Numerical Methods and Analysis discusses various root-finding methods including bisection, false position, and Newton-Raphson. Bisection uses interval halving to find a root between two values with opposite signs. False position uses the slope of a line between two points to estimate the next root. Newton-Raphson approximates the root using Taylor series expansion neglecting higher order terms. Interpolation uses forward difference tables to construct a polynomial approximation of a function.
Regula Falsi or False Position Method is one of the iterative (bracketing) Method for solving root(s) of nonlinear equation under Numerical Methods or Analysis.
Regula Falsi or False Position Method is one of the iterative (bracketing) Method for solving root(s) of nonlinear equation under Numerical Methods or Analysis.
Describes the mathematics of the Calculus of Variations.
For comments please contact me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website on http://www.solohermelin.com
Describes the mathematics of the Calculus of Variations.
For comments please contact me at solo.hermelin@gmail.com.
For more presentations on different subjects visit my website on http://www.solohermelin.com
Solutions Manual for College Algebra Concepts Through Functions 3rd Edition b...RhiannonBanksss
Full download : http://downloadlink.org/p/solutions-manual-for-college-algebra-concepts-through-functions-3rd-edition-by-sullivan-ibsn-9780321925725/ Solutions Manual for College Algebra Concepts Through Functions 3rd Edition by Sullivan IBSN 9780321925725
I have tried to discuss about the fundamental knowledge related to Irrigation and Flood Control in short. For more details anyone can visit the books that I have mentioned in my slide presentation. I have tried to cover major topics from books so that student can find it easy to understand and learn about irrigation and flood control. I hope it will help everyone who has interest to Irrigation Engineering.
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Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
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Final project report on grocery store management system..pdfKamal Acharya
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1. Numerical Methods and Analysis
Faruque Abdullah
Lecturer
Dept. of Civil Engineering
Dhaka International University
2. Roots of a equation
Roots of a equation is calculated by the following methods:
Bi-section method
Iteration method
Method of false position/ Interpolation method
Newton Raphson method, etc.
3. Bi-section/ Interval Halving Method(IVT)
IVT: If a function f(x) is continuous between a and b and f(a) and
f(b) is opposite sign, then there exists at least one root between a
and b.
f(x) = 𝑥4
- 2x;
f(2) = -ve
f(3) = +ve
x =
𝑎+𝑏
2
=
2+3
2
= 2.5
4. Procedure:
Step-1: Choose a real two numbers a and b such that f(a)f(b)<0
Step-2: Set, 𝑥 𝑟 =
𝑎+ 𝑏
2
Step-3: a) If f(a) f(𝑥 𝑟) < 0, the root lies in the interval (a,𝑥 𝑟), Then
set b = 𝑥 𝑟 and go to step 2.
b) If f(a) f(𝑥 𝑟) > 0, the root lies in the interval (𝑥 𝑟,b), Then
set a = 𝑥 𝑟 and go to step 2.
c) If f(a) f(𝑥 𝑟) = 0, it means that 𝑥 𝑟 is a root of the
equation f(x) = 0 and the computation may be terminated.
5. Problem-1: Find a real root of the equation 𝑥3- 2x-5 = 0 by bi-section method
up to an accuracy of 10−4.
Solution: Let, f(x) = 𝑥3 - 2x -5
f(2) = 23 - 2.2 – 5 = -1
f(3) = 33
- 2.3 – 5 = 16
As f(2) is –ve and f(3) is +ve .
So f(2) f(3) < 0.
Root lies between 2 and 3.
7. Problem-2: Figure shows a uniform subject to a linearly increasing distributed load.
The equation of the resulting elastic curve , y =
𝑤0
120𝐸𝐼𝐿
(-𝑥5
+ 2𝐿2
𝑥3
-𝐿4
𝑥). Use bi-
section method to determine point of maximum deflection up to an accuracy of 10−10
.
Use the following parameter. E = 50000 KN/𝑐𝑚2, I = 30000 𝑐𝑚4, 𝑤0 = 2.5 KN/cm, L
= 3 m.
3 m
𝑤0
8. Solution: For maximum deflection,
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
𝑤0
120𝐸𝐼𝐿
x
𝑑𝑦
𝑑𝑥
(-𝑥5 + 2 𝐿2 𝑥3-𝐿4 𝑥)
=
𝑤0
120𝐸𝐼𝐿
x (-5𝑥4 + 6 𝐿2 𝑥2-𝐿4)
=
2.5
120 X 50000 X 30000 X 300
x (-5𝑥4 + 6. 3002 𝑥2-3004)
= 4.63 x 10−14 x (- 5𝑥4 + 540000 𝑥2- 3004)
9. Let, f(x) = 4.63 x 10−14
x (- 5𝑥4
+ 540000 𝑥2
- 3004
)
f(134) = -7.34 x 10−7
f(135) = 3.73 x 10−6
As f(134) is –ve and f(135) is +ve .
So f(134) f(135) < 0.
Root lies between 134 and 135.
10. n a b 𝑥 𝑟 f(x)
(sign)
f(x)
1 134 135 134.5 +ve +1.5 x 10−6
2 134 134.5 134.25 +ve +3.84 10−7
3 134 134.25 134.125 -ve -1.75 x 10−7
4 134.125 134.25 134.1875 +ve +1.05 x 10−7
5 134.125 134.1875 134.15625 -ve -3.5 x 10−8
6 134.15625 134.1875 134.171875 +ve +3.5 x 10−8
7 134.15625 134.171875 134.1640625 -ve -7.22 x 10−11
11. Method of False Position
(0,0)
X
Y
[a, f(a)]
[b, f(b)]
f(x) 𝑦−𝑓(𝑎)
𝑥−𝑎
=
𝑓 𝑏 − 𝑓(𝑎)
𝑏−𝑎
Let, y = 0,
−𝑓(𝑎)
𝑥−𝑎
=
𝑓 𝑏 − 𝑓(𝑎)
𝑏−𝑎
x-a = -
(𝑏−𝑎)𝑓(𝑎)
𝑓 𝑏 −𝑓(𝑎)
x = a -
(𝑏−𝑎)𝑓(𝑎)
𝑓 𝑏 −𝑓(𝑎)
Derive the root of a curve f(x) by False Position method.
12. Newton Raphson Method
Let, 𝑥0 be a root of the function f(𝑥0) = 0 and assume that 𝑥1 = 𝑥0 + h be the correct
root of the function f(𝑥1) = 0.
𝑥1 = 𝑥0 + h -----------------(1)
f(𝑥1) = 0
f(𝑥0 + h) = 0-----------------(2)
Expanding eq (2) by Tailor’s series we get,
f(𝑥0) + h𝑓′(𝑥0) +
ℎ2
2!
𝑓′′(𝑥0) + ----- = 0
13. Neglecting 2nd and higher order derivatives,
f(𝑥0) + h𝑓′
(𝑥0) = 0
Or, h = -
f( 𝑥0)
𝑓′( 𝑥0)
From eq (1),
𝑥1 = 𝑥0 -
f( 𝑥0)
𝑓′( 𝑥0)
Similar approximation for 𝑥2, 𝑥3, ----, 𝑥 𝑛+1 we get,
𝑥2 = 𝑥1 -
f( 𝑥1)
𝑓′( 𝑥1)
14. 𝑥3 = 𝑥2 -
f( 𝑥2)
𝑓′( 𝑥2)
-------------
𝑥 𝑛+1 = 𝑥 𝑛 -
f( 𝑥 𝑛)
𝑓′( 𝑥 𝑛)
This is Newton Raphson Formula.
15. Problem-3: Find a real root of the equation 𝑥3- 2x-5 = 0 by Newton Raphson
method.
Solution: Let, f(x) = 𝑥3 - 2x -5
f(𝑥 𝑛) = 𝑥 𝑛
3 - 2𝑥 𝑛 -5
𝑓′(𝑥 𝑛) = 3𝑥 𝑛
2
- 2
From Newton Raphson Formula,
𝑥 𝑛+1 = 𝑥 𝑛 -
f( 𝑥 𝑛)
𝑓′( 𝑥 𝑛)
= 𝑥 𝑛 -
𝑥 𝑛
3 − 2 𝑥 𝑛 −5
3𝑥 𝑛
2 − 2
17. Problem-4: You are designing a spherical tank to hold water for a small village in a
developing country. The volume of liquid can be calculated as V = πℎ2 [3𝑅−ℎ]
3
. If R = 3
m, to what depth must the tank be filled so that it holds 30 𝑚3 water?
Solution: V = πℎ2 [3𝑅−ℎ]
3
=> 30 = πℎ2 [3.3−ℎ]
3
=> 90 = πℎ2(9-h)
=>
90
𝜋
= 9ℎ2 - ℎ3
=> ℎ3
- 9ℎ2
+ 28.65 = 0
23. Newton’s Formula for Forward Interpolation
Let, y = f(x) denote a function which takes the values from y0, y1, y2,
y3,……….., yn for the equidistance values x0, 𝑥1, 𝑥2, 𝑥3,……….., xn. Φ(x)
denote a function of polygonal for nth degree. As x is equidistance xi = x0 +
ih. Where, i = 1,2,3,…….,n.
x1 = x0 + h
=> (x1 - x0) = h
x2 = x0 + 2h
=> (x2 - x0) = 2h
27. Problem-5: Find the cubic polynomial which takes the following values
y(1) = 24
y(3) = 120
y(5) = 336
y(7) = 720
Hence, obtain the values of y(8) by Newton’s Forward Interpolation method.
28. Solution: We make the forward difference table,
h = (x1 - x0) = 3 – 1 = 2
x = 𝑥0 + ph => p =
𝑥 − 𝑥0
ℎ
=> p =
8 −1
2
= 3.5
x y ∆ ∆2 ∆3
1 24
96
3 120 120
216 48
5 336 168
384
7 720
29. From Newton’s formula for forward interpolation,
y(x) = 𝑦0 + p∆ 𝑦0 + p (p-1)
∆2 𝑦0
2!
+ p (p-1) (p-2)
∆3 𝑦0
3!
=> y(8)= 24 + 3.5 x 96 + 3.5 (3.5-1) x
120
2!
+ 3.5 (3.5-1) (3.5-2) x
48
3!
= 990.
30. Problem-6: Find the cubic polynomial which takes the following values
y(1) = 24
y(3) = 120
y(5) = 336
y(7) = 720
Hence, obtain the values of y(6) by Newton’s Backward Interpolation method.
Newton’s Formula for Backward Interpolation
31. Solution: We make the backward difference table,
h = (x1 - x0) = 3 – 1 = 2
x = 𝑥 𝑛 + ph => p =
𝑥 − 𝑥 𝑛
ℎ
=> p =
6 −7
2
= -0.5
x y ∇ ∇ 2
∇ 3
1 24
96
3 120 120
216 48
5 336 168
384
7 720
32. From Newton’s formula for backward interpolation,
y(x) = 𝑦𝑛 + p∆ 𝑦𝑛 + p (p+1)
∆2 𝑦 𝑛
2!
+ p (p+1)(p+2)
∆3 𝑦 𝑛
3!
=> y(6)= 720 + (-0.5) x 384 + (-0.5) (-0.5+1) x
168
2!
+ (-0.5) (-0.5+1) (-0.5+2) x
48
3!
= 576.
33. Problem-6: The table shows the value of tanx for 0.1 ≤ x ≤ 0.3
Find, (a) tan(0.12) by Forward Interpolation Method
(b) tan(0.26) by Backward Interpolation Method.
x Y = tanx
0.10 0.1745
0.15 0.2618
0.20 0.3491
0.25 0.4363
0.30 0.5236
34. Solution: We make the forward difference table,
x y ∆ ∆2
∆3
∆4
0.10 0.1745
0.0873
0.15 0.2618 0
0.0873 0
0.20 0.3491 0 0
0.0873 0
0.25 0.4364 0
0.0873
0.30 0.5237
35. (a) h = (x1 - x0) = 0.15 – 0.10 = 0.05
x = 𝑥0 + ph => p =
𝑥 − 𝑥0
ℎ
=> p =
0.12 −0.10
0.05
= 0.4
For Newton’s Forward formula,
tan(0.12) = 𝑦0 + p∆ 𝑦0 + p (p-1)
∆2 𝑦0
2!
+ p (p-1) (p-2)
∆3 𝑦0
3!
+ p (p-1) (p-2) (p-3)
∆4 𝑦0
4!
= 0.1745 + 0.4 x 0.0873 + 0.4 (0.4-1)
0
2!
+ 0.4 (0.4-1) (0.4-2)
0
3!
+
0.4 (0.4-1) (0.4-2)
0
4!
= 0.2094
37. Problem-7: The discharge through a hydraulic structure for different values of
head (H) is shown in table.
Calculate discharge, Q for H=3 ft using Lagrange formula.
H (ft) 1.2 2.1 2.7 4.0
Q (cusec) 25 60 90 155
39. Exercise: The load bearing capacity on different penetration level is given
below.
Calculate Load for D = 3.00 cm using Lagrange formula.
D (cm) 2.35 2.75 3.30 3.60
Load (KN) 25 43 66 70
46. Example: Solve the following equations by Gauss-Jordan Elimination method.
x + 3y + 6z = 12
2x + 3y = 16
x + 9z = 14
Solution: x = 8, y = 0, z = 2/3
47. Problem-9: Find whether the following system is consistent or not.
x - 4y + 5z = 8
3x + 7y - z = 3
x + 15y -11z = -14
Solution: Forming the matrix, A=
Now, A = 0 and ≠ 0.
r (A) = 2
1 -4 5
3 7 -1
1 15 -11
1 -4
3 7
48. Forming the augmented matrix, (A,b)=
Determinant of = 31≠ 0.
r (A,b) = 3
When, r (A,b) ≥ r(A), the equation will be inconsistence (No solution).
As, r(A) < r (A,b) ;
Hence the system is inconsistence.
1 -4 5 8
3 7 -1 3
1 15 -11 -14
-4 5 8
7 -1 3
15 -11 -14
50. Problem-10: In a laboratory the data of water demand with respect to population
is shown in the following table .
Population (thousand) 2.10 6.22 7.17 10.52 13.68
Water Demand (cumec) 2.90 3.83 5.98 5.71 7.74
Fit a straight line from this data to find the discharge for any corresponding area
using least square method.
Linear Curve Fitting
52. Non-Linear Curve Fitting
Problem-11: In a laboratory the deformation (mm) of a pre-stressed cantilever
beam with respect to distance (m) from the fixed support is shown below:
Fit a parabolic curve line from this data to find the deflection for any point from
the fixed support.
2
12
10
8
6
4
0.01
0.36
0.29
0.22
0.18
0.09
54. From eq (1), 6 𝑎0 + 42 𝑎1 + 364 𝑎2 = 1.15
From eq (2), 42 𝑎0 + 364 𝑎1 + 3528 𝑎2 = 10.44
From eq (3), 364 𝑎0 + 3528 𝑎1 + 36400 𝑎2 = 102.88
Solving the equation, 𝑎0 = -0.069, 𝑎1 = 0.042 & 𝑎2 = -5.803 x 10−4
⸫ y = 𝑎0 + 𝑎1x + 𝑎2 𝑥2 = -0.069 + 0.042 x + -5.803 x 10−4 𝑥2
Distance (m) 2 4 6 8 10 12
Deflection (mm)
y = -0.069 + 0.042 x + -5.803 x 10−4
𝑥2
0.01 0.10 0.18 0.27 0.35 0.43
Original Value(mm) 0.01 0.09 0.18 0.22 0.29 0.36
55. Problem-12: Use 4 segment to calculate the are of a function f(x) = 0.2 + 25 x +
3𝑥2 + 2𝑥4 from a = 0 to b = 2 by using Trapezoidal rule and Simpson's rule and
determine percentage of error.
Solution: For four segment n = 4. So, the value of x will be 0.5, 1.0, 1.5, 2.0.
Using Trapezoidal rules,
0
2
𝑦𝑑𝑥 =
ℎ
2
[𝑦0+ 𝑦𝑛 + 2 (𝑦1 +……..+ 𝑦(𝑛−1))]
=
0.5
2
[ 0.2 + 94.2 + 2(13.575 + 30.2 + 54.575)] = 72.775 unit
x 0 0.5 1.0 1.5 2.0
y 0.2 13.575 30.2 54.575 94.2
56. Using Simpson’s rules,
0
2
𝑦𝑑𝑥 =
ℎ
3
[𝑦0+ 𝑦𝑛 + 4 (𝑦1 + 𝑦3 + 𝑦5 +……..+ 𝑦(𝑛−1)) + 2 (𝑦2 + 𝑦4 + 𝑦6 +
……..+ 𝑦(𝑛−2))]
=
0.5
3
[ 0.2 + 94.2 + 4(13.575 + 54.575) + 2 (30.2)] = 71.23 unit
Actual area by integration method,
0
2
𝑦𝑑𝑥 = 0
2
(0.2 + 25 x + 3𝑥2 + 2𝑥4)𝑑𝑥
= 0.2x + 25
𝑥2
2
+ 3
𝑥3
3
+ 2
𝑥5
5 0
2
= 71.2 unit
Percentage of error for Trapezoidal rule =
72.775 −71.2
71.2
x 100% = 2.21%
Percentage of error for Simpson’s rule =
721.23 −71.2
71.2
x 100% = 0.042%
57. Problem-13: A irrigation reservoir discharging through sluice at a depth h, below the
water surface as a surface area A, for various value of h as given below:
If t denotes the time in min & the rate of fall of the surface is given by
𝑑ℎ
𝑑𝑡
= -
48
𝐴
ℎ , estimate the time taken for the water level to fall from 14 ft to 10 ft above
the sluice.
Solution: Area of the water reservoir, A =
ℎ
2
[ 950 + 1530 + 2 (1070 + 1200 + 1350)]
= 4860 𝑓𝑡2
h (ft) 10 11 12 13 14
A (𝑓𝑡2
) 950 1070 1200 1350 1530