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环、场和伽罗瓦理论|MATH4062 Rings, Fields and Galois Theory代写 Sydney代写

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这是一份Sydney悉尼大学MATH4062的成功案例

环、场和伽罗瓦理论|MATH4062 Rings, Fields and Galois Theory代写 Sydney代写


问题 1.

Suppose that $L: K$ is normal. To prove that $\operatorname{Gal}(M: L)$ is a normal subgroup of $\operatorname{Gal}(M: K)$, let $\varphi \in \operatorname{Gal}(M: K)$ and $\theta \in \operatorname{Gal}(M: L)$. We show that $\varphi^{-1} \theta \varphi \in \operatorname{Gal}(M: L)$, or equivalently,
$$
\varphi^{-1} \theta \varphi(\alpha)=\alpha \text { for all } \alpha \in L
$$
or equivalently, $\theta \varphi(\alpha)=\varphi(\alpha)$ for all $\alpha \in L$.
$\varphi(\alpha) \in L$ for all $\alpha \in L$, so $\theta(\varphi(\alpha))=\varphi(\alpha)$ since $\theta \in \operatorname{Gal}(M: L)$. This completes the proof that $\operatorname{Gal}(M: L) \leqslant \operatorname{Gal}(M: K)$.

证明 .

Finally, we prove the statement on quotients (still supposing that $L: K$ is a normal extension). Every automorphism $\varphi$ of $M$ over $K$ satisfies $\varphi L=L$, and therefore restricts to an automorphism $\hat{\varphi}$ of $L$. The function
$$
\begin{array}{rlc}
v: \quad \operatorname{Gal}(M: K) & \rightarrow & \operatorname{Gal}(L: K) \
\varphi & \mapsto & \hat{\varphi}
\end{array}
$$
is a group homomorphism, since it preserves composition. Its kernel is $\operatorname{Gal}(M$ : $L)$, by definition. If we can prove that $v$ is surjective then the last part of the theorem will follow from the first isomorphism theorem.





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MATH4062 COURSE NOTES ：



Then $W$ is defined by $n$ homogeneous linear equations in $n+1$ variables, so it is a nontrivial $M$-linear subspace of $M^{n+1}$.
Claim: let $\left(x_{0}, \ldots, x_{n}\right) \in W$ and $\varphi \in H$. Then $\left(\varphi\left(x_{0}\right), \ldots, \varphi\left(x_{n}\right)\right) \in W$.
Proof: For all $\theta \in H$, we have
$$
x_{0}\left(\varphi^{-1} \circ \theta\right)\left(\alpha_{0}\right)+\cdots+x_{n}\left(\varphi^{-1} \circ \theta\right)\left(\alpha_{n}\right)=0
$$
since $\varphi^{-1} \circ \theta \in H$. Applying $\varphi$ to both sides gives that for all $\theta \in H$,
$$
\varphi\left(x_{0}\right) \theta\left(\alpha_{0}\right)+\cdots+\varphi\left(x_{n}\right) \theta\left(\alpha_{n}\right)=0,
$$
proving the claim.


















度量空间|MATH4061 Metric Spaces代写 Sydney代写

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这是一份Sydney悉尼大学MATH4061的成功案例

度量空间|MATH4061 Metric Spaces代写 Sydney代写


问题 1.

set $U_{2}$ to get $x_{2} \in B\left(x_{1}, r_{1}\right) \cap U_{2}$. Again, we can find $r_{2}$ such that $0<r_{2}<2^{-2}$ and $B\left[x_{2}, r_{2}\right] \subset B\left(x_{1}, r_{1}\right) \cap U_{2}$. Proceeding this way, for each $n \in \mathbb{N}$, we get $x_{n} \in X$ and an $r_{n}$ with the properties
$$
B\left[x_{n}, r_{n}\right] \subset B\left(x_{n-1}, r_{n-1}\right) \cap U_{n} \text { and } 0<r_{n}<2^{-n}
$$

证明 .

Clearly, the sequence $\left(x_{n}\right)$ is Cauchy: if $m \leq n$,
$$
d\left(x_{m}, x_{n}\right) \leq d\left(x_{n}, x_{n-1}\right)+\cdots+d\left(x_{m+1}, x_{m}\right) \leq \sum_{k=m}^{n} 2^{-k}
$$
Since $\sum_{k} 2^{-k}$ is convergent, it follows that $\left(x_{n}\right)$ is Cauchy.





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MATH4061 COURSE NOTES ：


We shall sketch the argument. Let $m, M$ be such that
$$
m \leq \frac{\partial f}{\partial y}(x, y) \leq M, \text { for }(x, y) \in D
$$
Let $X:=\left(C[a, b], |_{\infty}\right)$. Consider
$$
T(\varphi)(x):=\varphi(x)-\frac{1}{M} f(x, \varphi(x)), \text { for } x \in[a, b]
$$
One shows that $T$ is a contraction by an obvious use of the mean value theorem.


















高级复杂系统|MATH3977 Complex Analysis (Advanced)代写 Sydney代写

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这是一份Sydney悉尼大学MATH3977的成功案例

高级复杂系统|MATH3977 Complex Analysis (Advanced)代写 Sydney代写


问题 1.

Let us define $r(x)$ by
$$
\sum_{p \leq x} \log p=x(1+r(x))
$$
so we have $r(x) \rightarrow 0$ for $x \rightarrow \infty$ by VII.4.4.
Trivially, the following holds:
$$
\sum_{p \leq x} \log p \leq \pi(x) \log x
$$
and hence
$$
\pi(x) \geq \frac{x}{\log x}(1+r(x))
$$

证明 .

$$
\pi(x) \leq \frac{x}{\log x}(1+r(x)) q^{-1}+x^{q} .
$$
This inequality can now be specified for a suitable value of $q$, namely for $q=1-$ $1 / \sqrt{\log x}(x \geq 2)$. Then we have
$$
\pi(x) \leq \frac{x}{\log x}(1+R(x))
$$
with
$$
R(x)=-1+(1+r(x))\left(1-\frac{1}{\sqrt{\log x}}\right)^{-1}+(\log x) x^{-1 / \sqrt{\log x}} .
$$
It obviously holds $R(x) \rightarrow 0$ for $x \rightarrow \infty$.





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MATH3977 COURSE NOTES ：


For $t \in \mathbb{R}$ we define
$$
\beta(t):=t-[t]-1 / 2 \quad([t]:=\max {n \in \mathbb{Z}, n \leq t})
$$
Then we have $\beta(t+1)=\beta(t)$ and $|\beta(t)| \leq \frac{1}{2}$.
The integral
$$
F(s):=\int_{1}^{\infty} t^{-s-1} \beta(t) d t
$$
converges absolutely for $R e(s)>0$, and represents in this right half-plane an analytic function $F$. For $\operatorname{Re}(s)>1$ it holds:
$$
\zeta(s)=\frac{1}{2}+\frac{1}{s-1}-s F(s)
$$


















高级拉格朗日和哈密尔顿动力学|MATH3977 Lagrangian and Hamiltonian Dynamics (Adv)代写 Sydney代写

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这是一份Sydney悉尼大学MATH3977的成功案例

高级拉格朗日和哈密尔顿动力学|MATH3977 Lagrangian and Hamiltonian Dynamics (Adv)代写 Sydney代写


问题 1.

We assumed we had a holonomic system and the $q$ ‘s were all independent, so this equation holds for arbitrary virtual displacements $\delta q_{j}$, and therefore
$$
\frac{d}{d t} \frac{\partial T}{\partial \dot{q}{j}}-\frac{\partial T}{\partial q{j}}-Q_{j}=0
$$
Now let us restrict ourselves to forces given by a potential, with $\vec{F}{i}=$ $-\vec{\nabla}{i} U({\vec{r}}, t)$, or
$$
Q_{j}=-\sum_{i} \frac{\partial \vec{r}{i}}{\partial q{j}} \cdot \vec{\nabla}{i} U=-\left.\frac{\partial \tilde{U}({q}, t)}{\partial q{j}}\right|_{t}
$$

证明 .

Notice that $Q_{j}$ depends only on the value of $U$ on the constrained surface.
Also, $U$ is independent of the $\dot{q}{i}$ ‘s, so $$ \frac{d}{d t} \frac{\partial T}{\partial \dot{q}{j}}-\frac{\partial T}{\partial q_{j}}+\frac{\partial U}{\partial q_{j}}=0=\frac{d}{d t} \frac{\partial(T-U)}{\partial \dot{q}{j}}-\frac{\partial(T-U)}{\partial q{j}}
$$
or
$$
\frac{d}{d t} \frac{\partial L}{\partial \dot{q}{j}}-\frac{\partial L}{\partial q{j}}=0
$$
This is Lagrange’s equation, which we have now derived in the more general context of constrained systems.





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MATH3977 COURSE NOTES ：


$$
\tilde{u}=\dot{i}=0 \quad \text { as } z \rightarrow \infty
$$
Hence
$$
\tilde{u}=C_{3} e^{-(1+i) z \delta_{E}}+C_{4} e^{-(1-i) z / \delta_{L}},
$$
while from (4.3.13)
$$
\tilde{v}=-i C_{3} e^{-(1+i) z^{\prime} \delta E}+i C_{4} e^{-(1-i) z / \delta_{E}} .
$$
On $z=0$,
$$
\begin{aligned}
&\tilde{v}=0 \
&\tilde{u}=-U
\end{aligned}
$$
so that
$$
C_{3}=C_{4}=-\frac{U}{2}
$$

















高级流体动力学 |MATH3974 Fluid Dynamics (Advanced)代写 Sydney代写

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这是一份Sydney悉尼大学MATH3974的成功案例

高级流体动力学 |MATH3974 Fluid Dynamics (Advanced)代写 Sydney代写


问题 1.

where $f=2 \Omega$. We may align the $x$-axis in the direction of the velocity at infinity, so that the boundary condition at $z \rightarrow \infty$ is simply
$$
\begin{aligned}
&u=U, \
&r=0, \quad z \rightarrow \infty \
&w=0
\end{aligned}
$$
At the rigid surface $z=0$ we suppose that the friction, in direct analogy with the role of molecular viscosity, inhibits the fluid motion to such an extent that both the normal and tangential velocities vanish on $z=0$, i.e.,
$$
u=v=w=0 \quad(z=0) .
$$

证明 .

The crucial condition is the condition of no slip on the tangential volocities $u$ and $v$. In the absence of retarding frictional forces, this boundary condition would be absent. Then, the uniform velocity $U$ would be an exact solution of $(4.3 .1 \mathrm{a}, \mathrm{b}, \mathrm{c})$, i.e.,
$$
\begin{aligned}
&u=U=-\frac{1}{\rho f} \frac{\hat{\partial p}}{\partial y}, \
&v=w=0,
\end{aligned}
$$





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MATH3974 COURSE NOTES ：


Since the fluid is homogeneous, it follows from that
or
$$
\begin{gathered}
\hat{\mathcal{c}}|\hat{\rho} p / \partial x| \
\hat{\partial z}|\hat{\hat{p} p / \partial y}|
\end{gathered}=0,
$$
so that the horizontal pressure gradient must be independent of $z$. Now as $z \rightarrow \infty$ both $u$ and $v$ become constant. so that for very large $z$, using
$$
\begin{gathered}
0=-\frac{1}{\rho} \frac{\hat{\partial} p}{\partial x} \
f U=-\frac{1 \hat{c} p}{\rho \hat{c} y}
\end{gathered}
$$

















测量理论和傅里叶分析 |MATH3969 Measure Theory and Fourier Analysis代写 Sydney代写

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这是一份Sydney悉尼大学MATH3969的成功案例

测量理论和傅里叶分析 |MATH3969 Measure Theory and Fourier Analysis代写 Sydney代写


The measures $\mu^{+}$and $\mu^{-}$constructed above are called the positive and negative parts of $\mu$, respectively. The measure
$$
|\mu|=\mu^{+}+\mu^{-}
$$
is called the total variation of $\mu$. The quantity
$$
|\mu|=|\mu|(X)
$$
is called the variation or the variation norm of $\mu$.
The decomposition $\mu=\mu^{+}-\mu^{-}$is called the Jordan or Jordan-Hahn decomposition.

One should not confuse the measure $|\mu|$ with the set function $A \mapsto|\mu(A)|$, which, typically, is not additive (e.g., if $|\mu|>\mu(X)=0$ ).

We observe that the measures $\mu^{+}$and $\mu^{-}$have the following properties that could be taken for their definitions:
$$
\begin{aligned}
&\mu^{+}(A)=\sup {\mu(B): B \subset A, B \in \mathcal{A}} \
&\mu^{-}(A)=\sup {-\mu(B): B \subset A, B \in \mathcal{A}}
\end{aligned}
$$





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MATH3969 COURSE NOTES ：


belongs to $\mathcal{A}^{n}$ and the function
$$
\left(x_{1}, \ldots, x_{n}\right) \mapsto \mu^{(n)}\left(A^{x_{1}, \ldots, x_{n}}\right)
$$
is measurable with respect to $\bigotimes_{i=1}^{n} \mathcal{A}{i}$. Denote by $B{1}^{k}$ the set of all points $x_{1}$ such that
$$
\mu^{(1)}\left(A_{k}^{x_{1}}\right)>\varepsilon / 2 .
$$
Then $B_{1}^{k} \in \mathcal{A}{1}$ and $\mu{1}\left(B_{1}^{k}\right)>\varepsilon / 2$, which follows by Fubini’s theorem for finite products and the inequality $\mu\left(A_{k}\right)>\epsilon$. Indeed, $A_{k}=C_{m} \times X_{m+1} \times \cdots$

















微分几何学|MATH3968 Differential Geometry代写 Sydney代写

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这是一份Sydney悉尼大学MATH3968的成功案例

微分几何学|MATH3968 Differential Geometry代写 Sydney代写


when expressed in the language of the symplectic form $\omega$, the Jacobi identity turns out to be equivalent to its closedness
$$
{f,{g, h}}+{h,{f, g}}+{g,{h, f}}=0 \quad \Leftrightarrow \quad d \omega=0
$$
when expressed in the language of the Poisson tensor $\mathcal{P}$, the Jacobi identity turns out to be equivalent to its invariance with respect to an arbitrary Hamiltonian field ${ }^{250}$
$$
{f,{g, h}}+{h,{f, g}}+{g,{h, f}}=0 \quad \Leftrightarrow \quad \mathcal{L}{\zeta \uparrow} \mathcal{P}=0, \psi \in \mathcal{F}(M) $$ and in components this gives the differential identity $$ \mathcal{L}{\zeta_{f}} \mathcal{P}=0 \text { for all } f \in \mathcal{F}(M) \quad \Leftrightarrow \quad \mathcal{P}^{r[i} \mathcal{P}_{, \gamma}^{j k]}=0
$$
So this is the identity (mentioned above) which each Poisson tensor is to satisfy.





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MATH3968 COURSE NOTES ：

The coordinates $\left(q^{a}, p_{a}\right)$, in which the symplectic form $\omega$ looks like this and whose existence is guaranteed by the Darboux theorem, are called the canonical coordinates in canonical coordinates the corresponding Poisson tensor is
$$
\mathcal{P} \equiv-\omega^{-1}=\frac{\partial}{\partial p_{a}} \wedge \frac{\partial}{\partial q^{a}}:=\frac{\partial}{\partial p_{a}} \otimes \frac{\partial}{\partial q^{a}}-\frac{\partial}{\partial q^{a}} \otimes \frac{\partial}{\partial p_{a}}
$$
in canonical coordinates the Hamiltonian field is
$$
\zeta_{f}=\left(\partial_{j} f\right) \mathcal{P}^{j /} \partial_{i} \equiv \frac{\partial f}{\partial p_{a}} \frac{\partial}{\partial q^{a}}-\frac{\partial f}{\partial q^{a}} \frac{\partial}{\partial p_{a}}
$$

















偏微分方程和波|MATH3078/MATH3978 PDEs and Waves代写 Sydney代写

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这是一份Sydney悉尼大学MATH3078/MATH3978的成功案例

偏微分方程和波|MATH3078/MATH3978 PDEs and Waves代写 Sydney代写


Note that $c=2, f(x)=\sin x$ and $g(x)=2 \cos x$. Proceeding as before, we set
$$
f^{(2 n)}(x)=(-1)^{n} \sin x, n=0,1,2, \cdots
$$
and
$$
g^{(2 n)}(x)=2(-1)^{n} \cos x, n=0,1,2, \cdots
$$
The solution in a series form is readily obtained by substituting and given by
$$
\begin{aligned}
u(x, t)=& \sin x\left(1-\frac{(2 t)^{2}}{2 !}+\frac{(2 t)^{4}}{4 !}+\cdots\right) \
&+\cos x\left((2 t)-\frac{(2 t)^{3}}{3 !}+\frac{(2 t)^{5}}{5 !}-\cdots\right)
\end{aligned}
$$
and in a closed form by
$$
\begin{aligned}
u(x, t) &=\sin x \cos (2 t)+\cos x \sin (2 t) \
&=\sin (x+2 t)
\end{aligned}
$$





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MATH3078/MATH3978 COURSE NOTES ：

To determine the function $F(x)$, we solve the second order linear ODE
$$
F^{\prime \prime}(x)+\lambda^{2} F(x)=0,
$$
to find that
$$
F(x)=A \cos (\lambda x)+B \sin (\lambda x),
$$
where $A$ and $B$ are constants. To determine the constants $A, B$, and $\lambda$, we use the homogeneous boundary conditions
$$
u(0, t)=0, u(L, t)=0,
$$
as given above gives
$$
F(0)=0, F(L)=0 .
$$
Using leads to
$$
A=0
$$
and
$$
\sin (\lambda L)=0 .
$$

















数学计算|MATH3076/MATH3976 Mathematical Computing代写 Sydney代写

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这是一份Sydney悉尼大学MATH3076/MATH3976 的成功案例

数学计算|MATH3076/MATH3976 Mathematical Computing代写 Sydney代写


Disjoint union axiom): Let $X_{1}$ and $X_{2}$ be compact oriented surfaces with boundaries whose labels are $\ell_{1}$ and $\ell_{2}$ respectively.
$$
F\left(X_{1} \text { I } X_{2}, \ell_{1} \text { Ш } \ell_{2}\right)=F\left(X_{1}, \ell_{1}\right) \otimes F\left(X_{2}, \ell_{2}\right) .
$$
(Gluing axiom): Let $\tilde{X}$ be a compact oriented surface obtained by sluing together a pair of boundary circles with dual labels in $X$.
$$
F(\tilde{X}, \ell)=\bigoplus_{x \in L} F(X, \ell \cup{x, \hat{x}})
$$
Duality axiom): Let $X^{}$ stnds for $X$ with the reversed orientation and labels applied $.$ $$ F\left(X^{}\right)=(F(X))^{} \text {, where we mean that } A^{}={ }^{t} \bar{A} .
$$





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MATH3076/MATH3976 COURSE NOTES ：


Let $X$ be a topological space. For any $p, q \in X$ if there is a continuous $\operatorname{map} f:[0,1] \rightarrow X$ such that $f(0)=p, f(1)=q$, then $X$ is called arcwise connected and such a map is called a path in $X$. In what follows, we always consider an arcwise connected topological space and so we refer it simply “a space” hereafter.

Let $X$ be a space, and for $p, q, r \in X$ let $f$ be a path connecting $p$ with $q$ and $g$ a path connecting $q$ with $r$. Then we define a product of paths, $f \cdot g$, as follows:
$$
f \cdot g(s)= \begin{cases}f(2 s) & (0 \leq s \leq 1 / 2) \ g(2 s-1) & (1 / 2 \leq s \leq 1)\end{cases}
$$
The $f \cdot g$ is a new path in $X$ connecting $p$ with $r$.
When there is a path $l$ such that the starting point and the end point coincide, i.e., a continuous map
$$
l:[0,1] \rightarrow X, l(0)=p, l(1)=p
$$
is called a loop in $X$ with a base point $p$.
















代数和逻辑|MATH3066 Algebra and Logic代写 Sydney代写

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这是一份Sydney悉尼大学MATH3066的成功案例

代数和逻辑|MATH3066 Algebra and Logic代写 Sydney代写


Therefore
$$
a b x+a b x^{\prime}<a x+b x^{\prime}
$$
or
$$
a b<a x+b x^{\prime}
$$
On the other hand,
$$
\begin{aligned}
&(a<a+b)<[a x<(a+b) x] \
&(b<a+b)<\left[b x^{\prime}<(a+b) x^{\prime}\right]
\end{aligned}
$$
Therefore
$$
a x+b x^{\prime}<(a+b)\left(x+x^{\prime}\right)
$$
or
$$
a x+b x^{\prime}<a+b
$$
To sum up,
$$
a b<a x+b x^{\prime}<a+b .
$$
Q. E. D.
Remark 1. This double inclusion may be expressed in the following form:
$$
f(b)<f(x)<f(a) .
$$
For
$$
\begin{gathered}
f(a)=a a+b a^{\prime}=a+b \
f(b)=a b+b b^{\prime}=a b
\end{gathered}
$$





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MATH3066 COURSE NOTES ：


The same conclusion may be reached by observing that abis the inferior limit of the function $a x+b x^{\prime}$, and that consequently the function can not vanish unless this limit is 0 .
$$
\left(a b<a x+b x^{\prime}\right)\left(a x+b x^{\prime}=0\right)<(a b=0) .
$$
We can express the resultant of elimination in other equivalent forms; for instance, if we write the equation in the form
$$
\left(a+x^{f}\right)(b+x)=0
$$
we observe that the resultant
$$
a b=0
$$
is obtained simply by dropping the unknown quantity (by suppressing the terms $x$ and $\left.x^{\prime}\right)$. A gain the equation may be written:
$$
a^{\prime} x+b^{\prime} x^{\prime}=1
$$