# 数学方法|MATH0016 Mathematical Methods 3代写2023

0

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0016 Mathematical Methods 3数学方法代写代考辅导服务！

## Instructions:

To summarize, the course covers the following topics:

(a) Fourier theory: This topic deals with decomposing a periodic function into a sum of sine and cosine functions, which can help analyze and understand the behavior of the function over time. The Fourier transform is a powerful tool used in signal processing, image analysis, and many other fields.

(b) The calculus of variations: This is a branch of mathematics that deals with finding functions that maximize or minimize an integral, subject to certain conditions. This topic has applications in many fields, including physics, engineering, and economics.

(c) Partial differential equations: These are equations that involve partial derivatives of a function, and they are commonly used to describe physical phenomena such as fluid dynamics and heat transfer. The course will cover both linear and quasilinear partial differential equations of the first and second order.

(d) Vector calculus: This topic deals with the manipulation of vector fields, including the divergence and curl operations. These operations are useful in many areas of physics and engineering, including electromagnetism and fluid dynamics. The course will also cover the divergence and Stokes’ theorem.

I hope this helps clarify what you’ll be studying in the course! Let me know if you have any questions or if there’s anything else I can help you with.

Solve the initial value problem
$$y^{\prime} y^{\prime \prime}-t=0, \quad y(1)=2, \quad y^{\prime}(1)=1 .$$

We can begin by using the method of integration by parts to integrate $y’y”$. Let $u = y’$ and $v’ = y”$, so that $u’ = y”$ and $v = y’$. Then we have

$y^{\prime} y^{\prime \prime}=u v^{\prime}=\frac{1}{2}\left(u^2\right)^{\prime}$.

Using this, we can rewrite the differential equation as

$\frac{1}{2}\left(y^{\prime 2}\right)^{\prime}-t=0$

Integrating once with respect to $t$, we get

$\frac{1}{2} y^{\prime 2}-\frac{1}{2} t^2=C_1$,

where $C_1$ is a constant of integration. We can then solve for $y’$ to get

$y^{\prime}= \pm \sqrt{t^2+2 C_1}$.

Using the initial condition $y(1) = 2$, we get

$y^{\prime}(1)= \pm \sqrt{1+2 C_1}=1$

Solving for $C_1$, we find that $C_1 = 0$. Thus, we have

$y^{\prime}= \pm \sqrt{t^2}, \quad y(1)=2$.

Since $y$ is increasing, we take the positive sign, giving

$y^{\prime}=t, \quad y(1)=2$.

Integrating with respect to $t$, we get

$y=\frac{1}{2} t^2+C_2$.

Using the initial condition $y(1) = 2$, we get $C_2 = \frac{3}{2}$, so the solution to the initial value problem is

$y=\frac{1}{2} t^2+\frac{3}{2}$

Consider the differential equation $y^{\prime}=y(5-y)(y-4)^2$. (a)Determine the critical points (stationary solutions).

To determine the critical points, we need to find the values of $y$ for which $y’=0$. Thus, we solve the equation:

$y^{\prime}=y(5-y)(y-4)^2=0$

This equation is satisfied when $y=0,5$, or $4$. Therefore, the critical points are $y=0,5,$ and $4$.

(b)Sketch the graph of $f(y)=y(5-y)(y-4)^2$.

To sketch the graph of $f(y)=y(5-y)(y-4)^2$, we can use the critical points found in part (a) and the behavior of $f(y)$ as $y$ approaches positive or negative infinity.

First, we note that $f(y)$ changes sign at $y=0,4,$ and $5$. We can determine the sign of $f(y)$ on each interval by testing a point in each interval. For example, for $y<0$, we can test $y=-1$, and we get:

$f(-1)=(-1)(5-(-1))(4-(-1))^2=-120<0$.

Therefore, $f(y)$ is negative on $(-\infty,0)$.

Next, we can consider the behavior of $f(y)$ as $y$ approaches infinity or negative infinity. Since the highest power of $y$ in $f(y)$ is $y^4$, we know that $f(y)$ approaches infinity as $y$ approaches positive or negative infinity.

Putting all of this together, we can sketch the graph of $f(y)$ as follows:

The graph has a local maximum at $y=4$ and a local minimum at $y=5$. The critical point $y=0$ is a double root.

# 数学方法|MATH0010 Mathematical Methods 1代写2023

0

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0010 Mathematical Methods 1数学方法代写代考辅导服务！

## Instructions:

Mathematical methods refer to the techniques and tools used in mathematics to solve problems and explore mathematical concepts. These methods can be used in various fields such as physics, engineering, economics, and computer science. Some of the key mathematical methods include:

1. Calculus: This is a branch of mathematics that deals with functions, limits, derivatives, integrals, and infinite series. Calculus is widely used in physics, engineering, and economics.
2. Linear algebra: This is a branch of mathematics that deals with linear equations, matrices, determinants, and vector spaces. Linear algebra is widely used in fields such as physics, engineering, computer science, and economics.
3. Probability theory: This is a branch of mathematics that deals with the study of random events and the likelihood of their occurrence. Probability theory is widely used in statistics, finance, and engineering.
4. Differential equations: This is a branch of mathematics that deals with equations that involve derivatives. Differential equations are widely used in physics, engineering, and economics.
5. Fourier analysis: This is a branch of mathematics that deals with the study of periodic functions and their representation as a sum of simpler functions. Fourier analysis is widely used in signal processing, image processing, and data analysis.
6. Numerical analysis: This is a branch of mathematics that deals with the development of algorithms and methods for solving mathematical problems using computers. Numerical analysis is widely used in scientific computing and engineering.

These methods are just a few examples of the many mathematical tools available to solve problems and explore concepts in various fields.

Assume that a velocity field, $u=u(x, t)$ exists. Show that the acceleration of an individual car is given by
$$\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}$$

To derive the acceleration of an individual car, we can start with the definition of acceleration, which is the rate of change of velocity with respect to time, i.e.,

$a=\frac{d u}{d t}$

Using the chain rule of differentiation, we can express this as

$a=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} \frac{d x}{d t}$.

Here, the second term on the right-hand side represents the change in velocity due to a change in position, which is given by the product of the velocity gradient $\partial u / \partial x$ and the car’s velocity $dx/dt$. We can rewrite this in terms of the velocity field $u(x,t)$ by using the chain rule again:

$\frac{d x}{d t}=u(x, t)$

Substituting this into the previous equation gives

$a=\frac{\partial u}{\partial t}+u(x, t) \frac{\partial u}{\partial x}$.

This is the desired expression for the acceleration of an individual car in terms of the velocity field.

In how many ways can a positive integer $n$ be written as a sum of positive integers, taking order into account? For instance, 4 can be written as a sum in the eight ways $4=3+1=$ $1+3=2+2=2+1+1=1+2+1=1+1+2=1+1+1+1$

The number of ways a positive integer $n$ can be written as a sum of positive integers, taking order into account, is given by the partition function $p(n)$. Unfortunately, there is no known closed-form expression for $p(n)$.

However, we can compute $p(n)$ recursively using the following recurrence relation:

$p(n)=\sum_{k=1}^n p(n-k)$

where the sum is taken over all positive integers $k$ such that $k \leq n$.

The base case for this recurrence relation is $p(0) = 1$, which represents the empty sum that adds up to $0$.

Using this recurrence relation, we can compute $p(n)$ for small values of $n$:

\begin{align*} p(1) &= 1 \ p(2) &= 2 \ p(3) &= 3 \ p(4) &= 5 \ p(5) &= 7 \ p(6) &= 11 \ p(7) &= 15 \ p(8) &= 22 \ p(9) &= 30 \ p(10) &= 42 \end{align*}

and so on.

For larger values of $n$, computing $p(n)$ using this recurrence relation becomes computationally expensive. However, there are more efficient algorithms for computing $p(n)$, such as the pentagonal number theorem and the Hardy-Ramanujan-Rademacher formula.

Let $p$ be a prime number and $1 \leq k \leq p-1$. How many $k$-element subsets $\left{a_1, \ldots, a_k\right}$ of ${1,2 \ldots, p}$ are there such that $a_1+\cdots+a_k \equiv 0(\bmod p)$ ?

Let $S_k$ denote the number of $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$. We will derive a recursive formula for $S_k$.

Consider a fixed integer $a_1$. Then the remaining $k-1$ integers must sum to $-a_1$ modulo $p$. If $a_1=0$, then we have $S_{k-1}$ choices for the remaining $k-1$ integers. Otherwise, there are $S_{k-1}$ choices for the remaining $k-1$ integers when we consider the $k-1$-element subset of ${1,2,\ldots,p}\setminus{a_1}$ that sums to $-a_1$ modulo $p$. Thus, we have the recursion

$S_k=S_{k-1}+(p-1) S_{k-1}=p S_{k-1}$

where the factor of $p-1$ accounts for the fact that there are $p-1$ choices for $a_1$.

Since $S_1 = p$, we have $S_k = p^k$ for all positive integers $k$. Therefore, there are $p^k$ $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$.

# 数理方法|Mathematical methods 1P4代写2023

0

Assignment-daixieTM为您提供剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写代考辅导服务！

## Instructions:

Mathematical methods are techniques used in mathematics to solve problems and analyze data. These methods are often used in scientific and engineering fields to model real-world phenomena and make predictions. Some examples of mathematical methods include:

1. Calculus: This branch of mathematics deals with the study of functions and their rates of change. It is used to solve problems in physics, engineering, and economics, among other fields.
2. Linear algebra: This is the study of systems of linear equations and their properties. It is used in fields such as computer graphics, physics, and economics.
3. Differential equations: These are mathematical equations that describe how a quantity changes over time. They are used in physics, engineering, and other sciences to model natural phenomena.
4. Statistics: This is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. It is used in fields such as psychology, sociology, and economics.
5. Numerical analysis: This is the study of algorithms and computational methods used to solve mathematical problems. It is used in fields such as engineering, finance, and science to make numerical predictions.

Overall, mathematical methods are essential tools for solving problems and understanding the world around us.

Consider the following equation $$\text { (c.a) } \mathbf{a}+(\mathbf{c} . \mathbf{b}) \mathbf{b}=\mathbf{p}$$ where $\mathbf{a}$ and $\mathbf{b}$ are known vectors which are not parallel. (i) What is the value of $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ ?

(i) We have $\mathbf{p} .(\mathbf{a} \times \mathbf{b})=\mathbf{p} \cdot \mathbf{n}$, where $\mathbf{n}=\mathbf{a} \times \mathbf{b}$ is the normal vector to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$. This dot product gives the scalar projection of $\mathbf{p}$ onto $\mathbf{n}$, which is the signed distance between $\mathbf{p}$ and the plane, multiplied by the magnitude of $\mathbf{n}$. Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the cross product $\mathbf{a} \times \mathbf{b}$ is a nonzero vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$, and therefore defines a plane that contains both vectors. Thus, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ gives the signed distance between $\mathbf{p}$ and the plane containing $\mathbf{a}$ and $\mathbf{b}$, multiplied by the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. In other words, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ is twice the volume of the parallelepiped defined by $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{p}$.

The response $y(t)$ of a linear system with input $f(t)$ is given by the convolution integral $$y(t)=\int_0^t g(t-\tau) f(\tau) d \tau$$ where $g(t)$ is the impulse response. Both $f(t)$ and $g(t)$ are zero for $t<0$. Show that $$\int_0^t g(t-\tau) f(\tau) d \tau=\int_0^t f(t-\tau) g(\tau) d \tau$$

We start by making a change of variables $\tau = t-s$ in the integral on the left-hand side of the equation: \begin{align*} \int_0^t g(t-\tau) f(\tau) d\tau &= \int_0^t g(t-(t-s))f(t-s)ds &&(\text{setting } s = t-\tau)\ &= \int_0^t g(s)f(t-s)ds. \end{align*} This is the convolution of $f(t-s)$ and $g(s)$, which by definition is:

$(f * g)(t)=\int_{-\infty}^{\infty} f(t-s) g(s) d s$.

However, since both $f(t)$ and $g(t)$ are zero for $t < 0$, we can restrict the integral to $s\in [0,t]$:

$(f * g)(t)=\int_0^t f(t-s) g(s) d s$.

Therefore, we have shown that

$\int_0^t g(t-\tau) f(\tau) d \tau=(f * g)(t)=\int_0^t f(t-\tau) g(\tau) d \tau$

as required.

A surface is defined by the equation $$\mathbf{r}=u^2 v \mathbf{i}+v^2 \mathbf{j}+u v \mathbf{k}$$ Find the normal to the surface at the point where $u=1$ and $v=2$.

To find the normal to the surface at a given point, we need to take the gradient of the surface equation at that point.

The gradient of the surface equation is given by:

$\nabla \mathbf{r}=\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}$

So, we need to find the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$, evaluate them at $u=1$ and $v=2$, and then take their cross product to find the normal vector.

Taking the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$ gives:

$\frac{\partial \mathbf{r}}{\partial u}=2 u v \mathbf{i}+v \mathbf{k}, \quad \frac{\partial \mathbf{r}}{\partial v}=u^2 \mathbf{i}+2 v \mathbf{j}+u \mathbf{k}$

Evaluating these partial derivatives at $u=1$ and $v=2$ gives:

$\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)}=4 \mathbf{i}+2 \mathbf{k},\left.\quad \frac{\partial \mathbf{r}}{\partial v}\right|{(1,2)}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$

Taking their cross product, we get:

$\left.\nabla \mathbf{r}\right|{(1,2)}=\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)} \times\left.\frac{\partial \mathbf{r}}{\partial v}\right|_{(1,2)}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 0 & 2 \ 1 & 4 & 2\end{array}\right|=-8 \mathbf{i}+6 \mathbf{j}-16 \mathbf{k}$

Therefore, the normal to the surface at the point $(1,2)$ is $\boxed{-8\mathbf{i} + 6\mathbf{j} – 16\mathbf{k}}$.