self learning activity ppt

1. SANJIVANI COLLEGE OF ENGINEERING, KOPARGAON
(An autonomous Institute affiliatedto SPPU Pune)
ENGINEERINGMATHEMATICSIIPRESENTATION
GUIDED BY:
PROF. A. S. AHER

2. 1.Solve :
ⅆ𝑦
ⅆ𝒙
= xy+ 𝒙𝟐
,
ⅆ𝒛
ⅆ𝒙
= xz , y(0)=1, z(0)=2, h=0.2
Find y , z at x=0.2.

3. 𝑦 0 = 1 , 𝑧 0 = 2
𝑥0 = 0 , 𝑦𝑜 = 1 𝑥0 = 0 , 𝑧0 = 2
ℎ = 0.2
Here,
𝑑𝑦
𝑑𝑥
= 𝑓 𝑥, 𝑦, 𝑧 = 𝑥𝑦 + 𝑥2 &
𝑑𝑧
𝑑𝑥
= 𝑔 𝑥, 𝑦, 𝑧 =
𝑥𝑧
Starting at (𝑥0,𝑦0,𝑧0) and taking the step for x,y,z to be
ℎ, 𝑘, 𝑢 respectively the RK method gives
By RK method,
𝑘1 = ℎ𝑓(𝑥0, 𝑦0, 𝑧0)
=0.2(𝑥0𝑦0 + 𝑥0
2)
=0.2(0*1+0)
=0
𝑢1 = ℎ𝑔 𝑥0, 𝑦0, 𝑧0
=ℎ(𝑥0𝑧0)
=0.2(0*2)
=0

4. 𝑘2 = ℎ𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
, 𝑧0 +
𝑢1
2
)
=ℎ𝑓(0 +
0.2
2
, 1 +
0
2
, 2 +
0
2
)
=ℎ𝑓(0.1 , 1 , 2)
=0.2[0.1*1 +(0.1)2
]
=0.2(0.1+0.01)
=0.2*0.11
=0.022
𝑘3 = ℎ𝑓(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
, 𝑧0 +
𝑢2
2
)
=ℎ𝑓(0 +
0.2
2
, 1 +
0.02
2
, 2 +
0.04
2
)
=ℎ𝑓(0.1,1.01,2.02)
= ℎ[0.1 ∗ 1.01 + (0.1)2
]
=0.2(0.1+0.01)
=0.2(0.11)
=0.02
𝑢2 = ℎ𝑔(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘1
2
, 𝑧0 +
𝑢1
2
)
=ℎ𝑔(0 +
0.2
2
, 1 +
0
2
, 2 +
0
2
)
=ℎ𝑔(0.1 , 1 , 2)
=0.2(0.1*2 )
=0.2(0.2)
=0.04
𝑢3 = ℎ𝑔(𝑥0 +
ℎ
2
, 𝑦0 +
𝑘2
2
, 𝑧0 +
𝑢2
2
)
=ℎ𝑔(0 +
0.2
2
, 1 +
0.02
2
, 2 +
0.04
2
)
= ℎ𝑔(0.1,1.01, 2.02)
=0.2(0.1*2.02)
=0.2(0.202)
= 0.04

5. 𝑘4= ℎ𝑓(𝑥0 + ℎ, 𝑦0 + 𝑘3, 𝑧0 + 𝑢3)
=ℎ𝑓(0 + 0.2,1 + 0.02 , 2 + 0.04)
=ℎ𝑓(0.2,1.02,2.04)
= ℎ[0.2 ∗ 1.02 + (0.2)2
]
=0.2(0.20+0.04)
=0.2(0.24)
=0.05
𝑘 =
1
6
(𝑘1 + 2𝑘2+ 2𝑘3+ 𝑘4)
=
1
6
[0+2*(0.02)+2*(0.02)+0.05]
=
1
6
0.04 + 0.04 + 0.05
=
1
6
0.13
k =0.02
Therefore ,
𝑦 at 𝑥=0 is 𝑦 = 𝑦0 + 𝑘
=1+0.02 and y=1.02
𝑢4 = ℎ𝑔(𝑥0 + ℎ, 𝑦0 + 𝑘3, 𝑧0 + 𝑢3)
=ℎ𝑔(0 + 0.2,1 + 0.02 , 2 + 0.04)
=ℎ𝑔(0.2,1.02,2.04)
= ℎ[0.2 ∗ 2.04]
=0.2(0)
=0.2(0.408)
=0.08
u=
1
6
(𝑢1 + 2𝑢2+2 𝑢3+ 𝑢4)
=
1
6
[0+2*(0.04)+2*(0.04)+0.08]
=
1
6
0.08 + 0.08 + 0.08
=
1
6
0.24
u = 0.04
𝑧 at 𝑥=0 is z= 𝑧0 + 𝑢
=2+0.04 and z=2.04

6. Q.2) Write a computer programming for backward difference formula.

7. Step 1 - Define a function backward_difference_table(x, y)
that takes two arguments, x and y.
Step 2 -Calculate the length of y and store it in n.
Step 3 - Create a 2D table of size n x n filled with zeros.
Step 4 - Initialize the first column of the table with the
values of y.
Step 5 - Iterate over the remaining columns of the table
and calculate the backward differences using the formula
table[i][j] = table[i+1][j-1] - table[i][j-1].
Step 6 -Return the table.
Step 7 - Define a function
newton_backward_interpolation(x, x_values, y_values)
that takes three arguments, x, x_values, and y_values.
Step 8 - Calculate the length of x_values and store it in n.
Step 9 - Calculate the value of h as the difference between
the first two values of x_values.
Step 10 - Calculate the value of u using the formula u = (x
- x_values[n-1]) / h.
Step 11 -Calculate the backward difference table using the
backward_difference_table function.
Step 12 -Initialize the result variable with the value at the
bottom of the last column of the table.
Algorithm:

8. STEP 13 - ITERATE OVER THE REMAINING COLUMNS OF THE TABLE AND
CALCULATE THE INTERPOLATION USING THE FORMULA RESULT += TERM,
WHERE TERM IS CALCULATED AS TERM *= (U + J) / (J + 1).
STEP 14 -RETURN THE RESULT.
STEP 15 -TAKE USER INPUT FOR X, X_VALUES, AND Y_VALUES.
STEP 16 -CALCULATE AND DISPLAY THE BACKWARD DIFFERENCE TABLE.
STEP 17 -CALCULATE AND DISPLAY THE INTERPOLATED VALUE AT X.

9. Source Code:
def backward_difference_table(x, y):
n = len(y)
table = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
table[i][0] = y[i]
for j in range(1, n):
for i in range(n-j):
table[i][j] = table[i+1][j-1] - table[i][j-1]
return table
def newton_backward_interpolation(x, x_values, y_values):
n = len(x_values)
h = x_values[1] - x_values[0]
u = (x - x_values[n-1]) / h
table = backward_difference_table(x_values, y_values) result = table[n-1][0]

10. Source Code:
# Taking user input for x, x_values, and y_values
x = float(input("Enter the value of x for which you want to find the value: "))
x_values = list(map(float, input("Enter the values of x (space-separated): ").split()))
# Calculating and displaying the backward difference table
table = backward_difference_table(x_values, y_values)
print("Newton's Backward Difference Table:")
for row in table:
print(row)
# Calculating and displaying the final output
output = newton_backward_interpolation(x, x_values, y_values)
printf("The interpolated value at x = {x} is: {output}")

11. Output:

12. Q.3) Using Taylor's series express 2𝒙𝟒+5𝒙𝟐+6 in powers of (x-1).

13. Answer:
Here,
f(𝑥)= 2𝑥^4+5𝑥^2+6
f’(𝑥) = 8𝑥^3+10𝑥
f’’(𝑥) =24𝑥^2+10
f’’’(𝑥) =48𝑥
f’’’’(𝑥) =48
f(1)= 2〖(1)〗^4+5〖(1)〗^2+6 =13
f’(1) = 8〖(1)〗^3+10(1) =18
f’’(1) =24〖(1)〗^2+10 =34
f’’’(1) =48(1)=48
f’’’’(1) =48

14. By Taylors Series:
f(𝑥)= f(a) + 𝑓′
(𝑎) 𝑥 − 𝑎 +
𝑥−𝑎 2
2!
𝑓′′
𝑎 +
𝑥−𝑎 3
3!
𝑓′′′
𝑎 +
𝑥−1 4
4!
𝑓′′′′ 𝑎 + ⋯⋯
Compare power 𝑥 − 1 𝑎𝑛𝑑 𝑥 − 𝑎
𝑥 − 1 = 𝑥 − 𝑎
𝑥 = a
f(𝑥)= f(1) + 𝑓′
(1) 𝑥 − 1 +
𝑥−1 2
2!
𝑓′′
1 +
𝑥−1 3
3!
𝑓′′′
1 +
𝑥−1 4
4!
𝑓′′′′ 1 + ⋯⋯
. ……equation(1)

15. Put the value in Equation (1)
f(x)= 13 + 18 𝑥 − 1 +
𝑥−1 2
2!
34 +
𝑥−1 3
3!
48 +
𝑥−1 4
4!
48
f(x)= 13 + 18 𝑥 − 1 +
𝑥−1 2
1×2
34 +
𝑥−1 3
1×2×3
48 +
𝑥−1 4
1×2×3×4
48
f(x)= 13 + 18 𝑥 − 1 +
𝑥−1 2
2
34 +
𝑥−1 3
6
48 +
𝑥−1 4
24
48
Thus, the Taylor series expansion becomes:
f(x)= 13 + 18 𝑥 − 1 + 17 𝑥 − 1 2
+ 8 𝑥 − 1 3
+ 2 𝑥 − 1 4

16. Q.4) Find the center of mass of a thin , uniform plate whose shape is the region between
y = cos x and the x - axis between x= ℿ/2 and
x = ℿ/2. Since the density is constant , we may takep(x, y)= 1.

17. It is clear that 𝑥 = 0
but for matrics lets compute it
First we compute the mass
m= −𝛱/2
𝛱/2
. 0
𝑐𝑜𝑠𝑥
. 1 dydx = −𝛱/2
𝛱/2
.cosxdx= sin 𝑥 −𝜋 2
𝛱 2
=sin
𝜋
2
− sin −
𝜋
2
= 1 + 1 = 2 … sin −
𝛱
2
= − sin
𝜋
2
mx= −𝛱/2
𝛱/2
. 0
𝑐𝑜𝑠𝑥
.ydydx = −𝛱/2
𝛱/2
.
1
2
cos2 𝑥 𝑑𝑥 =
1
2
−𝛱/2
𝛱/2
. 𝑐𝑜𝑠2 𝑥 𝑑𝑥.
=
𝝅
𝟒
.
My= −𝛱/2
𝛱/2
. 0
𝑐𝑜𝑠𝑥
. xdydx
= −𝛱/2
𝛱/2
.x cos x dx=0
So 𝑥 = 0 𝑎𝑠 𝑌 =
𝜋 4
2
=
𝜋
8

18. 1
2
cos2𝑥 𝑑𝑥
−
𝜋
2
𝜋
2
𝑐𝑜𝑠2𝑥 𝑑𝑥
𝑐𝑜𝑠2𝑥 + 1
2
𝑑𝑥 =
𝑐𝑜𝑠2𝑥
2
𝑑𝑥
1
2
𝑑𝑥
=
1
2
𝑥 𝑑𝑥
= 1
2
sin2x
𝜋
2
−
𝜋
2
x
𝜋
2
−
𝜋
2
1
2
1
2
sin 2 ×
𝜋
2
− Sin2 x −
𝜋
2
=
1
2
𝜋
2
+
𝜋
2
+
+
+
sin 𝜋 + sin 𝜋
1
2
= +
2𝜋
2
1
2
= 0 + 0
1
2
𝜋
2
+
𝜋
2
∗
1
2
=
𝜋
4
=
=
Mx =

19. −
𝜋
2
𝜋
2
𝑋 𝐶𝑂𝑆𝑋 𝐷𝑋 = 𝑥 .
−
𝜋
2
𝜋
2
𝐶𝑂𝑆 𝑋 𝐷𝑋 −
−
𝜋
2
𝜋
2 𝑆𝐼𝑁 𝑋 𝐷𝑋
= 𝑋 𝑆𝐼𝑁 𝑋 ]−𝛱
2
𝜋
2
- [- COSX ]−𝛱
2
𝜋
2
= [
𝜋
2
sin
𝛱
2
+
𝛱
2
sin −
𝜋
2
] + [𝐶𝑂𝑆
𝜋
2
− 𝐶𝑂𝑆
𝜋
2
]
= [
𝜋
2
sin
𝛱
2
−
𝛱
2
sin
𝜋
2
] + [𝐶𝑂𝑆
𝜋
2
− 𝐶𝑂𝑆
𝜋
2
]
= [ 1 - 1 ] + [0 – 0]
= 0
My=

20. References
 Books:
1. B. S. Grewal, Higher Engineering Mathematics, 42nd ed. Khanna Publishers, 2012. ISBN: 978-8174091154.
2. H. K. Das, Engineering Mathematics. S Chand, 2006. ISBN: 8121905209.
3. G. V. Davis, Numerical Methods in Engineering and Science. Springer, 1986. ISBN: 978-94-011-6958-5.
4. R. K. Jain and S. R. K. Iyengar, Advanced Engineering Mathematics, Narosa Publishing House, 2014. ISBN: 978-
1842653418.
5. N. P. Bali and M. Goyal, A Text Book of Engineering Mathematics, 8th ed. Lakshmi Publications, 2012.ISBN: 978-
8131808320.
6. E. Kreyszig, Advanced Engineering Mathematics, 9th ed. Wiley, 2013. ISBN: 978-0471488859.
7. E. B. Staff and A. D. Snider, Fundamentals of Complex Analysis with Application to Engineering and Science, 3rd ed.
ISBN: 0139078746.
8. P. Dawkins, Calculus III, Lamar University Texas, 2018. [Online]. Available: http://tutorial.math.lamar.edu.
 E-Resources
A. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/pages/lecture-notes/
B. https://www.codechef.com/
C. https://www.geeksforgeeks.org/courses?source=google&medium=cpc&device=m&keyword=geeksforgeeks&matchty
pe=e&campaignid=20039445781&adgroup=147845288105&gad_source=1&gclid=Cj0KCQjwk6SwBhDPARIsAJ59
GwegeTJbc6aGJqhfid7xlG892rYiAuuvZwkxULIPbtiDTYADMs8E3FEaAqF5EALw_wcB

21. Thank You!