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Page 1
NUMERICAL – METHODS
Numerical solution of algebraic equations
? Descartes Rule of sign:
An equation f(x) = 0 cannot have more positive roots then the number of sign
changes in f(x) & cannot have more negative roots then the number of sign
changes in f(-x).
? Bisection Method
If a function f(x) is continuous between a & b and f(a) & f(b) are of opposite sign,
then there exists at least one roots of f(x) between a & b.
Since root lies between a & b, we assume root
? ?
0
ab
x
2
?
?
If
? ?
0
f x 0 ? ;
0
x is the root
Else, if
? ?
0
fx has same sign as ? ? fa , then roots lies between
0
x & b and
we assume
0
1
xb
x
2
?
? , and follow same procedure otherwise if
? ?
0
fx has same sign
as ? ? fb , then
root lies between a &
0
x & we assume
0
1
ax
x
2
?
? & follow same procedure.
We keep on doing it, till ? ?
n
f x ?? , i.e., ? ?
n
fx is close to zero.
No. of step required to achieve an accuracy ?
e
e
ba
log
n
log 2
? ??
??
?
??
?
? Regula-Falsi Method
This method is similar to bisection method, as we assume two value
1 0
x & x
such that
? ? ? ?
1 0
f x f x 0 ? .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
If f(x 2)=0 then x 2 is the root , stop the process.
Page 2
NUMERICAL – METHODS
Numerical solution of algebraic equations
? Descartes Rule of sign:
An equation f(x) = 0 cannot have more positive roots then the number of sign
changes in f(x) & cannot have more negative roots then the number of sign
changes in f(-x).
? Bisection Method
If a function f(x) is continuous between a & b and f(a) & f(b) are of opposite sign,
then there exists at least one roots of f(x) between a & b.
Since root lies between a & b, we assume root
? ?
0
ab
x
2
?
?
If
? ?
0
f x 0 ? ;
0
x is the root
Else, if
? ?
0
fx has same sign as ? ? fa , then roots lies between
0
x & b and
we assume
0
1
xb
x
2
?
? , and follow same procedure otherwise if
? ?
0
fx has same sign
as ? ? fb , then
root lies between a &
0
x & we assume
0
1
ax
x
2
?
? & follow same procedure.
We keep on doing it, till ? ?
n
f x ?? , i.e., ? ?
n
fx is close to zero.
No. of step required to achieve an accuracy ?
e
e
ba
log
n
log 2
? ??
??
?
??
?
? Regula-Falsi Method
This method is similar to bisection method, as we assume two value
1 0
x & x
such that
? ? ? ?
1 0
f x f x 0 ? .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
If f(x 2)=0 then x 2 is the root , stop the process.
If f(x 2)>0 then
? ? ? ?
? ? ? ?
?
?
?
2 0
0
02
3
2
f x .x f x .x
x
f x f x
If f(x 2)<0 then
? ? ? ?
? ? ? ?
?
?
?
1 2
2
21
3
1
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Secant Method
In secant method, we remove the condition that
? ? ? ?
1 0
f x f x 0 ? and it doesn’t
provide the guarantee for existence of the root in the given interval , So it is called
un reliable method .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
and to compute x 3 replace every variable by its variable in x 2
? ? ? ?
? ? ? ?
?
?
?
2 1
1
12
3
2
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Newton-Raphson Method
? ?
? ?
n
n
n1
n
fx
xx
f ' x
?
??
Note : Since N.R. iteration method is quadratic convergence so to apply this formula
must exist.
Order of convergence
? Bisection = Linear
? Regula false = Linear
? Secant = superlinear
? Newton Raphson = quadratic
? ?
f " x
Page 3
NUMERICAL – METHODS
Numerical solution of algebraic equations
? Descartes Rule of sign:
An equation f(x) = 0 cannot have more positive roots then the number of sign
changes in f(x) & cannot have more negative roots then the number of sign
changes in f(-x).
? Bisection Method
If a function f(x) is continuous between a & b and f(a) & f(b) are of opposite sign,
then there exists at least one roots of f(x) between a & b.
Since root lies between a & b, we assume root
? ?
0
ab
x
2
?
?
If
? ?
0
f x 0 ? ;
0
x is the root
Else, if
? ?
0
fx has same sign as ? ? fa , then roots lies between
0
x & b and
we assume
0
1
xb
x
2
?
? , and follow same procedure otherwise if
? ?
0
fx has same sign
as ? ? fb , then
root lies between a &
0
x & we assume
0
1
ax
x
2
?
? & follow same procedure.
We keep on doing it, till ? ?
n
f x ?? , i.e., ? ?
n
fx is close to zero.
No. of step required to achieve an accuracy ?
e
e
ba
log
n
log 2
? ??
??
?
??
?
? Regula-Falsi Method
This method is similar to bisection method, as we assume two value
1 0
x & x
such that
? ? ? ?
1 0
f x f x 0 ? .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
If f(x 2)=0 then x 2 is the root , stop the process.
If f(x 2)>0 then
? ? ? ?
? ? ? ?
?
?
?
2 0
0
02
3
2
f x .x f x .x
x
f x f x
If f(x 2)<0 then
? ? ? ?
? ? ? ?
?
?
?
1 2
2
21
3
1
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Secant Method
In secant method, we remove the condition that
? ? ? ?
1 0
f x f x 0 ? and it doesn’t
provide the guarantee for existence of the root in the given interval , So it is called
un reliable method .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
and to compute x 3 replace every variable by its variable in x 2
? ? ? ?
? ? ? ?
?
?
?
2 1
1
12
3
2
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Newton-Raphson Method
? ?
? ?
n
n
n1
n
fx
xx
f ' x
?
??
Note : Since N.R. iteration method is quadratic convergence so to apply this formula
must exist.
Order of convergence
? Bisection = Linear
? Regula false = Linear
? Secant = superlinear
? Newton Raphson = quadratic
? ?
f " x
? Numerical Integration
Trapezoidal Rule
? ?
b
a
f x dx
?
, can be calculated as
Divide interval (a, b) into n sub-intervals such that width of each interval
? ? ba
h
n
?
?
we have (n + 1) points at edges of each intervals
? ?
12 n
0
x ,x ,x ,..........,x
? ? ? ? ? ?
nn
0 0 1 1
y f x ; y f x ,...................,y f x ? ? ?
? ? ? ?
12
b
n
0 n 1
a
h
f x dx y 2 y y .......... y y
2
?
??
? ? ? ? ? ?
??
?
Simpson’s
1
3
rd Rule
Here the number of intervals should be even
ba
h
n
?
??
?
??
??
? ? ? ? ? ?
54 1 3 2
b
n
0 n 1 n 2
a
h
f x dx y 4 y y y .......... y 2 y y ................ y y
3
??
??
? ? ? ? ? ? ? ? ? ? ?
??
?
Simpson’s
3
8
th Rule
Here the number of intervals should be even
? ?
??
?? ? ? ? ? ? ? ? ? ?
?? ?
b
n 4 5........ 0 1 2 n 1 3 6 9 n 3
a
3h
f x dx y 3(y y y y y ) 2(y y y ..............y ) y
8
Page 4
NUMERICAL – METHODS
Numerical solution of algebraic equations
? Descartes Rule of sign:
An equation f(x) = 0 cannot have more positive roots then the number of sign
changes in f(x) & cannot have more negative roots then the number of sign
changes in f(-x).
? Bisection Method
If a function f(x) is continuous between a & b and f(a) & f(b) are of opposite sign,
then there exists at least one roots of f(x) between a & b.
Since root lies between a & b, we assume root
? ?
0
ab
x
2
?
?
If
? ?
0
f x 0 ? ;
0
x is the root
Else, if
? ?
0
fx has same sign as ? ? fa , then roots lies between
0
x & b and
we assume
0
1
xb
x
2
?
? , and follow same procedure otherwise if
? ?
0
fx has same sign
as ? ? fb , then
root lies between a &
0
x & we assume
0
1
ax
x
2
?
? & follow same procedure.
We keep on doing it, till ? ?
n
f x ?? , i.e., ? ?
n
fx is close to zero.
No. of step required to achieve an accuracy ?
e
e
ba
log
n
log 2
? ??
??
?
??
?
? Regula-Falsi Method
This method is similar to bisection method, as we assume two value
1 0
x & x
such that
? ? ? ?
1 0
f x f x 0 ? .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
If f(x 2)=0 then x 2 is the root , stop the process.
If f(x 2)>0 then
? ? ? ?
? ? ? ?
?
?
?
2 0
0
02
3
2
f x .x f x .x
x
f x f x
If f(x 2)<0 then
? ? ? ?
? ? ? ?
?
?
?
1 2
2
21
3
1
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Secant Method
In secant method, we remove the condition that
? ? ? ?
1 0
f x f x 0 ? and it doesn’t
provide the guarantee for existence of the root in the given interval , So it is called
un reliable method .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
and to compute x 3 replace every variable by its variable in x 2
? ? ? ?
? ? ? ?
?
?
?
2 1
1
12
3
2
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Newton-Raphson Method
? ?
? ?
n
n
n1
n
fx
xx
f ' x
?
??
Note : Since N.R. iteration method is quadratic convergence so to apply this formula
must exist.
Order of convergence
? Bisection = Linear
? Regula false = Linear
? Secant = superlinear
? Newton Raphson = quadratic
? ?
f " x
? Numerical Integration
Trapezoidal Rule
? ?
b
a
f x dx
?
, can be calculated as
Divide interval (a, b) into n sub-intervals such that width of each interval
? ? ba
h
n
?
?
we have (n + 1) points at edges of each intervals
? ?
12 n
0
x ,x ,x ,..........,x
? ? ? ? ? ?
nn
0 0 1 1
y f x ; y f x ,...................,y f x ? ? ?
? ? ? ?
12
b
n
0 n 1
a
h
f x dx y 2 y y .......... y y
2
?
??
? ? ? ? ? ?
??
?
Simpson’s
1
3
rd Rule
Here the number of intervals should be even
ba
h
n
?
??
?
??
??
? ? ? ? ? ?
54 1 3 2
b
n
0 n 1 n 2
a
h
f x dx y 4 y y y .......... y 2 y y ................ y y
3
??
??
? ? ? ? ? ? ? ? ? ? ?
??
?
Simpson’s
3
8
th Rule
Here the number of intervals should be even
? ?
??
?? ? ? ? ? ? ? ? ? ?
?? ?
b
n 4 5........ 0 1 2 n 1 3 6 9 n 3
a
3h
f x dx y 3(y y y y y ) 2(y y y ..............y ) y
8
Truncation error
Trapezoidal Rule:
? ?
?
?
??
2
bound
(b a)
T h max f"
12
and order of error =2
Simpson’s
1
3
Rule:
? ?
? ?
?
?
??
4
iv
bound
(b a)
T h max f
180
and order of error =4
Simpson’s
3
8
th Rule:
? ?
? ?
?
?
??
4
iv
bound
3(b a)
T h max f
n80
and order of error =5
where ? ? ?
n 0
xx
Note : If truncation error occurs at n
th
order derivative then it gives exact result while
integrating the polynomial up to degree (n-1).
Numerical solution of Differential equation
Euler’s Method
? ?
dy
f x, y
dx
?
To solve differential equation by numerical method, we define a step size h
We can calculate value of y at
? ?
0 0 0
x h, x 2h,..........,x nh ? ? ? & not any
intermediate points.
? ?
i 1 i i i
y y hf x , y
?
??
? ?
?
i i
y y x ;
? ?
i 1 i 1
y y x
? ?
? ;
?
??
i
i 1
X X h
Modified Euler’s Method (Heun’s method)
? ? ? ?
??
? ? ? ? ?
??
0 1 0 0 0 0
h
y y f x ,y f x h, y h
2
Runge – Kutta Method
10
y y k ??
? ?
4 1 2 3
1
k k 2k 2k k
6
? ? ? ?
? ?
00 1
k hf x ,y ?
1
00 2
k h
k hf x ,y
22
??
? ? ?
??
??
Page 5
NUMERICAL – METHODS
Numerical solution of algebraic equations
? Descartes Rule of sign:
An equation f(x) = 0 cannot have more positive roots then the number of sign
changes in f(x) & cannot have more negative roots then the number of sign
changes in f(-x).
? Bisection Method
If a function f(x) is continuous between a & b and f(a) & f(b) are of opposite sign,
then there exists at least one roots of f(x) between a & b.
Since root lies between a & b, we assume root
? ?
0
ab
x
2
?
?
If
? ?
0
f x 0 ? ;
0
x is the root
Else, if
? ?
0
fx has same sign as ? ? fa , then roots lies between
0
x & b and
we assume
0
1
xb
x
2
?
? , and follow same procedure otherwise if
? ?
0
fx has same sign
as ? ? fb , then
root lies between a &
0
x & we assume
0
1
ax
x
2
?
? & follow same procedure.
We keep on doing it, till ? ?
n
f x ?? , i.e., ? ?
n
fx is close to zero.
No. of step required to achieve an accuracy ?
e
e
ba
log
n
log 2
? ??
??
?
??
?
? Regula-Falsi Method
This method is similar to bisection method, as we assume two value
1 0
x & x
such that
? ? ? ?
1 0
f x f x 0 ? .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
If f(x 2)=0 then x 2 is the root , stop the process.
If f(x 2)>0 then
? ? ? ?
? ? ? ?
?
?
?
2 0
0
02
3
2
f x .x f x .x
x
f x f x
If f(x 2)<0 then
? ? ? ?
? ? ? ?
?
?
?
1 2
2
21
3
1
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Secant Method
In secant method, we remove the condition that
? ? ? ?
1 0
f x f x 0 ? and it doesn’t
provide the guarantee for existence of the root in the given interval , So it is called
un reliable method .
? ? ? ?
? ? ? ?
?
?
?
1 0
0
01
2
1
f x .x f x .x
x
f x f x
and to compute x 3 replace every variable by its variable in x 2
? ? ? ?
? ? ? ?
?
?
?
2 1
1
12
3
2
f x .x f x .x
x
f x f x
Continue above process till required root not found
? Newton-Raphson Method
? ?
? ?
n
n
n1
n
fx
xx
f ' x
?
??
Note : Since N.R. iteration method is quadratic convergence so to apply this formula
must exist.
Order of convergence
? Bisection = Linear
? Regula false = Linear
? Secant = superlinear
? Newton Raphson = quadratic
? ?
f " x
? Numerical Integration
Trapezoidal Rule
? ?
b
a
f x dx
?
, can be calculated as
Divide interval (a, b) into n sub-intervals such that width of each interval
? ? ba
h
n
?
?
we have (n + 1) points at edges of each intervals
? ?
12 n
0
x ,x ,x ,..........,x
? ? ? ? ? ?
nn
0 0 1 1
y f x ; y f x ,...................,y f x ? ? ?
? ? ? ?
12
b
n
0 n 1
a
h
f x dx y 2 y y .......... y y
2
?
??
? ? ? ? ? ?
??
?
Simpson’s
1
3
rd Rule
Here the number of intervals should be even
ba
h
n
?
??
?
??
??
? ? ? ? ? ?
54 1 3 2
b
n
0 n 1 n 2
a
h
f x dx y 4 y y y .......... y 2 y y ................ y y
3
??
??
? ? ? ? ? ? ? ? ? ? ?
??
?
Simpson’s
3
8
th Rule
Here the number of intervals should be even
? ?
??
?? ? ? ? ? ? ? ? ? ?
?? ?
b
n 4 5........ 0 1 2 n 1 3 6 9 n 3
a
3h
f x dx y 3(y y y y y ) 2(y y y ..............y ) y
8
Truncation error
Trapezoidal Rule:
? ?
?
?
??
2
bound
(b a)
T h max f"
12
and order of error =2
Simpson’s
1
3
Rule:
? ?
? ?
?
?
??
4
iv
bound
(b a)
T h max f
180
and order of error =4
Simpson’s
3
8
th Rule:
? ?
? ?
?
?
??
4
iv
bound
3(b a)
T h max f
n80
and order of error =5
where ? ? ?
n 0
xx
Note : If truncation error occurs at n
th
order derivative then it gives exact result while
integrating the polynomial up to degree (n-1).
Numerical solution of Differential equation
Euler’s Method
? ?
dy
f x, y
dx
?
To solve differential equation by numerical method, we define a step size h
We can calculate value of y at
? ?
0 0 0
x h, x 2h,..........,x nh ? ? ? & not any
intermediate points.
? ?
i 1 i i i
y y hf x , y
?
??
? ?
?
i i
y y x ;
? ?
i 1 i 1
y y x
? ?
? ;
?
??
i
i 1
X X h
Modified Euler’s Method (Heun’s method)
? ? ? ?
??
? ? ? ? ?
??
0 1 0 0 0 0
h
y y f x ,y f x h, y h
2
Runge – Kutta Method
10
y y k ??
? ?
4 1 2 3
1
k k 2k 2k k
6
? ? ? ?
? ?
00 1
k hf x ,y ?
1
00 2
k h
k hf x ,y
22
??
? ? ?
??
??
2
00 3
k h
k hf x ,y
22
??
? ? ?
??
??
? ?
0 0 3 4
k hf x h, y k ? ? ?
Similar method for other iterations
Read More
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Short Notes: Numerical-Methods Electrical Engineering (EE) Questions
The "Short Notes: Numerical-Methods Electrical Engineering (EE) Questions" guide is a valuable resource for all aspiring students preparing for the
Electrical Engineering (EE) exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
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The content of Short Notes: Numerical-Methods is prepared as per the latest Electrical Engineering (EE) syllabus.
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