Q1. Evaluate $$\int_{-2}^3 f d \alpha$$ where $$f(x)=[|x|]$$ and $$\alpha(x)=|x|$$, [.] shows the greatest integer function.

The problem asks to evaluate the Stieltjes integral $$\int_{-2}^3 f \, d \alpha$$ where $$f(x) = [|x|]$$, the greatest integer function, and $$\alpha(x) = |x|$$, the absolute value function.

Introduction:
The Stieltjes integral is a generalization of the Riemann integral, where the integral is not necessarily with respect to $$x$$, but with respect to another function $$\alpha(x)$$. The greatest integer function, denoted by $$[x]$$, is the largest integer less than or equal to $$x$$.

Method/Approach:
To solve this problem, we need to analyze the behavior of the functions $$f(x)$$ and $$\alpha(x)$$ within the interval $$[-2,3]$$. We break the interval into segments where the greatest integer function changes its value. For this problem, since $$\alpha(x) = |x|$$, the increments of $$\alpha$$ occur at integer values of $$x$$. Thus, the Stieltjes integral is a sum of the products of the value of $$f$$ and the increment of $$\alpha$$ at these points.

Work/Calculations:
The given functions change values at integers. So we split the integration interval $$[-2,3]$$ into intervals of unit length.

$\int_{-2}^{3} f \, d \alpha = \int_{-2}^{-1} f \, d \alpha + \int_{-1}^{0} f \, d \alpha + \int_{0}^{1} f \, d \alpha + \int_{1}^{2} f \, d \alpha + \int_{2}^{3} f \, d \alpha$

Within each interval, $$f(x)$$ is a constant because the greatest integer function is constant between integers, and $$\alpha(x) = |x|$$ increments by one from the start to the end of the interval.

Thus, we can write each integral as the product of the value of $$f(x)$$ and the increment of $$\alpha(x)$$ within the interval:

$= f(-2) (\alpha(-1) – \alpha(-2)) + f(-1) (\alpha(0) – \alpha(-1)) + f(0) (\alpha(1) – \alpha(0)) + f(1) (\alpha(2) – \alpha(1)) + f(2) (\alpha(3) – \alpha(2))$

Substituting the values of $$f(x) = [|x|]$$ and $$\alpha(x) = |x|$$:

$= [-2] (|-1| – |-2|) + [-1] (|0| – |-1|) +  (|1| – |0|) +  (|2| – |1|) +  (|3| – |2|)$

The value of the greatest integer function is just the integer itself for integer values, so we can simplify further:

$= -2 (1 – 2) + -1 (0 – 1) + 0 (1 – 0) + 1 (2 – 1) + 2 (3 – 2)$

This simplifies to:

$= 2 +1 + 0 + 1 + 2 = 6$

Conclusion:
Hence, the value of the Stieltjes integral $$\int_{-2}^{3} f \, d \alpha$$ with $$f(x) = [|x|]$$ and $$\alpha(x) = |x|$$ is 6.

Q2. For the sequence of functions $$\left\{f_n\right\}$$ where $$f_n(x)=\frac{x^n}{1+x^n}, x \in[0,2]$$, show that $$\left\{f_n\right\}$$ is not uniformly convergent on $$[0,2]$$.

Introduction:
The sequence of functions $$\{f_n\}$$ where $$f_n(x) = \frac{x^n}{1 + x^n}$$ for $$x \in [0,2]$$ is given. We need to show that this sequence does not converge uniformly on the interval $$[0,2]$$.

Method/Approach:
A sequence of functions $$\{f_n\}$$ is said to converge uniformly on a set $$S$$ if for every $$\epsilon > 0$$, there exists a natural number $$N$$ such that for all $$n > N$$ and all $$x \in S$$, we have $$|f_n(x) – f(x)| < \epsilon$$, where $$f(x)$$ is the limiting function.

To prove that a sequence of functions does not converge uniformly, it suffices to show that there exists some $$\epsilon > 0$$ such that for any natural number $$N$$, there exists $$n > N$$ and some $$x \in S$$ where $$|f_n(x) – f(x)| \geq \epsilon$$.

Work/Calculations:
The pointwise limit of the sequence $$\{f_n\}$$ is $$f(x) = 0$$ for $$0 \leq x < 1$$ and $$f(x) = 1$$ for $$1 \leq x \leq 2$$.

For any natural number $$N$$, let’s take $$n = N+1$$ and $$x = \sqrt[n]{\frac{1}{2}}$$. Note that $$x \in [0, 2]$$ for all $$n$$.

Now, $$f_n(x) = \frac{x^n}{1 + x^n} = \frac{1/2}{1 + 1/2} = \frac{1}{3}$$.

So, $$|f_n(x) – f(x)| = |\frac{1}{3} – 0| = \frac{1}{3}$$.

We can see that for any $$N$$, we can find $$n > N$$ and $$x \in [0,2]$$ such that $$|f_n(x) – f(x)| = \frac{1}{3}$$. Hence, for $$\epsilon < \frac{1}{3}$$, we can’t find such $$N$$ that $$|f_n(x) – f(x)| < \epsilon$$ for all $$n > N$$ and $$x \in [0,2]$$.

Conclusion:
Therefore, the sequence of functions $$\{f_n\}$$ does not converge uniformly on the interval $$[0,2]$$.

Q3. Let $$f: R^3 \rightarrow R^3$$ be given by $$f(x, y, z)=\left(x^2+y+2, y^2+\cos x, z^2\right)$$. Then find the derivative of $$f$$ at $$(x, y, z)$$ in matrix form.

Introduction:
The task is to find the derivative of a function $$f: R^3 \rightarrow R^3$$ given by $$f(x, y, z) = (x^2 + y + 2, y^2 + \cos x, z^2)$$ at any point $$(x, y, z)$$ in matrix form. This is essentially a problem of multivariable calculus.

Method/Approach:
The derivative of a multivariable function at a specific point is given by the Jacobian matrix, a matrix comprising of all the first-order partial derivatives of the function. The $$i, j$$-th entry of the Jacobian matrix is the derivative of the $$i$$-th component function with respect to the $$j$$-th argument.

Work/Calculations:
For the given function, the Jacobian matrix $$J$$ at any point $$(x, y, z)$$ is:

$J(x, y, z) = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \\ \frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} & \frac{\partial f_3}{\partial z} \end{bmatrix}$

where $$f_1(x, y, z) = x^2 + y + 2$$, $$f_2(x, y, z) = y^2 + \cos x$$, and $$f_3(x, y, z) = z^2$$.

Differentiating each component function with respect to each variable gives us:

$J(x, y, z) = \begin{bmatrix} 2x & 1 & 0 \\ -\sin x & 2y & 0 \\ 0 & 0 & 2z \end{bmatrix}$

Conclusion:
So, the derivative of the given function $$f(x, y, z)$$ at any point $$(x, y, z)$$ in matrix form is given by the Jacobian matrix:

$\begin{bmatrix} 2x & 1 & 0 \\ -\sin x & 2y & 0 \\ 0 & 0 & 2z \end{bmatrix}$

Q4. Find the fundamental system of solutions of each of the following set of equation.
\begin{aligned} & x_1^{\prime}=-x_1+8 x_2, \\ & x_2^{\prime}=x_1+x_2 . \end{aligned}

Introduction:
The problem requires us to find the fundamental system of solutions for the given system of first-order homogeneous linear differential equations.

Method/Approach:
To solve such a system, we first write it in matrix form, find the eigenvalues of the corresponding matrix, and then find the eigenvectors corresponding to those eigenvalues. The solutions will be in the form of $$X(t) = c_i e^{\lambda_i t} v_i$$, where $$c_i$$ are constants, $$\lambda_i$$ are the eigenvalues, $$v_i$$ are the corresponding eigenvectors, and $$i$$ ranges over the number of dimensions (in this case, 2).

Work/Calculations:
The given system can be written in matrix form as:

$\begin{bmatrix} x_1′ \\ x_2′ \end{bmatrix} = \begin{bmatrix} -1 & 8 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$

So, our matrix $$A$$ is:

$A = \begin{bmatrix} -1 & 8 \\ 1 & 1 \end{bmatrix}$

We then find the eigenvalues $$\lambda$$ by solving the characteristic equation $$\det(A – \lambda I) = 0$$, where $$I$$ is the identity matrix.

The characteristic equation is:

$\begin{vmatrix} -1 – \lambda & 8 \\ 1 & 1 – \lambda \end{vmatrix} = 0$

Solving this gives us $$\lambda^2 – (Tr(A))\lambda + \det(A) = 0$$, where $$Tr(A)$$ is the trace of the matrix $$A$$ (the sum of the elements on the main diagonal), and $$\det(A)$$ is the determinant of $$A$$. Substituting the values, we get $$\lambda^2 + \lambda – 9 = 0$$.

The solutions to this equation are the eigenvalues of the matrix $$A$$, which are $$\lambda_1 = 3$$ and $$\lambda_2 = -3$$.

To find the corresponding eigenvectors, we substitute each eigenvalue back into the equation $$Av = \lambda v$$, where $$v$$ is the eigenvector, and solve for $$v$$.

For $$\lambda_1 = 3$$, we have the equation:

$\begin{bmatrix} -1 – 3 & 8 \\ 1 & 1 – 3 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0$

Solving this system of equations gives $$v_1 = (2, 1)$$.

For $$\lambda_2 = -3$$, we have the equation:

$\begin{bmatrix} -1 + 3 & 8 \\ 1 & 1 + 3 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0$

Solving this system of equations gives $$v_2 = (-4, 1)$$.

Conclusion:
The fundamental system of solutions of the given set of equations is:

$x_1(t) = c_1 e^{3t} (2, 1)$

and

$x_2(t) = c_2 e^{-3t} (-4, 1)$

Q5. Find the solution of the non-homogenous differential equation:
$\frac{d y}{d t}=\left[\begin{array}{cc} 2 & 3 \\ -1 & -2 \end{array}\right] \mathrm{y}+\left[\begin{array}{c} 4 e^{3 t} \\ -e^{3 t} \end{array}\right] \text { where } \mathrm{y}=\left[\begin{array}{l} y_1 \\ y_2 \end{array}\right]$

Introduction:
The task is to find the solution to the given non-homogeneous system of differential equations.

Method/Approach:
This is a first order non-homogeneous linear differential equation system. To find its solution, we first solve the associated homogeneous system. We then find a particular solution for the non-homogeneous system. The general solution is the sum of the general solution to the homogeneous system and the particular solution to the non-homogeneous system.

Work/Calculations:

1) The associated homogeneous system is given by:

$\frac{d y}{d t} = \begin{bmatrix} 2 & 3 \\ -1 & -2 \end{bmatrix} y$

The eigenvalues of the matrix are the roots of the characteristic polynomial:

$\begin{vmatrix} 2-\lambda & 3 \\ -1 & -2-\lambda \end{vmatrix} = 0$

which gives $$\lambda^{2} – (Tr(A))\lambda + \det(A) = 0$$, where $$Tr(A)$$ is the trace of the matrix (the sum of the elements on the main diagonal) and $$\det(A)$$ is the determinant. The roots are $$\lambda = 1, -1$$.

The eigenvectors corresponding to these eigenvalues can be found by substituting the eigenvalues back into the equation $$Av = \lambda v$$, where $$v$$ is the eigenvector. The corresponding eigenvectors are $$v_1 = (3, 1)$$ and $$v_2 = (-1, 1)$$.

So, the general solution of the homogeneous system is:

$y_{H}(t) = c_1 e^{t} \begin{bmatrix} -3 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

2) To find a particular solution for the non-homogeneous system, we guess a solution of the form $$y_{P}(t) = e^{3t} \begin{bmatrix} a \\ b \end{bmatrix}$$. Substituting $$y_{P}(t)$$ into the non-homogeneous differential equation, and comparing coefficients, we can solve for $$a$$ and $$b$$.

Doing this, we find that the particular solution is:

$y_{P}(t) = e^{3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Conclusion:
The general solution of the non-homogeneous system is the sum of the general solution of the homogeneous system and the particular solution, which is:

$y(t) = y_{H}(t) + y_{P}(t) = c_1 e^{t} \begin{bmatrix} -3 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -1 \\ 1 \end{bmatrix} + e^{3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

Q6. Show that for the initial value problem $$y^{\prime}=y, y(0)=1$$, the constant $$\mathrm{h}$$ in Picard’s Theorem must be smaller than unity.

Introduction:
The given initial value problem is $$y’ = y, y(0) = 1$$. We are required to show that for this problem, the constant $$h$$ in Picard’s Theorem must be smaller than unity.

Method/Approach:
Picard’s theorem is a fundamental theorem in the theory of ordinary differential equations which guarantees the existence and uniqueness of solutions to initial value problems. For a first order initial value problem of the form $$y’ = f(t, y)$$, $$y(t_0) = y_0$$, the theorem states that there exists a unique solution $$y(t)$$ for $$t$$ in the interval $$(t_0 – h, t_0 + h)$$ where $$h$$ is a positive constant.

For the given problem, $$f(t, y) = y$$. According to Picard’s theorem, $$h$$ must be smaller than both $$b – a$$ and $$\frac{M}{K}$$, where $$[a, b]$$ is the interval of $$t$$ for which $$f$$ is defined and continuous, $$M$$ is an upper bound of $$|f(t, y)|$$ on the rectangle $$[a, b] \times [y_0 – K, y_0 + K]$$, and $$K$$ is an upper bound of $$|y – y_0|$$ for $$y$$ in $$[y_0 – K, y_0 + K]$$.

Work/Calculations:
Here, we have $$y_0 = 1$$, $$a = -\infty$$, and $$b = \infty$$, so $$h < b – a$$ doesn’t give us a useful restriction. However, if we take $$K = 1$$, we see that for any $$y$$ in $$[y_0 – K, y_0 + K] = [0, 2]$$, we have $$|f(t, y)| = |y| \leq 2$$. Therefore, $$M = 2$$, and $$h < \frac{M}{K} = \frac{2}{1} = 2$$.

However, we can also see that $$f$$ and $$\frac{\partial f}{\partial y}$$ are both unbounded as $$y$$ approaches $$\pm\infty$$. Therefore, $$h$$ must be chosen small enough that the function and its derivative are bounded on the interval $$(t_0 – h, t_0 + h)$$. Since $$f(t, y) = y$$ and $$\frac{\partial f}{\partial y} = 1$$ are both unbounded as $$y$$ goes to $$\pm\infty$$, to keep them bounded, we must have $$h < 1$$.

Conclusion:
So, for the given initial value problem, the constant $$h$$ in Picard’s Theorem must be smaller than unity.

Q7. Discuss the existence and uniqueness the initial value problem $$y^{\prime}=(1+2 x+3 y) /\left(2+x^2+y^2\right), y(0)=0, R:|x| \leq 2,|y| \leq 1$$

Introduction:
The given initial value problem (IVP) is $$y^{\prime}=(1+2 x+3 y) /\left(2+x^2+y^2\right), y(0)=0$$. We are required to discuss the existence and uniqueness of a solution for this IVP on the rectangle $$R$$ defined by $$|x| \leq 2,|y| \leq 1$$.

Method/Approach:
Existence and uniqueness of solutions to an initial value problem are guaranteed by the Picard-Lindelöf theorem (also known as the Cauchy-Lipschitz theorem), which states that given a function $$f(t, y)$$ that is Lipschitz continuous in $$y$$, the initial value problem $$y’ = f(t, y)$$, $$y(t_0) = y_0$$ has a unique solution locally around $$t = t_0$$.

The Lipschitz condition is a measure of boundedness of the derivative of the function. A function $$f(t, y)$$ is said to be Lipschitz continuous in $$y$$ on a set $$D$$ if there exists a constant $$L$$ such that for all $$(t, y_1), (t, y_2) \in D$$, $$|f(t, y_1) – f(t, y_2)| \leq L|y_1 – y_2|$$.

Work/Calculations:

We consider the function $$f(x, y) = (1+2 x+3 y) /\left(2+x^2+y^2\right)$$. On the rectangle $$R: |x| \leq 2, |y| \leq 1$$, this function is continuous, and its partial derivative with respect to $$y$$, $$\frac{\partial f}{\partial y}$$, exists and is continuous, which means that $$f(x, y)$$ is continuously differentiable with respect to $$y$$ on $$R$$.

The Lipschitz condition requires that there exists a constant $$L$$ such that

$\left|\frac{\partial f}{\partial y}(x, y_1) – \frac{\partial f}{\partial y}(x, y_2)\right| \leq L|y_1 – y_2|$

for all $$(x, y_1), (x, y_2) \in R$$.

However, we can see that $$\frac{\partial f}{\partial y} = \frac{3(2 + x^2 + y^2) – 2y(1 + 2x + 3y)}{(2 + x^2 + y^2)^2}$$.

On the rectangle $$R: |x| \leq 2, |y| \leq 1$$, this function is continuous, and bounded. Hence, the Lipschitz condition holds, and there exists a constant $$L$$ that satisfies the inequality.

Conclusion:
Since the function $$f(x, y) = (1+2 x+3 y) /\left(2+x^2+y^2\right)$$ is continuously differentiable with respect to $$y$$ and satisfies the Lipschitz condition on the rectangle $$R: |x| \leq 2, |y| \leq 1$$, the Picard-Lindelöf theorem guarantees the existence and uniqueness of a solution to the initial value problem $$y^{\prime}=f(x,y), y(0)=0$$ in this region.

Q8. Give examples of polynomials over the field of real numbers with
(a) exactly two purely imaginary roots
(b) exactly one real root and two complex roots

(a) A polynomial with exactly two purely imaginary roots:

Consider the polynomial $$p(x) = x^2 + 4$$. The roots of this polynomial are given by $$x = \pm \sqrt{-4}$$, which are $$x = \pm 2i$$. These are purely imaginary roots.

(b) A polynomial with exactly one real root and two complex roots:

Consider the polynomial $$q(x) = x^3 – x^2 + 2$$. The roots of this polynomial are $$x = -1, 1 \pm i$$. Here, -1 is a real root and $$1 \pm i$$ are complex roots.

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