1. NUMERICAL METHODS IN ELECTRICAL ENGINEERING
CEE203
NUMERICAL METHODS IN ELECTRICAL ENGINEERING
Assist. Prof. Dr. Çağatay ULUIŞIK

2. OUTLINE Introducing MatLab Fundamental Terms, Basic Definitions
Solving Nonlinear Equations – Root Finding
Solving Systems of Linear Equations
Taylor Series
Numerical Differentiation
Numerical Integration
Curve Fitting and Interpolation
Ordinary Differential Equations

3. INTRODUCING MATLAB Vectors, Arrays and Matrices
Assign a value to a variable
(observe the effect of the semicolon):
Creating a row vector of size (14)
Creating a column vector of size (41)
Creating an array of size (19) using the colon sign (:)

4. INTRODUCING MATLAB
Creating an array of size (19) using the for loop :
Creating a matrix of size (33) :
Creating a matrix of size (33) using 2 for loops :

5. BASIC MATRIX OPERATIONS IN MATLAB

6. GRAPHIC UTILITIES OF MATLAB

7. GRAPHIC UTILITIES OF MATLAB

8. COMPLEX NUMBERS IN MATLAB
Defining a Complex Number :
Conjugate :
Magnitude :
Angle in Radians :
Angle in Degrees :
Addition :
Subtraction :
Multiplication :
Division :

9. LOADING DATA FROM A FILE AND SAVING DATA TO A FILE

10. STATISTICS IN MATLAB
Generation of Uniformly Distributed Random Numbers :
Generation of Uniformly Distributed Random Integers :
Mean Value :
Variance :
Standard Deviation :

11. INTRODUCING MATLAB Roots of a Polynom : Symbolic Integration :
Numeric Integration :
clc : To clear the command window
clear all : To clear all the variables in memory

12. FORMAT OF DATA round floor ceil Round towards nearest integer
Round towards minus infinity
ceil
Round towards plus infinity

13. EXAMPLES OF MATLAB SCRIPTS
Example 1 : Calculating of the sum of integers from 1 to a user-specifed value n a) using a formula b)using iterations
Example 2 : Creating and Calling a function

14. EXAMPLES OF MATLAB SCRIPTS
Example 3 : Error of the first n terms in the Taylor Series of the function ex a) Calculating the error for a given n value b) Calculating the minimum required number of terms n for a given error. a) b)

15. FUNDAMENTAL TERMS
Accuracy: Closeness of agreement between a measured / computed value and a true value.
Measurement accuracy: the ability of an instrument to measure the true value to within some stated error specifications.
Numerical accuracy: the degree to which the numerical solution to the approximate physical problem approximates the exact solution to the approximate physical problem.

16. MEASUREMENT ACCURACY
Comparison of a measured value with the real value
Meas. Value
Meas. frequency
Mean
Value
interval
Accuracy =
(%)
True val. – Meas. val
True value
True value ?

17. ERROR
It is the difference between a measured or calculated value of a quantity and its exact value.
Systematic error plagues experiments or calculations caused by negative factors. For example, a DC voltage component, which unintentionally is present, e.g., because of a failure on the blockage capacitor, is a systematic error. These can be removed once understood/discovered via controls and calibration.
Random error is an error which is always present, but varies unpredictably in size and direction. They are related to the scatter in the data obtained under fixed conditions which determine the repeatability (precision) of the measurement and follow well-behaved statistical rules.

18. ERROR
Absolute error:
It is an error that is expressed in physical units. It is the absolute value of the difference between the measured value and the true value (or the average value if the true value is not known) of a quantity.
Relative error:
An error expressed as a fraction of the absolute error to the true (or average) value of a quantity. It is always given as a percentage.

19. ERROR
A number is represented with a finite (fixed) number of digits called word length.
Precision of a measurement or the accuracy of a computation bounds the number of significant digits. All non-zero digits beyond the number of significant digits at the right of a number are removed in one of two ways; truncation or round-off.
For example, has 6 significant digits. If it is going to be represented only by 4 significant digits both truncation and round-off processes yield On the other hand, if the number of significant digits will be 3, then truncation and round-off processes yield, 53.0 and 53.1, respectively.

20. ERROR CALCULATIONS
A measurement/calculation result should be given as (a  a), which means the value may be anything between (a - a) and (a + a).
For example, if the measured speed is 98  3 km/hr, then the real value may be anything between 95 and 101 km/hr.
Total error is the sum of individual errors for the arithmetic combination of two measurements.
Total relative error is the sum of individual relative errors for the mulitlicative combination of two measurements.

21. ERROR CALCULATIONS
A parameter (say “A”) is going to be estimated / calculated from two measurements (say “B” and “C”), with the measurement errors of “b” and “c”, respectively.
If A = B + C then total error will be
If A = B  C then total error will be

22. ERROR CALCULATIONS
In general, the total (propagated) error is obtained from these two properties as
For a multi-variable function, the total error is calculated from

23. UNCERTAINTY
Uncertainty is a range that is likely to contain the true value of a quantity being measured or calculated. Uncertainty can be expressed in absolute or relative terms.
Modeling uncertainty is defined as the potential deficiency due to a lack of information.
Modeling error is the recognizable deficiency not due to a lack of information but due to the approximations and simplifications made there.
Measurement error is the difference between the measured and true values, while measurement uncertainty is an estimate of the error in a measurement.

24. UNCERTAINTY
Modeling and simulation uncertainties occur during the phase of
Conceptual modeling of the physical system
Mathematical modeling of the conceptual model
Discretization
Computer modeling of the mathematical model
Computer modeling of the discrete model
Numerical solution
Numerical uncertainties occur during computations due to round-off, turncation, non-convergence, artificial dissipations, etc.

25. ERROR / UNCERTAINTY Digital Voltmeter Accuracy:
(0.25 % Reading + 2 digit)
Meaning; multiply the value you read on the multimeter by % (i.e., by ) . This is called scaling error.
Add 2 times the value of the least significant digit to the result. This is called quantization error.
Example: You read V from a 20 V scale.
Measurement error :  = V
Quantization error : 2  0.01 = 0.02 V
Total error :  =  V

26. MODELING ERROR
Truncation error is also defined for mathematical modeling. For example, Taylor’s expansion or a Fourier-series representation are used to replace a function in terms of an infinite summation.
Taking only a given number of low-order terms, and neglecting the rest of the higher-order terms, introduces a truncation error.
For example, the Taylor expansion of the exponent function:
yields (for n=3)

27. 2Pt 3Pt + O(∆x) + O(∆x2) FDTD MODELING ERROR Derivative
f(x)
Mathematical definition
Finite difference approximation
2Pt
+ O(∆x)
3Pt
+ O(∆x2)

28. SOLVING NONLINEAR EQUATIONS
Fixed Point Iteration Method
Bisection Method
Secant Method
Newton-Raphson Method
MatLab Built-in Functions

29. FIXED POINT ITERATION METHOD
The equation f(x)=0 is rewritten in the form :
The intersection point of the graphs of the functions y=x and y=g(x) is the solution and is called the fixed point .
The numerical solution is obtained by an iterative process.
First a xi value near the fixed point is chosen and substituted into g(x). The solution is substituted back into g(x). So the iteration formula is given by:

30. FIXED POINT ITERATION METHOD
g(x1) x
g(x2)
x3=
g(x3)
x4=
g(x4)
x5=
g(x5)
x6=
x7=
g(x6)
g(x7)
g(x)
y=x

31. FIXED POINT ITERATION METHOD
Example:
Convergence Test: i xi 1
6.0000 2
0.8571 3
3.2308 4
1.4182 5
2.4812 6
1.7235 7
2.2030 8
1.8732 9
2.0882 10
1.9429 11
2.0388
y=x
g(x)=6/(x+1)

32. FIXED POINT ITERATION METHOD
Choosing the appropriate iteration function g(x)
The fixed point iteration method converges if in the neighborhood of the fixed point the derivative of g(x) has an absolute value smaller than 1. (also called Lipschitz continious)
Example:
The plot of the function shows that the equation has a solution between 1 and 2.
The actual solution is at x=
f(x)=xe0.5x+1.2x-5 x

33. FIXED POINT ITERATION METHOD
Case A
Case B
Case C
Divergent
Convergent i xi 1
1.0000 2
2.7927 3 4
4.4849 5 6
4.1667 7 8
4.1668 9 10 11 i xi 1
1.0000 2
1.7552 3
1.3869 4
1.5622 5
1.4776 6
1.5182 7
1.4987 8
1.5080 9
1.5035 10
1.5057 11
1.5046 i xi 1
1.0000 2
2.3048 3
0.7057 4
2.9183 5
0.3482 6
3.8500 7
0.0554 8
4.7986 9 10
5.2606 11

34. BISECTION METHOD
STEP 1: Choose two points a and b such that a solution exists between them : f(a) f(b) < 0
STEP 2: Find c=(a+b)/2
STEP 3: Determine whether the true solution is between a and c or between c and b. Then select the subinterval that contains the true solution
If f(a) f(c) < 0 , solution is between a and c anew=a, bnew=c
If f(b) f(c) < 0 , solution is between c and b anew=c, bnew=b
STEP 4: Repeat steps 2 and 3 until (b-a) is less than a specified tolerance

35. BISECTION METHOD f(a) f(b) f(a)  f(b) < 0 True Root f(a) f(b) f(b)
c=(a+b) /2
True Root
f(a)
f(b)
f(b)
f(a)
f(b)
f(a)

36. BISECTION METHOD a b c f(c) abs(b-a) 5.0000 12.0000 8.5000 18.7500
7.0000
6.7500
3.5000
7.6250
7.2656
1.7500
7.1875
2.0977
0.8750
6.9688
0.4375
7.0781
0.8655
0.2188
7.0234
0.2584
0.1094
6.9961
0.0547
7.0098
0.1075
0.0273
7.0029
0.0322
0.0137
6.9995
0.0068
7.0012
0.0134
0.0034
7.0004
0.0040
0.0017

37. NEWTON RAPSON METHOD f(x1) f(x) f(x2) Exact Root x x5 x4 x3 x2 x1
Slope: f’(x1)
f(x2)
Exact Root
Slope:
f’(x2)
Slope:
f’(x3) x x5 x4 x3 x2 x1
Slope:
f’(x4)

38. NEWTON RAPSON METHOD xi f(xi) xi -xi-1 6.00000000 31.00000000
1.3976
1.3239
1.1453
0.7849
0.3094
0.0383
0.0005

39. SECANT METHOD f(x1) f(x) f(x2) f(x3) Exact Root f(x4) f(x5) x x6 x5 x4

40. SECANT METHOD xi f(xi) xi -xi-1 6.00000000 31.00000000 5.00000000
1.0000
0.9375
0.9017
0.8062
0.6558
0.4375
0.2046
0.0517
0.0049
0.0001

41. MATLAB BUILT-IN FUNCTIONS
The roots(p) command finds the roots of a polynomial, whose coefficients are defined in the row vector p. The following MatLab script find the roots of the polynom P(x)=x3-3x2-13x+15
>> coef=[ ];
>> xi=roots(coef)
xi =
5.0000
1.0000
The fzero(‘function’,x0) command finds the roots of the ‘function’ near x0.
The following MatLab script find the roots of the function f(x)=x3-3x2-13x+15
>> x=fzero('x^3-3*x^2-13*x+15',7)
x = 5
>> x=fzero('x^3-3*x^2-13*x+15',3)
x = 1

42. SYSTEMS OF LINEAR EQUATIONS
Gauss Elimination Method
Gauss-Jordan Elimination Method
LU Decomposition Method

43. GAUSS ELIMINATION METHOD
Backward Substitution :

44. LU DECOMPOSITION METHOD
A is the coefficients matrix.
U is an upper triangular matrix and is obtained at the end of the Gauss elimination procedure.
L is a lower triangular matrix which has all 1’s on the diagonal and the elements below the diagonal are the multipliers mij that multiply the pivot equation when it is used to eliminate the elements below the pivot coefficient at the Gauss elimination procedure.
L is a lower triangular matrix and y is obtained from Ly=b using forward substitution.
U is an upper triangular matrix and x is obtained from Ux=y using backward substitution.

45. LU DECOMPOSITION METHOD
Forward Substitution :
Backward Substitution :

46. GAUSS-JORDAN ELIMINATION METHOD

47. GAUSS-JORDAN ELIMINATION METHOD

48. GAUSS-JORDAN ELIMINATION METHOD
MATLAB SCRIPT
A=[ ; ; ];
[n,col]=size(A); %
for i=1:n
A(i,:)=A(i,:)/A(i,i);
disp('A ='); disp(A); pause;
for k=i+1:n
dd=A(k,i);
A(k,:)=A(k,:)-dd*A(i,:); A
pause;
end
for i=n:-1:2
for k=i-1:-1:1
x=A(:,col);
disp(' The unknowns are :'); disp(x);
rref(A) % MatLab built-in function

49. TAYLOR SERIES
Taylor Series :
Maclaurin Series :
Example :

50. TAYLOR SERIES

51. NUMERICAL DIFFERENTIATION
Two-Point Forward Difference Method
Two-Point Backward Difference Method
Two-Point Central Difference Method
Three-Point Forward Difference Method
Three-Point Backward Difference Method

52. NUMERICAL DIFFERENTIATIONxi
xi+1
f(xi) h
f(xi+1)
True
derivative
Approximated
f(x)
True derivative xi
xi-1
f(xi) h
f(xi-1)
True
derivative
Approximated
f(x)
Forward Difference
Backward
Difference
Forward Difference:
xi+1
xi-1
f(xi+1) 2h
f(xi-1)
Approximated
derivative
f(x)
Central Difference
Backward Difference:
Central Difference:

53. NUMERICAL DIFFERENTIATION
Truncation
Error:
2 Point Forward Difference:
2 Point Backward Difference:
2 Point Central Difference:
3 Point Forward Difference:
3 Point Backward Difference:

54. NUMERICAL DIFFERENTIATION
Example : For f(x)=ln(x) estimate the value of f’(0.2) taking h=0.01.
xi-2
0.18
xi-1
0.19 xi
0.2
xi+1
0.21
xi+2
0.22
f(xi-2)
ln(0.18)
f(xi-1)
ln(0.19)
f(xi)
ln(0.2)
f(xi+1)
ln(0.21)
f(xi+2)
ln(0.22)
Method Used
f’(0.2)
Abs. Error
2 Point Forward Difference
4.879
0.121
2 Point Backward Difference
5.1293
0.1293
2 Point Central Difference
5.0042
0.0042
3 Point Forward Difference
4.9925
0.0075
3 Point Backward Difference
4.9906
0.0094

55. NUMERICAL DIFFERENTIATIONh
f’(0.8)
Abs. Error
0.7500
0.8819
0.3681
0.5000
0.9710
0.2790
0.4000
1.0137
0.2363
0.3000
1.0615
0.1885
0.2000
1.1157
0.1343
0.1000
1.1778
0.0722
0.0500
1.2125
0.0375
0.0100
1.2423
0.0077
0.0050
1.2461
0.0039
0.0010
1.2492
0.0008
0.0005
1.2496
0.0004
0.0001
1.2499 h
f’(0.8)
Abs. Error
0.7500
3.6968
2.4468
0.5000
1.9617
0.7117
0.4000
1.7329
0.4829
0.3000
1.5667
0.3167
0.2000
1.4384
0.1884
0.1000
1.3353
0.0853
0.0500
1.2908
0.0408
0.0100
1.2579
0.0079
0.0050
1.2539
0.0039
0.0010
1.2508
0.0008
0.0005
1.2504
0.0004
0.0001
1.2501
2 Point
Forward
Difference:
2 Point
Backward
Difference: h
f’(0.8)
Abs. Error
0.7500
2.2893
1.0393
0.5000
1.4663
0.2163
0.4000
1.3733
0.1233
0.3000
1.3141
0.0641
0.2000
1.2771
0.0271
0.1000
1.2566
0.0066
0.0500
1.2516
0.0016
0.0100
1.2501
6.51e-005
0.0050
1.2500
1.63e-005
0.0010
6.51e-007
0.0005
1.63e-007
0.0001
6.51e-009 h
f’(0.8)
Abs. Error
0.7500
1.0597
0.1903
0.5000
1.1311
0.1189
0.4000
1.1609
0.0891
0.3000
1.1903
0.0597
0.2000
1.2178
0.0322
0.1000
1.2399
0.0101
0.0500
1.2472
0.0028
0.0100
1.2499
0.0001
0.0050
3.21e-005
0.0010
1.298e-006
0.0005
3.25e-007
1.30e-008
2 Point
Central
Difference:
3 Point
Forward
Difference:

56. NUMERICAL DIFFERENTIATION

57. NUMERICAL DIFFERENTIATIONh
f’(1.5)
Abs. Error
1.0000
0.8000
0.2000
0.5000
0.1000
0.9000
0.0500
0.9500
0.0100
0.9900
0.0050
0.9950
0.0010
0.9990

58. NUMERICAL INTEGRATION
Rectangle Methods
Midpoint Method
Trapezoidal Method
Simpsons Methods
Simpsons 1/3 Method
Simpsons 3/8 Method

59. NUMERICAL INTEGRATION
1. Rectangle Method a
f(a) b
f(b)
f(x) a
f(a) b
f(b)
f(x)
2. Rectangle Method
Midpoint Method
Trapezoidal Method a
f(a) b
f(b)
f(x) a
f(a) b
f(b)
f(x)

60. NUMERICAL INTEGRATION
1. Rectangle Method
f(x)
f(x2)
f(x1) I1 I2 I3 Ii
Ii+1
IN-1 IN h h h h h h h   x1 a x2 x3 xi
xi+1
xN-1 xN
xN+1 b

61. NUMERICAL INTEGRATION
2. Rectangle Method x1 a x2  xi
xi+1
xN-1 xN
xN+1 b x3
f(x1) I1 I2 I3 Ii
Ii+1
IN-1 IN h
f(x2)
f(x)

62. NUMERICAL INTEGRATION
Midpoint Method x1 a x2  xi
xi+1
xN-1 xN
xN+1 b x3 I1 I2 I3 Ii
Ii+1
IN-1 IN h
f(x)
(x1+x2) 2
(x2+x3) 2

63. NUMERICAL INTEGRATION
Trapezoidal Method x1 a x2  xi
xi+1
xN-1 xN
xN+1 b x3
f(x1) I1 I2 I3 Ii
Ii+1
IN-1 IN h
f(x2)
f(x)

64. NUMERICAL INTEGRATION
Simpson 1/3 Method
A second order polynomial g(x) is used
to approximate the integrand f(x)
f(x)
g(x)
f(x1)
h/2=k
h/2=k x1 a x2 x3 b

65. NUMERICAL INTEGRATION
Simpson 3/8 Method
A third order polynomial g(x) is used
to approximate the integrand f(x)
f(x)
g(x)
f(x1)
h/3=k
h/3=k
h/3=k x1 a x2 x3 x4 b

66. NUMERICAL INTEGRATION

67. NUMERICAL INTEGRATION
1. Rectangular Method h
Num Int.
Abs. Error
Rel.
Error % 5
1115
325
22.569
2.5
1277.5
162.5
11.285 2
1310
130
9.0278 1
1375 65
4.5139
0.5
1407.5
32.5
2.2569
0.1
1433.5
6.5
0.05
1436.8
3.25
0.01
1439.3
0.65
0.005
1439.7
0.325
0.001
1439.9
0.065
0.0005
1440
0.0325
0.0001
0.0065
2. Rectangular Method h
Num Int.
Abs. Error
Rel. Error % 5
1765
325
22.569
2.5
1602.5
162.5
11.285 2
1570
130
9.0278 1
1505 65
4.5139
0.5
1472.5
32.5
2.2569
0.1
1446.5
6.5
0.05
1443.3
3.25
0.01
1440.6
0.65
0.005
1440.3
0.325
0.001
1440.1
0.065
0.0005
1440
0.0325
0.0001
0.0065

68. NUMERICAL INTEGRATION
1. Rect. Method
2. Rect. Method h
Num Int.
Abs.
Error 8
400
10.667
720
330.67 4
352
37.333
512
122.67 2
360
29.333
440
50.667 1
372
17.333
412
22.667
0.5
380
9.3333
0.1
387.36
1.9733
391.36
2.0267
0.05
388.34
390.34
1.0067
0.01
389.13
389.53
0.005
389.23
389.43
0.001
389.31
389.35
0.0005
389.32
389.34
0.0001
389.33
0.002
Midpoint Method h
Num Int.
Abs.
Error 8
304
85.333 4
368
21.333 2
384
5.3333 1
388
1.3333
0.5
389
0.1
389.32
0.05
389.33
0.01
0.005
3.3333e-005
0.001
1.3333e-006
0.0005
3.3333e-007
0.0001
1.3332e-008
Trapezoidal Method h
Num Int.
Abs.
Error 8
560
170.67 4
432
42.667 2
400
10.667 1
392
2.6667
0.5
390
0.1
389.36
0.05
389.34
0.01
389.33
0.005
6.6667e-005
0.001
2.6667e-006
0.0005
6.6667e-007
0.0001
2.6663e-008

69. NUMERICAL INTEGRATION
(h=0.1)
Numerical Integration
Absolute
Error
1. Rectangle Method
387.36
1.9733
2. Rectangle Method
391.36
2.0267
Midpoint Method
389.32
Trapezoidal Method
389.36
Simpsons 1/3 Method
389.33
5.6843e-014
Simpsons 3/8 Method

70. CURVE FITTING WITH A LINEAR FUNCTION
f(x)=a1x+a0
(xi ,yi) yi rN yN ri
(xN ,yN)
(x1 ,y1) y1 r1 r2 y2
(x2 ,y2) x1 x2 xi xN
A linear function f(x)=a1x+a0 is used to best fit the given data points (xi ,yi).
A residual ri at point (xi , yi) is the difference between the value yi of the data point and the value of the function f(xi) used to approximate the data points : ri= yi -f(xi)

71. CURVE FITTING WITH A LINEAR FUNCTION
The overall error E is defined as the sum of the squares of the residuals :
Linear least-squares regression :
The coefficients a1 and a0 of the linear function f(x)=a1x+a0 are determined such that the error E has the smallest possible value. E is minimum if:
The coefficients a1 and a0 are found as:

72. CURVE FITTING WITH A LINEAR FUNCTION
Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below. x 1 2 3 4 5 6 7 8 9 10 y
9.2 13
14.7
19.7
21.8
22.8
29.1
30.2
32.2

73. CURVE FITTING WITH A LINEAR FUNCTIONyi
f(xi)
ri= yi -f(xi)
(ri)2
6.0000
6.5145
0.2648
9.2000
9.4824
0.0798
0.5497
0.3022
0.5158
1.3139
1.7264
0.4461
0.1990
2.3159
1.8103
3.2772
0.0033
1.0516 E=
9.736
f(x)=2.9679x +

74. CURVE FITTING WITH A LINEAR FUNCTION
Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below. x 2 5 6 8 9 13 15 y 7 10 11 12 14

75. CURVE FITTING WITH A LINEAR FUNCTION
f(x)=0.64x yi
f(xi)
ri= yi -f(xi)
(ri)2
7.0000
6.9769
0.0231
0.0005
8.0000
8.8970
0.8046
9.5370
0.4630
0.2143
0.1829
0.0334
0.5428
0.2947
0.0003
0.0885 E=
1.4363 +

76. CURVE FITTING WITH A LINEAR FUNCTION
Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below. Use f(x) to determine the interpolated value n for x=4 and the extrapolated value m for x= -8. x -8 -7 -5 -1 2 4 5 6 y m 15 12 n -9

77. CURVE FITTING WITH A LINEAR FUNCTIONm
f(x)=1.7786x n
Interpolation : Using the data points for estimating the expexted values between the known points.
Extrapolation : Using the data points for estimating the expexted values beyond the known points.

78. LAGRANGE POLYNOMIALS
For n+1 data points (x0 , y0), (x1 , y1), (x2 , y2), ….. (xk , yk), ….. (xn , yn) a unique polynomial of order n can be found, which passes through these n data points. A Lagrange interpolating polynomial Pn(x) that passes through n points is defined as :

79. LAGRANGE POLYNOMIALS MATLAB SCRIPT
Example : Find the Lagrange interpolation polynomial that passes through the points : x 2
2.5 4 y
0.5
0.4
0.25
MATLAB SCRIPT
x=[ ];
y=[ ];
polyfit(x,y,2)

80. LAGRANGE POLYNOMIALS x 2
2.5 4 y
0.5
0.4
0.25
The Lagrange interpolation polynomial that passes through the points :

81. LAGRANGE POLYNOMIALS
Example : Find the Lagrange interpolation polynomial that passes through the points : x 1 2 y -1 7

82. LAGRANGE POLYNOMIALS x 1 2 y -1 7
The Lagrange interpolation polynomial that passes through the points :

83. LAGRANGE POLYNOMIALS
Example : Find the Lagrange interpolation polynomial that passes through the points : x 1 3 4 y 15

84. LAGRANGE POLYNOMIALS x 1 3 4 y 15
The Lagrange interpolation polynomial that passes through the points :

85. LAGRANGE POLYNOMIALS
Example : Find the Lagrange interpolation polynomial that passes through the points : x 3 4 5 y 8 -4 18

86. LAGRANGE POLYNOMIALS x 3 4 5 y 8 -4 18
The Lagrange interpolation polynomial that passes through the points :

87. ORDINARY DIFFERENTIAL EQUATIONS
A differential equation is an equation that contains derivatives of an unknown function.
A differential equation that has one independent variable is called an ordinary differential equation (ODE).
If an ODE involves only first derivatives of the dependent variable (y) with respect to the independent variable (x), it is a first – order ordinary differential equation.
A first –order ODE is linear, if it is a linear function of y and dy/dx.

88. EULER’S METHOD yi+1 yi h xi xi+1 Numerical Solution Exact Solution
Slope =f(xi, yi)
y(x)
Numerical Solution
Exact Solution

89. EULER’S METHOD
Example : Use Euler’s method to solve the Ordinary Differential Equation below from x=0 to x=1 with the initial conditions x=0 and y=2. (Take h=0.2) Compute the errors in each step for the exact solution y=3ex-x-1

90. EULER’S METHOD h=0.25 h=0.1 h=0.05 h=0.2 xi yi (Euler) (Exact) Abs.
Error
2.0000
0.2500
2.5000
2.6021
0.1021
0.5000
3.1875
3.4462
0.2587
0.7500
4.1094
4.6010
0.4916
1.0000
5.3242
6.1548
0.8306 xi yi
(Euler)
(Exact)
Abs.
Error
2.0000
0.1000
2.2000
2.2155
0.0155
0.2000
2.4300
2.4642
0.0342
0.3000
2.6930
2.7496
0.0566
0.4000
2.9923
3.0755
0.0832
0.5000
3.3315
3.4462
0.1146
0.6000
3.7147
3.8664
0.1517
0.7000
4.1462
4.3413
0.1951
0.8000
4.6308
4.8766
0.2459
0.9000
5.1738
5.4788
0.3050
1.0000
5.7812
6.1548
0.3736 xi yi
(Euler)
(Exact)
Abs.
Error
2.0000
0.0500
2.1000
2.1038
0.0038
0.1000
2.2075
2.2155
0.0080
0.1500
2.3229
2.3355
0.0126
0.2000
2.4465
2.4642
0.0177
0.2500
2.5788
2.6021
0.0232
0.3000
2.7203
2.7496
0.0293
0.3500
2.8713
2.9072
0.0359
0.4000
3.0324
3.0755
0.0431
0.4500
3.2040
3.2549
0.0510
0.5000
3.3867
3.4462
0.0595
0.5500
3.5810
3.6498
0.0687
0.6000
3.7876
3.8664
0.0788
0.6500
4.0069
4.0966
0.0897
0.7000
4.2398
4.3413
0.1015
0.7500
4.4868
4.6010
0.1142
0.8000
4.7486
4.8766
0.1280
0.8500
5.0261
5.1689
0.1429
0.9000
5.3199
5.4788
0.1590
0.9500
5.6309
5.8071
0.1763
1.0000
5.9599
6.1548
0.1950
h=0.2 xi yi
(Euler)
(Exact)
Abs.
Error
2.0000
0.2000
2.4000
2.4642
0.0642
0.4000
2.9200
3.0755
0.1555
0.6000
3.5840
3.8664
0.2824
0.8000
4.4208
4.8766
0.4558
1.0000
5.4650
6.1548
0.6899

91. EULER’S METHOD

92. EULER’S METHOD
Example : Use Euler’s method to solve the Ordinary Differential Equation below from x=0 to x=1.8 with the initial conditions x=0 and y=1. (Take h=0.6) Compute the errors in each step for the exact solution.

93. EULER’S METHOD h=0.6 h=0.1 h=0.3 xi yi (Euler) (Exact) Abs. Error xi
1.0000
0.6000
1.1628
0.1628
1.2000
1.2304
1.3856
0.1552
1.8000
1.0795
0.1869
0.8926 xi yi
(Euler)
(Exact)
Abs.
Error
1.0000
0.1000
1.0050
0.0050
0.2000
1.0099
1.0198
0.0099
0.3000
1.0293
1.0440
0.0147
0.4000
1.0575
1.0767
0.0192
0.5000
1.0934
1.1169
0.0235
0.6000
1.1355
1.1628
0.0272
0.7000
1.1821
1.2124
0.0303
0.8000
1.2305
1.2629
0.0323
0.9000
1.2778
1.3107
0.0329
1.3199
1.3513
0.0314
1.1000
1.3518
1.3787
0.0269
1.2000
1.3675
1.3856
0.0181
1.3000
1.3587
1.3620
0.0033
1.4000
1.3157
1.2955
0.0201
1.5000
1.2255
1.1698
0.0557
1.6000
1.0718
0.9634
0.1084
1.7000
0.8337
0.6481
0.1855
1.8000
0.4841
0.1869
0.2972
h=0.3 xi yi
(Euler)
(Exact)
Abs.
Error
1.0000
0.3000
1.0440
0.0440
0.6000
1.0819
1.1628
0.0809
0.9000
1.2118
1.3107
0.0989
1.2000
1.3203
1.3856
0.0652
1.5000
1.2773
1.1698
0.1075
1.8000
0.8395
0.1869
0.6526

94. EULER’S METHOD