Wednesday, July 4, 2012

Aptitude Question part 3

1. 6, 24, 60,120, 210
a) 336 b) 366 c) 330 d) 660
Answer : a) 336
Explanation : The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, ..... ( '.' means product)
2. 1, 5, 13, 25
Answer : 41
Explanation : The series is of the form 0^2+1^2, 1^2+2^2,...
3. 0, 5, 8, 17
Answer : 24
Explanation : 1^2-1, 2^2+1, 3^2-1, 4^2+1, 5^2-1
4. 1, 8, 9, 64, 25 (Hint : Every successive terms are related)
Answer : 216
Explanation : 1^2, 2^3, 3^2, 4^3, 5^2, 6^3
5. 8,24,12,36,18,54
Answer : 27
6. 71,76,69,74,67,72
Answer : 67
7. 5,9,16,29,54
Answer : 103
Explanation : 5*2-1=9; 9*2-2=16; 16*2-3=29; 29*2-4=54; 54*2-5=103
8. 1,2,4,10,16,40,64 (Successive terms are related)
Answer : 200
Explanation : The series is powers of 2 (2^0,2^1,..).
All digits are less than 8. Every second number is in octal number system.
128 should follow 64. 128 base 10 = 200 base 8.
9. 3,5,7,12,13,17,19
Answer : 12
Explanation : All but 12 are odd numbers
10. 2,5,10,17,26,37,50,64
Answer : 64
Explanation : 2+3=5; 5+5=10; 10+7=17; 17+9=26; 26+11=37; 37+13=50; 50+15=65;
11. 105,85,60,30,0,-45,-90
Answer : 0
Explanation : 105-20=85; 85-25=60; 60-30=30; 30-35=-5; -5-40=-45; -45-45=-90;
11. Hacking is the only vulnerability of the computers for the usage of the data
Ans. False
12. Hacking is done mostly due to the lack of computer knowledge
Ans. False
PASSAGE C: Alpine tunnels are closed tunnels. In the past 30 yrs not even a single accident has been recorded for there is one accident in the rail road system. Even in case of a fire accident it is possible to shift the passengers into adjacent wagons and even the live fire can be detected and extinguished with in the duration of 30 min.
Answer questions 13-16 based on passage C
13. No accident can occur in the closed tunnels
Ans. True
14. Fire is allowed to live for 30 min
Ans. False
16. All the care that travel in the tunnels will be carried by rail shutters.
Ans.True
PASSAGE D: In the past helicopters were forced to ground or crash because of the formation of the ice on the rotors and engines. A new electronic device has been developed which can detect the water content in the atmosphere and warns the pilot if the temperature is below freezing temperature about the formation of the ice on the rotors and wings.
Answer questions 17-20 based on passage D
17. The electronic device can avoid formation of the ice on the wings
Ans.False
18. There will be the malfunction of rotor & engine because of formation of ice
Ans.True
19. The helicopters were to be crashed or grounded
Ans.True
20. There is only one device that warn about the formation of ice
Ans.True
PASSAGE E: In the survey conducted in Mumbai out of 63 newly married house wives not a single house wife felt that the husbands should take equal part in the household work as they felt they loose their power over their husbands. In spite of their careers they opt to do the kitchen work themselves after coming back to home. The wives get half as much leisure time as the husbands get at the week ends.
Answer questions 21-23 based on passage E
21. Housewives want the husbands to take part equally in the household
Ans.False
22. Wives have half as much leisure time as the husbands have
Ans. False
23. 39% of the men will work equally in the house in cleaning and washing
Ans. False
PASSAGE F: Copernicus is the intelligent. In the days of Copernicus the transport and technology development was less & it took place weeks to communicate a message at that time, wherein we can send it through satellite with in no time. Even with this fast development it has become difficult to understand each other.
Answer questions 24-27 based on passage F
24. People were not intelligent during Copernicus days
Ans.False
25. Transport facilities are very much improved in now a day
Ans.Can't say
26. Even with the fast developments of the technology we can't live happily.
Ans. Can't say
27. We can understand the people very much with the development of communication
Ans. False.
PASSAGE G: Senior managers warned the workers that because of the introductory of Japanese industry in the car market. There is the threat to the workers. They also said that there will be the reduction in the purchase of the sales of car in public. the interest rates of the car will be increased with the loss in demand.
Answer questions 28-31 based on passage G
28. Japanese workers are taking over the jobs of Indian industry.
Ans.False
29. Managers said car interests will go down after seeing the raise in interest rates.
Ans.True
30. Japanese investments are ceasing to end in the car industry.
Ans. False
31. People are very interested to buy the cars.
Ans.False
PASSAGE H: In the Totalitarian days, the words have very much devalued. In the present day, they are becoming domestic that is the words will be much more devalued. In those days, the words will be very much affected in political area. but at present, the words came very cheap .We can say they come free at cost.
Answer questions 32-34 based on passage H
32. Totalitarian society words are devalued.
Ans.False
33. Totalitarians will have to come much about words
Ans.True
34. The art totalitarian society the words are used for the political speeches.
Ans. False
PASSAGE I: There should be copyright for all arts. The reel has came that all the arts has come under one copy right society, they were use the money that come from the arts for the developments . There may be a lot of money will come from the Tagore works. We have to ask the benifiters from Tagore work to help for the development of his works.
Answer questions 35-39 based on passage I
35. Tagore works are come under this copy right rule.
Ans. False
36. People are free to go to the public because of the copy right rule.
Ans: Can't say
37 People gives to theater and collect the money for development.
Ans: Can't say
38 We have asked the Tagore residents to help for the developments of art.
Ans: Can't say
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Mathematical Formulae

Numbers
-----------
1. A number is divisible by 2, if its unit’s place digit is 0, 2, 4, or 8

2. A number is divisible by 3, if the sum of its digits is divisible by 3

3. A number is divisible by 4, if the number formed by its last two digits is divisible by 4

4. A number is divisible by 8, if the number formed by its last three digits is divisible by 8

5. A number is divisible by 9, if the sum of its digits is divisible by 9

6. A number is divisible by 11, if, starting from the RHS,
(Sum of its digits at the odd place) – (Sum of its digits at even place) is equal to 0 or 11x

7. Product of two numbers = Their H. C. F. × Their L. C. M.

Simple Interest
-----------------
1. Let Principle = P, Rate = R% per annum and Time = T years. Then,

a. S.I. = ( P × R × T ) / 100

b. P = ( 100 × S.I. ) / ( R × T ),

c. R = ( 100 × S.I. ) / ( P × T ),

d. T = ( 100 × S.I. ) / ( P × R ).

Compound Interest
---------------------
1. Let Principle = P, Rate = R% per annum and Time = T years. Then,

I. When interest is compounded Annually,
Amount = P (1 + R/100)N

II. When interest is compounded Half-yearly:
Amount = P (1 + R/2/100)2N

III. When interest is compounded Quarterly:
Amount = P (1 + R/4/100)4N

2. When interest is compounded Annually, but the time is in fraction, say 3⅞ years.
Then, Amount = P (1 + R/100)3 × (1 + ⅞R/100)

3. When Rates are different for different years, say R1%, R2%, R3% for 1st, 2nd, and 3rd year
respectively,
Then, Amount = P (1 + R1/100) (1 + R2/100) (1 + R3/100)

4. Present worth of Rs. x due n years hence is given by:
Present Worth = x / (1 + R/100)n

Logarithms
-------------
1. Logarithm: If a is a positive real number, other than 1 and am = x, then we write m = loga x
and say that the value of log x to the base a is m.

2. Properties of Logarithms:

a. loga (xy) = loga x + loga y

b. loga (x/y) = loga x - loga y

c. logx x = 1 (i.e. Log of any number to its own base is 1)

d. loga 1 = 0 (i.e. Log of 1 to any base is 0)

e. loga (xp) = p loga x

f. loga x = 1 / logx a

g. loga x = logb x / logb a
= log x / log a (Change of base rule)

h. When base is not mentioned, it is taken as 10

i. Logarithms to the base 10 are known as common logarithms

Algebra
-----------
1. (a + b)2 = a2 + 2ab + b2

2. (a - b)2 = a2 - 2ab + b2

3. (a + b)2 - (a - b)2 = 4ab

4. (a + b)2 + (a - b)2 = 2(a2 + b2)

5. (a2 – b2) = (a + b)(a - b)

6. (a3 + b3) = (a + b)(a2 - ab + b2)

7. (a3 – b3) = (a - b)(a2 + ab + b2)

8. Results on Division:
Dividend = Quotient × Divisor + Remainder

9. An Arithmetic Progression (A. P.) with first term ‘a’ and Common Difference ‘d’ is given
by:
[a], [(a + d)], [(a + 2d)], … … …, [a + (n - 1)d]
nth term, Tn = a + (n - 1)d
Sum of first ‘n’ terms, Sn
= n/2 (First Term + Last Term)

10. A Geometric Progression (G. P.) with first term ‘a’ and Common Ratio ‘r’ is given by:
a, ar, ar2, ar3, … … …, arn-1
nth term, Tn = arn-1
Sum of first ‘n’ terms Sn = [a(1 - rn)] / [1 - r]

11. (1 + 2 + 3 + … … … + n) = [n(n + 1)] / 2

12. (12 + 22 + 32 + … … … + n2) = [n(n + 1)(2n + 1)] / 6

13. (13 + 23 + 33 + … … … + n3) = [n2(n + 1)2] / 4

Percentage
------------------
1. To express x% as a fraction, we have x% = x / 100

2. To express a / b as a percent, we have a / b = (a / b × 100) %

3. If ‘A’ is R% more than ‘B’, then ‘B’ is less than ‘A’ by
OR
If the price of a commodity increases by R%, then the reduction in consumption, not
to increase the expenditure is
{100R / [100 + R] } %

4. If ‘A’ is R% less than ‘B’, then ‘B’ is more than ‘A’ by
OR
If the price of a commodity decreases by R%, then the increase in consumption, not to
increase the expenditure is
{100R / [100 - R] } %

5. If the population of a town is ‘P’ in a year, then its population after ‘N’ years is
P (1 + R/100)N

6. If the population of a town is ‘P’ in a year, then its population ‘N’ years ago is
P / [(1 + R/100)N]

Profit & Loss
---------------

7. If the value of a machine is ‘P’ in a year, then its value after ‘N’ years at a depreciation of
‘R’ p.c.p.a is
P (1 - R/100)N

8. If the value of a machine is ‘P’ in a year, then its value ‘N’ years ago at a depreciation of
‘R’ p.c.p.a is
P / [(1 - R/100)N]

9. Selling Price = [(100 + Gain%) × Cost Price] / 100
= [(100 - Loss%) × Cost Price] / 100

Ratio & Proportion
-----------------------
1. The equality of two ratios is called a proportion. If a : b = c : d, we write a : b :: c : d and
we say that a, b, c, d are in proportion.
In a proportion, the first and fourth terms are known as extremes, while the second and
third are known as means.

2. Product of extremes = Product of means

3. Mean proportion between a and b is=(ab)^(1/2)

4. The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf)

5. a2 : b2 is a duplicate ratio of a : b

6. a^(1/2).b^(1/2) : is a sub-duplicate ration of a : b

7. a3 : b3 is a triplicate ratio of a : b

8. a1/3 : b1/3 is a sub-triplicate ratio of a : b

9. If a / b = c / d, then, (a + b) / b = (c + d) / d, which is called the componendo.

10. If a / b = c / d, then, (a - b) / b = (c - d) / d, which is called the dividendo.

11. If a / b = c / d, then, (a + b) / (a - b) = (c + d) / (c - d), which is called the componendo &
dividendo.

12. Variation: We say that x is directly proportional to y if x = ky for some constant k and we
write, x α y.

13. Also, we say that x is inversely proportional to y if x = k / y for some constant k and we
write x α 1 / y.

Area
------
1. Rectangle:

a. Area of a rectangle = (length × breadth)

b. Perimeter of a rectangle = 2 (length + breadth)

2. Square:

a. Area of square = (side)2

b. Area of a square = ½ (diagonal)2

3. Area of 4 walls of a room
= 2 (length + breadth) × height

4. Triangle:

a. Area of a triangle = ½ × base × height

b. Area of a triangle = , where
s = ½ (a + b + c), and a, b, c are the sides of the triangle

c. Area of an equilateral triangle = / 4 × (side)2

d. Radius of incircle of an equilateral triangle of side a = a / 2

e. Radius of circumcircle of an equilateral triangle of side a = a /

5. Parallelogram/Rhombus/Trapezium:

a. Area of a parallelogram = Base × Height

b. Area of a rhombus = ½ × (Product of diagonals)

c. The halves of diagonals and a side of a rhombus form a right angled triangle with
side as the hypotenuse.

d. Area of trapezium = ½ × (sum of parallel sides) × (distance between them)

6. Circle/Arc/Sector, where R is the radius of the circle:

a. Area of a circle = πR2

b. Circumference of a circle = 2πR

c. Length of an arc = Ө/360 × 2πR

d. Area of a sector = ½ (arc × R)
= Ө/360 × πR2

Volume & Surface Area
-------------------------
1. Cuboid:
Let length = l, breadth = b & height = h units Then,

a. Volume = (l × b × h) cu units

b. Surface Area = 2 (lb + bh + hl) sq. units

c. Diagonal = units

2. Cube:
Let each edge of a cube be of length a. Then,

a. Volume = a3 cu units

b. Surface Area = 6a2 sq. units

c. Diagonal = ( l^2 + b^2 +h^2)^1/2 units

3. Cylinder:
Let radius of base = r & height (or length) = h. Then,

a. Volume = (πr2h) cu. units

b. Curved Surface Area = (2πrh) sq. units
c. Total Surface Area = 2πr(r + h) sq. units

4. Cone:
Let radius of base = r & height = h. Then,

a. Slant height, l =(h^2 + R^2)^1/2 units

b. Volume = (⅓ πr2h) cu. units

c. Curved Surface Area = (πrl) sq. units

d. Total Surface Area = πr(r + l) sq. units

5. Sphere:
Let the radius of the sphere be r. Then,

a. Volume = (4/3 πr3) cu. units

b. Surface Area = (4πr2) sq. units

6. Hemi-sphere:
Let the radius of the sphere be r. Then,

a. Volume = (2/3 πr3) cu. units

b. Curved Surface Area = (2πr2) sq. units

c. Total Surface Area = (3πr2) sq. units

Stocks & Shares
-----------------
1. Brokerage: The broker’s charge is called brokerage.

2. When stock is purchased, brokerage is added to the cost price.

3. When the stock is sold, brokerage is subtracted from the selling price.

4. The selling price of a Rs. 100 stock is said to be:

a. at par, if S.P. is Rs. 100 exactly;

b. above par (or at premium), if S.P. is more than Rs. 100;

c. below par (or at discount), if S.P. is less than Rs. 100.

5. By ‘a Rs. 800, 9% stock at 95’, we mean a stock whose face value is Rs. 800, annual
interest is 9% of the face value and the market price of a Rs. 100 stock is Rs. 95.

True Discount
----------------
1. Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per
annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of
Rs. 100 now will clear off the debt of Rs. 156 due 4 years hence. We say that:
Sum due = Rs. 156 due 4 years hence;
Present Worth (P.W.) = Rs. 100;
True Discount (T.D.) = Rs. (156 - 100)
= (Sum due) – (P.W.)

2. T.D. = Interest on P.W.

3. Amount = (P.W.) + (T.D.)

4. Interest is reckoned on R.W. and true discount is reckoned on the amount

5. Let rate = R% per annum & time = T years. Then,

a. P.W. = (100 × Amount) / (100 + [R × T])
= (100 × T.D.) / (R × T)

b. T.D. = (P.W.) × R × T / 100
= ([Amount] × R × T) / (100 + [R × T])

c. Sum = ([S.I.] × [T.D.]) / ([S.I.] – [T.D.])

d. (S.I.) – (T.D.) = S.I. on T.D.

e. When the sum is put at compound interest, then
P.W. = Amount / (1 + R/100)T

Banker’s Discount
---------------------
1. Banker’s Discount (B.D.) is the S.I. on the face value for the period from the date on
which the bill was discounted and the legally due date.

2. Banker’s Gain (B.G.) = (B.D.) – (T.D.) for the unexpired time

3. When the date of the bill is not given, grace days are not to be added

4. B.D. = S.I. on bill for unexpired time

5. B.G. = (B.D.) – (T.D.)
= S.I. on T.D.
= (T.D.)2 / P.W.

6. T.D. =(P.W x B.G)^1/2

7. B.D. = (Amount × Rate × Time) / 100

8. T.D. = (Amount × Rate × Time) / (100 + [Rate × Time])

9. Amount = (B.D. × T.D.) / (B.D. – T.D.)

10. T.D. = (B.G. × 100) / (Rate × Time)

Partnership
--------------
1. If a number of partners have invested in a business and it has a profit, then
Share Of Partner = (Total_Profit × Part_Share / Total_Share)
Chain Rule

2. The cost of articles is directly proportional to the number of articles.

3. The work done is directly proportional to the number of men working at it.

4. The time (number of days) required to complete a job is inversely proportional to the
number of hours per day allocated to the job.

5. Time taken to cover a distance is inversely proportional to the speed of the car.
Time & Work

6. If A can do a piece of work in n days, then A’s 1 day’s work = 1/n.

7. If A’s 1 day’s work = 1/n, then A can finish the work in n days.

8. If A is thrice as good a workman as B, then:
Ratio of work done by A and B = 3 : 1,
Ratio of times taken by A & B to finish a work = 1 : 3

Pipes & Cisterns
-----------------
1. If a pipe can fill a tank in ‘x’ hours and another pipe can empty the full tank in ‘y’ hours
(where y > x), then on opening both the pipes, the net part of the tank filled in 1 hour is
(1/x – 1/y)

Time And Distance
----------------------
1. Suppose a man covers a distance at ‘x’ kmph and an equal distance at ‘y’ kmph, then
average speed during his whole journey is
[2xy / (x + y)] kmph

Trains
--------
1. Lengths of trains are ‘x’ km and ‘y’ km, moving at ‘u’ kmph and ‘v’ kmph (where, u > v) in
the same direction, then the time taken y the over-taker train to cross the slower train is
[(x + y) / (u - v)] hrs

2. Time taken to cross each other is
[(x + y) / (u + v)] hrs

3. If two trains start at the same time from two points A and B towards each other and after
crossing they take a and b hours in reaching B and A respectively.
Then, A’s speed : B’s speed = ( b^(1/2) : a^(1/2) ).

4. x kmph = (x × 5/18) m/sec.

5. y metres/sec = (y × 18/5) km/hr.

Boats & Streams
-------------------
1. If the speed of a boat in still water is u km/hr and the speed of the stream is v hm/hr,
then:
Speed downstream = (u + v) km/hr.
Speed upstream = (u - v) km/hr.

2. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:
Speed in still water = ½ (a + b) km/hr.
Rate of stream = ½ (a - b) km/hr.

Alligation or Mixture
----------------------
1. Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at
the given price must be mixed to produce a mixture at a given price.

2. Mean Price: The cost price of a quantity of the mixture is called the mean price.


Numerical solution of nonlinear equations
------------------------------------------
We want to develop numerical methods for solving nonlinear
equations of the form
f(x) = 0. (1)
Examples:
(i) f(x) = x2 − A = 0, A > 0 (⇒ x = ±√A).
(ii) f(x) = x3 − 3x2 + 2 = 0.
(iii) f(x) = 1
x − A = 0 (⇒ x = 1
A).
(iv) f(x) = sin x − 1
x = 0.
Numerical solution of nonlinear equations
Idea: Approximate Eqn. (1) by an iterative process (recursion)
xn+1 = F(xn), x1 is given, (2)
such that
lim
n→∞
xn = X, with f(X) = 0.
Easy to implement (1) on a computer.
Goal: find a systematic way for obtaining iterative process of the
form (2) for solving equations of the form (1).
Numerical solution of nonlinear equations
Consider Example (i), f(x) = x2 − A = 0, A > 0.
x2 − A = 0 ⇒ 2x2 − x2 = A
⇒ 2x2 = x2 + A ⇒ x =
1
2 x +
A
x .
Use the following iterative scheme:
xn+1 =
1
2 xn +
A
xn . (3)
Choose x1 (initial guess), use (3) to calculate approximately the
square root of A. Let A = 25 ⇒ X = ±5.
Numerical solution of nonlinear equations
Choose x1 = 1:
x1 = 1 ⇒ x2 = 13.0 ⇒ x3 = 7.4615
⇒ x4 = 5.406 ⇒ x5 = 5.0152 . . .
i.e. the iterative process converges to X = 5. Take now x1 = −1.
x1 = −1 ⇒ x2 = −13.0 ⇒ x3 = −7.4615
⇒ x4 = −5.406 ⇒ x5 = −5.0152 . . .
i.e. the iterative process converges to X = −5. Choose x1 = 8:
x1 = 8 ⇒ x2 = 5.5625 ⇒ x3 = 5.0284
⇒ x4 = 5.0001 . . .
The iteration process converges faster to the root of the equation.
Numerical solution of nonlinear equations
When we have an iterative procedure of the form (2) which we use
to calculate the roots of (1) we want to address the following issues.
1. Does the iterative process converge, i.e. is there a limit of xn as
n → ∞?
2. Does the iterative process converge to the right value (i.e. one
of the roots of the equation F(x) = 0)?
3. How does its accuracy increase with n (i.e. is it practical)?
4. How does the choice of x1 affect the convergence of the method
and the value to which it converges (in the case when the
equation has more than one roots)?
A iterative process of the form (2) for solving an equation of the
form (1) is called a fixed point method.
Numerical solution of nonlinear equations
Consider example (ii) x3 − 3x2 + 2 = 0:
x3 − 3x2 + 2 = 0 ⇒ x3 = 3x2 − 2
⇒ x = 3 −
2
x2 .
• Use the iterative process
xn+1 = 3 −
2
x2
n
.
• If this method converges, to which of the 3 roots of the
equation does it converge?
Numerical solution of nonlinear equations
If an iterative process is to be convergent, then we have
lim
n→∞xn+1 − xn = 0,
i.e. the distance between two subsequent elements of the sequence
{xn}∞
n=1 generated by (2) becomes smaller as n increases.
We expect that the starting value x1 affects the eventual
convergence of the iterative process to a limit. It might be that the
iterative process that we have chosen converges only for certain
values of x1.
If we sketch the curves y = x, y = F(x), their intersection satisfies
x = F(x). Start with x1. The iterates generated by (2) should give
x = X.
Numerical solution of nonlinear equations
Calculation of √A.
• Consider the iterative process
xn+1 =
1
2 xn +
A
xn ,
• which approximates √A.
• Assume 0 < x2 =" 1">
1
2 √A +
A
√A
= √A ⇒ x2 > √A.
Numerical solution of nonlinear equations
Now:
x3 =
1
2 x2 +
A
x2
< n =" 2,"> √A. Then xn+1 < n =" 1," a =" 1" a =" 1" a =" 1" 1 =" xn(2" x =" 0" 1 =" Ax" 1 =" 2" x =" x(2" xn =" 1" x1 =" 0" x1 =" 2" xn =" 0," n =" 1,">
2
A ⇒ lim
n→∞
xn = −∞.
Numerical solution of nonlinear equations
Summary
• An iterative process can converge or diverge, depending on the
initial choice x1.
• The root to which the iterative process converges depends on
the initial choice x1.
• The convergence can be fast (quadratic).
How to choose the iterative process for solving f(x) = 0.
A given equation f(x) = 0 may often be written in several different
ways as x = F(x).
Example: f(x) = x2 − 6x + 2 = 0 with roots X1,2 = 3 ± √7. It
can be written as x = F(x) in many different ways:
1. x = x
2
−2
2x−6 ⇒ the iterative process is xn+1 = x
2
−2
2xn−6 :
x2−6x+2 = 0 ⇒ 2x2−6x+2 = x2 ⇒ 2x2−6x = x2−2 ⇒ x =
x2 − 2
2x − 6
.
2. x = 6 − 2
x ⇒ the iterative process is xn+1 = 6 − 2
xn
:
x2 − 6x + 2 = 0 ⇒ x2 = 6x − 2 ⇒ x = 6 −
2
x
.
3. x = 1
6x2 + 1
3 ⇒ the iterative process is xn+1 = 1
6x2
n + 1
3 :
6x = x2 + 2 ⇒ x =
1
6
x2 +
1
3
.
How to choose the iterative process for solving f(x) = 0.
The resulting different iterative processes xn+1 = F(xn) have, in
general, different convergence properties. We want to choose the
iterative process that has the best convergence properties.
This means that not only does it converge to the roots of the
equation f(x) = 0, but also that it converges sufficiently fast. We
want to develop criteria that enable us to choose the best iterative
process for solving a nonlinear equation of the form f(x) = 0.
How to choose the iterative process for solving f(x) = 0.
Suppose that xn+1 = F(xn) with limn→∞ xn = X and
f(X) = 0 ⇒ X = F(X). Let us write
X = xn + ǫn, n = 1, 2, . . .
i.e.
X = approximation + error.
We want to calculate ǫn+1 as a function of ǫn. Assume that ǫn is
small so that we can use the Taylor series expansion:
xn+1 = F(xn) ≈ F(X) − ǫnF′(X) +
1
2
ǫ2
nF′′(X).
But xn+1 = X + ǫn+1 ⇒
ǫn+1 = X − xn+1 ≈ X − F(X) + ǫnF′(X) −
1
2
ǫ2
nF′′(X)
= ǫnF′(X) −
1
2
ǫ2
nF′′(X).
⇒ The size of ǫn+1, relative to the size of ǫn, depends on F′(X).
How to choose the iterative process for solving f(x) = 0.
Indeed:
ǫn+1
ǫn
= F′(X) + O(ǫn). (4)
DEFINITION 1 Assume that F′(X) 6= 0. Then the iterative
process xn+1 = F(xn) is called a first order process.
The convergence of a first order process is guranteed when
F′(X) <> 1. The case F′(X) = 1 requires further study.
How to choose the iterative process for solving f(x) = 0.
We can also define second order processes:
DEFINITION 2 Assume that F′(X) = 0 and that F′′(X) 6= 0.
Then the iterative process xn+1 = F(xn) is called a second order
process.
For a second order process we have that
ǫn+1 = −
1
2
ǫ2
nF′′(X) + O(ǫ3
n) ⇒ ǫn+1 = Cǫ2
n
and the convergence is much faster (quadratic as opposed to
linear). It is to our advantage to use a second order iterative
process to solve the equation f(x) = 0.
How to choose the iterative process for solving f(x) = 0.
Examples
1. The square root formula F(x) = 1
2 􀀀x + A
x .
F′(x) =
1
2 1 −
A
x2 ⇒ F′(√A) =
1
2 1 −
A
A = 0
⇒ We have to calculate the second derivative
at the root X = √A.
F′′(x) =
A
x3 ⇒ F′′(√A) =
1
√A
.
⇒ Second Order Process.
⇒ The convergence is quadratic.
How to choose the iterative process for solving f(x) = 0.
Examples
2. The reciprocal formula F(x) = x (2 − Ax).
F′(x) = (2 − Ax) + x(−A) = 2 − 2Ax
⇒ F′ 1
A = 2 − 2A 1
A = 0
⇒ We have to calculate the second derivative
at the root X =
1
A
.
F′′(x) = −2A ⇒ F′′ 1
A = −2A.
⇒ Second Order Process.
⇒ The convergence is quadratic.
How to choose the iterative process for solving f(x) = 0.
Examples
3. The quadratic equation f(x) = x2 − 6x + 2 = 0. The two roots
of this equation are R1 = 3 + √7, R2 = 3 − √7. Consider the
following 4 different iterative processes.
(a) xn+1 = 6 − 2
xn
.
(b) xn+1 = 1
6x2
n + 1
3 .
(c) xn+1 = √6xn − 2.
(d) xn+1 = x2
−2
2xn−6 .
How to choose the iterative process for solving f(x) = 0.
The performance of the the four different iterative processes is
summarized in the following table.
Iterative process R1 attainable R2 attainable order of process
xn+1 = 6 − 2
xn
. NO YES 1
xn+1 = 1
6x2
n + 1
3 . YES NO 1
xn+1 = √6xn − 2. NO YES 1
xn+1 = x2
−6
2xn−6 . YES YES 2
⇒ The fourth iterative process performs better: both roots are
attainable and it converges faster.
Q: Is there a systematic method for finding the best method for a
given nonlinear equation f(x) = 0? A: Yes, the
Newton–Raphson Method.
The Newton–Raphson Method
Consider the equation f(x) = 0 and let X be the root of this
equation: f(X) = 0. Let x1 be our initial approximation and write
X = x1 + ǫ1.
Use the Taylor series expansion to obtain
0 = f(X) = f(x1 + ǫ1) = f(x1) + ǫ1f′(x1) + . . .
Assuming that f′(x1) 6= 0, we solve this equation for ǫ1 to obtain
ǫ1 ≈ −
f(x1)
f′(x1)
.
Hence, a better approximation to X than x1 would appear to be
x2 = x1 −
f(x1)
f′(x1)
.
The Newton–Raphson Method
We can repeat this procedure to obtain the iterative process
xn+1 = xn −
f(xn)
f′(xn)
=: F(xn). (6)
This is the Newton–Raphson iterative process. The
Newton–Raphson process converges to a particular root of f(x) = 0
if x1 is suitably chosen, assuming that all the roots are attainable.
The process is usually second order convergent if X is a
simple root:
The Newton–Raphson Method
F(x) = x −
f(x)
f′(x) ⇒
F′(x) = 1 −
(f′(x))2 − f(x)f′′(x)
(f′(x))2 =
(f′(x))2 − (f′(x))2 + f(x)f′′(x)
(f′(x))2
=
f(x)f′′(x)
(f′(x))2 ⇒
F′(X) =
f(X)f′′(X)
(f′(X))2 = 0, since f(X) = 0.
If X is not a simple root then the Newton–Raphson process is
usually first order, but it still converges.
The Newton–Raphson Method: Examples
1. The square root formula:
f(x) = x2 − A = 0 ⇒ f′(x) = 2x ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
x2
n − A
2xn
= xn −
xn
2
+
A
2xn
=
1
2 xn +
A
xn .
The Newton–Raphson Method: Examples
2. The reciprocal formula:
f(x) =
1
x − A = 0 ⇒ f′(x) = −
1
x2 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
1
xn − A
− 1
x2
n
= xn + x2
n 1
xn − A
= xn + xn − Ax2
n
= xn(2 − Axn).
The Newton–Raphson Method: Examples
3. A quadratic equation:
f(x) = x2 − 6x + 2 = 0 ⇒ f′(x) = 2x − 6 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
x2
n − 6xn + 2
2xn − 6
=
xn(2xn − 6) − x2
n + 6xn − 2
2xn − 6
=
2x2
n − 6xn − x2
n + 6xn − 2
2xn − 6
=
x2
n − 2
2xn − 6
.
The Newton–Raphson process converges to R1 or R2 depending
on the initial guess:
• x1 < r1 =" 3"> 3 ⇒ xn → R2, R2 = 3 + √7.
• x1 = 3 ⇒ x2 = +∞, the method diverges.
The Newton–Raphson Method: Examples
4. The equation sin x − 1
x = 0.
f(x) = sin x −
1
x
= 0 ⇒ f′(x) = cos x +
1
x2 ⇒
xn+1 = xn −
f(xn)
f′(xn)
= xn −
sin xn − 1
xn
cos xn + 1
x2
n
= xn − x2
n
sin xn − 1
xn
x2
n cos xn + 1
= xn −
x2
n sin xn − xn
x2
n cos xn + 1
.
The equation sin x − 1/x = 0 has infinitely many solutions. Let
us find the first positive solution R = 1.11416. Start with
x1 = 0.5.
x1 = 0.5 ⇒ x2 = 0.8117 ⇒ x3 = 1.0413
⇒ x4 = 1.1095 ⇒ x5 = 1.1141.
The Newton–Raphson Method: Examples
0.5 1 1.5 2 2.5 3 3.5 4
−1
−0.5
0
0.5
1
1.5
2
sin(x)
1/x
0.5 1 1.5 2 2.5 3 3.5 4
−1.5
−1
−0.5
0
0.5
a. f1(x) = sin(x), f2(x) = 1/x. b. f(x) = sin(x) − 1/x.
The Newton–Raphson Method: Examples
0 5 10 15 20 25 30
−1
−0.5
0
0.5
1
1.5
2
sin(x)
1/x
f1(x) = sin(x), f2(x) = 1/x.
The Newton–Raphson Method: Examples
We want to show that a solution to the equation
f(x) = sin x − 1
x = 0 exists in the interval (
4 ,
2 ). We have that
f′(x) = cos x +
1
x2 > 0 for x ∈ hπ
4
,
π
2 i.
Consequently, the function f(x) is strictly increasing in the interval
[
4 ,
2 ]. Furthermore,
f(π/4) = sin(π/4) −
1
π/4
=
√2
2 −
4
π
< 2 =" 1"> 0.
Since f(x) is continuous and strictly increasing, there exists an
R ∈ (
4 ,
2 ) such that f(R) = 0. This shows that there exists a root
of the equation f(x) = 0 in the interval (
4 ,
2 ).
The Newton–Raphson Method: Examples
We want to show that the equation
f(θ) = sin(θ) − θ cos(θ) −
1
2
π = 0
in the interval (
2 , 2
3 ).
f′(θ) = cos θ − cos θ + θ sin θ = θ sin θ > 0 for θ ∈ π
2
,

3 .
Consequently, the function f(x) is strictly increasing in the interval
[
2 , 3
2 ]. Furthermore,
f(
π
2
) = sin(
π
2
) −
π
2
cos(
π
2
) −
π
2
= 1 −
π
2
< 3 =" 0.3424"> 0.
Since f(x) is continuous and strictly increasing, there exists an
R ∈ (
2 , 2
3 ) such that f(R) = 0. This shows that there exists a
root of the equation f(x) = 0 in the interval (
2 , 2
3 ).
The Newton–Raphson Method: Examples
Now we find the Newton–Raphson process for this equation. We
have that
f′(θ) = θ sin(θ).
The Newton–Raphson process is
θn+1 = θn −
sin θn − θn cos θn −
2
θn sin θn
.
The root of the equation f(θ) = sin(θ) − θ cos(θ) − 1
2π = 0 in the
interval (
2 , 2
3 ) is R = 1.9057. We calculate it using the
Newton-Raphson process with x1 =
2 :
x1 =
π
2 ⇒ x2 = 1.9342 ⇒ x3 = 1.9058 ⇒ x4 = 1.9057.
The Newton–Raphson Method: Examples
1.5 1.6 1.7 1.8 1.9 2 2.1 2.2
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0 2 4 6 8 10
−8
−6
−4
−2
0
2
4
6
8
a. θ ∈
2 , 2
3 . b. θ ∈ [0, 10].
f(θ) = sin(θ) − θ cos(θ) − 1
2π.
Numerical solution of nonlinear equations
Final Remarks
• One can also define higher order processes:
F′(X) = 0, F′′(X) = 0, . . . F(n)(X) 6= 0.
• This is an nth order process.
• There are many other methods for solving nonlinear equations
other than the Newton–Raphson method: Bisection method,
secant method etc.
• The Newton–Raphson methods works also for equations on the
complex plane:
f(z) = 0, z ∈ C.