. Thus, the complex number satisfy all Field axioms as below.

  1. Closer: \forall a,b \in G \implies a \ast b \in G
  2. Additive associative
    \forall a,b,c \in G \implies (a+ b) + c=a + (b + c)
  3. Additive identity
    \forall a \in G \implies a \ast e =e \ast a =a
    Additive identity (0,0)
  4. Additive Inverse
    \forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e
    Additive inverse of z=x+iy is -z=-x-iy
  5. Additive commutative
    \forall a,b \in G \implies a \ast b =b \ast a
  6. Distributive
    \forall a,b,c \in G \implies (a + b)\bullet c =a c+bc
  7. Multiplicative associative
    \forall a,b,c \in G \implies (a. b) . c=a . (b . c)
  8. Multiplicative identity
    \forall a \in G \implies a \ast e =e \ast a =a
    Multiplicative identity (1,0)
  9. Multiplicative inverse
    \forall a \in G \implies a \ast (a^{-1}) =(a^{-1}) \ast a =e
    Multiplicative inverse of non-zero complex number z=x+iy is
    z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}
    Proof
    Let z=x+iy be non-zero complex number and z^{-1} =u+iv be its multiplicative inverse, then
    zz^{-1} =( 1,0 )
    or ( x+iy )( u+iv )=( 1,0 )
    or ( xu-yv )+i( xv+uy )=( 1,0 )
    Comparing real and imaginary parts separately, we get
    ux-vy=1 and uy+vx=0
    Solving for u and v , we get
    u=\frac{x}{x^2 +y^2} and v=\frac{-y}{x^2 +y^2}

    Hence,
    z^{-1} =\frac{x}{x^2 +y^2} +i\frac{-y}{x^2 +y^2}
  10. Multiplicative commutative
    \forall a,b \in G \implies a \ast b =b \ast a

Algebric Structure

Axioms Structure Example
1-2 Semi Group Example
1-3 Monoid Example
1-4 Group Example
1-5 Abelian Group Example
1-6 Ring Example
1-7 Assocoative Ring Example
1-8 Assocoative Ring with Unity Example
1-9 Division Ring (Skew Field) Example
1-10 Field Example



Polar form

Drag the point for intaraction : 👇

Let z=x+iy be a complex number with magnitude r and amplitude \theta , then
x=r\cos \theta and y=r\sin \theta
Hence,
z=x+iy=r\cos \theta +ir \sin \theta
or z=r( \cos \theta +i\sin \theta )
Thus, a complex number z=x+iy is represented by polar coordinate (r,\theta) , as
z=r \cos \theta +i\sin \theta
Here,
r is the length of z , and \theta is argument of z with
r=| z |=\sqrt{x^2 +y^2} , and \theta =argz=\tan^{-1}( \frac{y}{x} )

Example

Find polar form of complex number z=-5+5i

Solution
Given that
z=-5+5i
or z=5( -1+i)
or z=5\sqrt{2}( -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i )
or z=5\sqrt{2}( \cos \frac{3\pi }{4}+\sin \frac{3\pi }{4}i )

The relevance of complex number in polar form is that multiplication and division are simpler with this form than the Cartesian form.

Let z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 ) and z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 ) be two complex numbers, then
z_1 z_2 =r_1 r_2 (\cos (\theta_1 +\theta_2 )+i\sin (\theta_1 +\theta_2 ))
Proof
Given that z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 ) and z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 ) . Thus
z_1 z_2 = r_1 (\cos \theta_1 +i\sin \theta_1 ) \times r_2 (\cos \theta_2 +i\sin \theta_2 )
or z_1 z_2 = r_1 r_2 (\cos \theta_1 +i\sin \theta_1 )(\cos \theta_2 +i\sin \theta_2 )
or z_1 z_2 = r_1 r_2 [\cos \theta_1 (\cos \theta_2 +i\sin \theta_2 )+i\sin \theta_1 (\cos \theta_2 +i\sin \theta_2 )
or z_1 z_2 = r_1 r_2 [\cos \theta_1 \cos \theta_2 +i\cos \theta_1\sin \theta_2 +i\sin \theta_1\cos \theta_2 -\cos \theta_1\sin \theta_2 ]
or z_1 z_2 = r_1 r_2 [(\cos \theta_1 \cos \theta_2-\cos \theta_1\sin \theta_2 ) +i(\cos \theta_1\sin \theta_2 +\sin \theta_1\cos \theta_2) ]
or z_1 z_2 = r_1 r_2 [\cos (\theta_1 +\theta_2) +i \sin (\theta_1+ \theta_2)]

Theorem

arg( z_1 z_2 )=arg z_1 +argz_2

Proof

Let z_1 =r_1 ( \cos \theta_1 +i \sin \theta_1 ) and z_2 =r_2 ( \cos \theta_2 +i\sin \theta_2 ) then
z_1 .z_2=r_1 .r_2 [ \cos ( \theta_1 +\theta_2 )+i\sin ( \theta_1 +\theta_2 ) ]
Thus,
arg( z_1 z_2 )=argz_1 +argz_2

Note:
Any complex number z has infinite arguments; all differ by multiple of 2\pi . The principal value is in the interval [ -\pi ,\pi ]

Some important property

  1. Let z_1 \text{ and } z_2 be two complex number then arg( z_1 .z_2 )=argz_1 +argz_2
    The argument of product of two complex number is sum of their arguments.
    Proof
    Let z_1 =r_1 e^{ i\theta_1} and z_2 =r_2 e^{ i\theta_2} then
    z_1 z_2 =( r_1 e^{ i\theta_1} )( r_2 e^{ i\theta_2} )
    or z_1 z_2 =( r_1 r_2 )e^{i( \theta_1 +\theta_2 )}
    Hence,
    arg( z_1 .z_2 )=\theta_1 +\theta_2 =argz_1 +argz_2
  2. Letz_1 \text{ and }z_2 be two complex number then,
    arg( \frac{z_1 }{z_2 } )=argz_1 -argz_2

    The argument of quotient of two complex number is difference of their arguments.
  3. Argument of complex number of the form z=a+i.0, a > 0 is 0
  4. Argument of complex number of the form z=a+i.0, a < 0 is \pi
  5. Argument of complex number of the form z=0+i.b, b > 0 is \frac{\pi }{2}
  6. Argument of complex number of the form z=0+i.b, b < 0 is -\frac{\pi }{2}

Drag the point for intaraction : 👇

If z =r( \cos \theta +i\sin \theta ) then show that z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )

Question

Let z=r( \cos \theta +i\sin \theta ) be a complex number and its inverse is z^{-1} =R( \cos \phi +i\sin \phi ) then
zz^{-1} =1
or r( \cos \theta +i\sin \theta )R( \cos \phi +i\sin \phi )=( 1,0 )
or rR\{ \cos ( \theta +\phi )+i\sin ( \theta +\phi ) \} =( 1,0 )
or rR\cos ( \theta +\phi )=1 \text{ and } rR\sin ( \theta +\phi )=0
or R=\frac{1}{r},\phi =-\theta
Thus,
z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )




Exponenrial Form

Euler’s formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. It states that for any real number x:
e^{i \theta}=\cos \theta+i\sin \theta
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions.

When x = π, Euler’s formula evaluates to e^{i\pi}+1=0, which is known as Euler’s identity.

Given Euler’s exponential form,
e^{i\theta}=\cos \theta +i\sin \theta
Thus, complex number z=r( \cos \theta +i\sin \theta ) is defined as
z=re^{i\theta}
The significance of exponential form of complex number is that we can easily compute conjugate and inverse.
For example,
\bar{z}=re^{- i\theta} and z^{-1} =\frac{1}{r} e^{- i\theta}

Proof of the Formula

  1. Function Method

    Consider the function f(θ) given by
    f(\theta )=\frac {\cos \theta +i\sin \theta }{e^{i\theta }}
    orf(\theta )= e^{- i\theta } ( \cos \theta +i\sin \theta)
    Differentiating gives by the product rule, we have
    f' (\theta )=0
    Thus, f(θ) is a constant.
    Since f(0) = 1, then f(θ) = 1 for all θ, and thus
    1= e^{- i\theta } ( \cos \theta +i\sin \theta)
    ore^{ i\theta }= \cos \theta +i\sin \theta
    This completes the proof.

  2. Series Method

    We know that
    e^x=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}
    \cos x= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}-
    \sin x= x-\frac{x^3}{3!} +\frac{x^5}{5!}-...
    Using power series expansion, we get
    e^{ i\theta }=                 \sum_{n=0}^\infty \frac{(i\theta)^n}{n!}
    or e^{ i\theta }=                 1+ \frac{(i\theta)^1}{1!}+ \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!}+ \frac{(i\theta)^4}{4!}+...
    or e^{ i\theta }=                 1+ i\theta- \frac{\theta^2}{2!} -i \frac{\theta^3}{3!}+ \frac{\theta^4}{4!}+...
    or e^{ i\theta }=                 \left ( 1- \frac{\theta^2}{2!}+ \frac{\theta^4}{4!}-...\right ) +i \left (\theta-\frac{\theta^3}{3!} +... \right )
    or e^{ i\theta }                 = \cos \theta +i\sin \theta
    This completes the proof.




De-Moivre’s theorem

Let z=r \cos \theta +i\sin \theta be a complex number then z^n =r^n (\cos n\theta +i\sin n\theta ) where n is a positive integer.
Proof

  1. Case 1: n=1
    Then
    z =r (\cos \theta +i\sin \theta )
    or z^1 =r^1 (\cos 1.\theta +i\sin 1.\theta )
    So,
    or z^n =r^n (\cos n\theta +i\sin n\theta ) when n=1
  2. Case 2: n=2
    z^2 =z.z
    or z^2 = r (\cos \theta +i\sin \theta ) \times r (\cos \theta +i\sin \theta )
    or z^2 = r^2 (\cos \theta +i\sin \theta )(\cos \theta +i\sin \theta )
    or z^2 = r^2 [\cos \theta (\cos \theta +i\sin \theta )+i\sin \theta (\cos \theta +i\sin \theta )
    or z^2 = r^2 [\cos \theta \cos \theta +i\cos \theta\sin \theta +i\sin \theta\cos \theta -\cos \theta\sin \theta ]
    or z^2 = r^2 [(\cos \theta \cos \theta-\sin \theta \sin \theta ) +i(\cos \theta\sin \theta +\sin \theta\cos \theta) ]
    or z^2 = r^2 [\cos (\theta +\theta) +i \sin (\theta+ \theta)]
    or z^2 = r^2 [\cos 2\theta +i \sin 2\theta]
    So,
    or z^n =r^n (\cos n\theta +i\sin n\theta ) when n=2
  3. Case 3: We assume the same formula is true for n = k, so we have
    (\cos\theta + i\sin\theta)^k = r^k(\cos(k\theta) + i\sin(k\theta))
    So,
    or z^n =r^n (\cos n\theta +i\sin n\theta ) when n=k
  4. Case 4: Now, we prove for n = k + 1,
    [r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos\theta + i\sin\theta)^k r (\cos\theta + i\sin\theta)
    or [r(\cos\theta + i\sin\theta)]^{k + 1} = r^k(\cos(k\theta) + i\sin(k\theta)) r(\cos\theta + i\sin\theta)
    or [r(\cos\theta + i\sin\theta)]^{k + 1} = r^{k+1}[(\cos(k\theta) \cos\theta-\sin(k\theta)\sin\theta )+i (\cos\theta\sin(k\theta) + \sin\theta\cos(k\theta))]
    or [r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k\theta+\theta)+i \sin(k\theta+\theta )]
    or [r(\cos\theta + i\sin\theta)]^{k + 1} =r^{k+1}[\cos(k+1)\theta+i \sin(k+1)\theta]
    So,
    or z^n =r^n (\cos n\theta +i\sin n\theta ) when n=k+1
  5. Using case 1-case 4, for any number n \in Z , we have
    [r(\cos\theta + i\sin\theta)]^n =r^n[\cos(n\theta)+i \sin(n\theta)]

Example 1

Compute (1+i)^6
Solution
Since
1+i= \sqrt{2} \left ( \frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}} \right )
or 1+i= \sqrt{2} \left ( \cos \frac{\pi }{4} +i\sin \frac{\pi }{4} \right )
We get
r=\sqrt{2} and \theta=\frac{\pi }{4}
Thus,
( 1+i )^6 =\sqrt{2}^6 \left [ \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right ]^6
or ( 1+i )^6 =\sqrt{2}^6[ \cos \left (6 \frac{\pi }{4} \right )+i\sin \left (6 \frac{\pi }{4} \right )]
or ( 1+i )^6 =8 (\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2})
or ( 1+i )^6 =-8i




nth root of Complex number

If Z=r(\cos \theta +i\sin \theta ) be a complex number then the nth root of z is
\sqrt[n]{Z}=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )
Proof
Given that Z is a complex number. Also let, nth root of Z is W such that W=R(\cos \phi +i\sin \phi )
Now we have
\sqrt[n]{Z}=W
or W^n=Z
or [R(\cos \phi +i\sin \phi )]^n=r(\cos \theta +i\sin \theta )
or R^n(\cos (n\phi) +i\sin (n\phi) )=r(\cos \theta +i\sin \theta )
Equating real and Imaginary parts, we get
R^n=r and \cos (n\phi)= \cos \theta and \sin (n\phi) =\sin \theta
or R=\sqrt[n]{r} and n\phi= \theta +2k \pi
or R=\sqrt[n]{r} and \phi= \frac{(\theta +2k\pi )}{n}
Thus, nth root of Z=r(\cos \theta +i\sin \theta ) is
W=R(\cos \phi +i\sin \phi )
or W=\sqrt[n]{r} \left ( \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right )

square roots of i

Find the square roots of i

We know that
\sqrt[n]{z}=\sqrt[n]{r} \left [ \cos \frac{(\theta +2k \pi )}{n} +i\sin \frac{(\theta +2k\pi )}{n} \right ]

Solution
Since i=\cos \frac{\pi }{2} +i\sin \frac{\pi }{2} , we get
r=1 and \theta=\frac{\pi }{2}
Hence the first square roots of i is
w_1[i=0] =\cos \frac{\frac{\pi }{2}+0}{2} +i\sin \frac{\frac{\pi }{2}+0}{2} =\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}
The second square root of i is

w_2[i=1] =\cos \frac{\frac{\pi }{2}+2\pi }{2} +i\sin \frac{\frac{\pi }{2}+2\pi }{2} =\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}

Conic Section

[latexpaage]

    Conic Section

    औपचारिक रूपमा, ग्रीकमा लगभग 500 to 200 BC को अवधिमा Conic sections पत्ता लगाइएको मानिन्छ। त्यतिखेर अपोलोनिस (200BC)ले Conic sections को बारेमा चर्चा गरेको पाईन्छ। यद्पि, सत्रौं शताब्दीको सुरुबाट मात्र Conic sections को व्यापक प्रयोग भएको पाईन्छ। आजका दिनहरूमा, प्रकृतिमा हुने धेरै प्रक्रियाहरूलाई मोडेल गर्न conic sections महत्त्वपूर्ण छन्। उदाहरण को लागी, universal bodies को locus कोनिक्स (सर्कल, अण्डाकार, प्याराबोला, र हाइपरबोला) हो ।

    The term “conic section” refers to the geometric shapes formed by the intersection of a plane with a cone. The cone is not necessarily a right circular cone. There are many conic sections on the basis of the angle of cutting.  In every cases, conic section is the section of cone by a plane.

    The ancient Greek mathematician Apollonius, also known as “Great Geometer,” made significant contributions to the understanding of conic sections around 200 BC. Apollonius explored the properties of ellipses, parabolas, and hyperbolas, categorizing them based on their unique characteristics.




    Cone

    In mathematics, cone is defined a three-dimensional surface
    traced out by a straight line
    passing through a fixed point and
    moving around a fixed line.
    In this definition of cone,
    The straight line is called generator
    The fixed point is called vertex
    The fixed line is called axis

    Conic भनेको डबल शंकुको cone लाई एउटा सतहले काट्दा बन्ने वक्र रेखा हो। सामान्यतया यसलाई right circular cone मा हेर्ने गरिन्छ, तर यो जुनसुकै cone मा पनि परिभाषित हुन्छ ।




    Conic Section: The Geometry

    Cone लाई एउटा plane ले काट्दा बन्ने plane curve (cross section) लाई conic section भनिन्छ ।

    Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
    Based on this intersection, there are seven types of conic section.
    These seven types of conic section are given below

    Conic Vertex Generator Axis
    1 Point Cuts outside
    2 Line Cuts touches
    3 Line Pair Cuts inside
    4 Circle Misses Right angle
    5 Ellipse Misses Not right angle
    6 Parabola Misses Parallel to
    7 Hyperbola Misses Not Parallel to

    According to this table, for example,
    Parabola: a parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which

    • the plane misses the vertex and
    • the plane is parallel to the generator.

    Hyperbola: Hyperbole is a conic section obtained by section of a cone by a plane in which

    • the plane misses the vertex of cone
    • the plane is not parallel to the generator of cone

    Similarly, we can define other conic sections

    Notice that, from the table above we see that in some intersection, plane does not pass through the vertex of the cone. When the plane does pass through the vertex, the resulting conic is a degenerate conic section.




    Conic Section: The Algebra

    In analytic geometry, conic section can be defined in algebraic expression. This algebraic forms of conic section is called analytic representation.

    In analytic geometry,
    Conic section can be defined based on the definition of circle.

    Circle

    Circle is defined a locus of point whose
    distance from a fixed point = constant

    In this definition of circle,

    • the constant distance is called radius.
    • the fixed point is called center

    Based on this definition of circle, we can define conic section.

    Conic Section

    Conic section is defined a locus of a point whose
    \( \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} \)= constant
    In this definition of conic section,

    • the constant ratio is called eccentricity, it is denoted by e.
    • the fixed point is called focus.
    • the fixed line is called directrix.

    Classification of Conic Section

    Based on the value of e, conic section can be classified into three standard types. These three standard types are

    1. Parabola (e =1)
    2. Ellipse (e <1)
    3. Hyperbola (e >1)
    Two more types are
    1. Circle (e =0)
    2. Straight Line (e =∞)



    General equation of Conic

    If the general equation of second degree \( ax^2 +2hxy + by^2 +2gx + 2fy + c = 0\) represents a pair of striaght lines then the discriminat must be perfect square, thus
    \((2hy+2g)^2-4(a)(by^2+2fy+c)\) is perfect square
    or \((hy+g)^2-a(by^2+2fy+c)\) is perfect square
    or \((h^2-ab)y^2+2(hg-af)y +(g^2-ac)\) is perfect square
    Again, \((h^2-ab)y^2+2(hg-af)y +(g^2-ac)\) is perfect square if its discriminant is zero
    Thus,
    \( 4(hg-af)^2-4(h^2-ab)(g^2-ac)=0\)
    or \( (hg-af)^2-(h^2-ab)(g^2-ac)=0\)
    or \( abc+2fgh-af^2-bg^2-ch^2=0\)

    Therefore, the discriminant is
    \( \Delta= abc+2fgh-af^2-bg^2-ch^2 \)

    Based on the value of the discriminant, following conics can be classified.
    1. \( Δ = 0, h^2=ab\) then the conic is pair of straight lines
    2. \( Δ = 0, \frac{a}{h}= \frac{h}{b}=\frac{g}{f} \) then the conic is parallel lines
    3. \( Δ \ne 0, a=b\ne 0, h=0\) then the conic is circle
    4. \( Δ \ne 0, h^2 < ab\) then the conic is ellipse
    5. \( Δ \ne 0, h^2=ab\) then the conic is parabola
    6. \( \Delta \ne 0, h^2 >ab\) then the conic is hyperbola
    7. \( Δ \ne 0, h^2 >ab, a+b=0\) then the conic is rectangular hyperbola

    The equation of the conic whose center is at the origin is of the form
    \( ax^2+by^2+2hxy+1=0 \)
    This conic is called central conic







    Parabola

    Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
    So,
    Parabola caan be defined as follows.

    1. Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which
      the plane misses the vertex and
      the plane is parallel to the generator

    2. Parabola is conic section defined as a plane curve obtained by intersection of a cone and a plane in which if
      α= angle between generator and axis and
      β= angle between plane and axis,
      and
      α=β
      then
      the section is a parabola,
      in which
      eccentricity = \(\frac{\cos \beta}{\cos \alpha}\)
      here
      the eccentricity is the measure of how far the conic deviates from being circular

    3. Parabola is a plane curve defined a locus of a point in which
      \( \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} \)= constant (e=1)

      In this definition of conic section,
      the constant ratio is called eccentricity, it is denoted by e.
      the fixed point is called focus.
      the fixed line is called directrix.

    4. Parabola is a plane curve defined a locus of a point in which the distance from a fixed point (or focus) and distance from a fixed line (or directrix) is always equal.
    5. Parabola is a plane curve defined a locus of a point which is always equidistant from a fixed point (or focus) to a fixed line (or directrix)
    In this definition
    • Focus: The fixed point of parabola is called focus
    • Directrix: The fixed line of parabola is called directrix
    • Axis: The straight line passing through focus and perpendicular to directrix is called axis
    • Vertex: The meeting point of axis and parabola is called vertex
    • Latus rectum: the chord passing through focus and perpendicular to axis is called latus rectum.
      The distance between the meeting points of latus rectum to the parabola is called length of latus rectum.

    Equation of Parabola

    Let C be a parabola whose
    Focus is F (a,0)
    Directrix is \(l: x = -a\)
    Vertex is O: (0,0)
    Take any point P(x,y) on parabola C,

    Then
    1. Draw PA ⊥ \(l\) then A (-a,y)
    2. Join F and P

    By the definition of parabola
    PA = PF
    or \( (x+a)^2=(x-a)^2+y^2 \)
    or \(x^2+2ax+a^2=x^2-2ax+a^2+y^2\)
    or \(2ax=-2ax+y^2 \)
    or \(y^2=4ax \)

    Summary on Equation of Parabola

    The basic parameters of parabola are summarized as below

    Parabola Parabola Parabola Parabola Parabola
    Equation \( y^2 =4 a x \) \( x^2 =4 a y \) \( (y-k)^2 =4 a (x-h) \) \( (x-h)^2 =4 a (y-k) \)
    Vertex (0,0) (0,0) (h,k) (h,k)
    Focus (a,0) (0,a) (h+a,k) (h,k+a)
    Directrix x=-a y=-a x=h-a y=k-a
    Axis y=0 x=0 y=k x=h
    Axis of Symmetry x-axis y-axis y=k x=h
    Endpoints of Latus Rectum (a,±2a) (±2a,a) (h+a,k±2a) (h±2a,k+a)



    Ellipse

    Conic section is a plane curve obtained by section (intersection) of a cone by a plane.
    So,
    Ellipse can be defined as follows.

    1. Ellipse is conic section defined as a plane curve obtained by intersection of a cone and a plane in which
      the plane misses the vertex and
      the plane is NOT at right angle with the axis

    2. Ellipse is conic section defined as a plane curve obtained by intersection of a cone and a plane in which if
      α= angle between generator and axis and
      β= angle between plane and axis,
      and
      α < β < 90
      then
      the section is a ellipse,
      in which
      eccentricity = \(\frac{\cos \beta}{\cos \alpha}\)
      here
      the eccentricity is the measure of how far the conic deviates from being circular

    3. Ellipse is a plane curve defined a locus of a point in which
      \( \frac{\text{distance from a fixed point}}{\text{distance from a fixed line}} \)= constant (e <1)

      In this definition of conic section,
      the constant ratio is called eccentricity, it is denoted by e.
      the fixed point is called focus.
      the fixed line is called directrix.

    4. Ellipse is a plane curve defined a locus of a point in which the distance from a fixed point (or focus) is always less than the distance from a fixed line (or directrix)
    5. Ellipse is a plane curve defined a locus of a point whose sum of distances from two fixed points (foci) is always a constant.

    In this definition

    • Foci: The two fixed points of ellipse
    • Directrix: The two fixed line of ellipse
    • Axis: The straight line passing through focus and perpendicular to directrix
    • Major axis: The straight line passing through the foci
    • Minor axis: The straight line passing through the center and perpendicular to the major axis
    • Length of major axis: The distance between the vertices on major axis
    • Length of minor axis: The distance between the meeting points of minor axis with ellipse
    • Vertices: The meeting points of the major axis with ellipse
    • Centre of ellipse: The middle point of the join of foci
    • Latus rectum: The chord passing through focus and perpendicular to major axis

    Please Note that

    1. Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis.
    2. Each endpoint of the major axis is the vertex of the ellipse, and each endpoint of the minor axis is a co-vertex of the ellipse.
    3. The center of an ellipse is the midpoint of both the major and minor axes.
      The axes are perpendicular at the center.
    4. The foci always lie on the major axis

    There are four variations of the standard form of the ellipse. These variations are categorized first by
    the location of the center (the origin or not the origin),
    and by
    the position (horizontal or vertical).

    Ellipse: Proof of Basic Facts

    1. c=ae

      If Z and Z’ are the directrix, F(c,0) and F'(-c,0) are the foci, and A(a,0) and A;(-a,0) are the vertices of ellipse, then by the definition of conic we have
      \(\frac{A’F}{A’Z}=e\)
      or \(A’F=A’Z e\) (1)
      Similarly, we have
      \(\frac{AF}{AZ}=e\)
      or \(AF=AZ e\) (2)
      Substracting (2) from (1), we get
      \(A’F-AF=(A’Z-AZ) e\)
      or\((A’O+OF)-(AO-OF)=(A’A) e\)
      or\(2OF=2a e\)
      or\(2c=2a e\)
      or\(c=a e\)

    2. sum of distance from Foci is 2a.

      the distance of a point A(a,0) from fous F(c,0) is
      a−(c)=a-c.
      the distance of a point A(a,0) from fous F'(-c,0) is
      a−(-c)=a+c
      The sum of the distances from Foci is
      (a+c)+(a−c)=2a

    3. Relation between a, b, c.

      \(2 \sqrt{b^2+c^2}=2a\)
      or\(b^2+c^2=a^2\)

    4. Equation of Ellipse

      Let C be an ellipse whose
      foci are (−c,0) and (c,0).
      center is O: (0,0)
      Take any point P(x,y) on ellipse C,

      Then,

      If (a,0) is a vertex of the ellipse, then the distance from (−c,0) to (a,0) is
      a−(−c)=a+c.
      The distance from (c,0) to (a,0) is
      a−c.
      The sum of the distances from the foci to the vertex is
      (a+c)+(a−c)=2a
      If (x,y) is a point on the ellipse, then we can define the following variables:
      d1=the distance from (−c,0)to (x,y)
      d2=the distance from (c,0)to (x,y)
      By the definition of an ellipse,
      d1+d2=2a

      Here
      \(d_1+d_2=2a\)
      or \(\sqrt{(x+c)^2+y^2}+\sqrt{(x−c)^2+y^2}=2a\)
      or \(\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x−c)^2+y^2}\)
      or \((x+c)^2+y^2=\left [2a-\sqrt{(x−c)^2+y^2} \right] ^2\)
      or \(x^2+2xc+c^2+y^2=4a^2-4a\sqrt{(x−c)^2+y^2} +(x−c)^2+y^2 \)
      or \(x^2+2xc+c^2+y^2=4a^2-4a\sqrt{(x−c)^2+y^2} +x^2-2xc+c^2+y^2 \)
      or \(2xc=4a^2-4a\sqrt{(x−c)^2+y^2} -2xc\)
      or \(4xc-4a^2=-4a\sqrt{(x−c)^2+y^2}\)
      or \(xc-a^2=-a\sqrt{(x−c)^2+y^2}\)
      or \(\left[xc-a^2\right]^2= \left[ -a\sqrt{(x−c)^2+y^2} \right ]^2\)
      or \(c^2x^2-2a^2cx+a^4= a^2 \left[(x−c)^2+y^2 \right ]\)
      or \(c^2x^2-2a^2cx+a^4= a^2 (x^2-2xc+c^2+y^2 ) \)
      or \(c^2x^2-2a^2cx+a^4= a^2x^2-a^22xc+a^2c^2+a^2y^2 \)
      or \(a^2x^2-c^2x^2+a^2y^2=a^4-a^2c^2 \)
      or \(x^2(a^2-c^2)+a^2y^2=a^2(a^2-c^2) \)
      or \(x^2b^2+a^2y^2=a^2b^2 \) \(a^2-c^2=b^2\)
      or \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 \)

      Thus, the standard equation of an ellipse is
      \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1 \)
      This equation defines an ellipse centered at the origin.

      NOTE
      1. If a>b, the ellipse is stretched further in the horizontal direction
      2. if b>a the ellipse is stretched further in the vertical direction.
      3. When c = 0, both foci merge together with the center of the ellipse and so the ellipse becomes a circle
      4. When c = a, then b = 0. The ellipse reduces to the line segment joining the two foci

      Summary of parameters in an Ellipse

      Ellipse Ellipse Ellipse Ellipse
      Equation \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\); a > b \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\); a < b \( \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\); a > b \( \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\); a < b
      Center (0,0) (0,0) (h,k) (h,k)
      Vertex \( (\pm a,0) \) \( (0,\pm b) \) \( (h\pm a,k) \) \( (h,k\pm b) \)
      Focus \( (\pm ae,0) \) \( (0,\pm be) \) \( (h\pm ae,k) \) \( (h,k\pm be) \)
      Directrix \( x=\pm \frac{a}{e} \) \( y=\pm \frac{b}{e} \) \( x=h\pm \frac{a}{e} \) \( y=k\pm \frac{b}{e} \)
      Length Rectum \( 2 \frac{b^2}{a} \) \( 2 \frac{a^2}{b}\) \( 2 \frac{b^2}{a} \) \( 2 \frac{a^2}{b}\)
      Eccentricity \( b^2=a^2(1-e^2) \) \( a^2=b^2(1-e^2) \) \( b^2=a^2(1-e^2) \) \( a^2=b^2(1-e^2) \)

Reflection

ज्यामितिमा स्थानान्तरण लामो समयदेखि व्यापक चासोको विषय बनेको थियो। युक्लिडको आधिकारिक वा स्वयंसिद्ध सिद्धान्त नभएपनि सामान्यतया स्थानान्तरणलाई सुपरपोजिसनको सिद्धान्त (उठाउर अर्कोमा राख्नु) मा आधारित गरिएको थियो । जब १९ औं शताब्दीमा (1872 मा) फ्लेक्स क्लेनले – एर्लान्जेन विश्वविद्यालयमा – ज्यामितिमा एक नयाँ परिप्रेक्ष्य प्रस्तावित गरे तबदेखी स्थानान्तरणलाई एर्लान्जेन कार्यक्रमको रूपमा चिनिन थालिएको छ।

Transformation Geometry

Transformation Geometry is a branch of mathematics that utilize analytic geometry and algebraic function to study geometric invariants. It was introduced (developed) by Flex Klein in19th century through Erlangen programme. According to him, theer are five Kinds of Transformations:

  1. Rigid motions: Plane geometry of congruent figures, preserve distances (e.g., Euclidean transformations)
  2. Similarity transformations: The familiar geometry with similar figures.
    Similarities preserve angles and ratios between distances (e.g., resizing)
  3. Affine (matrix) transformations: Geometry of computer animation. Rectangles and parallelograms the same
    Affine transformations preserve parallelism (e.g., scaling, shear)
  4. Projective Geeometry: Geometry of photo
    Projective transformations preserve collinearity
  5. Continuous (topological) transformations: Any loop is a circle. Any path is a segment

In transformation geometry, the idea of flips, slides, and turn are key features, which will be analyzed based on its mapping. This mapping are basically of two kinds.

  1. Isometric mapping (transformation)
  2. Non-isometric mapping (transformation)

Geometric transformation involves both analytic and algebraic geometry because it can be approached using the graphical perceptive, and algebraic computational theory.

Reflection

Reflection is very common topic in physics and mathematics. When an object is place before the mirror, the image is formed behind the mirror. This is called reflection.

Definition: Reflection

Let l be a line and A be a point.
Then, reflection about the line l is denoted by T_1, which is a transformation on a plane, where

  1. if A ∉l , then Tl (A) = A’
  2. if A ∈l , then Tl(A)=A
such that
  1. the mispoint of AA’ lies on l
  2. the segment AA’is perpendiculat to l

We start a short discussion for reflection in mathematics as below.

Let us take a geometrical figures like triangle. Now, we hold the triangle vertices, one-by-one, about a line source l in a grid. Then look at the image (shadow) of the triangle on the grid. The shadow have the same size as the original triangle has, but in opposite direction.

In the above figure, ∆XYZ is a triangle and its shadow is ∆X’Y’Z’.
By actual measurements it can be seen that
∡X = ∡X’, ∡Y = ∡Y’, ∡Z = ∡Z
XY=X’Y’, YZ=Y’Z’, XZ=X’Z’
Thus, the two triangles are congruent
The important note is that

  1. l ⊥ bisector of XX’
  2. l ⊥ bisector of YY’
  3. l ⊥ bisector of ZZ’

Here l is said to be line of reflection

In summary,
  1. The image is found opposite to the mirror line l .
  2. Mirror line l is the perpendicular bisector of the line joining the object and its image.
  3. X’ is called the reflection (image) of X and so on
  4. l is called the axis of reflection.

Thus, reflection can be seen in water, in a mirror, in glass. In mathematics, an object and its reflection have the same shape and size, but the figures face in opposite directions . Also, a reflection needs

  1. Object
  2. Line of Reflection