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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
Short Notes: Numerical-Methods Notes offer in-depth insights into the specific topic to help you master it with ease.
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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
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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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