Solution

Step 1: Rewrite the expression as an average

Define

We have to prove that T_n\to ab. Since both original sequences converge, they are bounded.

Step 2: Separate each sequence into its limit plus an error term

Write

where \alpha_k\to0 and \beta_k\to0. Then

Step 3: Deal with the two ordinary Cesàro averages

Because \alpha_k\to0, its arithmetic means also tend to zero:

Similarly, reversing the order of the \beta-terms does not change their average, so

Step 4: Show that the mixed error term also tends to zero

Since convergent sequences are bounded, choose M such that \lvert\alpha_k\rvert\le M and \lvert\beta_k\rvert\le M for all k. Given \varepsilon\gt0, choose N so that

In the sum \alpha_k\beta_{n+1-k}, only a fixed number of terms near the two ends can have one index below N. Their total contribution divided by n tends to zero. All remaining terms satisfy

Therefore

Step 5: Take the limit

All three error terms vanish. Hence only ab remains.

Final Answer: \boxed{\displaystyle \lim_{n\to\infty}\frac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n}=ab}

Solution

Step 1: Locate the singularity

The integrand has a pole at

Since \log2\approx0.693, this point lies inside the square whose vertices are 1,i,-1,-i. Therefore Cauchy’s integral formula for derivatives applies.

Step 2: Match the integral with Cauchy’s derivative formula

The generalized Cauchy integral formula is

Here

so n=3.

Step 3: Find the third derivative

At a=\log2,

hence

Step 4: Substitute into the formula

Final Answer: \boxed{72\pi i}

Solution

Step 1: Use the Laplace transform of an integral

Let

Then the required function is g(t)/t. First,

Therefore

Step 2: Apply the division-by-\(t\) property

If \mathcal L\{g(t)\}=G(s), then

Hence

Step 3: Decompose the integrand

Since

an antiderivative is

Step 4: Evaluate the upper limit

As v\to\infty, the logarithmic terms cancel in the limit and \tan^{-1}(v-1)\to\pi/2. Therefore

Final Answer:
\boxed{\displaystyle \mathcal L\left\{\frac1t\int_0^t e^u\sin u\,du\right\} = \frac{\pi}{4} -\frac12\log s +\frac14\log\big((s-1)^2+1\big) -\frac12\tan^{-1}(s-1)} for s\gt1.

Solution

Step 1: Write the Newton-Raphson iteration

Let the error at the n-th step be

Step 2: Expand \(f(\alpha)\) about \(x_n\)

By Taylor’s theorem, for some point \xi_n between x_n and \alpha,

Since f(\alpha)=0 and \alpha-x_n=-e_n,

Step 3: Express the next error in terms of the present error

Rearranging,

Using the Newton formula,

Step 4: Pass to the limiting error constant

As x_n\to\alpha, we also have \xi_n\to\alpha. Hence

provided the numerator is finite. Thus the next error is proportional to the square of the current error.

Final Conclusion: Newton-Raphson has quadratic convergence near a simple root: \boxed{e_{n+1}=O(e_n^2)}

Solution

Step 1: Observe that every automorphism is a permutation

An automorphism \phi:G\to G is, by definition, a bijective homomorphism. Since it is bijective, it is a permutation of the underlying set G. Therefore

Step 2: Check that the identity belongs to \(\operatorname{Aut}(G)\)

The identity map I_G(x)=x is bijective and satisfies

Hence I_G\in\operatorname{Aut}(G).

Step 3: Check closure under composition

Let \phi,\psi\in\operatorname{Aut}(G). Their composition is bijective. Moreover,

Thus \phi\circ\psi is again an automorphism.

Step 4: Check closure under inverses

Let \phi\in\operatorname{Aut}(G). Since \phi is bijective, \phi^{-1} exists. If a=\phi(x) and b=\phi(y), then

Hence \phi^{-1} is also an automorphism.

Final Conclusion: \operatorname{Aut}(G) contains the identity and is closed under composition and inverses. Therefore \boxed{\operatorname{Aut}(G)\le A(G)}

Solution

Step 1: Consider the increasing case

Assume first that f is increasing. A monotone function on a closed bounded interval is bounded, so upper and lower Darboux sums are defined.

Step 2: Choose a uniform partition

Divide [a,b] into n equal parts:

Because f is increasing, on [x_{i-1},x_i] the minimum is f(x_{i-1}) and the maximum is f(x_i).

Step 3: Compute the difference between upper and lower sums

Step 4: Make the difference arbitrarily small

Given \varepsilon\gt0, choose n sufficiently large so that

Then U(f,P)-L(f,P)\lt\varepsilon. By the Darboux criterion, f is Riemann integrable.

Step 5: Handle the decreasing case

If f is decreasing, the same argument applies after interchanging the upper and lower endpoint values, or equivalently by applying the increasing-case result to -f.

Final Conclusion: Every monotone real-valued function on a closed bounded interval is Riemann integrable: \boxed{f\in\mathcal R[a,b]}

Solution

Step 1: Rewrite the recurrence as a linear difference equation

Look for a solution of the form S_n=r^n. This gives the characteristic equation

Step 2: Solve the characteristic equation

so the roots are

Therefore

Step 3: Determine \(A\) from the first two terms

For n=1 and n=2,

Subtracting,

Then

Step 4: Take the limit

Since

we get

Final Answer: The sequence converges, and \boxed{\displaystyle \lim_{n\to\infty}S_n=\frac{S_1+2S_2}{3}}

Solution

Step 1: Understand the map \(f\)

For finite real x,

Together with the assigned values at \pm\infty, the map sends

It is strictly increasing and therefore one-to-one.

Step 2: Verify non-negativity and definiteness

For all x,y,

Also,

because f is injective.

Step 3: Verify symmetry

Step 4: Verify the triangle inequality

Using the ordinary triangle inequality in \mathbb R,

Thus d is a metric.

Step 5: Prove boundedness

Since every value of f lies in [-1,1],

In fact,

so the diameter is exactly 2.

Final Conclusion: (\mathbb R_\infty,d) is a bounded metric space with \boxed{\operatorname{diam}(\mathbb R_\infty)=2}

Solution

Step 1: Prove that closed \(Y\) implies complete \(Y\)

Assume Y is closed in X. Let (y_n) be a Cauchy sequence in Y. It is also a Cauchy sequence in X.

Since X is complete, there exists x\in X such that

Every y_n lies in Y, and Y is closed, so it contains all limits of its convergent sequences. Hence x\in Y. Therefore every Cauchy sequence in Y converges in Y.

Step 2: Prove that complete \(Y\) implies closed \(Y\)

Now assume Y is complete. Let x be a limit point of Y. Then there exists a sequence (y_n) in Y such that

Every convergent sequence is Cauchy, so (y_n) is Cauchy in Y. Since Y is complete, there exists y\in Y with

Step 3: Use uniqueness of limits

The same sequence converges to both x and y in the metric space X. Limits are unique, hence x=y. Therefore x\in Y.

So Y contains every one of its limit points and is closed.

Final Conclusion: \boxed{Y\text{ is complete }\iff Y\text{ is closed in }X}

Solution

Step 1: Use the geometric information about the image

A circle in the complex plane has empty interior: no open disk is contained entirely in the circumference of a circle.

Step 2: Assume, for contradiction, that \(f\) is non-constant

The Open Mapping Theorem states that a non-constant analytic function on a region maps open sets to open sets. Therefore, if f were non-constant, f(G) would be an open subset of \mathbb C.

Step 3: Obtain the contradiction

But the hypothesis says that f(G) is contained in a circle. A nonempty open subset of \mathbb C cannot be contained in a one-dimensional circle.

This contradicts the Open Mapping Theorem.

Final Conclusion: The assumption that f is non-constant is impossible. Hence \boxed{f\text{ is constant on }G.}

Solution

Step 1: Write Lagrange’s auxiliary equations

This is a Lagrange first-order PDE of the form Pp+Qq=R, with

Therefore the subsidiary equations are

Step 2: Find the first integral from differences

Introduce a parameter s along a characteristic:

Then

and similarly

Therefore

so

Step 3: Find a second independent integral

Consider

Along a characteristic,

Substituting the characteristic equations and simplifying gives

Hence

Step 4: Combine the two first integrals

The general solution of a Lagrange PDE is an arbitrary relation between two independent first integrals.

Final Answer: \boxed{\displaystyle \Phi\left(\frac{x-y}{\,y-z\,},\,xy+yz+zx\right)=0} where \Phi is an arbitrary differentiable function. Equivalent forms obtained from ratios of the pairwise differences are also valid.

Solution

Step 1: Write the PDE in standard second-order form

The equation is

It is the two-variable wave equation.

Step 2: Find the characteristic directions

For

the characteristic equation is

Here A=1, H=0 and B=-a^2. Hence

or

Step 3: Integrate the characteristic equations

The two independent characteristic variables are

Step 4: Write the general solution

For the wave equation, the general solution is an arbitrary function of each characteristic variable:

Step 5: Verify

Differentiating twice,

while

Thus z_{xx}=a^2z_{yy}.

Final Answer: \boxed{z=\phi(y+ax)+\psi(y-ax)} where \phi and \psi are arbitrary twice-differentiable functions.

Solution

Step 1: Find the extremal

The integrand is

The Euler-Lagrange equation is

Now

so

Hence

From y(0)=0, C_1=0. From y(a)=0,

Since the question assumes a\ne n\pi/2, \sin2a\ne0, so C_2=0. Thus the extremal is y=0.

Step 2: Form the Jacobi equation

Here

The Jacobi equation becomes

or

Step 3: Find points conjugate to \(x=0\)

A Jacobi field vanishing at x=0 is, up to a constant multiple,

Its positive zeros occur at

Thus the first conjugate point to 0 is \pi/2.

Step 4: Apply the Jacobi condition

For the Jacobi condition to hold throughout (0,a], there must be no conjugate point before or at a. Therefore, for positive a, the strengthened Jacobi condition holds when

If a\gt\pi/2, the interval already contains the conjugate point \pi/2, so the Jacobi condition fails even when a itself is not a multiple of \pi/2.

Final Answer: For a\gt0, the Jacobi condition is fulfilled on the whole interval precisely before the first conjugate point: \boxed{0\lt a\lt\frac{\pi}{2}} It fails for a\gt\pi/2.

Solution

Step 1: Write Euler’s iteration formula

For

Euler’s method is

Step 2: Compute the successive approximations

Starting with t_0=0, y_0=1:

At t_1=0.1:
y_1=1+0.1(1-0+4)=1.5.

At t_2=0.2:
y_2=1.5+0.1(1-0.1+4(1.5))=2.19.

At t_3=0.3:
y_3=2.19+0.1(1-0.2+4(2.19))=3.146.

At t_4=0.4:
y_4=3.146+0.1(1-0.3+4(3.146))=4.4744.

At t_5=0.5:
y_5=4.4744+0.1(1-0.4+4(4.4744))=6.32416.

Step 3: Verify the correct exact solution

Solving

gives

Using y(0)=1,

Step 4: Compute the error at \(t=0.5\)

The exact value is

The Euler approximation is

Therefore the absolute error is

Step 5: Compute the percentage error

Final Answer:
Euler approximation at t=0.5: \boxed{6.32416}
Correct exact value: \boxed{8.712004\text{ (approximately)}}
Absolute error: \boxed{2.387844\text{ (approximately)}}
Percentage error: \boxed{27.41\%\text{ (approximately)}}

Solution

Step 1: Start from the definition

Step 2: Interchange the order of integration

The region is 0\le t\le x\lt\infty. Hence

Step 3: Evaluate the remaining parameter integral

Define

Differentiate with respect to s:

Step 4: Integrate with respect to \(s\)

Since I(0)=0,

Step 5: Substitute back

Final Answer: \boxed{\displaystyle \mathcal L\{C(t)\}(s) =\frac{1}{2s}\log(1+s^2) =\frac1s\log\sqrt{1+s^2}} for s\gt0.

Solution

Step 1: State the condition for the Bisection method

If f is continuous on [a,b] and

then, by the Intermediate Value Theorem, at least one real root lies between a and b.

Step 2: Bisection algorithm

1. Choose a and b with f(a)f(b)\lt0.
2. Compute the midpoint c=(a+b)/2.
3. If \lvert f(c)\rvert is below the required tolerance, stop.
4. If f(a)f(c)\lt0, replace b by c.
5. Otherwise replace a by c.
6. Repeat until the interval or function-value error is below the tolerance.

Step 3: Select an interval for the given polynomial

Let

On [1,2],

so a root lies in (1,2). Bisection converges to the positive root

Step 4: C program

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

double f(double x)
{
    return x*x*x - 2*x - 1;
}

int main(void)
{
    double a = 1.0, b = 2.0, c;
    double tol = 1e-6;

    if (f(a) * f(b) > 0)
    {
        printf("Choose another interval.\n");
        return 0;
    }

    while ((b - a) / 2.0 > tol)
    {
        c = (a + b) / 2.0;

        if (fabs(f(c)) < tol)
            break;

        if (f(a) * f(c) < 0)
            b = c;
        else
            a = c;
    }

    c = (a + b) / 2.0;
    printf("Root = %.6f\n", c);

    return 0;
}

Step 5: State the numerical result

Running the program with the interval [1,2] gives approximately

Final Answer: A real root in (1,2) is \boxed{x\approx1.618034}.

Solution

Step 1: Evaluate the contour integral by Cauchy’s formula

The function e^{az} is entire. Since 0 lies inside the unit circle, Cauchy’s integral formula gives

Step 2: Parametrize the unit circle

Set

Then

Therefore

Step 3: Separate the real and imaginary parts

Using e^{i\theta}=\cos\theta+i\sin\theta,

Hence

Step 4: Use symmetry

The function

is odd in \theta, so its integral from -\pi to \pi is zero. Therefore

Step 5: Use evenness

The remaining integrand is even, hence

Final Answer: \boxed{\displaystyle \int_0^\pi e^{a\cos\theta}\cos(a\sin\theta)\,d\theta=\pi}

Solution

Step 1: Use boundedness of Riemann integrable functions

Since f and g are Riemann integrable, both are bounded. Choose constants M_f,M_g such that

Step 2: Control the oscillation of the product

For any two points x,y in the same subinterval,

Thus the oscillation of fg on each subinterval is at most M_f times the oscillation of g plus M_g times the oscillation of f.

Step 3: Apply the Darboux criterion

Because f and g are integrable, for any \varepsilon\gt0 we can choose a common refinement P for which their upper-minus-lower sums are sufficiently small. The inequality above then gives

Hence fg is Riemann integrable.

Step 4: Show that the absolute value of an integrable function is integrable

If h is integrable, then

Therefore the oscillation of \lvert h\rvert is no larger than that of h. So \lvert h\rvert is integrable.

Step 5: Express maximum and minimum algebraically

Since f-g is integrable, \lvert f-g\rvert is integrable. Now use

and

Sums and scalar multiples of integrable functions are integrable. Hence both the maximum and minimum are integrable.

Final Conclusion: \boxed{fg,\ \max(f,g),\ \min(f,g)\in\mathcal R[a,b].}