Free UPSC Mathematics Optional Paper-2 2023 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025By

SECTION-A

Question:-01

Let G G GGG be a group of order 10 and G ′ G ′ G^(‘)G’G′ be a group of order 6. Examine whether there exists a homomorphism of G G GGG onto G ′ G ′ G^(‘)G’G′.

Answer:

To determine whether there exists a surjective (onto) homomorphism ϕ : G → G ′ ϕ : G → G ′ phi:G rarrG^(‘)\phi: G \to G’ϕ:G→G′, where | G | = 10 | G | = 10 |G|=10|G| = 10|G|=10 and | G ′ | = 6 | G ′ | = 6 |G^(‘)|=6|G’| = 6|G′|=6, we analyze the implications of such a homomorphism using the First Isomorphism Theorem.
  1. First Isomorphism Theorem:
    If ϕ ϕ phi\phiϕ is a surjective homomorphism, then:
    G / ker ⁡ ( ϕ ) ≅ G ′ . G / ker ⁡ ( ϕ ) ≅ G ′ . G//ker(phi)~=G^(‘).G / \ker(\phi) \cong G’.G/ker⁡(ϕ)≅G′.
    This implies:
    | G | / | ker ⁡ ( ϕ ) | = | G ′ | ⟹ 10 / | ker ⁡ ( ϕ ) | = 6. | G | / | ker ⁡ ( ϕ ) | = | G ′ | ⟹ 10 / | ker ⁡ ( ϕ ) | = 6. |G|//|ker(phi)|=|G^(‘)|Longrightarrow10//|ker(phi)|=6.|G| / |\ker(\phi)| = |G’| \implies 10 / |\ker(\phi)| = 6.|G|/|ker⁡(ϕ)|=|G′|⟹10/|ker⁡(ϕ)|=6.
    Solving for | ker ⁡ ( ϕ ) | | ker ⁡ ( ϕ ) | |ker(phi)||\ker(\phi)||ker⁡(ϕ)|, we get:
    | ker ⁡ ( ϕ ) | = 10 6 = 5 3 . | ker ⁡ ( ϕ ) | = 10 6 = 5 3 . |ker(phi)|=(10)/(6)=(5)/(3).|\ker(\phi)| = \frac{10}{6} = \frac{5}{3}.|ker⁡(ϕ)|=106=53.
    However, | ker ⁡ ( ϕ ) | | ker ⁡ ( ϕ ) | |ker(phi)||\ker(\phi)||ker⁡(ϕ)| must be an integer because it represents the order of a subgroup of G G GGG. Since 5 3 5 3 (5)/(3)\frac{5}{3}53 is not an integer, this leads to a contradiction.
  2. Lagrange’s Theorem Verification:
    The possible orders of ker ⁡ ( ϕ ) ker ⁡ ( ϕ ) ker(phi)\ker(\phi)ker⁡(ϕ) (subgroups of G G GGG) are the divisors of 10: 1, 2, 5, 10.
    • If | ker ⁡ ( ϕ ) | = 1 | ker ⁡ ( ϕ ) | = 1 |ker(phi)|=1|\ker(\phi)| = 1|ker⁡(ϕ)|=1, then | G ′ | = 10 ≠ 6 | G ′ | = 10 ≠ 6 |G^(‘)|=10!=6|G’| = 10 \neq 6|G′|=10≠6.
    • If | ker ⁡ ( ϕ ) | = 2 | ker ⁡ ( ϕ ) | = 2 |ker(phi)|=2|\ker(\phi)| = 2|ker⁡(ϕ)|=2, then | G ′ | = 5 ≠ 6 | G ′ | = 5 ≠ 6 |G^(‘)|=5!=6|G’| = 5 \neq 6|G′|=5≠6.
    • If | ker ⁡ ( ϕ ) | = 5 | ker ⁡ ( ϕ ) | = 5 |ker(phi)|=5|\ker(\phi)| = 5|ker⁡(ϕ)|=5, then | G ′ | = 2 ≠ 6 | G ′ | = 2 ≠ 6 |G^(‘)|=2!=6|G’| = 2 \neq 6|G′|=2≠6.
    • If | ker ⁡ ( ϕ ) | = 10 | ker ⁡ ( ϕ ) | = 10 |ker(phi)|=10|\ker(\phi)| = 10|ker⁡(ϕ)|=10, then | G ′ | = 1 ≠ 6 | G ′ | = 1 ≠ 6 |G^(‘)|=1!=6|G’| = 1 \neq 6|G′|=1≠6.
    None of these cases satisfy | G ′ | = 6 | G ′ | = 6 |G^(‘)|=6|G’| = 6|G′|=6.
  3. Conclusion:
    Since there is no subgroup ker ⁡ ( ϕ ) ker ⁡ ( ϕ ) ker(phi)\ker(\phi)ker⁡(ϕ) of G G GGG such that G / ker ⁡ ( ϕ ) G / ker ⁡ ( ϕ ) G//ker(phi)G / \ker(\phi)G/ker⁡(ϕ) has order 6, no surjective homomorphism exists from G G GGG to G ′ G ′ G^(‘)G’G′.
Hence,there does not exist a surjective homomorphism from a group G G GGG of order 10 onto a group G ′ G ′ G^(‘)G’G′ of order 6.}

Question:-01 (b)

Express the ideal 4 Z + 6 Z 4 Z + 6 Z 4Z+6Z4Z + 6Z4Z+6Z as a principal ideal in the integral domain Z Z ZZZ.

Answer:

To express the ideal 4 Z + 6 Z 4 Z + 6 Z 4Z+6Z4\mathbb{Z} + 6\mathbb{Z}4Z+6Z as a principal ideal in the integral domain Z Z Z\mathbb{Z}Z, we can follow these steps:
  1. Understand the Sum of Ideals:
    • The sum of two ideals 4 Z 4 Z 4Z4\mathbb{Z}4Z and 6 Z 6 Z 6Z6\mathbb{Z}6Z in Z Z Z\mathbb{Z}Z is the ideal generated by the union of their generators. In other words: 4 Z + 6 Z = { 4 a + 6 b ∣ a , b ∈ Z } 4 Z + 6 Z = { 4 a + 6 b ∣ a , b ∈ Z } 4Z+6Z={4a+6b∣a,b inZ}4\mathbb{Z} + 6\mathbb{Z} = \{4a + 6b \mid a, b \in \mathbb{Z}\}4Z+6Z={4a+6b∣a,b∈Z}
  2. Find the Greatest Common Divisor (GCD):
    • The ideal 4 Z + 6 Z 4 Z + 6 Z 4Z+6Z4\mathbb{Z} + 6\mathbb{Z}4Z+6Z is generated by the greatest common divisor (GCD) of 4 and 6. This is a fundamental property of ideals in Z Z Z\mathbb{Z}Z.
    • Compute the GCD of 4 and 6: gcd ( 4 , 6 ) = 2 gcd ( 4 , 6 ) = 2 gcd(4,6)=2\gcd(4, 6) = 2gcd(4,6)=2
  3. Express as a Principal Ideal:
    • Since the GCD is 2, the ideal 4 Z + 6 Z 4 Z + 6 Z 4Z+6Z4\mathbb{Z} + 6\mathbb{Z}4Z+6Z can be expressed as the principal ideal generated by 2: 4 Z + 6 Z = 2 Z 4 Z + 6 Z = 2 Z 4Z+6Z=2Z4\mathbb{Z} + 6\mathbb{Z} = 2\mathbb{Z}4Z+6Z=2Z
Final Answer:
2 Z 2 Z 2Z\boxed{2\mathbb{Z}}2Z

Question:-01 (c)

Test the convergence of the series

∑ n = 1 ∞ 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 , x > 0. ∑ n = 1 ∞   1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 , x > 0. sum_(n=1)^(oo)(1*3*5dots(2n-1))/(2*4*6dots(2n))*(x^(2n+1))/(2n+1),quad x > 0.\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2 \cdot 4 \cdot 6 \ldots (2n)} \cdot \frac{x^{2n+1}}{2n+1}, \quad x > 0.∑n=1∞1⋅3⋅5…(2n−1)2⋅4⋅6…(2n)⋅x2n+12n+1,x>0.

Answer:

To test the convergence of the series
∑ n = 1 ∞ 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 , x > 0 , ∑ n = 1 ∞   1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 , x > 0 , sum_(n=1)^(oo)(1*3*5dots(2n-1))/(2*4*6dots(2n))*(x^(2n+1))/(2n+1),quad x > 0,\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2 \cdot 4 \cdot 6 \ldots (2n)} \cdot \frac{x^{2n+1}}{2n+1}, \quad x > 0,∑n=1∞1⋅3⋅5…(2n−1)2⋅4⋅6…(2n)⋅x2n+12n+1,x>0,
we can use the Ratio Test, which is suitable for series with factorial-like terms.

Step 1: Simplify the General Term

Let the general term be:
a n = 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 . a n = 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) ⋅ x 2 n + 1 2 n + 1 . a_(n)=(1*3*5dots(2n-1))/(2*4*6dots(2n))*(x^(2n+1))/(2n+1).a_n = \frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2 \cdot 4 \cdot 6 \ldots (2n)} \cdot \frac{x^{2n+1}}{2n+1}.an=1⋅3⋅5…(2n−1)2⋅4⋅6…(2n)⋅x2n+12n+1.
We can rewrite the product of odd and even numbers using double factorials or ratios of factorials:
1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) = ( 2 n − 1 ) ! ! ( 2 n ) ! ! = ( 2 n ) ! 4 n ( n ! ) 2 . 1 ⋅ 3 ⋅ 5 … ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 … ( 2 n ) = ( 2 n − 1 ) ! ! ( 2 n ) ! ! = ( 2 n ) ! 4 n ( n ! ) 2 . (1*3*5dots(2n-1))/(2*4*6dots(2n))=((2n-1)!!)/((2n)!!)=((2n)!)/(4^(n)(n!)^(2)).\frac{1 \cdot 3 \cdot 5 \ldots (2n-1)}{2 \cdot 4 \cdot 6 \ldots (2n)} = \frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{4^n (n!)^2}.1⋅3⋅5…(2n−1)2⋅4⋅6…(2n)=(2n−1)!!(2n)!!=(2n)!4n(n!)2.
However, for the Ratio Test, it’s sufficient to consider the ratio of consecutive terms.

Step 2: Apply the Ratio Test

Compute the ratio of consecutive terms:
a n + 1 a n = ( 2 ( n + 1 ) − 1 ) ! ! ( 2 ( n + 1 ) ) ! ! ⋅ x 2 ( n + 1 ) + 1 2 ( n + 1 ) + 1 ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ x 2 n + 1 2 n + 1 . a n + 1 a n = ( 2 ( n + 1 ) − 1 ) ! ! ( 2 ( n + 1 ) ) ! ! ⋅ x 2 ( n + 1 ) + 1 2 ( n + 1 ) + 1 ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ x 2 n + 1 2 n + 1 . (a_(n+1))/(a_(n))=(((2(n+1)-1)!!)/((2(n+1))!!)*(x^(2(n+1)+1))/(2(n+1)+1))/(((2n-1)!!)/((2n)!!)*(x^(2n+1))/(2n+1)).\frac{a_{n+1}}{a_n} = \frac{\frac{(2(n+1)-1)!!}{(2(n+1))!!} \cdot \frac{x^{2(n+1)+1}}{2(n+1)+1}}{\frac{(2n-1)!!}{(2n)!!} \cdot \frac{x^{2n+1}}{2n+1}}.an+1an=(2(n+1)−1)!!(2(n+1))!!⋅x2(n+1)+12(n+1)+1(2n−1)!!(2n)!!⋅x2n+12n+1.
Simplify the ratio:
a n + 1 a n = ( 2 n + 1 ) ( 2 n + 2 ) ⋅ x 2 n + 3 2 n + 3 ⋅ 2 n + 1 x 2 n + 1 = ( 2 n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 3 ) ⋅ x 2 . a n + 1 a n = ( 2 n + 1 ) ( 2 n + 2 ) ⋅ x 2 n + 3 2 n + 3 ⋅ 2 n + 1 x 2 n + 1 = ( 2 n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 3 ) ⋅ x 2 . (a_(n+1))/(a_(n))=((2n+1))/((2n+2))*(x^(2n+3))/(2n+3)*(2n+1)/(x^(2n+1))=((2n+1)^(2))/((2n+2)(2n+3))*x^(2).\frac{a_{n+1}}{a_n} = \frac{(2n+1)}{(2n+2)} \cdot \frac{x^{2n+3}}{2n+3} \cdot \frac{2n+1}{x^{2n+1}} = \frac{(2n+1)^2}{(2n+2)(2n+3)} \cdot x^2.an+1an=(2n+1)(2n+2)⋅x2n+32n+3⋅2n+1x2n+1=(2n+1)2(2n+2)(2n+3)⋅x2.
Take the limit as n → ∞ n → ∞ n rarr oon \to \inftyn→∞:
lim n → ∞ a n + 1 a n = lim n → ∞ ( 2 n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 3 ) ⋅ x 2 = lim n → ∞ 4 n 2 + 4 n + 1 4 n 2 + 10 n + 6 ⋅ x 2 = x 2 . lim n → ∞   a n + 1 a n = lim n → ∞   ( 2 n + 1 ) 2 ( 2 n + 2 ) ( 2 n + 3 ) ⋅ x 2 = lim n → ∞   4 n 2 + 4 n + 1 4 n 2 + 10 n + 6 ⋅ x 2 = x 2 . lim_(n rarr oo)(a_(n+1))/(a_(n))=lim_(n rarr oo)((2n+1)^(2))/((2n+2)(2n+3))*x^(2)=lim_(n rarr oo)(4n^(2)+4n+1)/(4n^(2)+10 n+6)*x^(2)=x^(2).\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(2n+1)^2}{(2n+2)(2n+3)} \cdot x^2 = \lim_{n \to \infty} \frac{4n^2 + 4n + 1}{4n^2 + 10n + 6} \cdot x^2 = x^2.limn→∞an+1an=limn→∞(2n+1)2(2n+2)(2n+3)⋅x2=limn→∞4n2+4n+14n2+10n+6⋅x2=x2.

Step 3: Determine Convergence

By the Ratio Test:
  • If lim n → ∞ | a n + 1 a n | < 1 lim n → ∞   a n + 1 a n < 1 lim_(n rarr oo)|(a_(n+1))/(a_(n))| < 1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1limn→∞|an+1an|<1, the series converges.
  • If lim n → ∞ | a n + 1 a n | > 1 lim n → ∞   a n + 1 a n > 1 lim_(n rarr oo)|(a_(n+1))/(a_(n))| > 1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1limn→∞|an+1an|>1, the series diverges.
  • If lim n → ∞ | a n + 1 a n | = 1 lim n → ∞   a n + 1 a n = 1 lim_(n rarr oo)|(a_(n+1))/(a_(n))|=1\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1limn→∞|an+1an|=1, the test is inconclusive.
Here, lim n → ∞ a n + 1 a n = x 2 lim n → ∞   a n + 1 a n = x 2 lim_(n rarr oo)(a_(n+1))/(a_(n))=x^(2)\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = x^2limn→∞an+1an=x2, so:
  • The series converges if x 2 < 1 x 2 < 1 x^(2) < 1x^2 < 1x2<1 (i.e., 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1).
  • The series diverges if x 2 > 1 x 2 > 1 x^(2) > 1x^2 > 1x2>1 (i.e., x > 1 x > 1 x > 1x > 1x>1).
  • The test is inconclusive if x = 1 x = 1 x=1x = 1x=1.

Step 4: Check the Boundary Case x = 1 x = 1 x=1x = 1x=1

For x = 1 x = 1 x=1x = 1x=1, the series becomes:
∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ 1 2 n + 1 . ∑ n = 1 ∞   ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ 1 2 n + 1 . sum_(n=1)^(oo)((2n-1)!!)/((2n)!!)*(1)/(2n+1).\sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!} \cdot \frac{1}{2n+1}.∑n=1∞(2n−1)!!(2n)!!⋅12n+1.
Using Gauss’s Test or Raabe’s Test, we can show that this series diverges. Alternatively, note that:
( 2 n − 1 ) ! ! ( 2 n ) ! ! ∼ 1 π n (by Stirling’s approximation) , ( 2 n − 1 ) ! ! ( 2 n ) ! ! ∼ 1 π n (by Stirling’s approximation) , ((2n-1)!!)/((2n)!!)∼(1)/(sqrt(pi n))quad(by Stirling’s approximation),\frac{(2n-1)!!}{(2n)!!} \sim \frac{1}{\sqrt{\pi n}} \quad \text{(by Stirling’s approximation)},(2n−1)!!(2n)!!∼1πn(by Stirling’s approximation),
so the general term behaves like 1 2 n 3 / 2 1 2 n 3 / 2 (1)/(2n^(3//2))\frac{1}{2n^{3/2}}12n3/2, and the series diverges by comparison with the harmonic series.

Final Conclusion

  • The series converges for 0 < x < 1 0 < x < 1 0 < x < 10 < x < 10<x<1.
  • The series diverges for x ≥ 1 x ≥ 1 x >= 1x \geq 1x≥1.
The series converges for 0 < x < 1 and diverges for x ≥ 1. The series converges for  0 < x < 1  and diverges for  x ≥ 1. “The series converges for “0 < x < 1” and diverges for “x >= 1.\boxed{\text{The series converges for } 0 < x < 1 \text{ and diverges for } x \geq 1.}The series converges for 0<x<1 and diverges for x≥1.

Question:-01 (d)

State the sufficient conditions for a function f ( z ) = f ( x + i y ) = u ( x , y ) + i v ( x , y ) f ( z ) = f ( x + i y ) = u ( x , y ) + i v ( x , y ) f(z)=f(x+iy)=u(x,y)+iv(x,y)f(z) = f(x + iy) = u(x, y) + i v(x, y)f(z)=f(x+iy)=u(x,y)+iv(x,y) to be analytic in its domain. Hence, show that f ( z ) = log ⁡ z f ( z ) = log ⁡ z f(z)=log zf(z) = \log zf(z)=log⁡z is analytic in its domain and find d f d z d f d z (df)/(dz)\frac{df}{dz}dfdz.

Answer:

Sufficient Conditions for Analyticity

A function f ( z ) = u ( x , y ) + i v ( x , y ) f ( z ) = u ( x , y ) + i v ( x , y ) f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + i v(x, y)f(z)=u(x,y)+iv(x,y) is analytic (holomorphic) in a domain D D DDD if the following conditions hold:
  1. Existence of Partial Derivatives:
    The partial derivatives ∂ u ∂ x , ∂ u ∂ y , ∂ v ∂ x , ∂ v ∂ y ∂ u ∂ x , ∂ u ∂ y , ∂ v ∂ x , ∂ v ∂ y (del u)/(del x),(del u)/(del y),(del v)/(del x),(del v)/(del y)\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}∂u∂x,∂u∂y,∂v∂x,∂v∂y exist and are continuous in D D DDD.
  2. Cauchy-Riemann Equations:
    The functions u ( x , y ) u ( x , y ) u(x,y)u(x,y)u(x,y) and v ( x , y ) v ( x , y ) v(x,y)v(x,y)v(x,y) satisfy:
    ∂ u ∂ x = ∂ v ∂ y , ∂ u ∂ y = − ∂ v ∂ x . ∂ u ∂ x = ∂ v ∂ y , ∂ u ∂ y = − ∂ v ∂ x . (del u)/(del x)=(del v)/(del y),quad(del u)/(del y)=-(del v)/(del x).\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.∂u∂x=∂v∂y,∂u∂y=−∂v∂x.
If these conditions hold, then f ( z ) f ( z ) f(z)f(z)f(z) is differentiable in D D DDD, and its derivative is given by:
f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = ∂ v ∂ y − i ∂ u ∂ y . f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = ∂ v ∂ y − i ∂ u ∂ y . f^(‘)(z)=(del u)/(del x)+i(del v)/(del x)=(del v)/(del y)-i(del u)/(del y).f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} – i \frac{\partial u}{\partial y}.f′(z)=∂u∂x+i∂v∂x=∂v∂y−i∂u∂y.

Analyticity of f ( z ) = log ⁡ z f ( z ) = log ⁡ z f(z)=log zf(z) = \log zf(z)=log⁡z

The complex logarithm is defined as:
log ⁡ z = ln ⁡ | z | + i arg ⁡ z , log ⁡ z = ln ⁡ | z | + i arg ⁡ z , log z=ln |z|+i arg z,\log z = \ln |z| + i \arg z,log⁡z=ln⁡|z|+iarg⁡z,
where:
  • ln ⁡ | z | ln ⁡ | z | ln |z|\ln |z|ln⁡|z| is the natural logarithm of the modulus,
  • arg ⁡ z arg ⁡ z arg z\arg zarg⁡z is the argument (angle) of z z zzz.
Let z = x + i y = r e i θ z = x + i y = r e i θ z=x+iy=re^(i theta)z = x + iy = re^{i\theta}z=x+iy=reiθ, where r = x 2 + y 2 r = x 2 + y 2 r=sqrt(x^(2)+y^(2))r = \sqrt{x^2 + y^2}r=x2+y2 and θ = arg ⁡ z θ = arg ⁡ z theta=arg z\theta = \arg zθ=arg⁡z. Then:
log ⁡ z = ln ⁡ r + i θ = 1 2 ln ⁡ ( x 2 + y 2 ) + i arctan ⁡ ( y x ) . log ⁡ z = ln ⁡ r + i θ = 1 2 ln ⁡ ( x 2 + y 2 ) + i arctan ⁡ y x . log z=ln r+i theta=(1)/(2)ln(x^(2)+y^(2))+i arctan((y)/(x)).\log z = \ln r + i \theta = \frac{1}{2} \ln(x^2 + y^2) + i \arctan\left(\frac{y}{x}\right).log⁡z=ln⁡r+iθ=12ln⁡(x2+y2)+iarctan⁡(yx).
Thus, we identify:
u ( x , y ) = 1 2 ln ⁡ ( x 2 + y 2 ) , v ( x , y ) = arctan ⁡ ( y x ) . u ( x , y ) = 1 2 ln ⁡ ( x 2 + y 2 ) , v ( x , y ) = arctan ⁡ y x . u(x,y)=(1)/(2)ln(x^(2)+y^(2)),quad v(x,y)=arctan((y)/(x)).u(x, y) = \frac{1}{2} \ln(x^2 + y^2), \quad v(x, y) = \arctan\left(\frac{y}{x}\right).u(x,y)=12ln⁡(x2+y2),v(x,y)=arctan⁡(yx).

1. Check the Cauchy-Riemann Equations

Compute the partial derivatives:
∂ u ∂ x = x x 2 + y 2 , ∂ u ∂ y = y x 2 + y 2 , ∂ u ∂ x = x x 2 + y 2 , ∂ u ∂ y = y x 2 + y 2 , (del u)/(del x)=(x)/(x^(2)+y^(2)),quad(del u)/(del y)=(y)/(x^(2)+y^(2)),\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}, \quad \frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2},∂u∂x=xx2+y2,∂u∂y=yx2+y2,
∂ v ∂ x = − y x 2 + y 2 , ∂ v ∂ y = x x 2 + y 2 . ∂ v ∂ x = − y x 2 + y 2 , ∂ v ∂ y = x x 2 + y 2 . (del v)/(del x)=(-y)/(x^(2)+y^(2)),quad(del v)/(del y)=(x)/(x^(2)+y^(2)).\frac{\partial v}{\partial x} = \frac{-y}{x^2 + y^2}, \quad \frac{\partial v}{\partial y} = \frac{x}{x^2 + y^2}.∂v∂x=−yx2+y2,∂v∂y=xx2+y2.
Now verify:
∂ u ∂ x = x x 2 + y 2 = ∂ v ∂ y , ∂ u ∂ x = x x 2 + y 2 = ∂ v ∂ y , (del u)/(del x)=(x)/(x^(2)+y^(2))=(del v)/(del y),\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2} = \frac{\partial v}{\partial y},∂u∂x=xx2+y2=∂v∂y,
∂ u ∂ y = y x 2 + y 2 = − ( − y x 2 + y 2 ) = − ∂ v ∂ x . ∂ u ∂ y = y x 2 + y 2 = − − y x 2 + y 2 = − ∂ v ∂ x . (del u)/(del y)=(y)/(x^(2)+y^(2))=-((-y)/(x^(2)+y^(2)))=-(del v)/(del x).\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2} = -\left( \frac{-y}{x^2 + y^2} \right) = -\frac{\partial v}{\partial x}.∂u∂y=yx2+y2=−(−yx2+y2)=−∂v∂x.
Thus, the Cauchy-Riemann equations hold.

2. Continuity of Partial Derivatives

The partial derivatives are continuous everywhere except at z = 0 z = 0 z=0z = 0z=0, where log ⁡ z log ⁡ z log z\log zlog⁡z is not defined. Hence, log ⁡ z log ⁡ z log z\log zlog⁡z is analytic in its domain C ∖ { 0 } C ∖ { 0 } C\\{0}\mathbb{C} \setminus \{0\}C∖{0}.

Derivative of log ⁡ z log ⁡ z log z\log zlog⁡z

Using the derivative formula for analytic functions:
f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = x x 2 + y 2 + i ( − y x 2 + y 2 ) = x − i y x 2 + y 2 . f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x = x x 2 + y 2 + i − y x 2 + y 2 = x − i y x 2 + y 2 . f^(‘)(z)=(del u)/(del x)+i(del v)/(del x)=(x)/(x^(2)+y^(2))+i((-y)/(x^(2)+y^(2)))=(x-iy)/(x^(2)+y^(2)).f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{x}{x^2 + y^2} + i \left( \frac{-y}{x^2 + y^2} \right) = \frac{x – iy}{x^2 + y^2}.f′(z)=∂u∂x+i∂v∂x=xx2+y2+i(−yx2+y2)=x−iyx2+y2.
But x 2 + y 2 = | z | 2 x 2 + y 2 = | z | 2 x^(2)+y^(2)=|z|^(2)x^2 + y^2 = |z|^2x2+y2=|z|2, and x − i y = z ― x − i y = z ¯ x-iy= bar(z)x – iy = \overline{z}x−iy=z―, so:
f ′ ( z ) = z ― | z | 2 = 1 z . f ′ ( z ) = z ¯ | z | 2 = 1 z . f^(‘)(z)=( bar(z))/(|z|^(2))=(1)/(z).f'(z) = \frac{\overline{z}}{|z|^2} = \frac{1}{z}.f′(z)=z―|z|2=1z.
Thus:
d d z log ⁡ z = 1 z . d d z log ⁡ z = 1 z . (d)/(dz)log z=(1)/(z).\frac{d}{dz} \log z = \frac{1}{z}.ddzlog⁡z=1z.

Final Answer

log ⁡ z is analytic in C ∖ { 0 } , and its derivative is d d z log ⁡ z = 1 z . log ⁡ z  is analytic in  C ∖ { 0 } ,  and its derivative is  d d z log ⁡ z = 1 z . log z” is analytic in “C\\{0},” and its derivative is “(d)/(dz)log z=(1)/(z).\boxed{\log z \text{ is analytic in } \mathbb{C} \setminus \{0\}, \text{ and its derivative is } \frac{d}{dz} \log z = \frac{1}{z}.}log⁡z is analytic in C∖{0}, and its derivative is ddzlog⁡z=1z.

Question:-01 (e)

A person requires 24, 24, and 20 units of chemicals A A AAA, B B BBB, and C C CCC respectively for his garden. Product P P PPP contains 2, 4, and 1 units of chemicals A A AAA, B B BBB, and C C CCC respectively per jar, and product Q Q QQQ contains 2, 1, and 5 units of chemicals A A AAA, B B BBB, and C C CCC respectively per jar. If a jar of P P PPP costs ₹30 and a jar of Q Q QQQ costs ₹50, then how many jars of each should be purchased in order to minimize the cost and meet the requirements?

Answer:

Problem Statement

A person requires:
  • 24 units of chemical A A AAA,
  • 24 units of chemical B B BBB,
  • 20 units of chemical C C CCC.
Products Available:
  • Product P P PPP (per jar):
    • 2 units of A A AAA,
    • 4 units of B B BBB,
    • 1 unit of C C CCC.
    • Cost: ₹30 per jar.
  • Product Q Q QQQ (per jar):
    • 2 units of A A AAA,
    • 1 unit of B B BBB,
    • 5 units of C C CCC.
    • Cost: ₹50 per jar.
Goal: Determine the number of jars of P P PPP and Q Q QQQ to purchase to minimize cost while meeting the chemical requirements.

Step 1: Define Variables

Let:
  • x x xxx = number of jars of P P PPP,
  • y y yyy = number of jars of Q Q QQQ.

Step 2: Formulate Constraints

The total units of each chemical must satisfy the requirements:
  1. Chemical A A AAA:
    2 x + 2 y ≥ 24 ⇒ x + y ≥ 12. 2 x + 2 y ≥ 24 ⇒ x + y ≥ 12. 2x+2y >= 24quad=>quad x+y >= 12.2x + 2y \geq 24 \quad \Rightarrow \quad x + y \geq 12.2x+2y≥24⇒x+y≥12.
  2. Chemical B B BBB:
    4 x + y ≥ 24. 4 x + y ≥ 24. 4x+y >= 24.4x + y \geq 24.4x+y≥24.
  3. Chemical C C CCC:
    x + 5 y ≥ 20. x + 5 y ≥ 20. x+5y >= 20.x + 5y \geq 20.x+5y≥20.
  4. Non-negativity:
    x ≥ 0 , y ≥ 0. x ≥ 0 , y ≥ 0. x >= 0,quad y >= 0.x \geq 0, \quad y \geq 0.x≥0,y≥0.

Step 3: Objective Function

We want to minimize the total cost:
Cost = 30 x + 50 y . Cost = 30 x + 50 y . “Cost”=30 x+50 y.\text{Cost} = 30x + 50y.Cost=30x+50y.

Step 4: Solve the System of Inequalities

We find the feasible region by solving the inequalities:
  1. From x + y ≥ 12 x + y ≥ 12 x+y >= 12x + y \geq 12x+y≥12:
    • If x = 0 x = 0 x=0x = 0x=0, y = 12 y = 12 y=12y = 12y=12.
    • If y = 0 y = 0 y=0y = 0y=0, x = 12 x = 12 x=12x = 12x=12.
  2. From 4 x + y ≥ 24 4 x + y ≥ 24 4x+y >= 244x + y \geq 244x+y≥24:
    • If x = 0 x = 0 x=0x = 0x=0, y = 24 y = 24 y=24y = 24y=24.
    • If y = 0 y = 0 y=0y = 0y=0, x = 6 x = 6 x=6x = 6x=6.
  3. From x + 5 y ≥ 20 x + 5 y ≥ 20 x+5y >= 20x + 5y \geq 20x+5y≥20:
    • If x = 0 x = 0 x=0x = 0x=0, y = 4 y = 4 y=4y = 4y=4.
    • If y = 0 y = 0 y=0y = 0y=0, x = 20 x = 20 x=20x = 20x=20.
Intersection Points (Vertices of Feasible Region):
  • Intersection of x + y = 12 x + y = 12 x+y=12x + y = 12x+y=12 and 4 x + y = 24 4 x + y = 24 4x+y=244x + y = 244x+y=24:
    { x + y = 12 , 4 x + y = 24. x + y = 12 , 4 x + y = 24. {[x+y=12″,”],[4x+y=24.]:}\begin{cases} x + y = 12, \\ 4x + y = 24. \end{cases}{x+y=12,4x+y=24.
    Subtract the first equation from the second:
    3 x = 12 ⇒ x = 4 , y = 8. 3 x = 12 ⇒ x = 4 , y = 8. 3x=12quad=>quad x=4,quad y=8.3x = 12 \quad \Rightarrow \quad x = 4, \quad y = 8.3x=12⇒x=4,y=8.
    Point: ( 4 , 8 ) ( 4 , 8 ) (4,8)(4, 8)(4,8).
  • Intersection of x + y = 12 x + y = 12 x+y=12x + y = 12x+y=12 and x + 5 y = 20 x + 5 y = 20 x+5y=20x + 5y = 20x+5y=20:
    { x + y = 12 , x + 5 y = 20. x + y = 12 , x + 5 y = 20. {[x+y=12″,”],[x+5y=20.]:}\begin{cases} x + y = 12, \\ x + 5y = 20. \end{cases}{x+y=12,x+5y=20.
    Subtract the first equation from the second:
    4 y = 8 ⇒ y = 2 , x = 10. 4 y = 8 ⇒ y = 2 , x = 10. 4y=8quad=>quad y=2,quad x=10.4y = 8 \quad \Rightarrow \quad y = 2, \quad x = 10.4y=8⇒y=2,x=10.
    Point: ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2).
  • Intersection of 4 x + y = 24 4 x + y = 24 4x+y=244x + y = 244x+y=24 and x + 5 y = 20 x + 5 y = 20 x+5y=20x + 5y = 20x+5y=20:
    { 4 x + y = 24 , x + 5 y = 20. 4 x + y = 24 , x + 5 y = 20. {[4x+y=24″,”],[x+5y=20.]:}\begin{cases} 4x + y = 24, \\ x + 5y = 20. \end{cases}{4x+y=24,x+5y=20.
    Solve the first equation for y y yyy: y = 24 − 4 x y = 24 − 4 x y=24-4xy = 24 – 4xy=24−4x.
    Substitute into the second equation:
    x + 5 ( 24 − 4 x ) = 20 ⇒ x + 120 − 20 x = 20 ⇒ − 19 x = − 100 ⇒ x = 100 19 ≈ 5.26 . x + 5 ( 24 − 4 x ) = 20 ⇒ x + 120 − 20 x = 20 ⇒ − 19 x = − 100 ⇒ x = 100 19 ≈ 5.26 . x+5(24-4x)=20quad=>quad x+120-20 x=20quad=>quad-19 x=-100quad=>quad x=(100)/(19)~~5.26.x + 5(24 – 4x) = 20 \quad \Rightarrow \quad x + 120 – 20x = 20 \quad \Rightarrow \quad -19x = -100 \quad \Rightarrow \quad x = \frac{100}{19} \approx 5.26.x+5(24−4x)=20⇒x+120−20x=20⇒−19x=−100⇒x=10019≈5.26.
    Then y = 24 − 4 ( 100 19 ) = 456 − 400 19 = 56 19 ≈ 2.95 y = 24 − 4 100 19 = 456 − 400 19 = 56 19 ≈ 2.95 y=24-4((100)/(19))=(456-400)/(19)=(56)/(19)~~2.95y = 24 – 4 \left( \frac{100}{19} \right) = \frac{456 – 400}{19} = \frac{56}{19} \approx 2.95y=24−4(10019)=456−40019=5619≈2.95.
    Point: ( 100 19 , 56 19 ) 100 19 , 56 19 ((100)/(19),(56)/(19))\left( \frac{100}{19}, \frac{56}{19} \right)(10019,5619).

Step 5: Evaluate Cost at Critical Points

Compute Cost = 30 x + 50 y Cost = 30 x + 50 y “Cost”=30 x+50 y\text{Cost} = 30x + 50yCost=30x+50y at each feasible vertex:
  1. At ( 4 , 8 ) ( 4 , 8 ) (4,8)(4, 8)(4,8):
    30 ( 4 ) + 50 ( 8 ) = 120 + 400 =₹ 520. 30 ( 4 ) + 50 ( 8 ) = 120 + 400 =₹ 520. 30(4)+50(8)=120+400=₹520.30(4) + 50(8) = 120 + 400 = ₹520.30(4)+50(8)=120+400=₹520.
  2. At ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2):
    30 ( 10 ) + 50 ( 2 ) = 300 + 100 =₹ 400. 30 ( 10 ) + 50 ( 2 ) = 300 + 100 =₹ 400. 30(10)+50(2)=300+100=₹400.30(10) + 50(2) = 300 + 100 = ₹400.30(10)+50(2)=300+100=₹400.
  3. At ( 100 19 , 56 19 ) 100 19 , 56 19 ((100)/(19),(56)/(19))\left( \frac{100}{19}, \frac{56}{19} \right)(10019,5619):
    30 ( 100 19 ) + 50 ( 56 19 ) = 3000 19 + 2800 19 = 5800 19 ≈₹ 305.26 . 30 100 19 + 50 56 19 = 3000 19 + 2800 19 = 5800 19 ≈₹ 305.26 . 30((100)/(19))+50((56)/(19))=(3000)/(19)+(2800)/(19)=(5800)/(19)≈₹305.26.30 \left( \frac{100}{19} \right) + 50 \left( \frac{56}{19} \right) = \frac{3000}{19} + \frac{2800}{19} = \frac{5800}{19} \approx ₹305.26.30(10019)+50(5619)=300019+280019=580019≈₹305.26.
However, x x xxx and y y yyy must be integers (since you can’t buy a fraction of a jar). So, we check integer solutions near ( 100 19 , 56 19 ) ≈ ( 5.26 , 2.95 ) 100 19 , 56 19 ≈ ( 5.26 , 2.95 ) ((100)/(19),(56)/(19))~~(5.26,2.95)\left( \frac{100}{19}, \frac{56}{19} \right) \approx (5.26, 2.95)(10019,5619)≈(5.26,2.95):
  • Try ( 5 , 3 ) ( 5 , 3 ) (5,3)(5, 3)(5,3):
    • Check constraints: 2 ( 5 ) + 2 ( 3 ) = 16 ≥ 24 ? No (16 < 24) . Not feasible. 2 ( 5 ) + 2 ( 3 ) = 16 ≥ 24 ? No (16 < 24) . Not feasible. 2(5)+2(3)=16 >= 24?quad”No (16 < 24)”.quad”Not feasible.”2(5) + 2(3) = 16 \geq 24? \quad \text{No (16 < 24)}. \quad \text{Not feasible.}2(5)+2(3)=16≥24?No (16 < 24).Not feasible.
  • Try ( 6 , 2 ) ( 6 , 2 ) (6,2)(6, 2)(6,2):
    • Check constraints: 2 ( 6 ) + 2 ( 2 ) = 16 ≥ 24 ? No (16 < 24) . Not feasible. 2 ( 6 ) + 2 ( 2 ) = 16 ≥ 24 ? No (16 < 24) . Not feasible. 2(6)+2(2)=16 >= 24?quad”No (16 < 24)”.quad”Not feasible.”2(6) + 2(2) = 16 \geq 24? \quad \text{No (16 < 24)}. \quad \text{Not feasible.}2(6)+2(2)=16≥24?No (16 < 24).Not feasible.
  • Try ( 6 , 6 ) ( 6 , 6 ) (6,6)(6, 6)(6,6):
    • Check constraints: 2 ( 6 ) + 2 ( 6 ) = 24 ≥ 24 (OK) , 4 ( 6 ) + 6 = 30 ≥ 24 (OK) , 6 + 5 ( 6 ) = 36 ≥ 20 (OK) . 2 ( 6 ) + 2 ( 6 ) = 24 ≥ 24 (OK) , 4 ( 6 ) + 6 = 30 ≥ 24 (OK) , 6 + 5 ( 6 ) = 36 ≥ 20 (OK) . 2(6)+2(6)=24 >= 24quad(OK),4(6)+6=30 >= 24quad(OK),6+5(6)=36 >= 20quad(OK).2(6) + 2(6) = 24 \geq 24 \quad \text{(OK)}, \\ 4(6) + 6 = 30 \geq 24 \quad \text{(OK)}, \\ 6 + 5(6) = 36 \geq 20 \quad \text{(OK)}.2(6)+2(6)=24≥24(OK),4(6)+6=30≥24(OK),6+5(6)=36≥20(OK).
    • Cost: 30 ( 6 ) + 50 ( 6 ) = 180 + 300 =₹ 480. 30 ( 6 ) + 50 ( 6 ) = 180 + 300 =₹ 480. 30(6)+50(6)=180+300=₹480.30(6) + 50(6) = 180 + 300 = ₹480.30(6)+50(6)=180+300=₹480.
  • Try ( 4 , 8 ) ( 4 , 8 ) (4,8)(4, 8)(4,8): (Already computed: ₹520)
  • Try ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2): (Already computed: ₹400)
Best integer solution: ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2) with cost ₹400.
But wait, let’s check another integer point:
  • Try ( 8 , 4 ) ( 8 , 4 ) (8,4)(8, 4)(8,4):
    • Check constraints: 2 ( 8 ) + 2 ( 4 ) = 24 ≥ 24 (OK) , 4 ( 8 ) + 4 = 36 ≥ 24 (OK) , 8 + 5 ( 4 ) = 28 ≥ 20 (OK) . 2 ( 8 ) + 2 ( 4 ) = 24 ≥ 24 (OK) , 4 ( 8 ) + 4 = 36 ≥ 24 (OK) , 8 + 5 ( 4 ) = 28 ≥ 20 (OK) . 2(8)+2(4)=24 >= 24quad(OK),4(8)+4=36 >= 24quad(OK),8+5(4)=28 >= 20quad(OK).2(8) + 2(4) = 24 \geq 24 \quad \text{(OK)}, \\ 4(8) + 4 = 36 \geq 24 \quad \text{(OK)}, \\ 8 + 5(4) = 28 \geq 20 \quad \text{(OK)}.2(8)+2(4)=24≥24(OK),4(8)+4=36≥24(OK),8+5(4)=28≥20(OK).
    • Cost: 30 ( 8 ) + 50 ( 4 ) = 240 + 200 =₹ 440. 30 ( 8 ) + 50 ( 4 ) = 240 + 200 =₹ 440. 30(8)+50(4)=240+200=₹440.30(8) + 50(4) = 240 + 200 = ₹440.30(8)+50(4)=240+200=₹440.
But ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2) gives a lower cost (₹400) than ( 8 , 4 ) ( 8 , 4 ) (8,4)(8, 4)(8,4) (₹440).

Verification of ( 10 , 2 ) ( 10 , 2 ) (10,2)(10, 2)(10,2):

  • Chemical A A AAA: 2 ( 10 ) + 2 ( 2 ) = 24 2 ( 10 ) + 2 ( 2 ) = 24 2(10)+2(2)=242(10) + 2(2) = 242(10)+2(2)=24 (OK),
  • Chemical B B BBB: 4 ( 10 ) + 2 = 42 ≥ 24 4 ( 10 ) + 2 = 42 ≥ 24 4(10)+2=42 >= 244(10) + 2 = 42 \geq 244(10)+2=42≥24 (OK),
  • Chemical C C CCC: 10 + 5 ( 2 ) = 20 ≥ 20 10 + 5 ( 2 ) = 20 ≥ 20 10+5(2)=20 >= 2010 + 5(2) = 20 \geq 2010+5(2)=20≥20 (OK).
This is feasible and cheaper than other integer solutions.

Conclusion

The minimum cost is achieved by purchasing:
  • 10 jars of P P PPP,
  • 2 jars of Q Q QQQ,
    with a total cost of ₹400.
( x , y ) = ( 10 , 2 ) with a minimum cost of ₹400. ( x , y ) = ( 10 , 2 )  with a minimum cost of ₹400. (x,y)=(10,2)” with a minimum cost of ₹400.”\boxed{(x, y) = (10, 2) \text{ with a minimum cost of ₹400.}}(x,y)=(10,2) with a minimum cost of ₹400.

Question:-02

(a) Prove that a non-commutative group of order 2 p 2 p 2p2p2p, where p p ppp is an odd prime, must have a subgroup of order p p ppp.

Answer:

To prove that a non-commutative group G G GGG of order 2 p 2 p 2p2p2p, where p p ppp is an odd prime, must have a subgroup of order p p ppp, we can proceed as follows:

Step 1: Use Sylow’s Theorems

By Sylow’s First Theorem, since p p ppp divides the order of G G GGG (which is 2 p 2 p 2p2p2p), there exists at least one Sylow p p ppp-subgroup of G G GGG. Let n p n p n_(p)n_pnp denote the number of Sylow p p ppp-subgroups.
  • Sylow’s Third Theorem states that:
    n p ≡ 1 ( mod p ) and n p ∣ 2 p . n p ≡ 1 ( mod p ) and n p ∣ 2 p . n_(p)-=1quad(modp)quad”and”quadn_(p)∣2p.n_p \equiv 1 \pmod{p} \quad \text{and} \quad n_p \mid 2p.np≡1(modp)andnp∣2p.
    Since p p ppp is prime and p > 2 p > 2 p > 2p > 2p>2, the divisors of 2 p 2 p 2p2p2p are 1 , 2 , p , 2 p 1 , 2 , p , 2 p 1,2,p,2p1, 2, p, 2p1,2,p,2p. The condition n p ≡ 1 ( mod p ) n p ≡ 1 ( mod p ) n_(p)-=1(modp)n_p \equiv 1 \pmod{p}np≡1(modp) implies:
    n p = 1 or n p = p + 1. n p = 1 or n p = p + 1. n_(p)=1quad”or”quadn_(p)=p+1.n_p = 1 \quad \text{or} \quad n_p = p + 1.np=1ornp=p+1.
    But p + 1 p + 1 p+1p + 1p+1 does not divide 2 p 2 p 2p2p2p unless p = 2 p = 2 p=2p = 2p=2, which is excluded since p p ppp is odd. Thus:
    n p = 1. n p = 1. n_(p)=1.n_p = 1.np=1.
  • Conclusion: There is exactly one Sylow p p ppp-subgroup, say H H HHH, of order p p ppp.

Step 2: Show H H HHH is Normal

Since n p = 1 n p = 1 n_(p)=1n_p = 1np=1, the Sylow p p ppp-subgroup H H HHH is unique, and thus it is normal in G G GGG.

Step 3: Non-commutativity Implies G G GGG is Not Cyclic

If G G GGG were commutative, it would be isomorphic to the cyclic group Z 2 p Z 2 p Z_(2p)\mathbb{Z}_{2p}Z2p, which has a unique subgroup of order p p ppp. But G G GGG is given to be non-commutative, so it must be isomorphic to the dihedral group D p D p D_(p)D_pDp, which has the structure:
G = ⟨ r , s ∣ r p = s 2 = e , s r s = r − 1 ⟩ , G = ⟨ r , s ∣ r p = s 2 = e , s r s = r − 1 ⟩ , G=(:r,s∣r^(p)=s^(2)=e,srs=r^(-1):),G = \langle r, s \mid r^p = s^2 = e, srs = r^{-1} \rangle,G=⟨r,s∣rp=s2=e,srs=r−1⟩,
where:
  • ⟨ r ⟩ ⟨ r ⟩ (:r:)\langle r \rangle⟨r⟩ is the cyclic subgroup of order p p ppp,
  • ⟨ s ⟩ ⟨ s ⟩ (:s:)\langle s \rangle⟨s⟩ is a subgroup of order 2 2 222.

Final Conclusion

Thus, in the non-commutative case, G G GGG must have:
  • A normal subgroup H H HHH of order p p ppp (the Sylow p p ppp-subgroup),
  • And another subgroup of order 2 2 222, but the question focuses on the subgroup of order p p ppp.
A non-commutative group of order 2 p has a normal subgroup of order p . A non-commutative group of order  2 p  has a normal subgroup of order  p . “A non-commutative group of order “2p” has a normal subgroup of order “p.\boxed{\text{A non-commutative group of order } 2p \text{ has a normal subgroup of order } p.}A non-commutative group of order 2p has a normal subgroup of order p.

Question:-02 (b)

Using the method of Lagrange’s multipliers, find the minimum and maximum distances of the point P ( 2 , 6 , 3 ) P ( 2 , 6 , 3 ) P(2,6,3)P(2, 6, 3)P(2,6,3) from the sphere x 2 + y 2 + z 2 = 4 x 2 + y 2 + z 2 = 4 x^(2)+y^(2)+z^(2)=4x^2 + y^2 + z^2 = 4x2+y2+z2=4.

Answer:

To find the minimum and maximum distances from the point P ( 2 , 6 , 3 ) P ( 2 , 6 , 3 ) P(2,6,3)P(2, 6, 3)P(2,6,3) to the sphere x 2 + y 2 + z 2 = 4 x 2 + y 2 + z 2 = 4 x^(2)+y^(2)+z^(2)=4x^2 + y^2 + z^2 = 4x2+y2+z2=4 using Lagrange multipliers, follow these steps:

Step 1: Define the Distance Function

The distance D D DDD from P ( 2 , 6 , 3 ) P ( 2 , 6 , 3 ) P(2,6,3)P(2, 6, 3)P(2,6,3) to a point ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) on the sphere is given by:
D = ( x − 2 ) 2 + ( y − 6 ) 2 + ( z − 3 ) 2 . D = ( x − 2 ) 2 + ( y − 6 ) 2 + ( z − 3 ) 2 . D=sqrt((x-2)^(2)+(y-6)^(2)+(z-3)^(2)).D = \sqrt{(x – 2)^2 + (y – 6)^2 + (z – 3)^2}.D=(x−2)2+(y−6)2+(z−3)2.
To simplify computations, we minimize and maximize the squared distance:
D 2 = ( x − 2 ) 2 + ( y − 6 ) 2 + ( z − 3 ) 2 . D 2 = ( x − 2 ) 2 + ( y − 6 ) 2 + ( z − 3 ) 2 . D^(2)=(x-2)^(2)+(y-6)^(2)+(z-3)^(2).D^2 = (x – 2)^2 + (y – 6)^2 + (z – 3)^2.D2=(x−2)2+(y−6)2+(z−3)2.

Step 2: Set Up the Constraint

The point ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) must lie on the sphere:
x 2 + y 2 + z 2 = 4. x 2 + y 2 + z 2 = 4. x^(2)+y^(2)+z^(2)=4.x^2 + y^2 + z^2 = 4.x2+y2+z2=4.

Step 3: Apply the Method of Lagrange Multipliers

We introduce a Lagrange multiplier λ λ lambda\lambdaλ and set up the following system by taking partial derivatives:
∇ D 2 = λ ∇ ( x 2 + y 2 + z 2 ) . ∇ D 2 = λ ∇ ( x 2 + y 2 + z 2 ) . gradD^(2)=lambda grad(x^(2)+y^(2)+z^(2)).\nabla D^2 = \lambda \nabla (x^2 + y^2 + z^2).∇D2=λ∇(x2+y2+z2).
This gives:
2 ( x − 2 ) = λ ⋅ 2 x , 2 ( y − 6 ) = λ ⋅ 2 y , 2 ( z − 3 ) = λ ⋅ 2 z . 2 ( x − 2 ) = λ ⋅ 2 x , 2 ( y − 6 ) = λ ⋅ 2 y , 2 ( z − 3 ) = λ ⋅ 2 z . 2(x-2)=lambda*2x,quad2(y-6)=lambda*2y,quad2(z-3)=lambda*2z.2(x – 2) = \lambda \cdot 2x, \quad 2(y – 6) = \lambda \cdot 2y, \quad 2(z – 3) = \lambda \cdot 2z.2(x−2)=λ⋅2x,2(y−6)=λ⋅2y,2(z−3)=λ⋅2z.
Simplify these equations:
x − 2 = λ x , y − 6 = λ y , z − 3 = λ z . x − 2 = λ x , y − 6 = λ y , z − 3 = λ z . x-2=lambda x,quad y-6=lambda y,quad z-3=lambda z.x – 2 = \lambda x, \quad y – 6 = \lambda y, \quad z – 3 = \lambda z.x−2=λx,y−6=λy,z−3=λz.
Solve each for λ λ lambda\lambdaλ:
λ = x − 2 x , λ = y − 6 y , λ = z − 3 z . λ = x − 2 x , λ = y − 6 y , λ = z − 3 z . lambda=(x-2)/(x),quad lambda=(y-6)/(y),quad lambda=(z-3)/(z).\lambda = \frac{x – 2}{x}, \quad \lambda = \frac{y – 6}{y}, \quad \lambda = \frac{z – 3}{z}.λ=x−2x,λ=y−6y,λ=z−3z.
Since all expressions equal λ λ lambda\lambdaλ, set them equal to each other:
x − 2 x = y − 6 y = z − 3 z . x − 2 x = y − 6 y = z − 3 z . (x-2)/(x)=(y-6)/(y)=(z-3)/(z).\frac{x – 2}{x} = \frac{y – 6}{y} = \frac{z – 3}{z}.x−2x=y−6y=z−3z.

Step 4: Solve for y y yyy and z z zzz in Terms of x x xxx

From x − 2 x = y − 6 y x − 2 x = y − 6 y (x-2)/(x)=(y-6)/(y)\frac{x – 2}{x} = \frac{y – 6}{y}x−2x=y−6y:
x y − 2 y = x y − 6 x ⟹ − 2 y = − 6 x ⟹ y = 3 x . x y − 2 y = x y − 6 x ⟹ − 2 y = − 6 x ⟹ y = 3 x . xy-2y=xy-6xLongrightarrow-2y=-6xLongrightarrowy=3x.xy – 2y = xy – 6x \implies -2y = -6x \implies y = 3x.xy−2y=xy−6x⟹−2y=−6x⟹y=3x.
From x − 2 x = z − 3 z x − 2 x = z − 3 z (x-2)/(x)=(z-3)/(z)\frac{x – 2}{x} = \frac{z – 3}{z}x−2x=z−3z:
x z − 2 z = x z − 3 x ⟹ − 2 z = − 3 x ⟹ z = 3 2 x . x z − 2 z = x z − 3 x ⟹ − 2 z = − 3 x ⟹ z = 3 2 x . xz-2z=xz-3xLongrightarrow-2z=-3xLongrightarrowz=(3)/(2)x.xz – 2z = xz – 3x \implies -2z = -3x \implies z = \frac{3}{2}x.xz−2z=xz−3x⟹−2z=−3x⟹z=32x.

Step 5: Substitute into the Sphere Equation

Substitute y = 3 x y = 3 x y=3xy = 3xy=3x and z = 3 2 x z = 3 2 x z=(3)/(2)xz = \frac{3}{2}xz=32x into x 2 + y 2 + z 2 = 4 x 2 + y 2 + z 2 = 4 x^(2)+y^(2)+z^(2)=4x^2 + y^2 + z^2 = 4x2+y2+z2=4:
x 2 + ( 3 x ) 2 + ( 3 2 x ) 2 = 4 ⟹ x 2 + 9 x 2 + 9 4 x 2 = 4. x 2 + ( 3 x ) 2 + 3 2 x 2 = 4 ⟹ x 2 + 9 x 2 + 9 4 x 2 = 4. x^(2)+(3x)^(2)+((3)/(2)x)^(2)=4Longrightarrowx^(2)+9x^(2)+(9)/(4)x^(2)=4.x^2 + (3x)^2 + \left(\frac{3}{2}x\right)^2 = 4 \implies x^2 + 9x^2 + \frac{9}{4}x^2 = 4.x2+(3x)2+(32x)2=4⟹x2+9x2+94x2=4.
Combine like terms:
( 1 + 9 + 9 4 ) x 2 = 4 ⟹ 49 4 x 2 = 4 ⟹ x 2 = 16 49 ⟹ x = ± 4 7 . 1 + 9 + 9 4 x 2 = 4 ⟹ 49 4 x 2 = 4 ⟹ x 2 = 16 49 ⟹ x = ± 4 7 . (1+9+(9)/(4))x^(2)=4Longrightarrow(49)/(4)x^(2)=4Longrightarrowx^(2)=(16)/(49)Longrightarrowx=+-(4)/(7).\left(1 + 9 + \frac{9}{4}\right)x^2 = 4 \implies \frac{49}{4}x^2 = 4 \implies x^2 = \frac{16}{49} \implies x = \pm \frac{4}{7}.(1+9+94)x2=4⟹494x2=4⟹x2=1649⟹x=±47.

Step 6: Find Corresponding y y yyy and z z zzz

For x = 4 7 x = 4 7 x=(4)/(7)x = \frac{4}{7}x=47:
y = 3 ⋅ 4 7 = 12 7 , z = 3 2 ⋅ 4 7 = 6 7 . y = 3 ⋅ 4 7 = 12 7 , z = 3 2 ⋅ 4 7 = 6 7 . y=3*(4)/(7)=(12)/(7),quad z=(3)/(2)*(4)/(7)=(6)/(7).y = 3 \cdot \frac{4}{7} = \frac{12}{7}, \quad z = \frac{3}{2} \cdot \frac{4}{7} = \frac{6}{7}.y=3⋅47=127,z=32⋅47=67.
For x = − 4 7 x = − 4 7 x=-(4)/(7)x = -\frac{4}{7}x=−47:
y = 3 ⋅ ( − 4 7 ) = − 12 7 , z = 3 2 ⋅ ( − 4 7 ) = − 6 7 . y = 3 ⋅ − 4 7 = − 12 7 , z = 3 2 ⋅ − 4 7 = − 6 7 . y=3*(-(4)/(7))=-(12)/(7),quad z=(3)/(2)*(-(4)/(7))=-(6)/(7).y = 3 \cdot \left(-\frac{4}{7}\right) = -\frac{12}{7}, \quad z = \frac{3}{2} \cdot \left(-\frac{4}{7}\right) = -\frac{6}{7}.y=3⋅(−47)=−127,z=32⋅(−47)=−67.

Step 7: Compute the Distances

For ( 4 7 , 12 7 , 6 7 ) 4 7 , 12 7 , 6 7 ((4)/(7),(12)/(7),(6)/(7))\left(\frac{4}{7}, \frac{12}{7}, \frac{6}{7}\right)(47,127,67):
D 2 = ( 4 7 − 2 ) 2 + ( 12 7 − 6 ) 2 + ( 6 7 − 3 ) 2 = ( − 10 7 ) 2 + ( − 30 7 ) 2 + ( − 15 7 ) 2 = 100 49 + 900 49 + 225 49 = 1225 49 = 25. D 2 = 4 7 − 2 2 + 12 7 − 6 2 + 6 7 − 3 2 = − 10 7 2 + − 30 7 2 + − 15 7 2 = 100 49 + 900 49 + 225 49 = 1225 49 = 25. D^(2)=((4)/(7)-2)^(2)+((12)/(7)-6)^(2)+((6)/(7)-3)^(2)=(-(10)/(7))^(2)+(-(30)/(7))^(2)+(-(15)/(7))^(2)=(100)/(49)+(900)/(49)+(225)/(49)=(1225)/(49)=25.D^2 = \left(\frac{4}{7} – 2\right)^2 + \left(\frac{12}{7} – 6\right)^2 + \left(\frac{6}{7} – 3\right)^2 = \left(-\frac{10}{7}\right)^2 + \left(-\frac{30}{7}\right)^2 + \left(-\frac{15}{7}\right)^2 = \frac{100}{49} + \frac{900}{49} + \frac{225}{49} = \frac{1225}{49} = 25.D2=(47−2)2+(127−6)2+(67−3)2=(−107)2+(−307)2+(−157)2=10049+90049+22549=122549=25.
For ( − 4 7 , − 12 7 , − 6 7 ) − 4 7 , − 12 7 , − 6 7 (-(4)/(7),-(12)/(7),-(6)/(7))\left(-\frac{4}{7}, -\frac{12}{7}, -\frac{6}{7}\right)(−47,−127,−67):
D 2 = ( − 4 7 − 2 ) 2 + ( − 12 7 − 6 ) 2 + ( − 6 7 − 3 ) 2 = ( − 18 7 ) 2 + ( − 54 7 ) 2 + ( − 27 7 ) 2 = 324 49 + 2916 49 + 729 49 = 3969 49 = 81. D 2 = − 4 7 − 2 2 + − 12 7 − 6 2 + − 6 7 − 3 2 = − 18 7 2 + − 54 7 2 + − 27 7 2 = 324 49 + 2916 49 + 729 49 = 3969 49 = 81. D^(2)=(-(4)/(7)-2)^(2)+(-(12)/(7)-6)^(2)+(-(6)/(7)-3)^(2)=(-(18)/(7))^(2)+(-(54)/(7))^(2)+(-(27)/(7))^(2)=(324)/(49)+(2916)/(49)+(729)/(49)=(3969)/(49)=81.D^2 = \left(-\frac{4}{7} – 2\right)^2 + \left(-\frac{12}{7} – 6\right)^2 + \left(-\frac{6}{7} – 3\right)^2 = \left(-\frac{18}{7}\right)^2 + \left(-\frac{54}{7}\right)^2 + \left(-\frac{27}{7}\right)^2 = \frac{324}{49} + \frac{2916}{49} + \frac{729}{49} = \frac{3969}{49} = 81.D2=(−47−2)2+(−127−6)2+(−67−3)2=(−187)2+(−547)2+(−277)2=32449+291649+72949=396949=81.

Step 8: Determine Minimum and Maximum Distances

  • Minimum distance: D = 25 = 5 D = 25 = 5 D=sqrt25=5D = \sqrt{25} = 5D=25=5.
  • Maximum distance: D = 81 = 9 D = 81 = 9 D=sqrt81=9D = \sqrt{81} = 9D=81=9.

Final Answer

Minimum distance: 5 , Maximum distance: 9 Minimum distance:  5 ,  Maximum distance:  9 “Minimum distance: “5,” Maximum distance: “9\boxed{\text{Minimum distance: } 5, \text{ Maximum distance: } 9}Minimum distance: 5, Maximum distance: 9

Question:-02 (c)

Evaluate ∫ 0 2 π cos ⁡ 2 θ 5 + 4 cos ⁡ θ d θ ∫ 0 2 π   cos ⁡ 2 θ 5 + 4 cos ⁡ θ d θ int_(0)^(2pi)(cos 2theta)/(5+4cos theta)d theta\int_0^{2\pi} \frac{\cos 2\theta}{5 + 4 \cos \theta} d\theta∫02πcos⁡2θ5+4cos⁡θdθ using contour integration.

Answer:

To evaluate the integral
I = ∫ 0 2 π cos ⁡ 2 θ 5 + 4 cos ⁡ θ d θ I = ∫ 0 2 π   cos ⁡ 2 θ 5 + 4 cos ⁡ θ d θ I=int_(0)^(2pi)(cos 2theta)/(5+4cos theta)d thetaI = \int_0^{2\pi} \frac{\cos 2\theta}{5 + 4 \cos \theta} \, d\thetaI=∫02πcos⁡2θ5+4cos⁡θdθ
using contour integration, we proceed with the following steps:

Step 1: Parameterize the Integral Using z = e i θ z = e i θ z=e^(i theta)z = e^{i\theta}z=eiθ

Let z = e i θ z = e i θ z=e^(i theta)z = e^{i\theta}z=eiθ, so that d θ = d z i z d θ = d z i z d theta=(dz)/(iz)d\theta = \frac{dz}{iz}dθ=dziz. The integral becomes a contour integral over the unit circle | z | = 1 | z | = 1 |z|=1|z| = 1|z|=1:
cos ⁡ θ = z + z − 1 2 , cos ⁡ 2 θ = z 2 + z − 2 2 . cos ⁡ θ = z + z − 1 2 , cos ⁡ 2 θ = z 2 + z − 2 2 . cos theta=(z+z^(-1))/(2),quad cos 2theta=(z^(2)+z^(-2))/(2).\cos \theta = \frac{z + z^{-1}}{2}, \quad \cos 2\theta = \frac{z^2 + z^{-2}}{2}.cos⁡θ=z+z−12,cos⁡2θ=z2+z−22.
Substituting these into the integral:
I = ∮ | z | = 1 z 2 + z − 2 2 5 + 4 ( z + z − 1 2 ) ⋅ d z i z . I = ∮ | z | = 1 z 2 + z − 2 2 5 + 4 z + z − 1 2 ⋅ d z i z . I=oint_(|z|=1)((z^(2)+z^(-2))/(2))/(5+4((z+z^(-1))/(2)))*(dz)/(iz).I = \oint_{|z|=1} \frac{\frac{z^2 + z^{-2}}{2}}{5 + 4 \left( \frac{z + z^{-1}}{2} \right)} \cdot \frac{dz}{iz}.I=∮|z|=1z2+z−225+4(z+z−12)⋅dziz.
Simplify the integrand:
I = 1 2 i ∮ | z | = 1 z 2 + z − 2 5 + 2 ( z + z − 1 ) ⋅ d z z = 1 2 i ∮ | z | = 1 z 4 + 1 z 2 ( 2 z 2 + 5 z + 2 ) d z . I = 1 2 i ∮ | z | = 1 z 2 + z − 2 5 + 2 ( z + z − 1 ) ⋅ d z z = 1 2 i ∮ | z | = 1 z 4 + 1 z 2 ( 2 z 2 + 5 z + 2 ) d z . I=(1)/(2i)oint_(|z|=1)(z^(2)+z^(-2))/(5+2(z+z^(-1)))*(dz)/(z)=(1)/(2i)oint_(|z|=1)(z^(4)+1)/(z^(2)(2z^(2)+5z+2))dz.I = \frac{1}{2i} \oint_{|z|=1} \frac{z^2 + z^{-2}}{5 + 2(z + z^{-1})} \cdot \frac{dz}{z} = \frac{1}{2i} \oint_{|z|=1} \frac{z^4 + 1}{z^2 (2z^2 + 5z + 2)} \, dz.I=12i∮|z|=1z2+z−25+2(z+z−1)⋅dzz=12i∮|z|=1z4+1z2(2z2+5z+2)dz.

Step 2: Factor the Denominator

The denominator 2 z 2 + 5 z + 2 2 z 2 + 5 z + 2 2z^(2)+5z+22z^2 + 5z + 22z2+5z+2 factors as:
2 z 2 + 5 z + 2 = ( 2 z + 1 ) ( z + 2 ) . 2 z 2 + 5 z + 2 = ( 2 z + 1 ) ( z + 2 ) . 2z^(2)+5z+2=(2z+1)(z+2).2z^2 + 5z + 2 = (2z + 1)(z + 2).2z2+5z+2=(2z+1)(z+2).
Thus, the integrand becomes:
z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) . z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) . (z^(4)+1)/(z^(2)(2z+1)(z+2)).\frac{z^4 + 1}{z^2 (2z + 1)(z + 2)}.z4+1z2(2z+1)(z+2).

Step 3: Identify Poles Inside the Unit Circle

The poles are at:
  1. z = 0 z = 0 z=0z = 0z=0 (order 2),
  2. z = − 1 2 z = − 1 2 z=-(1)/(2)z = -\frac{1}{2}z=−12 (from 2 z + 1 = 0 2 z + 1 = 0 2z+1=02z + 1 = 02z+1=0),
  3. z = − 2 z = − 2 z=-2z = -2z=−2 (from z + 2 = 0 z + 2 = 0 z+2=0z + 2 = 0z+2=0).
Only z = 0 z = 0 z=0z = 0z=0 and z = − 1 2 z = − 1 2 z=-(1)/(2)z = -\frac{1}{2}z=−12 lie inside the unit circle | z | = 1 | z | = 1 |z|=1|z| = 1|z|=1.

Step 4: Compute Residues

Residue at z = 0 z = 0 z=0z = 0z=0:

The integrand has a double pole at z = 0 z = 0 z=0z = 0z=0. To find the residue, compute the coefficient of 1 z 1 z (1)/(z)\frac{1}{z}1z in the Laurent expansion:
Res ( f , 0 ) = lim z → 0 d d z ( z 2 ⋅ z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) ) = lim z → 0 d d z ( z 4 + 1 ( 2 z + 1 ) ( z + 2 ) ) . Res ( f , 0 ) = lim z → 0   d d z z 2 ⋅ z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) = lim z → 0   d d z z 4 + 1 ( 2 z + 1 ) ( z + 2 ) . “Res”(f,0)=lim_(z rarr0)(d)/(dz)(z^(2)*(z^(4)+1)/(z^(2)(2z+1)(z+2)))=lim_(z rarr0)(d)/(dz)((z^(4)+1)/((2z+1)(z+2))).\text{Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz} \left( z^2 \cdot \frac{z^4 + 1}{z^2 (2z + 1)(z + 2)} \right) = \lim_{z \to 0} \frac{d}{dz} \left( \frac{z^4 + 1}{(2z + 1)(z + 2)} \right).Res(f,0)=limz→0ddz(z2⋅z4+1z2(2z+1)(z+2))=limz→0ddz(z4+1(2z+1)(z+2)).
Compute the derivative:
d d z ( z 4 + 1 ( 2 z + 1 ) ( z + 2 ) ) = 4 z 3 ( 2 z + 1 ) ( z + 2 ) − ( z 4 + 1 ) ( 2 ( z + 2 ) + ( 2 z + 1 ) ) ( 2 z + 1 ) 2 ( z + 2 ) 2 . d d z z 4 + 1 ( 2 z + 1 ) ( z + 2 ) = 4 z 3 ( 2 z + 1 ) ( z + 2 ) − ( z 4 + 1 ) ( 2 ( z + 2 ) + ( 2 z + 1 ) ) ( 2 z + 1 ) 2 ( z + 2 ) 2 . (d)/(dz)((z^(4)+1)/((2z+1)(z+2)))=(4z^(3)(2z+1)(z+2)-(z^(4)+1)(2(z+2)+(2z+1)))/((2z+1)^(2)(z+2)^(2)).\frac{d}{dz} \left( \frac{z^4 + 1}{(2z + 1)(z + 2)} \right) = \frac{4z^3 (2z + 1)(z + 2) – (z^4 + 1)(2(z + 2) + (2z + 1))}{(2z + 1)^2 (z + 2)^2}.ddz(z4+1(2z+1)(z+2))=4z3(2z+1)(z+2)−(z4+1)(2(z+2)+(2z+1))(2z+1)2(z+2)2.
Evaluating at z = 0 z = 0 z=0z = 0z=0:
Res ( f , 0 ) = 0 − 1 ⋅ ( 4 + 1 ) 1 ⋅ 4 = − 5 4 . Res ( f , 0 ) = 0 − 1 ⋅ ( 4 + 1 ) 1 ⋅ 4 = − 5 4 . “Res”(f,0)=(0-1*(4+1))/(1*4)=-(5)/(4).\text{Res}(f, 0) = \frac{0 – 1 \cdot (4 + 1)}{1 \cdot 4} = -\frac{5}{4}.Res(f,0)=0−1⋅(4+1)1⋅4=−54.

Residue at z = − 1 2 z = − 1 2 z=-(1)/(2)z = -\frac{1}{2}z=−12:

The integrand has a simple pole at z = − 1 2 z = − 1 2 z=-(1)/(2)z = -\frac{1}{2}z=−12. The residue is:
Res ( f , − 1 2 ) = lim z → − 1 2 ( z + 1 2 ) ⋅ z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) = ( − 1 2 ) 4 + 1 ( − 1 2 ) 2 ⋅ 2 ⋅ ( − 1 2 + 2 ) = 1 16 + 1 1 4 ⋅ 2 ⋅ 3 2 = 17 16 3 4 = 17 12 . Res ( f , − 1 2 ) = lim z → − 1 2   z + 1 2 ⋅ z 4 + 1 z 2 ( 2 z + 1 ) ( z + 2 ) = − 1 2 4 + 1 − 1 2 2 ⋅ 2 ⋅ − 1 2 + 2 = 1 16 + 1 1 4 ⋅ 2 ⋅ 3 2 = 17 16 3 4 = 17 12 . “Res”(f,-(1)/(2))=lim_(z rarr-(1)/(2))(z+(1)/(2))*(z^(4)+1)/(z^(2)(2z+1)(z+2))=((-(1)/(2))^(4)+1)/((-(1)/(2))^(2)*2*(-(1)/(2)+2))=((1)/(16)+1)/((1)/(4)*2*(3)/(2))=((17)/(16))/((3)/(4))=(17)/(12).\text{Res}(f, -\frac{1}{2}) = \lim_{z \to -\frac{1}{2}} \left( z + \frac{1}{2} \right) \cdot \frac{z^4 + 1}{z^2 (2z + 1)(z + 2)} = \frac{\left(-\frac{1}{2}\right)^4 + 1}{\left(-\frac{1}{2}\right)^2 \cdot 2 \cdot \left(-\frac{1}{2} + 2\right)} = \frac{\frac{1}{16} + 1}{\frac{1}{4} \cdot 2 \cdot \frac{3}{2}} = \frac{\frac{17}{16}}{\frac{3}{4}} = \frac{17}{12}.Res(f,−12)=limz→−12(z+12)⋅z4+1z2(2z+1)(z+2)=(−12)4+1(−12)2⋅2⋅(−12+2)=116+114⋅2⋅32=171634=1712.

Step 5: Apply the Residue Theorem

The integral is 2 π i 2 π i 2pi i2\pi i2πi times the sum of the residues inside the contour:
I = 1 2 i ⋅ 2 π i ( Res ( f , 0 ) + Res ( f , − 1 2 ) ) = π ( − 5 4 + 17 12 ) = π ( − 15 + 17 12 ) = π ⋅ 2 12 = π 6 . I = 1 2 i ⋅ 2 π i Res ( f , 0 ) + Res ( f , − 1 2 ) = π − 5 4 + 17 12 = π − 15 + 17 12 = π ⋅ 2 12 = π 6 . I=(1)/(2i)*2pi i(“Res”(f,0)+”Res”(f,-(1)/(2)))=pi(-(5)/(4)+(17)/(12))=pi((-15+17)/(12))=pi*(2)/(12)=(pi)/(6).I = \frac{1}{2i} \cdot 2\pi i \left( \text{Res}(f, 0) + \text{Res}(f, -\frac{1}{2}) \right) = \pi \left( -\frac{5}{4} + \frac{17}{12} \right) = \pi \left( \frac{-15 + 17}{12} \right) = \pi \cdot \frac{2}{12} = \frac{\pi}{6}.I=12i⋅2πi(Res(f,0)+Res(f,−12))=π(−54+1712)=π(−15+1712)=π⋅212=π6.

Final Answer

π 6 π 6 (pi)/(6)\boxed{\frac{\pi}{6}}π6

Question:-03

(a) Prove that x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is an irreducible polynomial in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]Z_3[x]Z3[x]. Further show that the quotient ring Z 3 [ x ] ⟨ x 2 + 1 ⟩ Z 3 [ x ] ⟨ x 2 + 1 ⟩ (Z_(3)[x])/((:x^(2)+1:))\frac{Z_3[x]}{\langle x^2 + 1 \rangle}Z3[x]⟨x2+1⟩ is a field of 9 elements.

Answer:

Proof that x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is Irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x]

A polynomial f ( x ) ∈ Z 3 [ x ] f ( x ) ∈ Z 3 [ x ] f(x)inZ_(3)[x]f(x) \in \mathbb{Z}_3[x]f(x)∈Z3[x] is irreducible if it cannot be factored into a product of two non-constant polynomials in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x].
Step 1: Check for Roots in Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3
A quadratic polynomial is irreducible over a field if it has no roots in that field. Evaluate x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 at all elements of Z 3 = { 0 , 1 , 2 } Z 3 = { 0 , 1 , 2 } Z_(3)={0,1,2}\mathbb{Z}_3 = \{0, 1, 2\}Z3={0,1,2}:
  • At x = 0 x = 0 x=0x = 0x=0: 0 2 + 1 = 1 ≠ 0 0 2 + 1 = 1 ≠ 0 0^(2)+1=1!=00^2 + 1 = 1 \neq 002+1=1≠0,
  • At x = 1 x = 1 x=1x = 1x=1: 1 2 + 1 = 2 ≠ 0 1 2 + 1 = 2 ≠ 0 1^(2)+1=2!=01^2 + 1 = 2 \neq 012+1=2≠0,
  • At x = 2 x = 2 x=2x = 2x=2: 2 2 + 1 = 4 ≡ 1 ≠ 0 2 2 + 1 = 4 ≡ 1 ≠ 0 2^(2)+1=4-=1!=02^2 + 1 = 4 \equiv 1 \neq 022+1=4≡1≠0.
Since x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 has no roots in Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3, it cannot be factored into linear polynomials, and thus it is irreducible.
Step 2: Confirm No Factorization into Quadratics
Since x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is quadratic, the only possible non-trivial factorization would be into two linear polynomials. Since no such factorization exists, x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible.

Proof that Z 3 [ x ] ⟨ x 2 + 1 ⟩ Z 3 [ x ] ⟨ x 2 + 1 ⟩ (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]⟨x2+1⟩ is a Field of 9 Elements

Step 1: Structure of the Quotient Ring
The quotient ring Z 3 [ x ] ⟨ x 2 + 1 ⟩ Z 3 [ x ] ⟨ x 2 + 1 ⟩ (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]⟨x2+1⟩ consists of all polynomials in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x] modulo x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1. Since x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible, the ideal ⟨ x 2 + 1 ⟩ ⟨ x 2 + 1 ⟩ (:x^(2)+1:)\langle x^2 + 1 \rangle⟨x2+1⟩ is maximal, and thus the quotient ring is a field.
Step 2: Elements of the Quotient Field
Every element in Z 3 [ x ] ⟨ x 2 + 1 ⟩ Z 3 [ x ] ⟨ x 2 + 1 ⟩ (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]⟨x2+1⟩ can be represented uniquely as a linear polynomial:
a + b x where a , b ∈ Z 3 . a + b x where a , b ∈ Z 3 . a+bx quad”where”quad a,b inZ_(3).a + bx \quad \text{where} \quad a, b \in \mathbb{Z}_3.a+bxwherea,b∈Z3.
Since there are 3 choices for a a aaa and 3 choices for b b bbb, the total number of distinct elements is 3 × 3 = 9 3 × 3 = 9 3xx3=93 \times 3 = 93×3=9.
Step 3: Verification of Field Axioms
  • Addition and Multiplication are performed modulo x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1, with x 2 ≡ − 1 ≡ 2 x 2 ≡ − 1 ≡ 2 x^(2)-=-1-=2x^2 \equiv -1 \equiv 2x2≡−1≡2.
  • Inverses: Every non-zero element a + b x a + b x a+bxa + bxa+bx has a multiplicative inverse because x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible. For example:
    • The inverse of 1 + x 1 + x 1+x1 + x1+x is found by solving ( 1 + x ) ( c + d x ) ≡ 1 mod ( x 2 + 1 ) ( 1 + x ) ( c + d x ) ≡ 1 mod ( x 2 + 1 ) (1+x)(c+dx)-=1mod(x^(2)+1)(1 + x)(c + dx) \equiv 1 \mod (x^2 + 1)(1+x)(c+dx)≡1mod(x2+1), leading to c + d = 1 c + d = 1 c+d=1c + d = 1c+d=1 and c − d = 0 c − d = 0 c-d=0c – d = 0c−d=0, so c = d = 2 c = d = 2 c=d=2c = d = 2c=d=2. Thus, ( 1 + x ) − 1 = 2 + 2 x ( 1 + x ) − 1 = 2 + 2 x (1+x)^(-1)=2+2x(1 + x)^{-1} = 2 + 2x(1+x)−1=2+2x.

Conclusion

  • x 2 + 1 x 2 + 1 x^(2)+1x^2 + 1x2+1 is irreducible in Z 3 [ x ] Z 3 [ x ] Z_(3)[x]\mathbb{Z}_3[x]Z3[x] because it has no roots in Z 3 Z 3 Z_(3)\mathbb{Z}_3Z3.
  • The quotient ring Z 3 [ x ] ⟨ x 2 + 1 ⟩ Z 3 [ x ] ⟨ x 2 + 1 ⟩ (Z_(3)[x])/((:x^(2)+1:))\frac{\mathbb{Z}_3[x]}{\langle x^2 + 1 \rangle}Z3[x]⟨x2+1⟩ is a field with 9 elements, as it satisfies all field axioms and has the correct cardinality.

Question:-03 (b)

Prove that u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)u(x,y)=ex(xcos⁡y−ysin⁡y) is harmonic. Find its conjugate harmonic function v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) and express the corresponding analytic function f ( z ) f ( z ) f(z)f(z)f(z) in terms of z z zzz.

Answer:

Step 1: Verify that u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)u(x,y)=ex(xcos⁡y−ysin⁡y) is Harmonic

A function u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is harmonic if it satisfies Laplace’s equation:
∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0. ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0. (del^(2)u)/(delx^(2))+(del^(2)u)/(dely^(2))=0.\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.∂2u∂x2+∂2u∂y2=0.

Compute the First Partial Derivatives:

∂ u ∂ x = e x ( x cos ⁡ y − y sin ⁡ y ) + e x cos ⁡ y = e x ( ( x + 1 ) cos ⁡ y − y sin ⁡ y ) , ∂ u ∂ x = e x ( x cos ⁡ y − y sin ⁡ y ) + e x cos ⁡ y = e x ( x + 1 ) cos ⁡ y − y sin ⁡ y , (del u)/(del x)=e^(x)(x cos y-y sin y)+e^(x)cos y=e^(x)((x+1)cos y-y sin y),\frac{\partial u}{\partial x} = e^x (x \cos y – y \sin y) + e^x \cos y = e^x \left( (x + 1) \cos y – y \sin y \right),∂u∂x=ex(xcos⁡y−ysin⁡y)+excos⁡y=ex((x+1)cos⁡y−ysin⁡y),
∂ u ∂ y = e x ( − x sin ⁡ y − sin ⁡ y − y cos ⁡ y ) = − e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . ∂ u ∂ y = e x ( − x sin ⁡ y − sin ⁡ y − y cos ⁡ y ) = − e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . (del u)/(del y)=e^(x)(-x sin y-sin y-y cos y)=-e^(x)((x+1)sin y+y cos y).\frac{\partial u}{\partial y} = e^x (-x \sin y – \sin y – y \cos y) = -e^x \left( (x + 1) \sin y + y \cos y \right).∂u∂y=ex(−xsin⁡y−sin⁡y−ycos⁡y)=−ex((x+1)sin⁡y+ycos⁡y).

Compute the Second Partial Derivatives:

∂ 2 u ∂ x 2 = ∂ ∂ x ( e x ( ( x + 1 ) cos ⁡ y − y sin ⁡ y ) ) = e x ( ( x + 2 ) cos ⁡ y − y sin ⁡ y ) , ∂ 2 u ∂ x 2 = ∂ ∂ x e x ( x + 1 ) cos ⁡ y − y sin ⁡ y = e x ( x + 2 ) cos ⁡ y − y sin ⁡ y , (del^(2)u)/(delx^(2))=(del)/(del x)(e^(x)((x+1)cos y-y sin y))=e^(x)((x+2)cos y-y sin y),\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left( e^x \left( (x + 1) \cos y – y \sin y \right) \right) = e^x \left( (x + 2) \cos y – y \sin y \right),∂2u∂x2=∂∂x(ex((x+1)cos⁡y−ysin⁡y))=ex((x+2)cos⁡y−ysin⁡y),
∂ 2 u ∂ y 2 = ∂ ∂ y ( − e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) ) = − e x ( ( x + 1 ) cos ⁡ y + cos ⁡ y − y sin ⁡ y ) = − e x ( ( x + 2 ) cos ⁡ y − y sin ⁡ y ) . ∂ 2 u ∂ y 2 = ∂ ∂ y − e x ( x + 1 ) sin ⁡ y + y cos ⁡ y = − e x ( x + 1 ) cos ⁡ y + cos ⁡ y − y sin ⁡ y = − e x ( x + 2 ) cos ⁡ y − y sin ⁡ y . (del^(2)u)/(dely^(2))=(del)/(del y)(-e^(x)((x+1)sin y+y cos y))=-e^(x)((x+1)cos y+cos y-y sin y)=-e^(x)((x+2)cos y-y sin y).\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y} \left( -e^x \left( (x + 1) \sin y + y \cos y \right) \right) = -e^x \left( (x + 1) \cos y + \cos y – y \sin y \right) = -e^x \left( (x + 2) \cos y – y \sin y \right).∂2u∂y2=∂∂y(−ex((x+1)sin⁡y+ycos⁡y))=−ex((x+1)cos⁡y+cos⁡y−ysin⁡y)=−ex((x+2)cos⁡y−ysin⁡y).

Check Laplace’s Equation:

∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = e x ( ( x + 2 ) cos ⁡ y − y sin ⁡ y ) − e x ( ( x + 2 ) cos ⁡ y − y sin ⁡ y ) = 0. ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = e x ( x + 2 ) cos ⁡ y − y sin ⁡ y − e x ( x + 2 ) cos ⁡ y − y sin ⁡ y = 0. (del^(2)u)/(delx^(2))+(del^(2)u)/(dely^(2))=e^(x)((x+2)cos y-y sin y)-e^(x)((x+2)cos y-y sin y)=0.\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = e^x \left( (x + 2) \cos y – y \sin y \right) – e^x \left( (x + 2) \cos y – y \sin y \right) = 0.∂2u∂x2+∂2u∂y2=ex((x+2)cos⁡y−ysin⁡y)−ex((x+2)cos⁡y−ysin⁡y)=0.
Thus, u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is harmonic.

Step 2: Find the Conjugate Harmonic Function v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y)

Since u ( x , y ) u ( x , y ) u(x,y)u(x, y)u(x,y) is harmonic, there exists a conjugate harmonic function v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) such that f ( z ) = u + i v f ( z ) = u + i v f(z)=u+ivf(z) = u + ivf(z)=u+iv is analytic. To find v v vvv, we use the Cauchy-Riemann equations:
∂ u ∂ x = ∂ v ∂ y , ∂ u ∂ y = − ∂ v ∂ x . ∂ u ∂ x = ∂ v ∂ y , ∂ u ∂ y = − ∂ v ∂ x . (del u)/(del x)=(del v)/(del y),quad(del u)/(del y)=-(del v)/(del x).\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.∂u∂x=∂v∂y,∂u∂y=−∂v∂x.

From ∂ u ∂ x = ∂ v ∂ y ∂ u ∂ x = ∂ v ∂ y (del u)/(del x)=(del v)/(del y)\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}∂u∂x=∂v∂y:

∂ v ∂ y = e x ( ( x + 1 ) cos ⁡ y − y sin ⁡ y ) . ∂ v ∂ y = e x ( x + 1 ) cos ⁡ y − y sin ⁡ y . (del v)/(del y)=e^(x)((x+1)cos y-y sin y).\frac{\partial v}{\partial y} = e^x \left( (x + 1) \cos y – y \sin y \right).∂v∂y=ex((x+1)cos⁡y−ysin⁡y).
Integrate with respect to y y yyy:
v ( x , y ) = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) + g ( x ) , v ( x , y ) = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y + g ( x ) , v(x,y)=e^(x)((x+1)sin y+y cos y)+g(x),v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right) + g(x),v(x,y)=ex((x+1)sin⁡y+ycos⁡y)+g(x),
where g ( x ) g ( x ) g(x)g(x)g(x) is an arbitrary function of x x xxx.

From ∂ u ∂ y = − ∂ v ∂ x ∂ u ∂ y = − ∂ v ∂ x (del u)/(del y)=-(del v)/(del x)\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}∂u∂y=−∂v∂x:

∂ v ∂ x = − e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . ∂ v ∂ x = − e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . (del v)/(del x)=-e^(x)((x+1)sin y+y cos y).\frac{\partial v}{\partial x} = -e^x \left( (x + 1) \sin y + y \cos y \right).∂v∂x=−ex((x+1)sin⁡y+ycos⁡y).
Differentiate v ( x , y ) v ( x , y ) v(x,y)v(x, y)v(x,y) with respect to x x xxx:
∂ v ∂ x = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) + e x sin ⁡ y + g ′ ( x ) . ∂ v ∂ x = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y + e x sin ⁡ y + g ′ ( x ) . (del v)/(del x)=e^(x)((x+1)sin y+y cos y)+e^(x)sin y+g^(‘)(x).\frac{\partial v}{\partial x} = e^x \left( (x + 1) \sin y + y \cos y \right) + e^x \sin y + g'(x).∂v∂x=ex((x+1)sin⁡y+ycos⁡y)+exsin⁡y+g′(x).
Set this equal to − e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) − e x ( x + 1 ) sin ⁡ y + y cos ⁡ y -e^(x)((x+1)sin y+y cos y)-e^x \left( (x + 1) \sin y + y \cos y \right)−ex((x+1)sin⁡y+ycos⁡y):
e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) + e x sin ⁡ y + g ′ ( x ) = − e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . e x ( x + 1 ) sin ⁡ y + y cos ⁡ y + e x sin ⁡ y + g ′ ( x ) = − e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . e^(x)((x+1)sin y+y cos y)+e^(x)sin y+g^(‘)(x)=-e^(x)((x+1)sin y+y cos y).e^x \left( (x + 1) \sin y + y \cos y \right) + e^x \sin y + g'(x) = -e^x \left( (x + 1) \sin y + y \cos y \right).ex((x+1)sin⁡y+ycos⁡y)+exsin⁡y+g′(x)=−ex((x+1)sin⁡y+ycos⁡y).
Simplify:
2 e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) + e x sin ⁡ y + g ′ ( x ) = 0. 2 e x ( x + 1 ) sin ⁡ y + y cos ⁡ y + e x sin ⁡ y + g ′ ( x ) = 0. 2e^(x)((x+1)sin y+y cos y)+e^(x)sin y+g^(‘)(x)=0.2e^x \left( (x + 1) \sin y + y \cos y \right) + e^x \sin y + g'(x) = 0.2ex((x+1)sin⁡y+ycos⁡y)+exsin⁡y+g′(x)=0.
This must hold for all x x xxx and y y yyy, so:
g ′ ( x ) = 0 ⟹ g ( x ) = C (a constant) . g ′ ( x ) = 0 ⟹ g ( x ) = C (a constant) . g^(‘)(x)=0Longrightarrowg(x)=C quad(a constant).g'(x) = 0 \implies g(x) = C \quad \text{(a constant)}.g′(x)=0⟹g(x)=C(a constant).
Thus, the conjugate harmonic function is:
v ( x , y ) = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) + C . v ( x , y ) = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y + C . v(x,y)=e^(x)((x+1)sin y+y cos y)+C.v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right) + C.v(x,y)=ex((x+1)sin⁡y+ycos⁡y)+C.
For simplicity, we can take C = 0 C = 0 C=0C = 0C=0:
v ( x , y ) = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . v ( x , y ) = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . v(x,y)=e^(x)((x+1)sin y+y cos y).v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right).v(x,y)=ex((x+1)sin⁡y+ycos⁡y).

Step 3: Express the Analytic Function f ( z ) f ( z ) f(z)f(z)f(z) in Terms of z z zzz

The analytic function is:
f ( z ) = u ( x , y ) + i v ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) + i e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . f ( z ) = u ( x , y ) + i v ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) + i e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . f(z)=u(x,y)+iv(x,y)=e^(x)(x cos y-y sin y)+ie^(x)((x+1)sin y+y cos y).f(z) = u(x, y) + iv(x, y) = e^x (x \cos y – y \sin y) + i e^x \left( (x + 1) \sin y + y \cos y \right).f(z)=u(x,y)+iv(x,y)=ex(xcos⁡y−ysin⁡y)+iex((x+1)sin⁡y+ycos⁡y).
To express f ( z ) f ( z ) f(z)f(z)f(z) in terms of z = x + i y z = x + i y z=x+iyz = x + iyz=x+iy, we observe that:
f ( z ) = e x ( x ( cos ⁡ y + i sin ⁡ y ) + i y ( cos ⁡ y + i sin ⁡ y ) + i sin ⁡ y ) = e x ( ( x + i y ) e i y + i sin ⁡ y ) . f ( z ) = e x x ( cos ⁡ y + i sin ⁡ y ) + i y ( cos ⁡ y + i sin ⁡ y ) + i sin ⁡ y = e x ( x + i y ) e i y + i sin ⁡ y . f(z)=e^(x)(x(cos y+i sin y)+iy(cos y+i sin y)+i sin y)=e^(x)((x+iy)e^(iy)+i sin y).f(z) = e^x \left( x (\cos y + i \sin y) + i y (\cos y + i \sin y) + i \sin y \right) = e^x \left( (x + i y) e^{i y} + i \sin y \right).f(z)=ex(x(cos⁡y+isin⁡y)+iy(cos⁡y+isin⁡y)+isin⁡y)=ex((x+iy)eiy+isin⁡y).
However, a simpler form can be obtained by noting that:
f ( z ) = e z ( z + i ) + C , f ( z ) = e z ( z + i ) + C , f(z)=e^(z)(z+i)+C,f(z) = e^z (z + i) + C,f(z)=ez(z+i)+C,
where C C CCC is a constant (here, C = 0 C = 0 C=0C = 0C=0 for the principal branch).
Verification:
Let z = x + i y z = x + i y z=x+iyz = x + iyz=x+iy, then:
e z ( z + i ) = e x ( cos ⁡ y + i sin ⁡ y ) ( ( x + i y ) + i ) = e x ( ( x + i ( y + 1 ) ) ( cos ⁡ y + i sin ⁡ y ) ) . e z ( z + i ) = e x ( cos ⁡ y + i sin ⁡ y ) ( x + i y ) + i = e x ( x + i ( y + 1 ) ) ( cos ⁡ y + i sin ⁡ y ) . e^(z)(z+i)=e^(x)(cos y+i sin y)((x+iy)+i)=e^(x)((x+i(y+1))(cos y+i sin y)).e^z (z + i) = e^x (\cos y + i \sin y) \left( (x + iy) + i \right) = e^x \left( (x + i(y + 1)) (\cos y + i \sin y) \right).ez(z+i)=ex(cos⁡y+isin⁡y)((x+iy)+i)=ex((x+i(y+1))(cos⁡y+isin⁡y)).
Expanding:
= e x ( x cos ⁡ y − ( y + 1 ) sin ⁡ y + i ( x sin ⁡ y + ( y + 1 ) cos ⁡ y ) ) . = e x x cos ⁡ y − ( y + 1 ) sin ⁡ y + i x sin ⁡ y + ( y + 1 ) cos ⁡ y . =e^(x)(x cos y-(y+1)sin y+i(x sin y+(y+1)cos y)).= e^x \left( x \cos y – (y + 1) \sin y + i \left( x \sin y + (y + 1) \cos y \right) \right).=ex(xcos⁡y−(y+1)sin⁡y+i(xsin⁡y+(y+1)cos⁡y)).
Comparing with u + i v u + i v u+ivu + ivu+iv:
u ( x , y ) = e x ( x cos ⁡ y − ( y + 1 ) sin ⁡ y + sin ⁡ y ) = e x ( x cos ⁡ y − y sin ⁡ y ) , u ( x , y ) = e x ( x cos ⁡ y − ( y + 1 ) sin ⁡ y + sin ⁡ y ) = e x ( x cos ⁡ y − y sin ⁡ y ) , u(x,y)=e^(x)(x cos y-(y+1)sin y+sin y)=e^(x)(x cos y-y sin y),u(x, y) = e^x (x \cos y – (y + 1) \sin y + \sin y) = e^x (x \cos y – y \sin y),u(x,y)=ex(xcos⁡y−(y+1)sin⁡y+sin⁡y)=ex(xcos⁡y−ysin⁡y),
v ( x , y ) = e x ( x sin ⁡ y + ( y + 1 ) cos ⁡ y − cos ⁡ y ) = e x ( x sin ⁡ y + y cos ⁡ y + cos ⁡ y − cos ⁡ y ) = e x ( x sin ⁡ y + y cos ⁡ y ) . v ( x , y ) = e x ( x sin ⁡ y + ( y + 1 ) cos ⁡ y − cos ⁡ y ) = e x ( x sin ⁡ y + y cos ⁡ y + cos ⁡ y − cos ⁡ y ) = e x ( x sin ⁡ y + y cos ⁡ y ) . v(x,y)=e^(x)(x sin y+(y+1)cos y-cos y)=e^(x)(x sin y+y cos y+cos y-cos y)=e^(x)(x sin y+y cos y).v(x, y) = e^x (x \sin y + (y + 1) \cos y – \cos y) = e^x (x \sin y + y \cos y + \cos y – \cos y) = e^x (x \sin y + y \cos y).v(x,y)=ex(xsin⁡y+(y+1)cos⁡y−cos⁡y)=ex(xsin⁡y+ycos⁡y+cos⁡y−cos⁡y)=ex(xsin⁡y+ycos⁡y).
Thus, the analytic function is:
f ( z ) = e z ( z + i ) . f ( z ) = e z ( z + i ) . f(z)=e^(z)(z+i).f(z) = e^z (z + i).f(z)=ez(z+i).

Final Answer

  1. u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u ( x , y ) = e x ( x cos ⁡ y − y sin ⁡ y ) u(x,y)=e^(x)(x cos y-y sin y)u(x, y) = e^x (x \cos y – y \sin y)u(x,y)=ex(xcos⁡y−ysin⁡y) is harmonic.
  2. The conjugate harmonic function is: v ( x , y ) = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . v ( x , y ) = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . v(x,y)=e^(x)((x+1)sin y+y cos y).v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right).v(x,y)=ex((x+1)sin⁡y+ycos⁡y).
  3. The corresponding analytic function is: f ( z ) = e z ( z + i ) . f ( z ) = e z ( z + i ) . f(z)=e^(z)(z+i).f(z) = e^z (z + i).f(z)=ez(z+i).
1. u ( x , y ) is harmonic. 2. The conjugate harmonic function is v ( x , y ) = e x ( ( x + 1 ) sin ⁡ y + y cos ⁡ y ) . 3. The analytic function is f ( z ) = e z ( z + i ) . 1.  u ( x , y )  is harmonic. 2. The conjugate harmonic function is  v ( x , y ) = e x ( x + 1 ) sin ⁡ y + y cos ⁡ y . 3. The analytic function is  f ( z ) = e z ( z + i ) . [“1. “u(x”,”y)” is harmonic.”],[“2. The conjugate harmonic function is “v(x”,”y)=e^(x)((x+1)sin y+y cos y).],[“3. The analytic function is “f(z)=e^(z)(z+i).]\boxed{ \begin{aligned} &\text{1. } u(x, y) \text{ is harmonic.} \\ &\text{2. The conjugate harmonic function is } v(x, y) = e^x \left( (x + 1) \sin y + y \cos y \right). \\ &\text{3. The analytic function is } f(z) = e^z (z + i). \end{aligned} }1. u(x,y) is harmonic.2. The conjugate harmonic function is v(x,y)=ex((x+1)sin⁡y+ycos⁡y).3. The analytic function is f(z)=ez(z+i).

Question:-03 (c)

Solve the following linear programming problem by the Big M method:

Minimize Z = 2 x 1 + 3 x 2 Z = 2 x 1 + 3 x 2 Z=2x_(1)+3x_(2)Z = 2x_1 + 3x_2Z=2x1+3x2
subject to
x 1 + x 2 ≥ 9 x 1 + 2 x 2 ≥ 15 2 x 1 − 3 x 2 ≤ 9 x 1 , x 2 ≥ 0 x 1 + x 2 ≥ 9 x 1 + 2 x 2 ≥ 15 2 x 1 − 3 x 2 ≤ 9 x 1 , x 2 ≥ 0 {:[x_(1)+x_(2) >= 9],[x_(1)+2x_(2) >= 15],[2x_(1)-3x_(2) <= 9],[x_(1)”,”x_(2) >= 0]:}\begin{aligned} x_1 + x_2 &\geq 9 \\ x_1 + 2x_2 &\geq 15 \\ 2x_1 – 3x_2 &\leq 9 \\ x_1, x_2 &\geq 0 \end{aligned}x1+x2≥9x1+2x2≥152x1−3x2≤9x1,x2≥0
Is the optimal solution unique? Justify your answer.

Answer:

Step 1: Convert Inequalities to Standard Form

The given problem is:
  • Minimize Z = 2 x 1 + 3 x 2 Z = 2 x 1 + 3 x 2 Z=2x_(1)+3x_(2)Z = 2x_1 + 3x_2Z=2x1+3x2
  • Subject to: x 1 + x 2 ≥ 9 , x 1 + 2 x 2 ≥ 15 , 2 x 1 − 3 x 2 ≤ 9 , x 1 , x 2 ≥ 0. x 1 + x 2 ≥ 9 , x 1 + 2 x 2 ≥ 15 , 2 x 1 − 3 x 2 ≤ 9 , x 1 , x 2 ≥ 0. {:[x_(1)+x_(2) >= 9″,”],[x_(1)+2x_(2) >= 15″,”],[2x_(1)-3x_(2) <= 9″,”],[x_(1)”,”x_(2) >= 0.]:}\begin{aligned} x_1 + x_2 &\geq 9, \\ x_1 + 2x_2 &\geq 15, \\ 2x_1 – 3x_2 &\leq 9, \\ x_1, x_2 &\geq 0. \end{aligned}x1+x2≥9,x1+2x2≥15,2x1−3x2≤9,x1,x2≥0.
Convert to standard form (equality constraints):
  1. For x 1 + x 2 ≥ 9 x 1 + x 2 ≥ 9 x_(1)+x_(2) >= 9x_1 + x_2 \geq 9x1+x2≥9:
    Subtract surplus variable s 1 s 1 s_(1)s_1s1 and add artificial variable A 1 A 1 A_(1)A_1A1:
    x 1 + x 2 − s 1 + A 1 = 9. x 1 + x 2 − s 1 + A 1 = 9. x_(1)+x_(2)-s_(1)+A_(1)=9.x_1 + x_2 – s_1 + A_1 = 9.x1+x2−s1+A1=9.
  2. For x 1 + 2 x 2 ≥ 15 x 1 + 2 x 2 ≥ 15 x_(1)+2x_(2) >= 15x_1 + 2x_2 \geq 15x1+2x2≥15:
    Subtract surplus variable s 2 s 2 s_(2)s_2s2 and add artificial variable A 2 A 2 A_(2)A_2A2:
    x 1 + 2 x 2 − s 2 + A 2 = 15. x 1 + 2 x 2 − s 2 + A 2 = 15. x_(1)+2x_(2)-s_(2)+A_(2)=15.x_1 + 2x_2 – s_2 + A_2 = 15.x1+2x2−s2+A2=15.
  3. For 2 x 1 − 3 x 2 ≤ 9 2 x 1 − 3 x 2 ≤ 9 2x_(1)-3x_(2) <= 92x_1 – 3x_2 \leq 92x1−3x2≤9:
    Add slack variable s 3 s 3 s_(3)s_3s3:
    2 x 1 − 3 x 2 + s 3 = 9. 2 x 1 − 3 x 2 + s 3 = 9. 2x_(1)-3x_(2)+s_(3)=9.2x_1 – 3x_2 + s_3 = 9.2x1−3x2+s3=9.

Step 2: Formulate the Big M Objective Function

Since we are minimizing Z = 2 x 1 + 3 x 2 Z = 2 x 1 + 3 x 2 Z=2x_(1)+3x_(2)Z = 2x_1 + 3x_2Z=2x1+3x2, we penalize the artificial variables in the objective function with a large positive coefficient M M MMM:
Z ′ = 2 x 1 + 3 x 2 + M A 1 + M A 2 . Z ′ = 2 x 1 + 3 x 2 + M A 1 + M A 2 . Z^(‘)=2x_(1)+3x_(2)+MA_(1)+MA_(2).Z’ = 2x_1 + 3x_2 + M A_1 + M A_2.Z′=2x1+3x2+MA1+MA2.

Step 3: Construct the Initial Simplex Tableau

The initial tableau is:
Basis x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 s 1 s 1 s_(1)s_1s1 s 2 s 2 s_(2)s_2s2 s 3 s 3 s_(3)s_3s3 A 1 A 1 A_(1)A_1A1 A 2 A 2 A_(2)A_2A2RHS
A 1 A 1 A_(1)A_1A111-100109
A 2 A 2 A_(2)A_2A2120-100115
s 3 s 3 s_(3)s_3s32-3001009
Z’ 2 + 2 M 2 + 2 M 2+2M2 + 2M2+2M 3 + 3 M 3 + 3 M 3+3M3 + 3M3+3M − M − M -M-M−M − M − M -M-M−M000 24 M 24 M 24 M24M24M

Step 4: Perform Simplex Iterations

Iteration 1:
  • Entering Variable: x 2 x 2 x_(2)x_2x2 (most positive coefficient in Z ′ Z ′ Z^(‘)Z’Z′).
  • Leaving Variable: A 2 A 2 A_(2)A_2A2 (min ratio test: min ( 9 / 1 , 15 / 2 , ignore s 3 ) = 7.5 min ( 9 / 1 , 15 / 2 , ignore  s 3 ) = 7.5 min(9//1,15//2,”ignore “s_(3))=7.5\min(9/1, 15/2, \text{ignore } s_3) = 7.5min(9/1,15/2,ignore s3)=7.5).
Pivot on x 2 x 2 x_(2)x_2x2 in the A 2 A 2 A_(2)A_2A2 row:
New A 2 row = Old A 2 row 2 . New  A 2  row = Old  A 2  row 2 . “New “A_(2)” row”=(“Old “A_(2)” row”)/(2).\text{New } A_2 \text{ row} = \frac{\text{Old } A_2 \text{ row}}{2}.New A2 row=Old A2 row2.
New A 1 row = Old A 1 row − New A 2 row . New  A 1  row = Old  A 1  row − New  A 2  row . “New “A_(1)” row”=”Old “A_(1)” row”-“New “A_(2)” row”.\text{New } A_1 \text{ row} = \text{Old } A_1 \text{ row} – \text{New } A_2 \text{ row}.New A1 row=Old A1 row−New A2 row.
New s 3 row = Old s 3 row + 3 × New A 2 row . New  s 3  row = Old  s 3  row + 3 × New  A 2  row . “New “s_(3)” row”=”Old “s_(3)” row”+3xx”New “A_(2)” row”.\text{New } s_3 \text{ row} = \text{Old } s_3 \text{ row} + 3 \times \text{New } A_2 \text{ row}.New s3 row=Old s3 row+3×New A2 row.
Updated Tableau:
Basis x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 s 1 s 1 s_(1)s_1s1 s 2 s 2 s_(2)s_2s2 s 3 s 3 s_(3)s_3s3 A 1 A 1 A_(1)A_1A1 A 2 A 2 A_(2)A_2A2RHS
A 1 A 1 A_(1)A_1A10.50-10.501-0.51.5
x 2 x 2 x_(2)x_2x20.510-0.5000.57.5
s 3 s 3 s_(3)s_3s33.500-1.5101.531.5
Z’ 0.5 + 0.5 M 0.5 + 0.5 M 0.5+0.5 M0.5 + 0.5M0.5+0.5M0 − M − M -M-M−M 1.5 + 1.5 M 1.5 + 1.5 M 1.5+1.5 M1.5 + 1.5M1.5+1.5M00 − 1.5 − 1.5 M − 1.5 − 1.5 M -1.5-1.5 M-1.5 – 1.5M−1.5−1.5M 22.5 − 1.5 M 22.5 − 1.5 M 22.5-1.5 M22.5 – 1.5M22.5−1.5M
Iteration 2:
  • Entering Variable: x 1 x 1 x_(1)x_1x1.
  • Leaving Variable: A 1 A 1 A_(1)A_1A1 (min ratio test: min ( 1.5 / 0.5 , 7.5 / 0.5 , 31.5 / 3.5 ) = 3 min ( 1.5 / 0.5 , 7.5 / 0.5 , 31.5 / 3.5 ) = 3 min(1.5//0.5,7.5//0.5,31.5//3.5)=3\min(1.5/0.5, 7.5/0.5, 31.5/3.5) = 3min(1.5/0.5,7.5/0.5,31.5/3.5)=3).
Pivot on x 1 x 1 x_(1)x_1x1 in the A 1 A 1 A_(1)A_1A1 row:
New A 1 row = Old A 1 row 0.5 . New  A 1  row = Old  A 1  row 0.5 . “New “A_(1)” row”=(“Old “A_(1)” row”)/(0.5).\text{New } A_1 \text{ row} = \frac{\text{Old } A_1 \text{ row}}{0.5}.New A1 row=Old A1 row0.5.
New x 2 row = Old x 2 row − 0.5 × New A 1 row . New  x 2  row = Old  x 2  row − 0.5 × New  A 1  row . “New “x_(2)” row”=”Old “x_(2)” row”-0.5 xx”New “A_(1)” row”.\text{New } x_2 \text{ row} = \text{Old } x_2 \text{ row} – 0.5 \times \text{New } A_1 \text{ row}.New x2 row=Old x2 row−0.5×New A1 row.
New s 3 row = Old s 3 row − 3.5 × New A 1 row . New  s 3  row = Old  s 3  row − 3.5 × New  A 1  row . “New “s_(3)” row”=”Old “s_(3)” row”-3.5 xx”New “A_(1)” row”.\text{New } s_3 \text{ row} = \text{Old } s_3 \text{ row} – 3.5 \times \text{New } A_1 \text{ row}.New s3 row=Old s3 row−3.5×New A1 row.
Updated Tableau:
Basis x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 s 1 s 1 s_(1)s_1s1 s 2 s 2 s_(2)s_2s2 s 3 s 3 s_(3)s_3s3 A 1 A 1 A_(1)A_1A1 A 2 A 2 A_(2)A_2A2RHS
x 1 x 1 x_(1)x_1x110-2102-13
x 2 x 2 x_(2)x_2x2011-10-116
s 3 s 3 s_(3)s_3s3007-51-7521
Z’00 1 − 2 M 1 − 2 M 1-2M1 – 2M1−2M 1 + M 1 + M 1+M1 + M1+M0 − 1 + 2 M − 1 + 2 M -1+2M-1 + 2M−1+2M − 1 − M − 1 − M -1-M-1 – M−1−M21
Iteration 3:
  • Optimality Check: All coefficients in Z ′ Z ′ Z^(‘)Z’Z′ are non-positive (since M M MMM is large and positive).
  • Artificial Variables: A 1 A 1 A_(1)A_1A1 and A 2 A 2 A_(2)A_2A2 are non-basic (coefficients are zero in Z ′ Z ′ Z^(‘)Z’Z′), so they can be removed.

Step 5: Extract the Optimal Solution

From the final tableau:
x 1 = 3 , x 2 = 6 , s 3 = 21. x 1 = 3 , x 2 = 6 , s 3 = 21. x_(1)=3,quadx_(2)=6,quads_(3)=21.x_1 = 3, \quad x_2 = 6, \quad s_3 = 21.x1=3,x2=6,s3=21.
The minimum value of Z Z ZZZ is:
Z = 2 ( 3 ) + 3 ( 6 ) = 24. Z = 2 ( 3 ) + 3 ( 6 ) = 24. Z=2(3)+3(6)=24.Z = 2(3) + 3(6) = 24.Z=2(3)+3(6)=24.

Step 6: Check Uniqueness of the Optimal Solution

  • Uniqueness Condition: If all non-basic variables in the final tableau have strictly negative coefficients in the Z Z ZZZ-row, the solution is unique.
  • Final Z Z ZZZ-row Coefficients: For s 1 : 1 − 2 M ( Negative for large M ) , For s 2 : 1 + M ( Positive ) , For A 1 : − 1 + 2 M ( Positive for large M ) , For A 2 : − 1 − M ( Negative ) . For  s 1 : 1 − 2 M ( Negative for large  M ) , For  s 2 : 1 + M ( Positive ) , For  A 1 : − 1 + 2 M ( Positive for large  M ) , For  A 2 : − 1 − M ( Negative ) . “For “s_(1):1-2M quad(“Negative for large “M),”For “s_(2):1+M quad(“Positive”),”For “A_(1):-1+2M quad(“Positive for large “M),”For “A_(2):-1-M quad(“Negative”).\text{For } s_1: 1 – 2M \quad (\text{Negative for large } M), \\ \text{For } s_2: 1 + M \quad (\text{Positive}), \\ \text{For } A_1: -1 + 2M \quad (\text{Positive for large } M), \\ \text{For } A_2: -1 – M \quad (\text{Negative}).For s1:1−2M(Negative for large M),For s2:1+M(Positive),For A1:−1+2M(Positive for large M),For A2:−1−M(Negative).
  • Conclusion: Since the coefficient for s 2 s 2 s_(2)s_2s2 is positive, there exists an alternative optimal solution. Thus, the optimal solution is not unique.

Final Answer

  • Optimal Solution: x 1 = 3 x 1 = 3 x_(1)=3x_1 = 3x1=3, x 2 = 6 x 2 = 6 x_(2)=6x_2 = 6x2=6, with Z = 24 Z = 24 Z=24Z = 24Z=24.
  • Uniqueness: The optimal solution is not unique because the coefficient of s 2 s 2 s_(2)s_2s2 in the final Z Z ZZZ-row is positive.
Optimal Solution: x 1 = 3 , x 2 = 6 , with Z = 24. The optimal solution is not unique. Optimal Solution:  x 1 = 3 , x 2 = 6 ,  with  Z = 24. The optimal solution is not unique. [“Optimal Solution: “x_(1)=3″,”x_(2)=6″,”” with “Z=24.],[“The optimal solution is not unique.”]\boxed{ \begin{aligned} &\text{Optimal Solution: } x_1 = 3, x_2 = 6, \text{ with } Z = 24. \\ &\text{The optimal solution is not unique.} \end{aligned} }Optimal Solution: x1=3,x2=6, with Z=24.The optimal solution is not unique.

Question:-04

(a) Prove that the oscillation of a real-valued bounded function f f fff defined on [ a , b ] [ a , b ] [a,b][a, b][a,b] is the supremum of the set { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } {|f(x_(1))-f(x_(2))|:x_(1),x_(2)in[a,b]}\{ |f(x_1) – f(x_2)| : x_1, x_2 \in [a, b] \}{|f(x1)−f(x2)|:x1,x2∈[a,b]}.

Answer:

Proof: Oscillation of a Bounded Function on [ a , b ] [ a , b ] [a,b][a, b][a,b]

We aim to prove that the oscillation of a bounded real-valued function f f fff on [ a , b ] [ a , b ] [a,b][a, b][a,b] is equal to the supremum of the set of absolute differences of f f fff over [ a , b ] [ a , b ] [a,b][a, b][a,b].

Definitions:

  1. Oscillation of f f fff on [ a , b ] [ a , b ] [a,b][a, b][a,b]:
    osc ( f , [ a , b ] ) = sup x ∈ [ a , b ] f ( x ) − inf x ∈ [ a , b ] f ( x ) . osc ( f , [ a , b ] ) = sup x ∈ [ a , b ]   f ( x ) − inf x ∈ [ a , b ]   f ( x ) . “osc”(f,[a,b])=s u p_(x in[a,b])f(x)-i n f_(x in[a,b])f(x).\text{osc}(f, [a, b]) = \sup_{x \in [a, b]} f(x) – \inf_{x \in [a, b]} f(x).osc(f,[a,b])=supx∈[a,b]f(x)−infx∈[a,b]f(x).
  2. Set of Absolute Differences:
    S = { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } . S = { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } . S={|f(x_(1))-f(x_(2))|:x_(1),x_(2)in[a,b]}.S = \{ |f(x_1) – f(x_2)| : x_1, x_2 \in [a, b] \}.S={|f(x1)−f(x2)|:x1,x2∈[a,b]}.

Goal:

Prove that:
osc ( f , [ a , b ] ) = sup S . osc ( f , [ a , b ] ) = sup S . “osc”(f,[a,b])=s u p S.\text{osc}(f, [a, b]) = \sup S.osc(f,[a,b])=supS.

Step 1: Show osc ( f , [ a , b ] ) ≥ sup S osc ( f , [ a , b ] ) ≥ sup S “osc”(f,[a,b]) >= s u p S\text{osc}(f, [a, b]) \geq \sup Sosc(f,[a,b])≥supS

  • By definition of supremum and infimum, for any x 1 , x 2 ∈ [ a , b ] x 1 , x 2 ∈ [ a , b ] x_(1),x_(2)in[a,b]x_1, x_2 \in [a, b]x1,x2∈[a,b]: f ( x 1 ) ≤ sup x ∈ [ a , b ] f ( x ) , f ( x 2 ) ≥ inf x ∈ [ a , b ] f ( x ) . f ( x 1 ) ≤ sup x ∈ [ a , b ]   f ( x ) , f ( x 2 ) ≥ inf x ∈ [ a , b ]   f ( x ) . f(x_(1)) <= s u p_(x in[a,b])f(x),quad f(x_(2)) >= i n f_(x in[a,b])f(x).f(x_1) \leq \sup_{x \in [a, b]} f(x), \quad f(x_2) \geq \inf_{x \in [a, b]} f(x).f(x1)≤supx∈[a,b]f(x),f(x2)≥infx∈[a,b]f(x).
  • Therefore: | f ( x 1 ) − f ( x 2 ) | ≤ sup f − inf f = osc ( f , [ a , b ] ) . | f ( x 1 ) − f ( x 2 ) | ≤ sup f − inf f = osc ( f , [ a , b ] ) . |f(x_(1))-f(x_(2))| <= s u p f-i n f f=”osc”(f,[a,b]).|f(x_1) – f(x_2)| \leq \sup f – \inf f = \text{osc}(f, [a, b]).|f(x1)−f(x2)|≤supf−inff=osc(f,[a,b]).
  • Since this holds for all x 1 , x 2 x 1 , x 2 x_(1),x_(2)x_1, x_2x1,x2, it follows that: sup S ≤ osc ( f , [ a , b ] ) . sup S ≤ osc ( f , [ a , b ] ) . s u p S <= “osc”(f,[a,b]).\sup S \leq \text{osc}(f, [a, b]).supS≤osc(f,[a,b]).

Step 2: Show osc ( f , [ a , b ] ) ≤ sup S osc ( f , [ a , b ] ) ≤ sup S “osc”(f,[a,b]) <= s u p S\text{osc}(f, [a, b]) \leq \sup Sosc(f,[a,b])≤supS

  • Let sup f = f ( x ∗ ) sup f = f ( x ∗ ) s u p f=f(x^(**))\sup f = f(x^*)supf=f(x∗) and inf f = f ( x ∗ ) inf f = f ( x ∗ ) i n f f=f(x_(**))\inf f = f(x_*)inff=f(x∗) for some x ∗ , x ∗ ∈ [ a , b ] x ∗ , x ∗ ∈ [ a , b ] x^(**),x_(**)in[a,b]x^*, x_* \in [a, b]x∗,x∗∈[a,b] (since f f fff is bounded and [ a , b ] [ a , b ] [a,b][a, b][a,b] is compact, the supremum and infimum are attained).
  • Then: osc ( f , [ a , b ] ) = f ( x ∗ ) − f ( x ∗ ) = | f ( x ∗ ) − f ( x ∗ ) | ∈ S . osc ( f , [ a , b ] ) = f ( x ∗ ) − f ( x ∗ ) = | f ( x ∗ ) − f ( x ∗ ) | ∈ S . “osc”(f,[a,b])=f(x^(**))-f(x_(**))=|f(x^(**))-f(x_(**))|in S.\text{osc}(f, [a, b]) = f(x^*) – f(x_*) = |f(x^*) – f(x_*)| \in S.osc(f,[a,b])=f(x∗)−f(x∗)=|f(x∗)−f(x∗)|∈S.
  • Since sup S sup S s u p S\sup SsupS is the least upper bound of S S SSS, and osc ( f , [ a , b ] ) ∈ S osc ( f , [ a , b ] ) ∈ S “osc”(f,[a,b])in S\text{osc}(f, [a, b]) \in Sosc(f,[a,b])∈S, we have: osc ( f , [ a , b ] ) ≤ sup S . osc ( f , [ a , b ] ) ≤ sup S . “osc”(f,[a,b]) <= s u p S.\text{osc}(f, [a, b]) \leq \sup S.osc(f,[a,b])≤supS.

Step 3: Conclusion

From Steps 1 and 2, we have:
osc ( f , [ a , b ] ) ≤ sup S ≤ osc ( f , [ a , b ] ) . osc ( f , [ a , b ] ) ≤ sup S ≤ osc ( f , [ a , b ] ) . “osc”(f,[a,b]) <= s u p S <= “osc”(f,[a,b]).\text{osc}(f, [a, b]) \leq \sup S \leq \text{osc}(f, [a, b]).osc(f,[a,b])≤supS≤osc(f,[a,b]).
Thus:
osc ( f , [ a , b ] ) = sup S . osc ( f , [ a , b ] ) = sup S . “osc”(f,[a,b])=s u p S.\text{osc}(f, [a, b]) = \sup S.osc(f,[a,b])=supS.

Final Answer

osc ( f , [ a , b ] ) = sup { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } osc ( f , [ a , b ] ) = sup { | f ( x 1 ) − f ( x 2 ) | : x 1 , x 2 ∈ [ a , b ] } “osc”(f,[a,b])=s u p{|f(x_(1))-f(x_(2))|:x_(1),x_(2)in[a,b]}\boxed{\text{osc}(f, [a, b]) = \sup \{ |f(x_1) – f(x_2)| : x_1, x_2 \in [a, b] \}}osc(f,[a,b])=sup{|f(x1)−f(x2)|:x1,x2∈[a,b]}

Question:-04 (b)

Classify the singular point z = 0 z = 0 z=0z = 0z=0 of the function f ( z ) = e z z − sin ⁡ z f ( z ) = e z z − sin ⁡ z f(z)=(e^(z))/(z-sin z)f(z) = \frac{e^z}{z – \sin z}f(z)=ezz−sin⁡z and obtain the principal part of its Laurent series expansion.

Answer:

To classify the singular point z = 0 z = 0 z=0z = 0z=0 of the function f ( z ) = e z z − sin ⁡ z f ( z ) = e z z − sin ⁡ z f(z)=(e^(z))/(z-sin z)f(z) = \frac{e^z}{z – \sin z}f(z)=ezz−sin⁡z and find the principal part of its Laurent series expansion, we proceed with the following steps:

Step 1: Analyze the Denominator z − sin ⁡ z z − sin ⁡ z z-sin zz – \sin zz−sin⁡z Near z = 0 z = 0 z=0z = 0z=0

First, expand sin ⁡ z sin ⁡ z sin z\sin zsin⁡z in a Taylor series around z = 0 z = 0 z=0z = 0z=0:
sin ⁡ z = z − z 3 6 + z 5 120 − ⋯ sin ⁡ z = z − z 3 6 + z 5 120 − ⋯ sin z=z-(z^(3))/(6)+(z^(5))/(120)-cdots\sin z = z – \frac{z^3}{6} + \frac{z^5}{120} – \cdotssin⁡z=z−z36+z5120−⋯
Thus, the denominator becomes:
z − sin ⁡ z = z 3 6 − z 5 120 + ⋯ = z 3 6 ( 1 − z 2 20 + ⋯ ) . z − sin ⁡ z = z 3 6 − z 5 120 + ⋯ = z 3 6 1 − z 2 20 + ⋯ . z-sin z=(z^(3))/(6)-(z^(5))/(120)+cdots=(z^(3))/(6)(1-(z^(2))/(20)+cdots).z – \sin z = \frac{z^3}{6} – \frac{z^5}{120} + \cdots = \frac{z^3}{6} \left(1 – \frac{z^2}{20} + \cdots \right).z−sin⁡z=z36−z5120+⋯=z36(1−z220+⋯).
This shows that z − sin ⁡ z z − sin ⁡ z z-sin zz – \sin zz−sin⁡z has a zero of order 3 at z = 0 z = 0 z=0z = 0z=0.

Step 2: Analyze the Numerator e z e z e^(z)e^zez Near z = 0 z = 0 z=0z = 0z=0

The Taylor series for e z e z e^(z)e^zez around z = 0 z = 0 z=0z = 0z=0 is:
e z = 1 + z + z 2 2 + z 3 6 + ⋯ e z = 1 + z + z 2 2 + z 3 6 + ⋯ e^(z)=1+z+(z^(2))/(2)+(z^(3))/(6)+cdotse^z = 1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \cdotsez=1+z+z22+z36+⋯
The numerator e z e z e^(z)e^zez is non-zero and analytic at z = 0 z = 0 z=0z = 0z=0.

Step 3: Determine the Nature of the Singularity

The function f ( z ) f ( z ) f(z)f(z)f(z) can be written as:
f ( z ) = e z z − sin ⁡ z = 1 + z + z 2 2 + z 3 6 + ⋯ z 3 6 ( 1 − z 2 20 + ⋯ ) . f ( z ) = e z z − sin ⁡ z = 1 + z + z 2 2 + z 3 6 + ⋯ z 3 6 1 − z 2 20 + ⋯ . f(z)=(e^(z))/(z-sin z)=(1+z+(z^(2))/(2)+(z^(3))/(6)+cdots)/((z^(3))/(6)(1-(z^(2))/(20)+cdots)).f(z) = \frac{e^z}{z – \sin z} = \frac{1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \cdots}{\frac{z^3}{6} \left(1 – \frac{z^2}{20} + \cdots \right)}.f(z)=ezz−sin⁡z=1+z+z22+z36+⋯z36(1−z220+⋯).
Since the denominator has a zero of order 3 and the numerator is non-zero, f ( z ) f ( z ) f(z)f(z)f(z) has a pole of order 3 at z = 0 z = 0 z=0z = 0z=0.

Step 4: Find the Laurent Series Expansion Near z = 0 z = 0 z=0z = 0z=0

To find the principal part of the Laurent series, we perform long division or expand e z z − sin ⁡ z e z z − sin ⁡ z (e^(z))/(z-sin z)\frac{e^z}{z – \sin z}ezz−sin⁡z in powers of z z zzz.
First, express f ( z ) f ( z ) f(z)f(z)f(z) as:
f ( z ) = 6 e z z 3 ( 1 − z 2 20 + ⋯ ) . f ( z ) = 6 e z z 3 1 − z 2 20 + ⋯ . f(z)=(6e^(z))/(z^(3)(1-(z^(2))/(20)+cdots)).f(z) = \frac{6 e^z}{z^3 \left(1 – \frac{z^2}{20} + \cdots \right)}.f(z)=6ezz3(1−z220+⋯).
Using the geometric series expansion for the denominator:
1 1 − z 2 20 + ⋯ = 1 + z 2 20 + ⋯ , 1 1 − z 2 20 + ⋯ = 1 + z 2 20 + ⋯ , (1)/(1-(z^(2))/(20)+cdots)=1+(z^(2))/(20)+cdots,\frac{1}{1 – \frac{z^2}{20} + \cdots} = 1 + \frac{z^2}{20} + \cdots,11−z220+⋯=1+z220+⋯,
we get:
f ( z ) = 6 z 3 ( 1 + z + z 2 2 + z 3 6 + ⋯ ) ( 1 + z 2 20 + ⋯ ) . f ( z ) = 6 z 3 1 + z + z 2 2 + z 3 6 + ⋯ 1 + z 2 20 + ⋯ . f(z)=(6)/(z^(3))(1+z+(z^(2))/(2)+(z^(3))/(6)+cdots)(1+(z^(2))/(20)+cdots).f(z) = \frac{6}{z^3} \left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \cdots \right) \left(1 + \frac{z^2}{20} + \cdots \right).f(z)=6z3(1+z+z22+z36+⋯)(1+z220+⋯).
Multiplying the series:
f ( z ) = 6 z 3 ( 1 + z + ( 1 2 + 1 20 ) z 2 + ( 1 6 + 1 20 ) z 3 + ⋯ ) . f ( z ) = 6 z 3 1 + z + 1 2 + 1 20 z 2 + 1 6 + 1 20 z 3 + ⋯ . f(z)=(6)/(z^(3))(1+z+((1)/(2)+(1)/(20))z^(2)+((1)/(6)+(1)/(20))z^(3)+cdots).f(z) = \frac{6}{z^3} \left(1 + z + \left(\frac{1}{2} + \frac{1}{20}\right) z^2 + \left(\frac{1}{6} + \frac{1}{20}\right) z^3 + \cdots \right).f(z)=6z3(1+z+(12+120)z2+(16+120)z3+⋯).
Simplifying:
f ( z ) = 6 z 3 + 6 z 2 + 21 5 z + 23 10 + ⋯ . f ( z ) = 6 z 3 + 6 z 2 + 21 5 z + 23 10 + ⋯ . f(z)=(6)/(z^(3))+(6)/(z^(2))+(21)/(5z)+(23)/(10)+cdots.f(z) = \frac{6}{z^3} + \frac{6}{z^2} + \frac{21}{5z} + \frac{23}{10} + \cdots.f(z)=6z3+6z2+215z+2310+⋯.

Step 5: Identify the Principal Part

The principal part of the Laurent series consists of the terms with negative powers of z z zzz:
Principal Part = 6 z 3 + 6 z 2 + 21 5 z . Principal Part = 6 z 3 + 6 z 2 + 21 5 z . “Principal Part”=(6)/(z^(3))+(6)/(z^(2))+(21)/(5z).\text{Principal Part} = \frac{6}{z^3} + \frac{6}{z^2} + \frac{21}{5z}.Principal Part=6z3+6z2+215z.

Final Answer

  • Classification: z = 0 z = 0 z=0z = 0z=0 is a pole of order 3.
  • Principal Part of Laurent Series:
6 z 3 + 6 z 2 + 21 5 z 6 z 3 + 6 z 2 + 21 5 z (6)/(z^(3))+(6)/(z^(2))+(21)/(5z)\boxed{ \frac{6}{z^3} + \frac{6}{z^2} + \frac{21}{5z} }6z3+6z2+215z

Question:-04 (c)

A department head has 5 subordinates and 5 jobs to be performed. The time (in hours) that each subordinate will take to perform each job is given in the matrix below:

Jobs
ABCDE
I494124
II15112058
III177151218
IV91311914
V61112914
How should the jobs be assigned, one to each subordinate, so as to minimize the total time? Also, obtain the total minimum time to perform all the jobs if subordinate IV cannot be assigned job C.

Answer:

This is the original cost matrix:
4  9  4  12  4
15 11 20  5  8
17  7  15 12 18
6  13 11  9  14
Subtract row minima
We subtract the row minimum from each row:
0  5  0  8   0  (-4)
10 6  15 0   3  (-5)
10 0  8  5   11 (-7)
0  7  5  3   8  (-6)
Subtract column minima
Because each column contains a zero, subtracting column minima has no effect.
Cover all zeros with a minimum number of lines
There are 4 lines required to cover all zeros:
10  5  0  0   3  x
10  0  8  5   11 x
0   7  5  3   8  x
x   0  0  0   3  x
Create additional zeros
The number of lines is smaller than 5. The smallest uncovered number is 2. We subtract this number from all uncovered elements and add it to all elements that are covered twice:
2  5  0  10  0
10 4  13 0   1
12 0  8  7   11
0  2  0  0   3
0  3  4  3   6
Cover all zeros with a minimum number of lines
There are 5 lines required to cover all zeros:
2  5  0  10 0  x
12 0  8  0  1  x
0  0  0  7  11 x
0  2  0  0  3  x
x  3  4  3  6  x
The optimal assignment
Because there are 5 lines required, the zeros cover an optimal assignment:
2  5  0  10 0
12 0  8  0  1
0  0  0  7  11
0  2  0  0  3
This corresponds to the following 4 to 4 assignment in the original cost matrix:
4  9  4  12  4
15 11 20 5   8
17 7  15 12  18
6  13 11 9   14
The optimal value equals 33.

Question:-05 (c)

Evaluate, using binary arithmetic, the following numbers in their given system:

(i) ( 634.235 ) 8 − ( 132.223 ) 8 ( 634.235 ) 8 − ( 132.223 ) 8 (634.235)_(8)-(132.223)_(8)(634.235)_8 – (132.223)_8(634.235)8−(132.223)8
(ii) ( 7 A B .432 ) 16 − ( 5 C A . D 61 ) 16 ( 7 A B .432 ) 16 − ( 5 C A . D 61 ) 16 (7AB.432)_(16)-(5CA.D 61)_(16)(7AB.432)_{16} – (5CA.D61)_{16}(7AB.432)16−(5CA.D61)16

Answer:

Solution:

(i) ( 634.235 ) 8 − ( 132.223 ) 8 ( 634.235 ) 8 − ( 132.223 ) 8 (634.235)_(8)-(132.223)_(8)(634.235)_8 – (132.223)_8(634.235)8−(132.223)8)

Step 1: Convert both octal numbers to binary.
  • ( 634.235 ) 8 ( 634.235 ) 8 (634.235)_(8)(634.235)_8(634.235)8 to binary:
    6 → 110 , 3 → 011 , 4 → 100 , . 2 → 010 , 3 → 011 , 5 → 101 6 → 110 , 3 → 011 , 4 → 100 , . 2 → 010 , 3 → 011 , 5 → 101 6rarr110,quad3rarr011,quad4rarr100,quad.quad2rarr010,quad3rarr011,quad5rarr1016 \rightarrow 110, \quad 3 \rightarrow 011, \quad 4 \rightarrow 100, \quad . \quad 2 \rightarrow 010, \quad 3 \rightarrow 011, \quad 5 \rightarrow 1016→110,3→011,4→100,.2→010,3→011,5→101
    ( 634.235 ) 8 = ( 110 011 100.010 011 101 ) 2 ( 634.235 ) 8 = ( 110 011 100.010 011 101 ) 2 (634.235)_(8)=(110011100.010011101)_(2)(634.235)_8 = (110\,011\,100.010\,011\,101)_2(634.235)8=(110011100.010011101)2
  • ( 132.223 ) 8 ( 132.223 ) 8 (132.223)_(8)(132.223)_8(132.223)8 to binary:
    1 → 001 , 3 → 011 , 2 → 010 , . 2 → 010 , 2 → 010 , 3 → 011 1 → 001 , 3 → 011 , 2 → 010 , . 2 → 010 , 2 → 010 , 3 → 011 1rarr001,quad3rarr011,quad2rarr010,quad.quad2rarr010,quad2rarr010,quad3rarr0111 \rightarrow 001, \quad 3 \rightarrow 011, \quad 2 \rightarrow 010, \quad . \quad 2 \rightarrow 010, \quad 2 \rightarrow 010, \quad 3 \rightarrow 0111→001,3→011,2→010,.2→010,2→010,3→011
    ( 132.223 ) 8 = ( 001 011 010.010 010 011 ) 2 ( 132.223 ) 8 = ( 001 011 010.010 010 011 ) 2 (132.223)_(8)=(001011010.010010011)_(2)(132.223)_8 = (001\,011\,010.010\,010\,011)_2(132.223)8=(001011010.010010011)2
Step 2: Perform binary subtraction.
− 110 011 100.010 011 101 − 001 011 010.010 010 011 − 110 011 100.010 011 101 − 001 011 010.010 010 011 -110011100.010011101-001011010.010010011\begin{array}{r} \phantom{-}110\,011\,100.010\,011\,101 \\ -\;001\,011\,010.010\,010\,011 \\ \hline \end{array}−110011100.010011101−001011010.010010011
  • Subtraction process (using two’s complement for negative results): 110 011 100.010 011 101 − 001 011 010.010 010 011 = 101 000 010.000 001 010 110 011 100.010 011 101 − 001 011 010.010 010 011 = 101 000 010.000 001 010 110011100.010011101-001011010.010010011=101000010.000001010110\,011\,100.010\,011\,101 – 001\,011\,010.010\,010\,011 = 101\,000\,010.000\,001\,010110011100.010011101−001011010.010010011=101000010.000001010
Step 3: Convert the result back to octal.
101 000 010.000 001 010 → 5 0 2.0 1 2 101 000 010.000 001 010 → 5 0 2.0 1 2 101000010.000001010 rarr502.012101\,000\,010.000\,001\,010 \rightarrow 5\,0\,2.0\,1\,2101000010.000001010→502.012
( 502.012 ) 8 ( 502.012 ) 8 (502.012)_(8)(502.012)_8(502.012)8
Final Answer:
( 502.012 ) 8 ( 502.012 ) 8 (502.012)_(8)\boxed{(502.012)_8}(502.012)8

(ii) ( 7 A B .432 ) 16 − ( 5 C A . D 61 ) 16 ( 7 A B .432 ) 16 − ( 5 C A . D 61 ) 16 (7AB.432)_(16)-(5CA.D 61)_(16)(7AB.432)_{16} – (5CA.D61)_{16}(7AB.432)16−(5CA.D61)16)

Step 1: Convert both hexadecimal numbers to binary.
  • ( 7 A B .432 ) 16 ( 7 A B .432 ) 16 (7AB.432)_(16)(7AB.432)_{16}(7AB.432)16 to binary:
    7 → 0111 , A → 1010 , B → 1011 , . 4 → 0100 , 3 → 0011 , 2 → 0010 7 → 0111 , A → 1010 , B → 1011 , . 4 → 0100 , 3 → 0011 , 2 → 0010 7rarr0111,quad A rarr1010,quad B rarr1011,quad.quad4rarr0100,quad3rarr0011,quad2rarr00107 \rightarrow 0111, \quad A \rightarrow 1010, \quad B \rightarrow 1011, \quad . \quad 4 \rightarrow 0100, \quad 3 \rightarrow 0011, \quad 2 \rightarrow 00107→0111,A→1010,B→1011,.4→0100,3→0011,2→0010
    ( 7 A B .432 ) 16 = ( 0111 1010 1011.0100 0011 0010 ) 2 ( 7 A B .432 ) 16 = ( 0111 1010 1011.0100 0011 0010 ) 2 (7AB.432)_(16)=(011110101011.010000110010)_(2)(7AB.432)_{16} = (0111\,1010\,1011.0100\,0011\,0010)_2(7AB.432)16=(011110101011.010000110010)2
  • ( 5 C A . D 61 ) 16 ( 5 C A . D 61 ) 16 (5CA.D 61)_(16)(5CA.D61)_{16}(5CA.D61)16 to binary:
    5 → 0101 , C → 1100 , A → 1010 , . D → 1101 , 6 → 0110 , 1 → 0001 5 → 0101 , C → 1100 , A → 1010 , . D → 1101 , 6 → 0110 , 1 → 0001 5rarr0101,quad C rarr1100,quad A rarr1010,quad.quad D rarr1101,quad6rarr0110,quad1rarr00015 \rightarrow 0101, \quad C \rightarrow 1100, \quad A \rightarrow 1010, \quad . \quad D \rightarrow 1101, \quad 6 \rightarrow 0110, \quad 1 \rightarrow 00015→0101,C→1100,A→1010,.D→1101,6→0110,1→0001
    ( 5 C A . D 61 ) 16 = ( 0101 1100 1010.1101 0110 0001 ) 2 ( 5 C A . D 61 ) 16 = ( 0101 1100 1010.1101 0110 0001 ) 2 (5CA.D 61)_(16)=(010111001010.110101100001)_(2)(5CA.D61)_{16} = (0101\,1100\,1010.1101\,0110\,0001)_2(5CA.D61)16=(010111001010.110101100001)2
Step 2: Perform binary subtraction.
− 0111 1010 1011.0100 0011 0010 − 0101 1100 1010.1101 0110 0001 − 0111 1010 1011.0100 0011 0010 − 0101 1100 1010.1101 0110 0001 -011110101011.010000110010-010111001010.110101100001\begin{array}{r} \phantom{-}0111\,1010\,1011.0100\,0011\,0010 \\ -\;0101\,1100\,1010.1101\,0110\,0001 \\ \hline \end{array}−011110101011.010000110010−010111001010.110101100001
  • Subtraction process (using two’s complement for negative results): 0111 1010 1011.0100 0011 0010 − 0101 1100 1010.1101 0110 0001 = 0001 1101 1100.0110 1101 0001 0111 1010 1011.0100 0011 0010 − 0101 1100 1010.1101 0110 0001 = 0001 1101 1100.0110 1101 0001 011110101011.010000110010-010111001010.110101100001=000111011100.0110110100010111\,1010\,1011.0100\,0011\,0010 – 0101\,1100\,1010.1101\,0110\,0001 = 0001\,1101\,1100.0110\,1101\,0001011110101011.010000110010−010111001010.110101100001=000111011100.011011010001
Step 3: Convert the result back to hexadecimal.
0001 1101 1100.0110 1101 0001 → 1 D C .6 D 1 0001 1101 1100.0110 1101 0001 → 1 D C .6 D 1 000111011100.011011010001 rarr1DC.6D10001\,1101\,1100.0110\,1101\,0001 \rightarrow 1\,D\,C.6\,D\,1000111011100.011011010001→1DC.6D1
( 1 D C .6 D 1 ) 16 ( 1 D C .6 D 1 ) 16 (1DC.6 D1)_(16)(1DC.6D1)_{16}(1DC.6D1)16
Final Answer:
( 1 D C .6 D 1 ) 16 ( 1 D C .6 D 1 ) 16 (1DC.6 D1)_(16)\boxed{(1DC.6D1)_{16}}(1DC.6D1)16

Question:-05 (d)

A planet of mass m m mmm is revolving around the sun of mass M M MMM. The kinetic energy T T TTT and the potential energy V V VVV of the planet are given by T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) T=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))T = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2)T=12m(r˙2+r2θ˙2) and V = G M m ( 1 2 a − 1 r ) V = G M m 1 2 a − 1 r V=GMm((1)/(2a)-(1)/(r))V = G M m \left( \frac{1}{2a} – \frac{1}{r} \right)V=GMm(12a−1r), where ( r , θ ) ( r , θ ) (r,theta)(r, \theta)(r,θ) are the polar coordinates of the planet at time t t ttt, G G GGG is the gravitational constant, and 2 a 2 a 2a2a2a is the major axis of the ellipse (the path of the planet). Find the Hamiltonian and the Hamilton equations of the planet’s motion.

Answer:

To find the Hamiltonian and the Hamilton equations for the planet’s motion, we follow these steps:

1. Lagrangian of the System

The Lagrangian L L LLL is given by:
L = T − V L = T − V L=T-VL = T – VL=T−V
where:
  • Kinetic Energy (T): T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) T = 1 2 m r ˙ 2 + r 2 θ ˙ 2 T=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))T = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right)T=12m(r˙2+r2θ˙2)
  • Potential Energy (V): V = G M m ( 1 2 a − 1 r ) V = G M m 1 2 a − 1 r V=GMm((1)/(2a)-(1)/(r))V = G M m \left( \frac{1}{2a} – \frac{1}{r} \right)V=GMm(12a−1r)
Thus:
L = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 ) − G M m ( 1 2 a − 1 r ) L = 1 2 m r ˙ 2 + r 2 θ ˙ 2 − G M m 1 2 a − 1 r L=(1)/(2)m(r^(˙)^(2)+r^(2)theta^(˙)^(2))-GMm((1)/(2a)-(1)/(r))L = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 \right) – G M m \left( \frac{1}{2a} – \frac{1}{r} \right)L=12m(r˙2+r2θ˙2)−GMm(12a−1r)

2. Generalized Momenta

The generalized momenta p r p r p_(r)p_rpr and p θ p θ p_( theta)p_\thetapθ are obtained from:
p r = ∂ L ∂ r ˙ = m r ˙ p r = ∂ L ∂ r ˙ = m r ˙ p_(r)=(del L)/(del(r^(˙)))=mr^(˙)p_r = \frac{\partial L}{\partial \dot{r}} = m \dot{r}pr=∂L∂r˙=mr˙
p θ = ∂ L ∂ θ ˙ = m r 2 θ ˙ p θ = ∂ L ∂ θ ˙ = m r 2 θ ˙ p_( theta)=(del L)/(del(theta^(˙)))=mr^(2)theta^(˙)p_\theta = \frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta}pθ=∂L∂θ˙=mr2θ˙

3. Hamiltonian

The Hamiltonian H H HHH is defined as:
H = ∑ i p i q ˙ i − L H = ∑ i   p i q ˙ i − L H=sum _(i)p_(i)q^(˙)_(i)-LH = \sum_i p_i \dot{q}_i – LH=∑ipiq˙i−L
Substituting the momenta and Lagrangian:
H = p r r ˙ + p θ θ ˙ − L H = p r r ˙ + p θ θ ˙ − L H=p_(r)r^(˙)+p_( theta)theta^(˙)-LH = p_r \dot{r} + p_\theta \dot{\theta} – LH=prr˙+pθθ˙−L
Express r ˙ r ˙ r^(˙)\dot{r}r˙ and θ ˙ θ ˙ theta^(˙)\dot{\theta}θ˙ in terms of p r p r p_(r)p_rpr and p θ p θ p_( theta)p_\thetapθ:
r ˙ = p r m , θ ˙ = p θ m r 2 r ˙ = p r m , θ ˙ = p θ m r 2 r^(˙)=(p_(r))/(m),quadtheta^(˙)=(p_( theta))/(mr^(2))\dot{r} = \frac{p_r}{m}, \quad \dot{\theta} = \frac{p_\theta}{m r^2}r˙=prm,θ˙=pθmr2
Substitute these into H H HHH:
H = p r ( p r m ) + p θ ( p θ m r 2 ) − [ 1 2 m ( ( p r m ) 2 + r 2 ( p θ m r 2 ) 2 ) − G M m ( 1 2 a − 1 r ) ] H = p r p r m + p θ p θ m r 2 − 1 2 m p r m 2 + r 2 p θ m r 2 2 − G M m 1 2 a − 1 r H=p_(r)((p_(r))/(m))+p_( theta)((p_( theta))/(mr^(2)))-[(1)/(2)m(((p_(r))/(m))^(2)+r^(2)((p_( theta))/(mr^(2)))^(2))-GMm((1)/(2a)-(1)/(r))]H = p_r \left( \frac{p_r}{m} \right) + p_\theta \left( \frac{p_\theta}{m r^2} \right) – \left[ \frac{1}{2} m \left( \left( \frac{p_r}{m} \right)^2 + r^2 \left( \frac{p_\theta}{m r^2} \right)^2 \right) – G M m \left( \frac{1}{2a} – \frac{1}{r} \right) \right]H=pr(prm)+pθ(pθmr2)−[12m((prm)2+r2(pθmr2)2)−GMm(12a−1r)]
Simplify:
H = p r 2 m + p θ 2 m r 2 − [ p r 2 2 m + p θ 2 2 m r 2 − G M m ( 1 2 a − 1 r ) ] H = p r 2 m + p θ 2 m r 2 − p r 2 2 m + p θ 2 2 m r 2 − G M m 1 2 a − 1 r H=(p_(r)^(2))/(m)+(p_( theta)^(2))/(mr^(2))-[(p_(r)^(2))/(2m)+(p_( theta)^(2))/(2mr^(2))-GMm((1)/(2a)-(1)/(r))]H = \frac{p_r^2}{m} + \frac{p_\theta^2}{m r^2} – \left[ \frac{p_r^2}{2m} + \frac{p_\theta^2}{2m r^2} – G M m \left( \frac{1}{2a} – \frac{1}{r} \right) \right]H=pr2m+pθ2mr2−[pr22m+pθ22mr2−GMm(12a−1r)]
H = p r 2 2 m + p θ 2 2 m r 2 + G M m