Q1. Evaluate \(\int_{-2}^3 f d \alpha\) where \(f(x)=[|x|]\) and \(\alpha(x)=|x|\), [.] shows the greatest integer function.
Answer:
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 \(\alpha(x) = |x|\):
\[
= [-2] (|-1| – |-2|) + [-1] (|0| – |-1|) + [0] (|1| – |0|) + [1] (|2| – |1|) + [2] (|3| – |2|)
\]
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 \([0,2]\).
Answer:
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.
Answer:
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.
\[
\begin{aligned}
& x_1^{\prime}=-x_1+8 x_2, \\
& x_2^{\prime}=x_1+x_2 .
\end{aligned}
\]
Answer:
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:
\[
\frac{d y}{d t}=\left[\begin{array}{cc}
2 & 3 \\
-1 & -2
\end{array}\right] \mathrm{y}+\left[\begin{array}{c}
4 e^{3 t} \\
-e^{3 t}
\end{array}\right] \text { where } \mathrm{y}=\left[\begin{array}{l}
y_1 \\
y_2
\end{array}\right]
\]
Answer:
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.
Answer:
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 \(y^{\prime}=(1+2 x+3 y) /\left(2+x^2+y^2\right), y(0)=0, R:|x| \leq 2,|y| \leq 1\)
Answer:
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
Answer:
(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.
