# 牛顿力学|MATH009 Newtonian Mechanics代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH009 Newtonian Mechanics牛顿力学代写代考辅导服务！

## Instructions:

Newtonian mechanics, which is a branch of classical physics that deals with the motion of objects under the influence of forces. The course covers essential concepts such as force, torque, momentum, angular momentum, and energy.

The study of point particles is a common starting point in mechanics, and the course likely covers the dynamics of point particles in great detail, including the classic problem of a central force with the inverse square law. Vector methods are likely emphasized in this course, and you may encounter vector differential equations as well.

Overall, this course will provide a solid foundation in the fundamental principles of classical mechanics and equip you with the tools needed to analyze and solve problems in this area of physics.

A pendulum consists of a ball of mass $\mathrm{M}$ attached to the end of a rigid bar of length $2 \mathrm{~d}$ which is pivoted at the center. At the other end of the bar is a container (“catch”). A second ball of mass $\mathrm{M} / 2$ is thrown into the catch at a velocity $\mathrm{v}$ where it sticks. For this problem, ignore the mass of the pendulum bar and catch, and treat the balls as if they were point masses (i.e., neglect their individual moments of inertia).
(a) What is the initial angular rotation rate of the pendulum after the incoming ball is caught?

(a) The angular momentum of the system relative to the pivot point just prior to the upper ball being caught is:
$$L=\frac{M}{2} v d$$
As there are no external torques acting on the system relative to the pivot point during the time of the collision, the angular momentum can be determined as:
$$L=I \omega=\left(M d^2+\frac{M}{2} d^2\right) \omega=\frac{3}{2} M d^2 \omega$$
hence,
$$\omega=\frac{v}{3 d}$$

(b) How much total mechanical energy is lost when the incoming ball gets stuck in the catch?

(b) The energy of the upper ball prior to sticking to the catch is:
$$E_i=\frac{1}{2} \frac{M}{2} v^2=\frac{1}{4} M v^2$$
After the collision, all of the energy can be expressed as pure rotation about the stationary pivot point, hence:
$$E_f=\frac{1}{2} I \omega^2=\frac{1}{2}\left(\frac{3}{2} M d^2\right)\left(\frac{v}{3 d}\right)^2=\frac{1}{12} M v^2$$
so
$$\Delta E=E_f-E_i=\frac{1}{6} M v^2$$

\begin{prob}

(c) What minimum velocity does the incoming ball need in order to invert the pendulum (i.e., rotate it by $180^{\circ}$ )?

\end{prob}

\begin{proof}

(c) Because there are only conservative forces acting (the pivot does not move so it
does no work) total mechanical energy is conserved. Assuming that the bottom of
the swing sets the zeropoint for gravitational potential, at the point that the ball
sticks the total energy is:

$$E_i=\frac{M}{2}(2 d) g+\frac{1}{12} M v^2$$
The point at which the pendulum just comes to rest inverted, the bottom mass is at height $2 \mathrm{~d}$ :
$$E_f=M(2 d) g$$
With $\Delta \mathrm{E}=0$ this gives:
\begin{aligned} & 2 M d g=M d g+\frac{1}{12} M v^2 \ & \Rightarrow v=\sqrt{12 d g} \end{aligned}

# 几何学代写 Geometry代考2023

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## 几何学代写Geometry

### 点（几何学）Point (geometry)代写

• Line (geometry)线（几何学）
• Plane (geometry)平（几何学）

## 几何学的历史

The birth of geometry can be traced back to the time of the ancient Egyptians. Herodotus recounts that due to the erosion and deposition of the Nile floods, the extent of the Egyptians’ land changed from year to year, making it necessary to recalculate taxes. Therefore, it was necessary to invent land surveying techniques (geometry, in the original meaning of the term).

The development of practical geometry is very ancient, and since it could have many applications and was developed for those applications, in ancient times, practical geometry was sometimes reserved for a class of wise men with priestly attributes. In ancient Greece, mainly due to the influence of the Athenian philosopher Plato, and even before him, Anaximander of Miletus [citation needed], the use of rules and compasses became common (although these tools seem to have been invented elsewhere), and above all, new ideas using presentation techniques were born. Greek geometry became the basis for the development of geography, astronomy, optics, mechanics and other sciences, as well as various technologies (such as navigation technology).

In Greek civilization, in addition to Euclidean geometry and conic theory, which were still studied in schools, spherical geometry and trigonometry (plane and sphere) were born.

## 几何学相关课后作业代写

Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.

To show that $S^4$ has no symplectic structure, we will use the following fact from symplectic geometry: a compact symplectic manifold has even dimension.

Suppose that $S^4$ has a symplectic structure. Then $S^4$ is a compact symplectic manifold, so its dimension must be even. However, the dimension of $S^4$ is $4$, which is not even. Therefore, $S^4$ cannot have a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we will use the following fact: the product of two symplectic manifolds is symplectic if and only if both factors have even dimension.

Suppose that $S^2 \times S^4$ has a symplectic structure. Then both $S^2$ and $S^4$ are symplectic manifolds, so their dimensions must both be even. However, the dimension of $S^2$ is $2$, which is not even. Therefore, $S^2 \times S^4$ cannot have a symplectic structure.

# 应用数学|MATH0008 Applied Mathematics代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0008 Applied Mathematics应用数学代写代考辅导服务！

## Instructions:

Qualitative and analytic approaches to differential equations are essential tools for modeling a wide range of physical systems. Qualitative approaches involve using geometric and graphical techniques to understand the behavior of solutions to differential equations, while analytic approaches involve using mathematical techniques to derive explicit formulas for these solutions. By combining these two approaches, you’ll be able to develop a deep understanding of the behavior of mechanical, biological, and other systems.

Stability is another important concept that you’ll be studying. In the context of differential equations, stability refers to the tendency of a system to return to a state of equilibrium after experiencing a disturbance. Understanding the stability of a system is crucial for predicting its behavior over time.

You’ll also be learning about waves and oscillations, which are ubiquitous in mechanical and biological systems. Waves are disturbances that propagate through a medium, while oscillations are periodic movements around a point of equilibrium. Understanding these phenomena is essential for modeling a wide range of systems, from sound waves to the behavior of the heart.

Finally, the course will also introduce you to discrete dynamical systems. These are mathematical models that describe systems that change over time in a stepwise fashion, rather than continuously. Discrete dynamical systems are important for modeling systems that evolve in discrete time steps, such as population growth models or financial models.

Suppose one has two coins $C_1$ and $C_2$. Coin $C_1$ gives head with probability $1 / 2$, while coin $C_2$ gives head with probability $1 / 3$. We pick one of the coins uniformly at random and toss it twice. We get twice the same result. Compute the probability the it was coin $C_1$ being used.

Let $H$ be the event that we get heads twice in a row, and let $C_1$ and $C_2$ be the events that we picked coin $C_1$ and coin $C_2$, respectively. We want to compute the probability of $C_1$ given $H$.

Using Bayes’ theorem, we have:

$P\left(C_1 \mid H\right)=\frac{P\left(H \mid C_1\right) P\left(C_1\right)}{P(H)}$

We can compute the probabilities on the right-hand side as follows:

• $P(H | C_1)$ is the probability of getting heads twice in a row given that we picked coin $C_1$. Since $C_1$ gives head with probability $1/2$, this probability is $(1/2)^2 = 1/4$.
• $P(C_1)$ is the probability of picking coin $C_1$ in the first place, which is $1/2$.
• $P(H)$ is the probability of getting heads twice in a row, which we can compute using the law of total probability:

\begin{aligned} P(H) & =P\left(H \mid C_1\right) P\left(C_1\right)+P\left(H \mid C_2\right) P\left(C_2\right) \ & =\frac{1}{4} \cdot \frac{1}{2}+\left(\frac{1}{3}\right)^2 \cdot \frac{1}{2} \ & =\frac{1}{12}+\frac{1}{18} \ & =\frac{5}{36}\end{aligned}

Substituting these values into Bayes’ theorem, we get:

$P\left(C_1 \mid H\right)=\frac{(1 / 4) \cdot(1 / 2)}{5 / 36}=\frac{9}{20} \approx 0.45$

Therefore, the probability that we picked coin $C_1$ given that we got heads twice in a row is approximately $0.45$.

You have a biased coin that comes up heads with probability $1 / 3$. Show that the probability of obtaining 80 heads or more from 90 throws is not more than 0.16 .

Let $X$ be the number of heads obtained from 90 throws of the biased coin. We have $X \sim \mathrm{Bin}(n=90,p=1/3)$, where $\mathrm{Bin}(n,p)$ denotes a binomial distribution with $n$ trials and success probability $p$. We want to find $P(X \geq 80)$.

We will use the Chebyshev’s inequality, which states that for any random variable $X$ with finite mean $\mu$ and variance $\sigma^2$, and for any positive constant $k$, we have:

$P(|X-\mu| \geq k \sigma) \leq \frac{1}{k^2}$

Applying this inequality to our problem, we have:

$P(X \geq 80)=P(X-\mu \geq 80-\mu) \leq \frac{\operatorname{Var}(X)}{(80-\mu)^2}$

Note that $\mu = np = 90 \cdot 1/3 = 30$ and $\mathrm{Var}(X) = np(1-p) = 90 \cdot 1/3 \cdot 2/3 = 20$. Substituting these values, we get:

$P(X \geq 80) \leq \frac{20}{(80-30)^2}=\frac{1}{25}=0.04$

Thus, the probability of obtaining 80 heads or more from 90 throws is not more than 0.04. However, this is an upper bound obtained by Chebyshev’s inequality, and it may be loose in practice. To get a tighter bound, we can use the normal approximation to the binomial distribution, which is valid when $np$ and $n(1-p)$ are both large enough. In our case, $np=30$ and $n(1-p)=60$, so the normal approximation is reasonable. Using the continuity correction, we have:

$P(X \geq 80) \approx P\left(Z \geq \frac{80+0.5-n p}{\sqrt{n p(1-p)}}\right)$

where $Z$ is a standard normal random variable. Substituting the values, we get:

$P(X \geq 80) \approx P(Z \geq 4.45) \approx 2.4 \times 10^{-6}$

Thus, the probability of obtaining 80 heads or more from 90 throws is extremely small, and certainly not more than 0.16.

Prove that if $C$ is any subset of ${100,101, \ldots, 199}$ with $|C|=51$, then $C$ contains two consecutive integers.

We will prove the statement using the Pigeonhole Principle, which states that if $n$ pigeons are placed into $m$ pigeonholes, and $n>m$, then there exists at least one pigeonhole with more than one pigeon.

Consider the set $S={1,2,\ldots,100}$ and the set $T={101,102,\ldots,199}$. Note that $S$ and $T$ are disjoint, and their union is ${1,2,\ldots,199}$.

Suppose for the sake of contradiction that $C$ is a subset of $T$ with $|C|=51$ that does not contain two consecutive integers. Then, $C$ and $C+1={c+1: c\in C}$ are disjoint subsets of $S$ with $|C|=|C+1|=51$. This means that $S$ contains two disjoint subsets $A$ and $B$ with $|A|=|B|=51$.

Now, consider the set of differences $D={b-a: b\in B, a\in A}$, which consists of all the possible differences between an element in $B$ and an element in $A$. Note that $D$ is a subset of ${-99,-98,\ldots,99}$, since the largest difference between an element in $B$ and an element in $A$ is $99$ (when the largest element in $B$ is subtracted from the smallest element in $A$).

Since $|D|=51^2>199$, there must be two distinct pairs $(a_1,b_1)$ and $(a_2,b_2)$ in $A\times B$ with the same difference $b_1-a_1=b_2-a_2$. But then, $b_2-a_1=b_2-a_2+(a_2-a_1)$ is a difference between an element in $B$ and an element in $A$, and $b_2-a_1$ is equal to $b_1-a_1$, which means that $C$ contains two consecutive integers, namely $b_1$ and $a_1+1$. This is a contradiction, so we conclude that any subset $C$ of $T$ with $|C|=51$ must contain two consecutive integers.

# 力学代写 mechanics代考2023

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## 力学代写mechanics

### 统计力学Statistical mechanics代写

• Theory of relativity相对论
• Quantum mechanics量子力学

## 力学的相关

The history and development of differential geometry as a discipline can be traced back at least to the ancient classics. It is closely related to the development of geometry, the concepts of space and form, and the study of topology, particularly manifolds. In this section we focus on the history of the application of infinitesimal methods to geometry, and then on the idea of tangent spaces, and finally on the development of the modern formalism of the discipline in terms of tensors and tensor fields.

## 力学相关课后作业代写

Hooke’s law, a constitutive equation for a linear, elastic material, can be written in general form as:
$$\sigma_{i j}=\lambda \varepsilon_{k k} \delta_{i j}+2 \mu \varepsilon_{i j} \text { where } \lambda \text { and } \mu \text { are Làme constants. }$$
a) Expand Hooke’s Law. How many independent equations are there?

a) Hooke’s law
$$\sigma_{i j}=\lambda \varepsilon_{k k} \delta_{i j}+2 \mu \varepsilon_{i j}$$
where
$$\delta_{i j}=\left{\begin{array}{l} 1, \mathrm{i}=\mathrm{j} \ 0, \mathrm{i} \neq \mathrm{j} \end{array}\right.$$
For $i=1$
\begin{aligned} & \sigma_{11}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{11} \ & \sigma_{12}=2 \mu \varepsilon_{12} \ & \sigma_{13}=2 \mu \varepsilon_{13} \end{aligned}

For $\mathrm{i}=2$
\begin{aligned} & \sigma_{21}=2 \mu \varepsilon_{21} \ & \sigma_{22}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{22} \ & \sigma_{23}=2 \mu \varepsilon_{23} \end{aligned}
For $\mathrm{i}=3$
\begin{aligned} & \sigma_{31}=2 \mu \varepsilon_{31} \ & \sigma_{32}=2 \mu \varepsilon_{32} \ & \sigma_{33}=\lambda\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)+2 \mu \varepsilon_{33} \end{aligned}

# 代数|MATH0005 Algebra 1代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0005 Algebra 1代数代写代考辅导服务！

## Instructions:

Linear equations are fundamental to many areas of mathematics, science, and engineering, so it’s important to have a strong understanding of them. And it’s great that the course also covers other important topics in modern mathematics, such as logic, set theory, and functions.

By studying these topics in depth, students can develop important problem-solving skills and critical thinking abilities that will serve them well in future courses and in their careers. Additionally, understanding these foundational concepts will allow students to approach more advanced topics with greater ease and confidence.

Overall, it sounds like MATH0005 is an excellent course for anyone looking to build a strong foundation in modern mathematics.

(i) Exhibit a proper normal subgroup $V$ of $A_4$. To which group is $V$ isomorphic to?

(i) The subgroup $V = {(), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}$ is a proper normal subgroup of $A_4$. To see this, note that $V$ is a subgroup of $A_4$ since it is closed under multiplication and inverses. Moreover, it is normal since it is the kernel of the sign homomorphism from $A_4$ to $\mathbb{Z}/2\mathbb{Z}$, which maps even permutations to $0$ and odd permutations to $1$. To see that $V$ is proper, note that it has index $2$ in $A_4$ since $A_4$ has order $12$ and $V$ has order $4$, so $A_4/V$ has order $2$. But there is no subgroup of order $2$ in $A_4$.

The group $V$ is isomorphic to the Klein four-group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

(ii) Give the left cosets of $V$ inside $A_4$.

(ii) The left cosets of $V$ inside $A_4$ are $V$ and $\tau V$, where $\tau$ is any transposition not in $V$. For example, if we take $\tau = (1,2)$, then we have

$V={(),(12)(34),(13)(24),(14)(23)}$

and

$\tau V={(12),(12)(34)(12),(13)(24)(12),(14)(23)(12)}$

\begin{prob}

(iii) To which group is $A_4 / V$ isomorphic to?

\end{prob}

\begin{proof}

(iii) The group $A_4/V$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, the cyclic group of order $2$. To see this, note that $A_4/V$ has order $2$, so it is either isomorphic to $\mathbb{Z}/2\mathbb{Z}$ or to the trivial group ${1}$. But if $A_4/V \cong {1}$, then $A_4 = V$, which is false. Therefore, $A_4/V \cong \mathbb{Z}/2\mathbb{Z}$.

# 流体力学|Fluid mechanics II 3A3代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Fluid mechanics II 3A3流体力学代写代考辅导服务！

## Instructions:

Fluids are generally considered to be those materials that have the ability to constantly change their shape by adapting to the container, which is why liquids, vapors and gases are considered to be fluids. Fluid mechanics consists of two main branches:

Fluid mechanics deals with fluids that are stationary in an inertial system, i.e. with constant velocity in time and homogeneity in space. Historically, it was the first step towards the study of mechanics.
Fluid dynamics or fluid mechanics (including specifically aerodynamics, hydrodynamics, and oil dynamics), deals with fluids in motion.
Fluids are characterized by having their own volume and a density very similar to that of solids, which means that at the microscopic level, the distances between molecules remain small and the interaction forces are high. This is a fundamental difference from gaseous substances, which have a low density and therefore low intermolecular interactions, allowing them to expand at any volume.

(a) Supersonic flow enters a straight pipe of constant cross-sectional area. Heat transfer is negligible, but the pipe wall is rough. Draw a labelled graph to show how the Mach number distribution along the pipe evolves as the skin-friction coefficient increases from zero. You may assume that the exit pressure is low enough to ensure that the inlet conditions are always the same.

(a) The graph below shows the variation of the Mach number with distance along the pipe as the skin friction coefficient increases from zero. As the skin friction coefficient increases, the velocity near the pipe wall decreases, causing the boundary layer to thicken. This results in a reduction in the effective cross-sectional area available for flow, which in turn reduces the mass flow rate and increases the Mach number. The Mach number gradually increases until it reaches the sonic condition at the throat of the pipe, after which it remains constant until the exit.

(b) Air flows in a pipe of length $5.9 \mathrm{~m}$ and inside diameter $0.2 \mathrm{~m}$. The inlet stagnation pressure is 2.7 bar, and the static pressure at the pipe exit is 1 bar. If the exit is choked, and there are no shocks in the pipe, find: (i) the two possible values of the Mach number at the inlet; (ii) the skin-friction coefficient $c_f$ corresponding to each.

(b) From the given data, we can use the choked flow condition to find the Mach number at the inlet. The choked flow condition occurs when the flow velocity at the throat of the pipe reaches the local speed of sound. At the throat, the Mach number is therefore 1.

Using the isentropic relations for a perfect gas, we can relate the Mach number to the pressure ratio across the throat:

$\frac{P_{02}}{P_{01}}=\left(\frac{1+\frac{\gamma-1}{2} M_1^2}{\frac{\gamma+1}{2}}\right)^{\frac{\gamma}{\gamma-1}}=\frac{P_{02}}{P_e}=\left(\frac{A_e}{A_{02}}\right)^2=1$,

where $P_{01}$ is the stagnation pressure at the inlet, $P_{02}$ is the pressure at the throat, $P_e$ is the static pressure at the exit, $A_{02}$ is the area of the throat, and $A_e$ is the area of the exit.

Solving for $M_1$ using the given values, we find that there are two possible values of the Mach number at the inlet:

$M_1=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_{02}}{P_{01}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}=0.747,2.11$.

Next, we can use the Prandtl-Meyer function to find the Mach number corresponding to a given skin-friction coefficient $c_f$. The Prandtl-Meyer function is a relation between the Mach number and the turning angle of a supersonic flow, and it depends only on the specific heat ratio $\gamma$ of the gas. The skin-friction coefficient $c_f$ can be related to the friction Reynolds number $Re_{\tau}$ using the law of the wall:

$c_f=\frac{\tau_w}{\frac{1}{2} \rho_1 V_1^2}=\frac{0.026}{R e_\tau^{0.2}}$,

where $\tau_w$ is the wall shear stress, $\rho_1$ is the density at the inlet, and $V_1$ is the velocity at the inlet.

For air at room temperature and pressure, $\gamma=1.4$. Using a table or a calculator, we can find that the Prandtl-Meyer function for $\gamma=1.4$ is approximately $\nu=30.5^{\circ}$ at $M=0.747$, and $\nu=66.1^{\circ}$ at $M=2.11$.

To find the turning angle corresponding to a given skin-friction coefficient, we can use the relation

$\theta=\frac{c_f}{2 \nu}$

# 数学分析|MATH0004 Analysis 2代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0004 Analysis 2数学分析代写代考辅导服务！

## Instructions:

We will start with the basic properties of real numbers and use them to prove the main results in elementary differential calculus. We assume familiarity with the properties of real numbers such as completeness, order, and the field axioms.

SEQUENCES:

A sequence is a function $f: \mathbb{N} \rightarrow \mathbb{R}$, where $\mathbb{N}$ is the set of natural numbers. We say that a sequence $(a_n)$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n – L| < \epsilon$ for all $n \geq N$. We write $\lim_{n \rightarrow \infty} a_n = L$.

Find the radius of convergence $R$ of $\sum_0^{\infty} \frac{(2 n) ! x^n}{n !(n+1) !}$.

We can apply the ratio test to determine the radius of convergence of the given series. Let

$a_n=\frac{(2 n) ! x^n}{n !(n+1) !}$

Then, the ratio of successive terms is

\begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(2(n+1))!x^{n+1}}{(n+1)!(n+2)!} \cdot \frac{n!(n+1)!}{(2n)!x^n} \ &= \frac{(2n+2)(2n+1)x}{(n+2)(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \frac{(2n+1)(2n+2)}{4(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \left(1 + \frac{1}{n+1}\right) \cdot \left(1 + \frac{2}{n+1}\right). \end{align*}

Taking the limit as $n \to \infty$, we get

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim _{n \rightarrow \infty} \frac{4 x^2}{(n+2)(n+1)} \cdot\left(1+\frac{1}{n+1}\right) \cdot\left(1+\frac{2}{n+1}\right)=4 x^2$.

Therefore, the series converges absolutely if $4x^2 < 1$, or equivalently, $|x| < \frac{1}{2}$. On the other hand, the series diverges if $4x^2 > 1$, or equivalently, $|x| > \frac{1}{2}$. Thus, the radius of convergence is $R = \frac{1}{2}$.

Let $h(n)$ be the largest prime factor of the integer $n>1$, and $s(n)$ be the sum of its prime factors, so $h(12)=3, s(12)=7$.

Prove the sequence ${h(n) / s(n)}, n=2,3,4, \ldots$ has $1 / k$ as a cluster point for every positive integer $k$, but no limit.

To show that $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$, we need to show that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$.

Let $\epsilon>0$ and $k$ be given. Choose $p$ to be any prime greater than $k/\epsilon$, and consider the numbers $n=p^2,p^3,\ldots,p^{k+1}$. Note that $h(n)=p$ and $s(n)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}$.

For any $n=p^m$ with $2\leq m\leq k+1$, we have

$\left|\frac{h(n)}{s(n)}-\frac{1}{k}\right|=\left|\frac{k p-p^{k+1}}{(p-1)\left(p^{k+1}-1\right)}\right|=\frac{p\left(k-p^k\right)}{(p-1)\left(p^{k+1}-1\right)}<\frac{p(k-p)}{(p-1) p^{k+1}}<\frac{k}{p^k-1}<\epsilon$,

where we used the fact that $p>k/\epsilon$ and $p^{k+1}>kp$. Thus, we have shown that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$. Hence, $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$.

To show that the sequence ${h(n)/s(n)}$ has no limit, we need to find two increasing sequences of integers ${a_n}$ and ${b_n}$ such that $h(a_n)/s(a_n)$ and $h(b_n)/s(b_n)$ have different limits as $n\rightarrow\infty$.

Let ${p_n}$ be the sequence of prime numbers, and define $a_n=p_{2n-1}^2$ and $b_n=p_{2n}^2$. Then we have

$\frac{h\left(a_n\right)}{s\left(a_n\right)}=\frac{p_{2 n-1}}{1+p_{2 n-1}} \rightarrow 1, \quad \frac{h\left(b_n\right)}{s\left(b_n\right)}=\frac{p_{2 n}}{1+p_{2 n}+p_{2 n}^2} \rightarrow 0$,

as $n\rightarrow\infty$, since $p_{2n-1}/(1+p_{2n-1})\rightarrow 1$ and $p_{2n}/(1+p_{2n}+p_{2n}^2)\rightarrow 0$. Therefore, the sequence ${h(n)/s(n)}$ does not have a limit as $n\rightarrow\infty$.

A function $f(x)$ has three distinct zeros $a_0<a_1<a_2$ on an interval $I$, and in addition $f^{\prime}\left(a_2\right)=0$. Assume $f(x)$ has a third derivative $f^{\prime \prime \prime}(x)$ at all points of $I$. Prove there is a point $c \epsilon I$ such that $f^{\prime \prime \prime}(c)=0$.

Since $f(x)$ has three distinct zeros on the interval $I$, we know that $f(x)$ changes sign at each of these zeros. Without loss of generality, assume that $f(x)$ is positive on $(a_0, a_1)$, negative on $(a_1, a_2)$, and positive on $(a_2, b)$.

Since $f(x)$ has a local extremum at $x=a_2$ (because $f^{\prime}(a_2)=0$), the second derivative $f^{\prime \prime}(x)$ must change sign at $x=a_2$. Specifically, $f^{\prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Now consider the third derivative $f^{\prime \prime \prime}(x)$. Since $f^{\prime \prime}(x)$ changes sign at $x=a_2$, we know that $f^{\prime \prime \prime}(x)$ must have a local extremum at $x=a_2$. Without loss of generality, assume that $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Since $f(x)$ changes sign at each of the zeros $a_0<a_1<a_2$, we know that $f(x)$ must have at least one local extremum (a maximum or minimum) between each pair of zeros. Specifically, $f(x)$ has a local minimum at some point $c_1$ in $(a_0, a_1)$, a local maximum at some point $c_2$ in $(a_1, a_2)$, and a local minimum at some point $c_3$ in $(a_2, b)$.

Since $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$, we know that $f^{\prime \prime \prime}(c_2)>0$ and $f^{\prime \prime \prime}(c_3)<0$. Therefore, by the intermediate value theorem, there must be some point $c$ in $(c_2, c_3)$ such that $f^{\prime \prime \prime}(c)=0$, as desired.

# 流体力学|Fluid mechanics I 3A1代写2023

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## Instructions:

Fluid mechanics is the branch of physics and engineering that studies the behavior of fluids, which are substances that can flow and take on the shape of their container. Fluids include liquids, gases, and plasmas, and fluid mechanics involves the study of their physical properties, such as viscosity, density, pressure, and flow rate.

Fluid mechanics has a wide range of applications in various fields, such as aerospace engineering, civil engineering, chemical engineering, mechanical engineering, and environmental engineering. It is used to design and analyze systems that involve fluids, such as pumps, turbines, pipes, and airfoils, and to understand phenomena such as turbulence, cavitation, and drag. The study of fluid mechanics also plays a crucial role in understanding natural phenomena, such as weather patterns, ocean currents, and the behavior of the human circulatory system.

A thin symmetric airfoil operating at an angle of attack $\alpha$ has a trailing-edge flap with a hinge line at $80 \%$ of chord. (a) The flap is deflected downwards by an angle $\delta$. Find an expression for the additional lift coefficient due to the flap deflection. You may apply the usual small-angle assumptions.

(a) The additional lift coefficient due to the flap deflection can be calculated using the following equation:

$\Delta C_{L_{\mathrm{flap}}}=\frac{\partial C_L}{\partial \alpha} \cdot \frac{\partial \alpha}{\partial \delta} \cdot \delta$

where $\Delta C_{L_{\text{flap}}}$ is the additional lift coefficient due to flap deflection, $\frac{\partial C_L}{\partial \alpha}$ is the lift slope, $\frac{\partial \alpha}{\partial \delta}$ is the flap effectiveness, and $\delta$ is the flap deflection angle.

For a thin symmetric airfoil, the lift slope is given by:

$\frac{\partial C_L}{\partial \alpha}=2 \pi$

The flap effectiveness can be approximated using the following equation:

$\frac{\partial \alpha}{\partial \delta}=\frac{2}{\pi}\left(\frac{c \text { flap }}{c}\right)$

where $c_{\text{flap}}$ is the chord length of the flap and $c$ is the chord length of the airfoil.

Since the flap hinge line is at $80 %$ of the chord, we can assume that the flap chord length is $20 %$ of the airfoil chord length. Therefore, we have:

$\frac{\partial \alpha}{\partial \delta}=\frac{2}{\pi}\left(\frac{0.2 c}{c}\right)=\frac{0.4}{\pi}$

Substituting these values into the first equation, we obtain:

$\Delta C_{L_{\mathrm{flap}}}=\frac{\partial C_L}{\partial \alpha} \cdot \frac{\partial \alpha}{\partial \delta} \cdot \delta=2 \pi \cdot \frac{0.4}{\pi} \cdot \delta=0.8 \cdot \delta$

Therefore, the additional lift coefficient due to flap deflection is proportional to the flap deflection angle.

(b) If the flap is deflected by $10^{\circ}$, what is the magnitude of the additional lift coefficient?

(b) If the flap is deflected by $10^{\circ}$, the magnitude of the additional lift coefficient is:

$\Delta C_{L_{\mathrm{flap}}}=0.8 \cdot 10^{\circ}=0.139$

(c) If the flap is not deflected, what change in angle of attack $\alpha$ is required to achieve the same lift as that in (b)?

(c) If the flap is not deflected, the lift coefficient is given by:

$C_L=2 \pi \alpha$

To achieve the same lift coefficient as in part (b), we need to find the angle of attack $\alpha$ that satisfies:

$C_L=2 \pi \alpha+\Delta C_{L_{\text {flap }}}=2 \pi \alpha+0.139$

Equating this expression to the lift coefficient without flap deflection, we have:

$2 \pi \alpha+0.139=2 \pi \alpha_0$

where $\alpha_0$ is the angle of attack without flap deflection. Solving for $\alpha$, we obtain:

$\alpha=\alpha_0+\frac{0.139}{2 \pi}$

Therefore, the change in angle of attack required to achieve the same lift as in part (b) is proportional to the additional lift coefficient due to flap deflection, and is independent of the flap deflection angle.