Solution:

Step 1: Define the Centre of the Group

The group G is the General Linear Group, GL(2, \mathbb{R}). The centre of a group G, denoted as Z(G), is the set of all elements in G that commute with every element of G.

Let an arbitrary matrix A in the centre be: A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. Since A \in G, its determinant ad - bc \neq 0.

Step 2: Test with an upper triangular elementary matrix

Since A must commute with all matrices in G, we can choose specific matrices to test. Let’s choose X_1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. Notice that \det(X_1) = 1 \neq 0, so X_1 \in G.

We must have AX_1 = X_1A.

Calculate AX_1:
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & a+b \\ c & c+d \end{pmatrix}

Calculate X_1A:
\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ c & d \end{pmatrix}

Equating the corresponding elements of the two resulting matrices:

• a = a + c \implies c = 0
• c + d = d \implies c = 0
• a + b = b + d \implies a = d

Substituting c = 0 and d = a back into A, we know A must be of the form: \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}.

Step 3: Test with a lower triangular elementary matrix

Now, let’s test A with another matrix X_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}. Since \det(X_2) = 1 \neq 0, X_2 \in G.

We must have AX_2 = X_2A.

Calculate AX_2 using our updated A:
\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} a+b & b \\ a & a \end{pmatrix}

Calculate X_2A:
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} = \begin{pmatrix} a & b \\ a & a+b \end{pmatrix}

Equating the (1,1) elements of the resulting matrices:

Step 4: Finalize the form of the matrix

Substituting b = 0 into our matrix, we get A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} = aI, where I is the identity matrix.

For A to remain in G (meaning it must be non-singular), its determinant a^2 must not be zero. Thus, a can be any non-zero real number (a \in \mathbb{R} \setminus \{0\}).

Conversely, it is a known property that scalar matrices commute with all matrices of the same order.

Final Answer:

The centre of the group G consists of all non-zero scalar matrices:

Solution:

Step 1: Set up Charpit’s Auxiliary Equations

Let the given partial differential equation be denoted by f(x, y, z, p, q) = 0:

We first calculate the partial derivatives of f with respect to all five variables:

• f_x = p - q^2
• f_y = q
• f_z = 1
• f_p = x
• f_q = y - 2qx

Charpit’s auxiliary equations are given by:

Substituting our partial derivatives into the fractions for dx and dq:

Step 2: Solve the selected auxiliary equation for q

We can easily integrate this relation between x and q:

Integrating both sides:

(where a is an arbitrary constant).

Step 3: Find p from the original PDE

Substitute q = \frac{a}{x^2} back into the original differential equation px + qy + z = xq^2:

Solve for p:

Step 4: Integrate the total differential dz

The total derivative of z is dz = p\,dx + q\,dy. Substitute our expressions for p and q:

Bring the z term to the left side:

Multiply the entire equation by x to create exact differentials:

Recognize the exact differentials on both sides (d(xz) = x\,dz + z\,dx and d(y/x) = \frac{x\,dy - y\,dx}{x^2}):

Integrate directly:

Final Answer:

Dividing by x, the complete integral of the partial differential equation is:

(where a and b are arbitrary constants).

Solution:

To prove that the sequence converges to 3/2, we will use the Monotone Convergence Theorem.
This requires us to show that the sequence is monotonically increasing and bounded above.

Step 1: Show that the sequence is monotonically increasing

We need to prove that y_{n} < y_{n+1} for all n \ge 1. We will use mathematical induction.

Base Case (n=1):
We are given y_1 = 1.
Calculate y_2:
y_2 = \frac{2y_1 + 3}{4} = \frac{2(1) + 3}{4} = \frac{5}{4}
Since 1 < 5/4, we have y_1 < y_2. The base case holds.

Inductive Step:
Assume the statement is true for some arbitrary positive integer k, that is, y_k < y_{k+1}.
We must show it holds for k+1 (i.e., y_{k+1} < y_{k+2}).

From our assumption: y_k < y_{k+1}
Multiply by 2: 2y_k < 2y_{k+1}
Add 3: 2y_k + 3 < 2y_{k+1} + 3
Divide by 4:
\frac{2y_k + 3}{4} < \frac{2y_{k+1} + 3}{4}

By the inductive definition of the sequence, the left side is y_{k+1} and the right side is y_{k+2}.
Therefore, y_{k+1} < y_{k+2}.

By the Principle of Mathematical Induction, y_n < y_{n+1} for all n \ge 1.
The sequence is monotonically strictly increasing.

Step 2: Show that the sequence is bounded above

We will guess the upper bound is the expected limit, 3/2, and prove y_n < 3/2 for all n \ge 1 using induction.

Base Case (n=1):
y_1 = 1 < 3/2. The base case holds.

Inductive Step:
Assume y_k < 3/2 for some integer k \ge 1.
We must show y_{k+1} < 3/2.

From our assumption: y_k < \frac{3}{2}
Multiply by 2: 2y_k < 3
Add 3: 2y_k + 3 < 6
Divide by 4:
\frac{2y_k + 3}{4} < \frac{6}{4}

Since the left side is y_{k+1} and 6/4 = 3/2, we have y_{k+1} < 3/2.

By induction, y_n < 3/2 for all n \ge 1.
The sequence is bounded above.

Step 3: Evaluate the Limit

Result Used: The Monotone Convergence Theorem states that every bounded monotonic sequence of real numbers is convergent.
Since (y_n) is monotonically increasing and bounded above, it converges to some limit L.

Let \lim_{n\rightarrow\infty} y_n = L.
Since the sequence converges, the shifted sequence (y_{n+1}) also converges to the same limit L.

Taking the limit as n \to \infty on both sides of the recursive relation:
\lim_{n\rightarrow\infty} y_{n+1} = \lim_{n\rightarrow\infty} \frac{2y_n + 3}{4}
L = \frac{2L + 3}{4}

Now, solve for L:
4L = 2L + 3
2L = 3 \implies L = \frac{3}{2}

Final Answer:

Hence, it is shown that \lim_{n\rightarrow\infty} Y = \frac{3}{2}.

Solution:

Part 1: Prove f(z) = 1/z^2 is NOT uniformly continuous in |z| < 1

By definition, a function is uniformly continuous on a set if for every \epsilon > 0,
there exists a \delta > 0 such that for all z_1, z_2 in the set,
|z_1 - z_2| < \delta \implies |f(z_1) - f(z_2)| < \epsilon.

To show a function is not uniformly continuous, we can find two sequences z_n and z'_n in the region
such that |z_n - z'_n| \to 0 as n \to \infty,
but |f(z_n) - f(z'_n)| does not approach 0 (it remains bounded away from 0 or diverges).

Let us choose two real sequences within the region |z| < 1:
z_n = \frac{1}{n} \quad \text{and} \quad z'_n = \frac{1}{n+1}
For n \ge 2, both sequences lie strictly inside the unit disk |z| < 1.

Calculate the distance between the points:
|z_n - z'_n| = \left| \frac{1}{n} - \frac{1}{n+1} \right| = \frac{1}{n(n+1)}
As n \to \infty, \frac{1}{n(n+1)} \to 0.
This means we can make the points arbitrarily close to each other.

Now, calculate the distance between their function values:
|f(z_n) - f(z'_n)| = \left| \frac{1}{(1/n)^2} - \frac{1}{(1/(n+1))^2} \right| = |n^2 - (n+1)^2|
|f(z_n) - f(z'_n)| = |n^2 - (n^2 + 2n + 1)| = |-2n - 1| = 2n + 1

As n \to \infty, |f(z_n) - f(z'_n)| = 2n + 1 \to \infty.

Therefore, for any given \delta > 0, we can choose a large enough n
such that |z_n - z'_n| < \delta, yet |f(z_n) - f(z'_n)| will be greater than any \epsilon.
Thus, f(z) is not uniformly continuous in |z| < 1.

Part 2: Prove f(z) = 1/z^2 IS uniformly continuous in 1/2 \le |z| \le 1

Let S be the closed annulus defined by 1/2 \le |z| \le 1.

Method 1 (Theoretical):
The region S is a closed and bounded subset of the complex plane, which makes it a compact set.
The function f(z) = 1/z^2 is continuous everywhere except at z=0.
Since 0 \notin S, f(z) is continuous on S.
By the Heine-Cantor Theorem, any continuous function on a compact metric space is uniformly continuous.
Hence, f(z) is uniformly continuous on S.

Method 2 (Direct Proof using Lipschitz Condition):
Let z_1, z_2 \in S. Let’s evaluate |f(z_1) - f(z_2)|:
|f(z_1) - f(z_2)| = \left| \frac{1}{z_1^2} - \frac{1}{z_2^2} \right| = \left| \frac{z_2^2 - z_1^2}{z_1^2 z_2^2} \right| = \frac{|z_2 - z_1||z_2 + z_1|}{|z_1|^2 |z_2|^2}

Since both points are in S, we know 1/2 \le |z| \le 1. We can establish bounds for the terms:

• Upper bound for numerator factor: By the triangle inequality,
  |z_1 + z_2| \le |z_1| + |z_2| \le 1 + 1 = 2.
• Lower bound for denominator: Since |z| \ge 1/2,
  we have |z_1|^2 \ge 1/4 and |z_2|^2 \ge 1/4.
  Thus, |z_1|^2 |z_2|^2 \ge (1/4)(1/4) = 1/16.

Substitute these bounds back into the equation:
|f(z_1) - f(z_2)| \le \frac{2}{1/16} |z_1 - z_2| = 32 |z_1 - z_2|

Given any \epsilon > 0, we can choose \delta = \epsilon / 32.
Then for all z_1, z_2 \in S satisfying |z_1 - z_2| < \delta, we have:
|f(z_1) - f(z_2)| \le 32|z_1 - z_2| < 32\left(\frac{\epsilon}{32}\right) = \epsilon
This strictly satisfies the definition of uniform continuity.

Final Conclusion:
The function is not uniformly continuous in the open disk because it diverges near the singularity at zero,
but it is uniformly continuous on the closed annulus because the domain is strictly bounded away from that singularity.

Solution:

Step 1: Write the Augmented Matrix

The system of linear equations can be represented as an augmented matrix [A|B]:

Step 2: Forward Elimination (Row Operations)

To simplify our calculations and avoid fractions early on, let’s swap Row 1 (R_1) and Row 2 (R_2) so that the pivot element in the first row is 1.

Operation: R_1 \leftrightarrow R_2

Now, eliminate the entries below the first pivot (the x_1 column).

Operations: R_2 \to R_2 - 2R_1 and R_3 \to R_3 - 2R_1

• New R_2: (2-2, 1 - 2(-1), 1 - 2(2) \mid 4 - 2(2)) = (0, 3, -3 \mid 0)
• New R_3: (2-2, 2 - 2(-1), -1 - 2(2) \mid 3 - 2(2)) = (0, 4, -5 \mid -1)

Simplify Row 2 by dividing it by 3.

Operation: R_2 \to \frac{1}{3}R_2

Eliminate the entry below the second pivot (the x_2 column).

Operation: R_3 \to R_3 - 4R_2

• New R_3: (0-0, 4-4, -5 - 4(-1) \mid -1 - 4(0)) = (0, 0, -1 \mid -1)

The matrix is now in Row Echelon Form (upper triangular form).

Step 3: Back Substitution

Convert the augmented matrix back into algebraic equations:

1. -x_3 = -1
2. x_2 - x_3 = 0
3. x_1 - x_2 + 2x_3 = 2

Solve from the bottom up:

From Equation (1):
x_3 = 1

Substitute x_3 = 1 into Equation (2):
x_2 - 1 = 0 \implies x_2 = 1

Substitute x_2 = 1 and x_3 = 1 into Equation (3):
x_1 - 1 + 2(1) = 2
x_1 + 1 = 2 \implies x_1 = 1

Final Answer:

The solution to the system of equations by the Gauss elimination method is:
x_1 = 1, x_2 = 1, x_3 = 1.

Solution:

Let G be a group of order p^2, where p is a prime number. Let e be the identity element of G.

Step 1: Possible orders of elements

By Lagrange’s Theorem, the order of any element g \in G must divide the order of the group, |G| = p^2.
Since p is prime, the only possible orders for the elements in G are 1, p, or p^2.
Only the identity element e has an order of 1.

Step 2: Case 1 – The group contains an element of order p^2

Suppose there exists an element a \in G such that the order of a is p^2 (i.e., |a| = p^2).
This means the cyclic subgroup generated by a, denoted \langle a \rangle, contains p^2 distinct elements.
Since |\langle a \rangle| = p^2 = |G|, it must be that G = \langle a \rangle.
Therefore, in this case, G is a cyclic group of order p^2 (isomorphic to \mathbb{Z}_{p^2}).

Step 3: Case 2 – The group does not contain an element of order p^2

If no element has order p^2, then every non-identity element in G must have an order exactly equal to p.

Sub-step 3.1: Show that G must be Abelian
It is a known theorem that the center Z(G) of a p-group is non-trivial (meaning |Z(G)| > 1).
Thus, the order of Z(G) can be p or p^2.
If |Z(G)| = p^2, then Z(G) = G, which means G is abelian.
If |Z(G)| = p, then the quotient group G/Z(G) has order p^2 / p = p.
Since any group of prime order is cyclic, G/Z(G) is cyclic.
However, if G/Z(G) is cyclic, a fundamental group theory result states that G must be abelian. If G is abelian, its center is the whole group (|Z(G)| = p^2), which contradicts |Z(G)| = p.
Therefore, |Z(G)| must be p^2, and G is an abelian group.

Sub-step 3.2: Construct the product of two cyclic groups
Since all non-identity elements have order p, choose any element x \in G where x \neq e.
Let H = \langle x \rangle. Then H is a cyclic subgroup of order p.

Now, choose another element y \in G such that y \notin H.
Let K = \langle y \rangle. Then K is also a cyclic subgroup of order p.

Consider the intersection H \cap K. By Lagrange’s theorem, the order of H \cap K must divide |H| = p.
So |H \cap K| is either 1 or p. Since y \notin H, H and K are not the same subgroup, so their intersection cannot have order p.
Thus, |H \cap K| = 1 \implies H \cap K = \{e\}.

Since G is abelian, both H and K are normal subgroups of G.
We can form the internal direct product HK. The order of this product is:
|HK| = \frac{|H| \cdot |K|}{|H \cap K|} = \frac{p \cdot p}{1} = p^2

Since |HK| = p^2 = |G|, it follows that G = HK.

Final Answer:

Because G = HK, H and K are normal, and H \cap K = \{e\}, G is the internal direct product of H and K.
Therefore, if G is not cyclic, it is the product of two cyclic groups of order p (isomorphic to \mathbb{Z}_p \times \mathbb{Z}_p).

Solution:

This statement is precisely the Heine-Borel Theorem for \mathbb{R}^n.
Definition Used: A subset K is compact if every open cover of K has a finite subcover. (i.e., If K \subset \bigcup_{\alpha} U_\alpha for open sets U_\alpha, then there exist finitely many indices \alpha_1, \dots, \alpha_k such that K \subset \bigcup_{i=1}^k U_{\alpha_i}).

Because this is an “if and only if” proof, we must prove both directions.

Part 1: Forward Proof (Assume K is compact \implies K is closed and bounded)

Step 1.1: Prove K is bounded.
Consider the collection of all open balls centered at the origin 0 with integer radii m:
\mathcal{C} = \{ B_m(0) \mid m \in \mathbb{N} \}
Since the union of all such balls is the entire space \mathbb{R}^n, the collection \mathcal{C} forms an open cover of K.
Because K is compact, this open cover must have a finite subcover.
Let the finite subcover be B_{m_1}(0), B_{m_2}(0), \dots, B_{m_k}(0).
Let M = \max(m_1, m_2, \dots, m_k).
Then, K \subset B_M(0). Since K is contained within a ball of finite radius M, K is bounded.

Step 1.2: Prove K is closed.
To prove K is closed, we will show its complement K^c = \mathbb{R}^n \setminus K is open.
Let y \in K^c. We must find an open ball around y that does not intersect K.
For every x \in K, the distance |x - y| > 0. Let r_x = \frac{|x - y|}{2}.
The open balls B_{r_x}(x) for all x \in K form an open cover of K.
Since K is compact, there exists a finite subcover: B_{r_{x_1}}(x_1), \dots, B_{r_{x_m}}(x_m).
Now, consider the corresponding open balls around y: B_{r_{x_i}}(y).
Let R = \min(r_{x_1}, \dots, r_{x_m}). Since there are finitely many positive radii, R > 0.
The ball B_R(y) is completely disjoint from every B_{r_{x_i}}(x_i), and therefore disjoint from K.
This implies B_R(y) \subset K^c. Thus, y is an interior point, making K^c open.
Because its complement is open, K is closed.

Part 2: Reverse Proof (Assume K is closed and bounded \implies K is compact)

Step 2.1: Enclose K in an n-cell.
Since K is bounded, there exists some real number M > 0 such that for all x = (x_1, \dots, x_n) \in K, we have |x_i| \le M for all coordinates i.
Therefore, K is a subset of the closed n-cell (hypercube):
I = [-M, M] \times [-M, M] \times \dots \times [-M, M]

Step 2.2: Apply Standard Topological Lemmas.
To complete the proof elegantly, we invoke two foundational results in real analysis:
Result 1: Every closed n-cell in \mathbb{R}^n is compact. (This is traditionally proved using Cantor’s Intersection Theorem or the Nested Interval Property by recursively subdividing the cell).
Therefore, I is a compact set.

Result 2: A closed subset of a compact set is compact.
Brief proof of Result 2: If F is closed and C is compact, let \{U_\alpha\} be an open cover of F. Then \{U_\alpha\} \cup \{F^c\} is an open cover of C. Since C is compact, this has a finite subcover. Removing F^c (if present) from this finite subcover leaves a finite collection of U_\alpha that covers F.

Step 2.3: Conclusion.
We are given that K is closed. We established that K is a subset of the compact n-cell I.
By Result 2, since K is a closed subset of the compact set I, K itself must be compact.

Final Conclusion:

We have shown that if K is compact, it is necessarily closed and bounded. Conversely, if K is closed and bounded, it is necessarily compact. Thus, the statement is proven.

Solution:

Definition: A group G is called simple if its only normal subgroups are the trivial subgroup \{e\} and G itself.
Let N be a normal subgroup of A_n such that N \neq \{e\}. We must prove that N = A_n.

Step 1: Foundational Lemmas regarding 3-cycles

We rely on two well-known results in permutation groups:

1. Lemma 1: For n \ge 3, the alternating group A_n is generated by 3-cycles. (Any even permutation can be written as a product of 3-cycles).
2. Lemma 2: For n \ge 5, all 3-cycles are conjugate within A_n.
   Consequence: If a normal subgroup N \triangleleft A_n contains even a single 3-cycle, it must contain all of its conjugates. Since all 3-cycles are conjugate, N contains all 3-cycles. By Lemma 1, since it contains all generators of A_n, N = A_n.

Goal: We only need to prove that N contains at least one 3-cycle.

Step 2: Case Analysis to isolate a 3-cycle

Since N \neq \{e\}, choose a non-identity permutation \sigma \in N. Decompose \sigma into disjoint cycles.
Since N is normal, for any \tau \in A_n, the commutator \tau \sigma \tau^{-1} \sigma^{-1} must belong to N.
We analyze the possible cycle structures of \sigma:

Case 1: \sigma contains a cycle of length \ge 4.
Let \sigma = (1\ 2\ 3\ 4 \dots) \mu, where \mu is disjoint from (1\ 2\ 3\ 4).
Choose \tau = (1\ 2\ 3) \in A_n.
Then \tau \sigma \tau^{-1} = (2\ 3\ 1\ 4 \dots) \mu.
Calculate the commutator: \sigma^{-1} (\tau \sigma \tau^{-1}) \in N.
\sigma^{-1} (\tau \sigma \tau^{-1}) = (\dots 4\ 3\ 2\ 1) \mu^{-1} (2\ 3\ 1\ 4 \dots) \mu = (1\ 2\ 4)
We have isolated a 3-cycle (1\ 2\ 4) \in N.

Case 2: \sigma consists of 3-cycles and disjoint 2-cycles, with at least two 3-cycles.
Let \sigma = (1\ 2\ 3)(4\ 5\ 6) \mu.
Choose \tau = (1\ 2\ 4) \in A_n.
Then \tau \sigma \tau^{-1} = (2\ 4\ 3)(1\ 5\ 6) \mu.
The commutator \sigma^{-1} (\tau \sigma \tau^{-1}) \in N simplifies to:
(1\ 3\ 2)(4\ 6\ 5)(2\ 4\ 3)(1\ 5\ 6) = (1\ 2\ 5\ 3\ 4)
Now N contains a 5-cycle, which reduces back to Case 1.

Case 3: \sigma contains exactly one 3-cycle.
Let \sigma = (1\ 2\ 3) \mu, where \mu is a product of disjoint 2-cycles.
Since disjoint cycles commute, and 2-cycles squared yield the identity:
\sigma^2 = (1\ 2\ 3)^2 \mu^2 = (1\ 3\ 2) \cdot e = (1\ 3\ 2)
Since \sigma \in N and N is a subgroup, \sigma^2 \in N. Thus, N contains the 3-cycle (1\ 3\ 2).

Case 4: \sigma is a product of disjoint 2-cycles.
Since \sigma \in A_n (even permutation), it must contain at least two disjoint 2-cycles.
Let \sigma = (1\ 2)(3\ 4) \mu.
Choose \tau = (1\ 2\ 3) \in A_n.
Then \tau \sigma \tau^{-1} = (2\ 3)(1\ 4) \mu.
The element \alpha = \sigma^{-1} (\tau \sigma \tau^{-1}) \in N becomes:
\alpha = (1\ 2)(3\ 4) (2\ 3)(1\ 4) = (1\ 3)(2\ 4)
Now, since n \ge 5, there is a fifth element, say 5.
Choose \delta = (1\ 3\ 5) \in A_n.
Compute the commutator of \delta with \alpha:
\alpha^{-1} (\delta \alpha \delta^{-1}) = (1\ 3)(2\ 4) \cdot (3\ 5)(2\ 4) = (1\ 3\ 5)
We have successfully produced a 3-cycle (1\ 3\ 5) \in N.

Step 3: Conclusion

In every possible case for the cycle structure of a non-identity element in N, we have demonstrated that N must contain a 3-cycle.
By Lemma 2, containing one 3-cycle means N contains all 3-cycles.
By Lemma 1, containing all 3-cycles implies N = A_n.

Final Answer:

Since the only non-trivial normal subgroup of A_n is A_n itself, it follows that for all n \ge 5, A_n is a simple group.

Solution:

Step 1: State the Estimation Lemma (ML-Inequality)

Result Used: If f(z) is a continuous complex-valued function on a contour C, then the absolute value of the integral is bounded by:
\left| \int_C f(z) dz \right| \le M \cdot L
where M is the maximum value of |f(z)| on the path C, and L is the arc length of the path C.

Step 2: Calculate the Length of the Path (L)

The path C is a straight line segment connecting the origin (0,0) to the point (1, 2\sqrt{2}).
We can use the standard Euclidean distance formula to find the length L:

So, the length of the contour is L = 3.

Step 3: Calculate the Maximum Modulus (M)

The function being integrated is f(z) = e^z. We need to find the maximum value of |e^z| along the path C.
Let z = x + iy. We can evaluate the modulus of e^z:

Since |e^{iy}| = |\cos(y) + i\sin(y)| = 1 for any real y, this simplifies to:

Now, we look at the constraints of the path C. The line segment starts at x = 0 and ends at x = 1. Therefore, along the path C, the real part x is bounded by 0 \le x \le 1.

Because the real exponential function e^x is strictly increasing, it reaches its maximum on the interval [0, 1] at the upper endpoint, x = 1.

So, the maximum modulus is M = e.

Step 4: Compute the Upper Bound

Now, apply the values of M and L to the ML-inequality:

Final Answer:

The upper bound for the value of the integral \int_{C}e^{z}dz is 3e.

Solution:

Step 1: Define the partial sums of the original series

Let the given convergent series be \sum_{n=1}^{\infty} x_n.
Let S_n denote the n-th partial sum of this series:
S_n = x_1 + x_2 + x_3 + \dots + x_n
Because the series is given as convergent to some value, say S, the sequence of its partial sums (S_n) must converge to S.
Mathematically: \lim_{n \to \infty} S_n = S

Step 2: Define the new series obtained by grouping terms

Grouping the terms without changing their order means creating a new series \sum_{k=1}^{\infty} y_k, where each y_k consists of a finite sum of consecutive terms of the original series.
Let n_1 < n_2 < n_3 < \dots < n_k < \dots be a strictly increasing sequence of natural numbers.
We define the grouped terms as follows:

• y_1 = x_1 + x_2 + \dots + x_{n_1}
• y_2 = x_{n_1+1} + x_{n_1+2} + \dots + x_{n_2}
• y_3 = x_{n_2+1} + x_{n_2+2} + \dots + x_{n_3}
• …
• y_k = x_{n_{k-1}+1} + \dots + x_{n_k}

Step 3: Relate the partial sums of the new series to the original

Let T_k denote the k-th partial sum of the new grouped series \sum y_k. We evaluate the first few terms of T_k:

• T_1 = y_1 = S_{n_1}
• T_2 = y_1 + y_2 = S_{n_1} + (x_{n_1+1} + \dots + x_{n_2}) = S_{n_2}
• T_k = y_1 + y_2 + \dots + y_k = S_{n_k}

This shows a crucial relationship: The sequence of partial sums of the new series, (T_k), is exactly the sequence (S_{n_k}).
Since n_1 < n_2 < n_3 \dots is strictly increasing, (S_{n_k}) is a subsequence of the original sequence of partial sums (S_n).

Step 4: Invoke the Subsequence Theorem

Result Used: A fundamental theorem in real analysis states that if a sequence converges to a limit S, then every subsequence of that sequence must also converge to the exact same limit S.

Since (S_n) \to S, its subsequence (S_{n_k}) must also converge to S.
Because T_k = S_{n_k}, it immediately follows that:
\lim_{k \to \infty} T_k = S

Final Conclusion:

Because the sequence of partial sums T_k of the grouped series converges to S, the grouped series \sum y_k is convergent.
Furthermore, it converges to S, which is the exact same value as the original series \sum x_n.

Solution:

Example 1: A Pole of Order 2 (Simplest Example)

Consider the function:

1. Verify the singularity is non-removable:
A singularity at z=0 is removable if \lim_{z \to 0} f(z) exists and is finite. For f(z) = 1/z^2, as z \to 0, |f(z)| \to \infty. Therefore, it is a pole of order 2, which is a non-removable singularity.

2. Verify the residue is zero:
The residue of a function at a singularity z=z_0 is the coefficient a_{-1} of the \frac{1}{z-z_0} term in its Laurent series expansion.
The Laurent series expansion of f(z) = 1/z^2 around z=0 is already in its final form:
f(z) = \frac{1}{z^2} + 0 \cdot \frac{1}{z} + 0 + 0 \cdot z + \dots
Because the coefficient of the \frac{1}{z} term is exactly 0, the residue is:
\text{Res}(f, 0) = a_{-1} = 0

Example 2: An Essential Singularity (Alternative Example)

If we want an example with a more severe non-removable singularity (an essential singularity), consider:

1. Verify the singularity is non-removable:
There is no integer n such that \lim_{z \to 0} z^n e^{1/z^2} is finite. The function oscillates wildly and diverges near zero, making it an essential (non-removable) singularity.

2. Verify the residue is zero:
We find the Laurent series by substituting u = 1/z^2 into the Maclaurin series for e^u:
e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots
g(z) = e^{\frac{1}{z^2}} = 1 + \frac{1}{z^2} + \frac{1}{2! z^4} + \frac{1}{3! z^6} + \dots
Looking at this expansion, there are only even powers of 1/z. The term \frac{1}{z} does not appear, meaning its coefficient a_{-1} is zero.
Therefore:
\text{Res}(g, 0) = 0

Final Answer:

A simple and valid example is f(z) = \frac{1}{z^2}. It has a non-removable singularity (a pole of order 2) at z=0, and its residue at that point is 0.

Solution:

Step 1: Define the metric and the Cauchy sequence

To discuss completeness, we must establish the metric on C[0,1]. The standard metric for this space is the supremum metric (or uniform metric), defined as:
d(f, g) = \sup_{x \in [0,1]} |f(x) - g(x)|
A metric space is complete if every Cauchy sequence in the space converges to a point that is also in the space.

Let (f_n) be an arbitrary Cauchy sequence in C[0,1]. By definition, for every \epsilon > 0, there exists an integer N such that for all m, n \ge N:
d(f_n, f_m) = \sup_{x \in [0,1]} |f_n(x) - f_m(x)| < \epsilon \quad \dots \text{(Equation 1)}

Step 2: Establish Pointwise Convergence

Fix any point x \in [0,1]. From Equation 1, it follows that for all m, n \ge N:
|f_n(x) - f_m(x)| \le \sup_{x \in [0,1]} |f_n(x) - f_m(x)| < \epsilon
This implies that for each fixed x, the sequence of real numbers (f_n(x)) is a Cauchy sequence in \mathbb{R}.

Result Used: The set of real numbers \mathbb{R} is a complete metric space. Therefore, every Cauchy sequence in \mathbb{R} must converge to a real number.

Let the limit of this sequence be denoted by f(x). This defines a new function f:[0,1] \rightarrow \mathbb{R} such that:
\lim_{n \to \infty} f_n(x) = f(x) \quad \text{for all } x \in [0,1]
This proves that f_n converges pointwise to f.

Step 3: Upgrade to Uniform Convergence

We must show that (f_n) converges to f under our specific supremum metric (i.e., uniformly).
Return to the Cauchy condition: For a given \epsilon > 0, choose N such that for all m, n \ge N and all x \in [0,1]:
|f_n(x) - f_m(x)| < \epsilon
Hold n fixed (where n \ge N) and let m \to \infty. Since f_m(x) \to f(x) and the absolute value function is continuous, we get:
\lim_{m \to \infty} |f_n(x) - f_m(x)| = |f_n(x) - f(x)| \le \epsilon
Notice that this choice of N depends only on \epsilon, not on x. Because |f_n(x) - f(x)| \le \epsilon is true for all x \in [0,1], we can take the supremum over [0,1]:
\sup_{x \in [0,1]} |f_n(x) - f(x)| \le \epsilon \implies d(f_n, f) \le \epsilon
This proves that the sequence of functions (f_n) converges uniformly to the limit function f.

Step 4: Prove the limit function is in C[0,1]

To conclude completeness, the limit function f must belong to the original space C[0,1]. This means f must be continuous on [0,1].

Result Used (Uniform Limit Theorem): The uniform limit of a sequence of continuous functions is continuous.

Since each f_n is continuous on [0,1] (because f_n \in C[0,1]), and (f_n) converges uniformly to f, it strictly follows that f is also continuous on [0,1].
Therefore, f \in C[0,1].

Final Conclusion:

We started with an arbitrary Cauchy sequence in C[0,1] and rigorously proved that it converges (in the supremum metric) to a limit function f which is also in C[0,1]. Therefore, C[0,1] is a complete metric space.

Solution:

Part 1: Determine the points where the mapping is not conformal

Result Used: A complex mapping w = f(z) is conformal at all points z where the function is analytic and its derivative is non-zero (i.e., f'(z) \neq 0). It fails to be conformal at critical points where f'(z) = 0 and at singular points where it is not analytic.

Given the mapping: w = f(z) = z + \frac{1}{z}
First, note that f(z) has a singularity at z = 0, so it is not conformal there.

Next, find the derivative f'(z):
f'(z) = 1 - \frac{1}{z^2}

Set the derivative to zero to find the critical points:
1 - \frac{1}{z^2} = 0 \implies z^2 = 1 \implies z = \pm 1

Therefore, the mapping is not conformal at the points z = 1, z = -1, and z = 0.

Part 2: Demonstrate using the image of the real axis

We need to show practically that conformality (angle preservation) fails at these critical points by mapping the real axis of the z-plane.

Let z lie on the real axis, so z = x + iy with y = 0. Thus, z = x (where x \neq 0).
Substitute this into the mapping w = u + iv:
w = x + \frac{1}{x}
Since x is purely real, w is also purely real. This means v = 0 and u = x + \frac{1}{x}.
So, the image of the real axis in the z-plane lies entirely on the real axis in the w-plane.

Demonstrating the failure of conformality at z = 1:
Conformality means the angle between any two curves intersecting at a point is preserved in magnitude and sense. Let’s look at what happens at z = 1.

Consider two segments of the real axis meeting at z = 1:

• Segment A: The path from z = 1 to z \to \infty (i.e., x > 1).
• Segment B: The path from z = 1 to z \to 0^+ (i.e., 0 < x < 1).

In the z-plane, the angle between Segment A and Segment B at the point z=1 is exactly 180^\circ (or \pi radians) because they lie on a straight line in opposite directions.

Now, let’s find their images in the w-plane (u = x + 1/x):

• Image of Segment A (x > 1): At x=1, u=2. As x \to \infty, u \to \infty. The image is the ray from u=2 moving to the right towards +\infty.
• Image of Segment B (0 < x < 1): At x=1, u=2. As x \to 0^+, the term 1/x \to \infty, so u \to \infty. The image is also the ray from u=2 moving to the right towards +\infty.

In the w-plane, the images of both segments overlap perfectly, originating from w=2 and pointing in the exact same direction. The angle between the two image curves at w=2 is 0^\circ.

Final Conclusion:

Since the angle between the two segments was \pi in the z-plane but became 0 in the w-plane, the mapping “folded” the real axis back onto itself at z=1 (and similarly at z=-1). Because the angle was not preserved, we have successfully demonstrated that the mapping is not conformal at z = 1.

Solution:

Step 1: Set up the partial differential equation for the orthogonal surface

First, isolate the parameter a to define the given system of surfaces f(x, y, z) = a:
f(x, y, z) = \frac{2xz + 3yz}{z + 2} = a

Result Used: A surface cutting this system orthogonally must satisfy the linear partial differential equation P p + Q q = R, where P = \frac{\partial f}{\partial x}, Q = \frac{\partial f}{\partial y}, and R = \frac{\partial f}{\partial z}.

Calculate the partial derivatives:

• P = \frac{\partial f}{\partial x} = \frac{2z}{z+2}
• Q = \frac{\partial f}{\partial y} = \frac{3z}{z+2}
• R = \frac{\partial f}{\partial z} = \frac{(z+2)(2x+3y) - (2xz+3yz)(1)}{(z+2)^2} = \frac{2xz+3yz+4x+6y-2xz-3yz}{(z+2)^2} = \frac{2(2x+3y)}{(z+2)^2}

Step 2: Formulate Lagrange’s Auxiliary Equations

The auxiliary equations are given by \frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}.
Substituting our derivatives:
\frac{dx}{\frac{2z}{z+2}} = \frac{dy}{\frac{3z}{z+2}} = \frac{dz}{\frac{2(2x+3y)}{(z+2)^2}}
Multiply all denominators by (z+2)^2 to simplify:
\frac{dx}{2z(z+2)} = \frac{dy}{3z(z+2)} = \frac{dz}{2(2x+3y)}

Step 3: Find the first independent integral

Take the first two fractions:
\frac{dx}{2z(z+2)} = \frac{dy}{3z(z+2)}
Cancel the common term z(z+2):
\frac{dx}{2} = \frac{dy}{3} \implies 3dx - 2dy = 0
Integrating gives our first constant of integration, c_1:
u = 3x - 2y = c_1

Step 4: Find the second independent integral

Using the properties of proportions, apply multipliers 2, 3, 0 to the fractions to construct a new ratio:
\frac{2dx + 3dy}{2[2z(z+2)] + 3[3z(z+2)]} = \frac{2dx + 3dy}{13z(z+2)}
Equate this new ratio to the third fraction \frac{dz}{R}:
\frac{d(2x+3y)}{13z(z+2)} = \frac{dz}{2(2x+3y)}
Cross-multiply to separate variables:
2(2x+3y)d(2x+3y) = 13(z^2+2z)dz
Integrate both sides:
(2x+3y)^2 = 13\left(\frac{z^3}{3} + z^2\right) + c_2
This gives our second constant of integration, c_2:
v = (2x+3y)^2 - 13\left(\frac{z^3}{3} + z^2\right) = c_2

Step 5: Apply the boundary condition to find the specific surface

The general orthogonal surface is given by \phi(u, v) = 0.
We are given that it passes through the circle z=0 and x^2+y^2=9.

At z=0, our constants become:
1. u = 3x - 2y
2. v = (2x+3y)^2 - 0 = (2x+3y)^2

We need to eliminate x and y to find the relationship between u and v.
Square u and add it to v:
u^2 + v = (3x-2y)^2 + (2x+3y)^2
u^2 + v = (9x^2 - 12xy + 4y^2) + (4x^2 + 12xy + 9y^2)
u^2 + v = 13x^2 + 13y^2 = 13(x^2 + y^2)

Since we are on the circle x^2+y^2=9, substitute this value:
u^2 + v = 13(9) = 117

Now substitute the full expressions for u and v (including z) back into this relationship:
(3x-2y)^2 + (2x+3y)^2 - 13\left(\frac{z^3}{3} + z^2\right) = 117
Expand and simplify the left side:
13(x^2+y^2) - 13\left(\frac{z^3}{3} + z^2\right) = 117
Divide the entire equation by 13:
x^2 + y^2 - \left(\frac{z^3}{3} + z^2\right) = 9

Final Answer:

Multiplying by 3 to clear the fraction, the equation of the required orthogonal surface is:
3x^2 + 3y^2 - z^3 - 3z^2 = 27

Solution:

Step 1: Apply the Laplace Transform to the given equation

Let Y(s) denote the Laplace transform of y(t), i.e., \mathcal{L}\{y(t)\} = Y(s).

Standard Results Used:

• Transform of a derivative: \mathcal{L}\{y^{\prime}(t)\} = sY(s) - y(0)
• Transform of an integral: \mathcal{L}\left\{\int_{0}^{t}y(\tau)d\tau\right\} = \frac{1}{s}Y(s)
• Transform of t: \mathcal{L}\{t\} = \frac{1}{s^2}

Taking the Laplace transform of both sides of the given integro-differential equation:

Step 2: Substitute the initial condition and solve for Y(s)

We are given the initial condition y(0) = 0. Substituting this into the equation:

Factor out Y(s):

Find a common denominator for the terms inside the parentheses:

Isolate Y(s) by multiplying both sides by \frac{s}{s^2 + 3s + 2}:

Factor the quadratic expression in the denominator (s^2 + 3s + 2 = (s+1)(s+2)):

Step 3: Perform Partial Fraction Decomposition

To find the inverse Laplace transform, we decompose Y(s) into partial fractions:

Multiply through by the common denominator s(s+1)(s+2):

We can solve for A, B, and C by substituting strategic values for s:

• Let s = 0: 1 = A(1)(2) + 0 + 0 \implies 2A = 1 \implies <strong>A = 1/2</strong>
• Let s = -1: 1 = 0 + B(-1)(1) + 0 \implies -B = 1 \implies <strong>B = -1</strong>
• Let s = -2: 1 = 0 + 0 + C(-2)(-1) \implies 2C = 1 \implies <strong>C = 1/2</strong>

Substitute these constants back into the partial fraction expansion:

Step 4: Take the Inverse Laplace Transform

Apply the inverse Laplace operator \mathcal{L}^{-1} to each term:

Using the standard inverse transform formulas (\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at} and \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1):

Final Answer:

The solution to the given initial value problem is:
y(t) = \frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t}

Solution:

Step 1: Identify the functional and apply the Beltrami Identity

The given functional is F(x, y, y^{\prime}) = y(y^{\prime})^3.

Result Used: We observe that the functional F does not explicitly depend on the independent variable x (i.e., \frac{\partial F}{\partial x} = 0). Therefore, the Euler-Lagrange equation reduces to the Beltrami identity:
F - y^{\prime} \frac{\partial F}{\partial y^{\prime}} = C
where C is a constant of integration.

Step 2: Compute the partial derivative and substitute

Calculate the partial derivative of F with respect to y^{\prime}:
\frac{\partial F}{\partial y^{\prime}} = \frac{\partial}{\partial y^{\prime}} [y(y^{\prime})^3] = 3y(y^{\prime})^2

Substitute F and \frac{\partial F}{\partial y^{\prime}} into the Beltrami identity:
y(y^{\prime})^3 - y^{\prime} [3y(y^{\prime})^2] = C
y(y^{\prime})^3 - 3y(y^{\prime})^3 = C
-2y(y^{\prime})^3 = C

Step 3: Solve the resulting differential equation

Rearrange the equation to isolate y^{\prime}:
(y^{\prime})^3 = -\frac{C}{2y}

Take the cube root of both sides. Let k = \left(-\frac{C}{2}\right)^{1/3}, which is just another arbitrary constant:
y^{\prime} = \frac{k}{y^{1/3}}

Write y^{\prime} as \frac{dy}{dx} and separate the variables:
y^{1/3} dy = k \, dx

Integrate both sides:
\int y^{1/3} dy = \int k \, dx
\frac{3}{4} y^{4/3} = kx + C_2

To simplify, multiply the entire equation by \frac{4}{3} and absorb the constants into new constants A and B:
y^{4/3} = Ax + B
This is the general equation for the extremal.

Step 4: Apply the boundary conditions

We are given two boundary conditions: y(0) = 0 and y(1) = 1.

Apply the first condition, y(0) = 0:
0^{4/3} = A(0) + B \implies B = 0

Substitute B = 0 into the general equation, giving y^{4/3} = Ax.
Now apply the second condition, y(1) = 1:
1^{4/3} = A(1) \implies A = 1

Step 5: Final Conclusion

Substitute A = 1 and B = 0 back into the general equation:
y^{4/3} = x

Raise both sides to the power of \frac{3}{4} to solve explicitly for y(x):
y = x^{3/4}

Final Answer:

The extremal of the given functional subject to the boundary conditions is:
y(x) = x^{3/4}

Solution:

Part 1: Algorithm for the Regula Falsi (False Position) Method

The Regula Falsi method is a bracketed root-finding algorithm that estimates the root by connecting the endpoints of the interval with a straight line (a secant line) and finding where it crosses the x-axis.

Step 1: Choose two initial guesses, a and b, such that the function changes sign over the interval [a, b]. That is, ensure f(a) \cdot f(b) < 0.

Step 2: Choose a tolerance limit \epsilon > 0 for the stopping criterion and set the maximum number of iterations N.

Step 3: Loop for i = 1, 2, \dots, N:
Calculate the next root approximation c using the formula based on similar triangles:
c = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}

Step 4: Check for convergence:
If |f(c)| < \epsilon or the interval |b - a| < \epsilon, then c is the required root. Stop the algorithm.

Step 5: Update the interval based on the Intermediate Value Theorem:
If f(a) \cdot f(c) < 0, the root lies in [a, c]. Update b = c.
Else if f(b) \cdot f(c) < 0, the root lies in [c, b]. Update a = c.

Step 6: End loop. If the loop completes N iterations without converging, output a failure message.

Part 2: C Program Implementation

Below is the C program to find the root of f(x) = 4e^{-x}\sin(x) - 1 = 0. By observation, f(0) = -1 and f(1) \approx 1.24, so the root lies in the interval [0, 1].

#include <stdio.h>
#include <math.h>

/* Define the given function f(x) */
double f(double x) {
    return 4.0 * exp(-x) * sin(x) - 1.0;
}

int main() {
    /* Initial guesses for the interval [0, 1] */
    double a = 0.0; 
    double b = 1.0; 
    double c;
    
    /* Tolerance for stopping criterion */
    double tol = 0.0001; 
    int max_iter = 100, iter = 1;

    /* Step 1 check */
    if (f(a) * f(b) >= 0) {
        printf("Invalid interval. f(a) and f(b) must have opposite signs.\n");
        return 1;
    }

    printf("Iter\t a\t\t b\t\t c\t\t f(c)\n");
    printf("----------------------------------------------------------------------\n");

    do {
        /* Step 3: Regula Falsi formula */
        c = (a * f(b) - b * f(a)) / (f(b) - f(a));
        
        printf("%d\t %lf\t %lf\t %lf\t %lf\n", iter, a, b, c, f(c));

        /* Step 4: Check if root is found within tolerance */
        if (fabs(f(c)) < tol) {
            break;
        }

        /* Step 5: Update the interval */
        if (f(a) * f(c) < 0) {
            b = c;
        } else {
            a = c;
        }
        
        iter++;
    } while (iter <= max_iter);

    printf("\nApproximated Root found: %lf\n", c);
    
    return 0;
}

Final Note: This program includes the standard math library <math.h> for the exp() and sin() functions, and fabs() for the absolute value check against the tolerance.

Solution:

Step 1: Calculate the first few values of S_n

Let S_n be the sum of the cubes of the first n natural numbers. We evaluate S_n for n = 1, 2, 3, 4, 5:

• S_1 = 1^3 = 1
• S_2 = 1^3 + 2^3 = 1 + 8 = 9
• S_3 = 9 + 3^3 = 9 + 27 = 36
• S_4 = 36 + 4^3 = 36 + 64 = 100
• S_5 = 100 + 5^3 = 100 + 125 = 225

Step 2: Construct the Forward Difference Table

We calculate the finite differences (\Delta, \Delta^2, \Delta^3, \Delta^4) for the sequence S_n.

Since the general term is a cubic polynomial being summed, S_n must be a polynomial of degree 4. Thus, the 4th forward differences are constant (=6), and all higher-order differences are zero. The leading forward differences from the first row are:
S_1 = 1, \Delta S_1 = 8, \Delta^2 S_1 = 19, \Delta^3 S_1 = 18, \Delta^4 S_1 = 6.

Step 3: Apply Newton’s Forward Difference Formula

Newton’s formula for a function evaluated at a + uh is:
f(a + uh) = f(a) + u\Delta f(a) + \frac{u(u-1)}{2!}\Delta^2 f(a) + \frac{u(u-1)(u-2)}{3!}\Delta^3 f(a) + \frac{u(u-1)(u-2)(u-3)}{4!}\Delta^4 f(a)

Here, our starting point is a = 1 and the step size is h = 1.
We want to find the formula for general n, so we set n = a + uh = 1 + u(1).
This means u = n - 1.

Substitute u = n - 1 and our leading differences into the formula:

Step 4: Simplify the expression

Let’s expand and simplify each term individually:

• Term 1: 1
• Term 2: 8n - 8
• Term 3: \frac{19}{2}(n^2 - 3n + 2) = \frac{19}{2}n^2 - \frac{57}{2}n + 19
• Term 4: 3(n^3 - 6n^2 + 11n - 6) = 3n^3 - 18n^2 + 33n - 18
• Term 5: \frac{1}{4}(n^4 - 10n^3 + 35n^2 - 50n + 24) = \frac{1}{4}n^4 - \frac{5}{2}n^3 + \frac{35}{4}n^2 - \frac{25}{2}n + 6

Now, collect the like terms by powers of n:

• Coefficient of n^4: \frac{1}{4}
• Coefficient of n^3: 3 - \frac{5}{2} = \frac{1}{2}
• Coefficient of n^2: \frac{19}{2} - 18 + \frac{35}{4} = \frac{38}{4} - \frac{72}{4} + \frac{35}{4} = \frac{1}{4}
• Coefficient of n: 8 - \frac{57}{2} + 33 - \frac{25}{2} = 41 - \frac{82}{2} = 41 - 41 = 0
• Constant term: 1 - 8 + 19 - 18 + 6 = 26 - 26 = 0

Adding them all together yields:

Factor out \frac{n^2}{4}:

Final Answer:

Using Newton’s forward difference formula, the value of the sum is:
S_{n} = \left[ \frac{n(n+1)}{2} \right]^2

Solution:

Step 1: Set up the Laplace Transform for Convolution

Result Used: The Laplace transform of the Bessel function of order zero, J_0(t), is well-known:
\mathcal{L}\{J_0(t)\} = \frac{1}{\sqrt{s^2+1}}

To use the convolution theorem, we must express this transform as the product of two simpler transforms. We factor the denominator over complex numbers:
\frac{1}{\sqrt{s^2+1}} = \frac{1}{\sqrt{(s-i)(s+i)}} = \frac{1}{\sqrt{s-i}} \cdot \frac{1}{\sqrt{s+i}}

Let F(s) = \frac{1}{\sqrt{s-i}} and G(s) = \frac{1}{\sqrt{s+i}}.

Step 2: Find the Inverse Laplace Transforms of the factors

Result Used: The standard inverse Laplace transform involving the square root is \mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s}}\right\} = \frac{1}{\sqrt{\pi t}}. By applying the First Shifting Theorem (\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)), we find:

• f(t) = \mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s-i}}\right\} = \frac{e^{it}}{\sqrt{\pi t}}
• g(t) = \mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s+i}}\right\} = \frac{e^{-it}}{\sqrt{\pi t}}

Step 3: Apply the Convolution Theorem

The convolution theorem states that \mathcal{L}^{-1}\{F(s)G(s)\} = \int_{0}^{t} f(\tau)g(t-\tau)d\tau. Applying this to our functions gives the inverse transform of \frac{1}{\sqrt{s^2+1}}, which is J_0(t):

Simplify the constants and the exponents:
e^{i\tau} \cdot e^{-it + i\tau} = e^{-it} e^{2i\tau}
J_0(t) = \frac{e^{-it}}{\pi} \int_{0}^{t} \frac{e^{2i\tau}}{\sqrt{\tau(t-\tau)}} d\tau

Step 4: Substitute variables to change the limits of integration

We need to change the limits from [0, t] to something symmetrical, like [-1, 1].
Let \tau = \frac{t}{2}(1 + x).

• When \tau = 0, x = -1.
• When \tau = t, x = 1.
• The differential is d\tau = \frac{t}{2} dx.

Now, evaluate the term inside the square root in the denominator:
\tau(t-\tau) = \frac{t}{2}(1+x) \left( t - \frac{t}{2}(1+x) \right) = \frac{t}{2}(1+x) \cdot \frac{t}{2}(1-x) = \frac{t^2}{4}(1-x^2)
Taking the square root gives \frac{t}{2}\sqrt{1-x^2}.

Substitute these into the integral:
J_0(t) = \frac{e^{-it}}{\pi} \int_{-1}^{1} \frac{e^{2i\left(\frac{t}{2}(1+x)\right)}}{\frac{t}{2}\sqrt{1-x^2}} \left( \frac{t}{2} \right) dx

The \frac{t}{2} terms cancel out nicely:
J_0(t) = \frac{e^{-it}}{\pi} \int_{-1}^{1} \frac{e^{it(1+x)}}{\sqrt{1-x^2}} dx

Step 5: Simplify using Euler’s Formula and Parity Properties

Expand the exponential in the numerator:
J_0(t) = \frac{e^{-it}}{\pi} \int_{-1}^{1} \frac{e^{it} \cdot e^{itx}}{\sqrt{1-x^2}} dx = \frac{1}{\pi} \int_{-1}^{1} \frac{e^{itx}}{\sqrt{1-x^2}} dx

Using Euler’s formula, e^{itx} = \cos(tx) + i\sin(tx), we split the integral into real and imaginary parts:
J_0(t) = \frac{1}{\pi} \int_{-1}^{1} \frac{\cos(tx)}{\sqrt{1-x^2}} dx + \frac{i}{\pi} \int_{-1}^{1} \frac{\sin(tx)}{\sqrt{1-x^2}} dx

Now, evaluate the parity (even/odd properties) of the integrands:

• The function \frac{\sin(tx)}{\sqrt{1-x^2}} is an odd function of x. Integrating an odd function over the symmetric interval [-1, 1] yields 0.
• The function \frac{\cos(tx)}{\sqrt{1-x^2}} is an even function of x. Integrating an even function over [-1, 1] is equal to twice the integral over [0, 1].

Applying these properties eliminates the imaginary part and doubles the real part:

Final Answer:

Hence, it is shown that:

Solution:

Step 1: Identify the function and parameters

The integrand is y = f(x) = \frac{1}{1+x}.
The limits of integration are a = 0 and b = 1.
The step size is given as h = 0.25.

We can find the number of sub-intervals n by the formula:
n = \frac{b - a}{h} = \frac{1 - 0}{0.25} = 4
Since n = 4 is an even number, we can validly apply Simpson’s 1/3 rule.

Step 2: Construct the table of values

We evaluate y_i = \frac{1}{1+x_i} at each node x_i = a + i \cdot h for i = 0, 1, 2, 3, 4:

Step 3: Apply Simpson’s 1/3 Rule

Result Used: The formula for Simpson’s 1/3 rule is:
\int_a^b f(x) dx \approx \frac{h}{3} \left[ (y_0 + y_n) + 4(y_1 + y_3 + \dots) + 2(y_2 + y_4 + \dots) \right]

Substitute the values from our table into the formula:
I \approx \frac{0.25}{3} \left[ (y_0 + y_4) + 4(y_1 + y_3) + 2(y_2) \right]
I \approx \frac{0.25}{3} \left[ (1.00000 + 0.50000) + 4(0.80000 + 0.57143) + 2(0.66667) \right]
I \approx \frac{0.25}{3} \left[ 1.50000 + 4(1.37143) + 1.33334 \right]
I \approx \frac{0.25}{3} \left[ 1.50000 + 5.48572 + 1.33334 \right]
I \approx \frac{0.25}{3} \left[ 8.31906 \right] = 0.693255

Step 4: Compute the Exact Analytical Value

Evaluate the integral analytically using standard calculus:
I_{\text{exact}} = \int_{0}^{1} \frac{dx}{1+x} = \left[ \ln(1+x) \right]_0^1
I_{\text{exact}} = \ln(1+1) - \ln(1+0) = \ln(2) - \ln(1)
Since \ln(1) = 0, the exact value is:
I_{\text{exact}} = \ln(2) \approx 0.693147

Step 5: Calculate the Absolute Error

The absolute error is the difference between the exact value and the approximate value:
\text{Error} = |I_{\text{exact}} - I_{\text{approx}}|
\text{Error} = |0.693147 - 0.693255| = 0.000108

Final Answer:

The value of the integral evaluated by Simpson’s 1/3 rule is 0.693255. The absolute error when compared to the exact value (\ln 2) is 0.000108.

Solution:

Step 1: Find the Complementary Function (C.F.)

The given PDE is a homogeneous linear partial differential equation of the third order. First, we write the auxiliary equation by substituting D = m and D^{\prime} = 1:

By inspection, m = -1 is a root since (-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0. Using synthetic division or factorization, we find the other roots:

The roots are m_1 = -1, m_2 = -2, and m_3 = 3.

The Complementary Function is:
\text{C.F.} = \phi_1(y - x) + \phi_2(y - 2x) + \phi_3(y + 3x)

Step 2: Find Particular Integral 1 (P.I.₁) for the polynomial part

The first part of the Particular Integral deals with x^2 + xy^2 + y^3:

Factor out D^3 from the denominator to use binomial expansion:

Expand using (1 - \theta)^{-1} = 1 + \theta + \dots:

Now, apply the operators in the brackets to the polynomial (note that D^{\prime} represents differentiation with respect to y, and 1/D represents integration with respect to x):

• Term 1: 1 \cdot (x^2 + xy^2 + y^3) = x^2 + xy^2 + y^3
• Term 2: 7\frac{D^{\prime 2}}{D^2} (x^2 + xy^2 + y^3) = \frac{7}{D^2} (2x + 6y) = 7 \left( \frac{x^3}{3} + 3x^2y \right) = \frac{7}{3}x^3 + 21x^2y
• Term 3: 6\frac{D^{\prime 3}}{D^3} (x^2 + xy^2 + y^3) = \frac{6}{D^3} (6) = \frac{36}{D^3}(1) = 36 \left(\frac{x^3}{6}\right) = 6x^3

Summing these gives the intermediate result inside the bracket:

Now apply \frac{1}{D^3} (integrate three times with respect to x):

Step 3: Find Particular Integral 2 (P.I.₂) for the trigonometric part

For \cos(ax+by), we substitute D^2 = -a^2, DD^{\prime} = -ab, and D^{\prime 2} = -b^2. Here, a = 1 and b = -1, so D^2 = -1, DD^{\prime} = 1, and D^{\prime 2} = -1.

First, rewrite the denominator in terms of squares:
D^3 - 7DD^{\prime 2} - 6D^{\prime 3} = D(D^2) - 7D(D^{\prime 2}) - 6D^{\prime}(D^{\prime 2})
Substitute the values: D(-1) - 7D(-1) - 6D^{\prime}(-1) = 6D + 6D^{\prime} = 6(D + D^{\prime}).

Multiply numerator and denominator by (D - D^{\prime}):
\text{P.I.}_2 = \frac{D - D^{\prime}}{6(D^2 - D^{\prime 2})} \cos(x-y)
Substituting D^2 = -1 and D^{\prime 2} = -1 yields 6(-1 - (-1)) = 0. This is a case of failure.

To resolve this, return to the factored form of the operator: F(D, D^{\prime}) = (D+D^{\prime})(D-3D^{\prime})(D+2D^{\prime}).
We apply the non-failing part (D-3D^{\prime})(D+2D^{\prime}) = D^2 - DD^{\prime} - 6D^{\prime 2} first:
D^2 - DD^{\prime} - 6D^{\prime 2} \xrightarrow{\text{subst}} -1 - 1 - 6(-1) = 4
\text{P.I.}_2 = \frac{1}{4(D + D^{\prime})} \cos(x-y)

For the failing factor (D - mD^{\prime}) where m = -1, use the general integration formula \int \phi(x, c-mx) dx with y = c + x \implies x-y = -c:
\text{P.I.}_2 = \frac{1}{4} \int \cos(-c) dx = \frac{1}{4} \cos(c) \int 1 dx = \frac{x}{4} \cos(c)
Substitute back c = y - x (which means \cos(c) = \cos(x-y)):
\text{P.I.}_2 = \frac{x}{4} \cos(x-y)

Step 4: Final Complete Solution

The complete solution is z = \text{C.F.} + \text{P.I.}_1 + \text{P.I.}_2.

Final Answer:

Solution:

Step 1: Standard Form and Identities

The given equation is pt - qs = q^3. Comparing this with the standard Monge’s form Rr + Ss + Tt = V (where r = \frac{\partial^2 z}{\partial x^2}), we identify the coefficients:

We recall the fundamental total differential relations for p and q:

Substitute r and t into the original partial differential equation:

Multiply the entire equation by dy to clear the fraction:

Rearrange the terms to group those involving s together:

Step 2: Establish Monge’s Subsidiary Equations

To solve the system, Monge’s method requires setting both the independent group and the s-dependent group to zero simultaneously. This produces the following two subsidiary equations:

Equation 1: p\,dx + q\,dy = 0

Equation 2: p\,dq - q^3\,dy = 0

Step 3: Integrate the Subsidiary Equations

We know from total probability/calculus that the total differential of the surface function is dz = p\,dx + q\,dy. Substituting this relationship directly into Equation 1 gives:

Now, rewrite Equation 1 explicitly as p = -q\frac{dy}{dx} and substitute it into Equation 2:

Divide through by -q (assuming q \neq 0):

Factor out dy:

This splits into two branches: dy = 0 or dq + q^2\,dx = 0. Let’s solve the non-trivial differential branch:

Integrating both sides yields:

Step 4: Form the Intermediate Integral

According to Monge’s method, the two independent constants are connected via an arbitrary function \phi, such that b = \phi(a). Substituting our constants a and b creates our first intermediate integral relation:

We solve this relation explicitly for q:

Step 5: Final Integration for the General Solution

Invert the partial differential relation to express it in terms of the independent variable y:

Integrating both sides with respect to z treates x as a parameter and introduces a new arbitrary function of integration, \psi(x):

Let f(z) = \int \phi(z)\,dz, where f(z) remains an entirely arbitrary function since \phi(z) is arbitrary.

Final Answer:

The general solution of the non-linear partial differential equation is:

y = xz + f(z) + \psi(x)

where f and \psi are arbitrary structural functions.

Solution:

Step 1: Steady State Condition (Initial State)

Before the change, the rod is at steady state u_s(x). The heat equation at steady state is \frac{d^2 u_s}{dx^2} = 0.
The solution is u_s(x) = Ax + B.
Using boundary conditions u(0) = 0 and u(L) = 100:
u_s(0) = 0 \implies B = 0
u_s(L) = 100 \implies AL = 100 \implies A = 100/L
So, the initial temperature distribution at t=0 is f(x) = \frac{100}{L}x.

Step 2: The Heat Equation and Separation of Variables

We solve the transient heat equation \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} with new boundary conditions u(0, t) = 0 and u(L, t) = 0.
Using separation of variables u(x, t) = X(x)T(t), we get:
X^{\prime\prime} + \lambda^2 X = 0 and T^{\prime} + k\lambda^2 T = 0.

Applying the boundary conditions X(0) = 0 and X(L) = 0:
X(x) = C \sin\left(\frac{n\pi x}{L}\right) where \lambda_n = \frac{n\pi}{L}.
The time-dependent part is T(t) = D e^{-k(\frac{n\pi}{L})^2 t}.

By the Principle of Superposition, the general solution is:
u(x, t) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) e^{-k\left(\frac{n\pi}{L}\right)^2 t}

Step 3: Determine Fourier Coefficients (b_n)

At t = 0, we must have u(x, 0) = f(x) = \frac{100x}{L}.
\frac{100x}{L} = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)
This is a Fourier Sine Series, where:
b_n = \frac{2}{L} \int_{0}^{L} \frac{100x}{L} \sin\left(\frac{n\pi x}{L}\right) dx = \frac{200}{L^2} \int_{0}^{L} x \sin\left(\frac{n\pi x}{L}\right) dx

Use integration by parts (\int u \, dv = uv - \int v \, du) with u = x and dv = \sin\left(\frac{n\pi x}{L}\right) dx:

b_n = \frac{200}{L^2} \left[ x \left( -\frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right) - \int \left( -\frac{L}{n\pi} \cos\left(\frac{n\pi x}{L}\right) \right) dx \right]_0^L
b_n = \frac{200}{L^2} \left[ -\frac{xL}{n\pi} \cos\left(\frac{n\pi x}{L}\right) + \frac{L^2}{n^2\pi^2} \sin\left(\frac{n\pi x}{L}\right) \right]_0^L

Evaluating at the limits 0 and L:
The sine term is 0 at both ends.
At x = L: -\frac{L^2}{n\pi} \cos(n\pi) = -\frac{L^2}{n\pi}(-1)^n
At x = 0: The term is 0.

Final Answer:

Substituting b_n back into the general solution, the temperature distribution is: