液体和等离子体|PHYS3202 Fluids and Plasma代写

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“Fluids and Plasmas” is an applied physics course designed to give the students experience in working with, predicting and measuring the behaviour of fluid flows and plasmas. The course begins with an outline of the fluid equations of motion, which lead to solutions for waves in fluids, convection and buoyancy-driven flows. 

这是一份anu澳大利亚国立大学PHYS3202的成功案例

液体和等离子体|PHYS3202 Fluids and Plasma代写


问题 1.

Here we consider some special cases of $(100)$ obtained by specializing $a, b, c$, and $d$ in $H$ of (77). Our choices for these four functions will determine the structure of the first integrals $x_{0}$ and $y_{0}$ through (99). For the cases we consider, their structure will be easy to discern and will give some insight into the behavior of $\xi$. How $\mathbf{B}_{p}$ will propagate in each case is pointed out to make the discussion more physically concrete. To conclude, a physical interpretation for the terms of $H$ and the role they play in determining how solutions propagate are discussed as well.

证明 .

The first case we consider is a rather drastic simplification of the general result (99): we take $a, b, c$, and $d$ all to be zero, getting rid of Hentirely. Then we are simply left with
$$
x_{0}=x \quad \text { and } \quad y_{0}=y .
$$
Thus, in this case, the general solution for $\xi$ is of the form
$$
\xi=\xi(x, y, z-\gamma \tau),
$$
which corresponds to a structure propagating toroidally with specd $\gamma$.

$$
\psi=(1 / \gamma)\left(\xi-\alpha \nabla_{1}^{2} \xi\right)(x, y, z,-\gamma \tau) .
$$
The arguments in parentheses stress that $\psi$ moves in exactly the same way as $\xi$ : surfaces of constant poloidal flux simply propagate in the $z$ direction with constant velocity $\gamma$. Applying $\mathbf{B}{p}=-\epsilon B{T} \hat{\mathbf{z}} \times \nabla_{1} \psi$ to (105) shows that the disturbance $\mathbf{B}{p}$ also propagates in the same way: if we follow a point moving along a characteristic curve, $\mathbf{B}{p}$ at the point will be a constant vector. However, from $(105)$ and the arguments given at the end of Sec. III F, the solution is not necessarily an Alfvén-like wave because, in general, $\mathbf{B}{p}$ will not be proportional to $\mathbf{v}{t}$ for this case.






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PHYS3202 COURSE NOTES :

Having introduced the fluid equations, we next discuss a method for arriving at exact solutions of them.

We denote the partial derivative of a quantity by a subscript, e.g., $\partial U / \partial \tau \equiv U_{\tau}$. Then, after rearranging the terms of $(9)$ and $(10)$ and subtracting (14) from (9), we can write
$$
\begin{aligned}
&U_{r}+[\phi, U]+J_{z}+[J, \psi]=0, \
&\psi_{r}+(\phi-\alpha \chi){z}+[\phi-\alpha \chi, \psi]=0, \end{aligned} $$ This is the nonlinear system we will study. Note that we are taking $\hat{\eta}=0$ in (16); the resistivity of the plasma is neglected for all that follows. To satisfy (17) we take $$ \chi=g(z)+U, $$ where $g$ is an arbitrary function of $z$. This is by no means the general solution to (17); it is simply a special case that satisfies (17) with little effort. Defining $$ \begin{aligned} &\xi \xi \phi-\alpha g(z), \ &\text { and recasting (15) and (16) in terms of } \xi \text { gives } \ &\qquad U{+}+[\xi, U]+J_{z}+[J, \psi]=0 \
&\text { and } \
&\qquad \psi_{\tau}+(\xi-\alpha U){z}+[\xi-\alpha U, \psi]=0, \end{aligned} $$ where (18) has been used. We note in passing that from (19) and (6), the definition of $U$, we have $$ U=\nabla{1}^{2} \xi,
$$
a relation that will be used often in what follows.
Now we have to find solutions to (20) and (21). Let us first consider the simpler case of axisymmetric equilibrium.




















物理基础1A| Physics 1A: Foundations PHYS08016代写

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This is an introductory-level course, covering the classical physics of kinematics, dynamics, oscillations, forces and fields, and touching on aspects of contemporary physics, including relativity and chaos. The course is designed for those with qualifications in physics and mathematics at SCE-H level or equivalent.

这是一份anu澳大利亚国立大学PHYS08016的成功案例

公司财务|FINM2001 Corporate Finance代写


问题 1.

During an IFR flight, a passenger looks out the window while relaxing in his seat. He observes a turn and estimates the bank angle to be $30^{\circ}$. At the same time, the passenger observes the free surface of the orange juice in his glass: it is parallel to the tray.
a) The passenger assumes the turn being flown as a rate one turn. Explain the term rate one turn. Why is it correct to assume a rate one turn.
b) The passenger assumes the turn being flown as a coordinated turn. Explain the term coordinated turn. Why is it correct to assume a coordinated turn.
c) Calculate the aircraft’s true airspeed.

证明 .

a) A rate one turn is a turn with a heading change of $180^{\circ}$ in 60 seconds. During IFR flights turns are performed as rate one turns.
b) In a coordinated turn (correctly banked turn)

  • the lift force lies in the aircraft plane of symmetry,
  • the ball in the turn and slip indicator is centered,
  • there is no acceleration along the $y$-axis of the aircraft.
    This phenomenon is also shown by the free surface of the orange juice in the glass which is parallel to the tray.
    c) $\tan \Phi=\frac{V \cdot \Omega}{g}$
    $$
    V=\frac{g}{\Omega} \cdot \tan \Phi=\frac{9.81 \mathrm{~m} \cdot 60 \mathrm{~s}}{\mathrm{~s}^{2} \cdot \pi} \cdot \tan 30^{\circ}=108.2 \frac{\mathrm{m}}{\mathrm{s}}=210 \mathrm{kt}
    $$
    Answer: The aircraft’s true airspeed is $210 \mathrm{kt}$.






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FINM2001 COURSE NOTES :

The category of effect of a failure is judged to be hazardous. Following $A C J N o$. 1 . to $J A R$ $25.1309$
a) What is the largest permissible failure probability?
b) What is the mean time to failure $M T T F$ ?
Solution
$F(t) \quad$ probability of failure,
$\lambda \quad$ failure rate
MTTF mean time to failure
FH flight hour
a) hazardous : $F(t=1 \mathrm{FH}) \leq 10^{-7}$
b) For small probabilities of failure: $\quad \lambda \approx F / t=10^{-7} \cdot \frac{1}{\mathrm{FH}}$
$$
\mathrm{MTTF}=1 / \lambda=\frac{1}{10^{-7}} \mathrm{FH}=10000000 \mathrm{FH}
$$
Answer: If a failure has a hazardous effect, the mean time to this failure may not be less than $10000000 \mathrm{FH}$.




















公司财务|FINM2001 Corporate Finance代写

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This course focuses on tools and techniques used in modern financial management. Material in the course has an applied focus and is designed to provide students with the knowledge and skills required for understanding, exploring and analysing financial management issues. The course draws upon topical material in order to contextualise theoretical discussion, and present students with examples in practice.

这是一份anu澳大利亚国立大学FINM1000的成功案例

公司财务|FINM2001 Corporate Finance代写


问题 1.

Throughout the late 1950 ‘s, Myron J. Gordon (initially working with Ezra Shapiro) formalised the impact of distribution policies and their associated returns on current share price using the derivation of a constant growth formula, the mathematics for which are fully explained in the CVT text.

Required:

  1. Present a mathematical summary of the Gordon Growth Model under conditions of certainty.
  2. Comment on its hypothetical implications for corporate management seeking to maximise shareholder wealth.

证明 .

These questions not only provide an opportunity to test your understanding of the companion text, but also to practise your written skills and ability to editorialise source material.

  1. The Gordon Model
    According to Gordon (1962) movements in ex-div share price $\left(\mathrm{P}_{2}\right.$ ) under conditions of certainty relate to the profitability of corporate investment and not dividend policy.

Using Gordon’s original notation and our Equation numbering from $C V T$ (Chapter Three) where $\mathrm{K}{e}$ represents the equity capitalisation rate; $E{1}$ equals next year’s post-tax earnings; $b$ is the proportion retained; (1-b) $E_{1}$ is next year’s dividend; $r$ is the return on reinvestment and r.b equals the constant annual growth in dividends:
(16) $\quad \mathrm{P}{0}=(1-\mathrm{b}) \mathrm{E}{1} / \mathrm{K}{\varepsilon}-\mathrm{rb} \quad$ subject to the proviso that $\mathrm{K}{\varepsilon}>$ r.b for share price to be finite.
You will also recall that in many Finance texts today, the equation’s notation is simplified with $D_{1}$ and g representing the dividend term and growth rate, subject to the constraint that $\mathrm{K}{\varepsilon}>\mathrm{g}$ (17) $\mathrm{P}{\mathrm{b}}=\mathrm{D}{1} / \mathrm{K}{\mathrm{c}}-\mathrm{g}$

  1. The Implications
    In a world of certainty, Gordon’s analysis of share price behaviour confirms the importance of Fisher’s relationship between a company’s return on reinvestment $(\mathrm{r})$ and its shareholders’ opportunity cost of capital rate $(\mathrm{K})$ ).






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FINM2001 COURSE NOTES :

Moving into a world of uncertainty, Gordon (op cit) explains why rational-risk averse investors are no longer indifferent to managerial decisions to pay a dividend or reinvest earnings on their behalf, which therefore impacts on share price.

Required:

  1. Present a mathematical summary of the difference between the Gordon Growth Model under conditions of certainty and uncertainty.
  2. Comment on its hypothetical implications for corporate management seeking to maximise shareholder wealth.
    An Indicative Outline Solution
    Again, these questions provide opportunities to test your understanding of the companion text and practise your written and editorial skills.
  3. The Gordon Model and Uncertainty
    According to Gordon (ibid) movements in share price under conditions of uncertainty relate to dividend policy, rather than investment policy and the profitability of corporate investment. He begins with the basic mathematical growth model:
    (16) $\mathrm{P}{0}=(1-\mathrm{b}) \mathrm{E}{1} / \mathrm{K}{c}-\mathrm{rb} \quad$ subject to the proviso that $\mathrm{K}{\varepsilon}>$ r.b for share price to be finite.
    This again simplifies to:
    (17) $\mathrm{P}{0}=\mathrm{D}{1} / \mathrm{K}{c}-\mathrm{g}$ subject to the constraint that $\mathrm{K}{c}>\mathrm{g}$
    But now, the overall shareholder return (equity capitalisation rate) is no longer a constant but a function of the timing and size of the dividend payout. Moreover, an increase in the retention ratio also results in a further rise in the periodic capitalisation rate. Expressed mathematically:
    $$
    \mathrm{K}{\varepsilon}=f\left(\mathrm{~K}{\mathrm{el}}<\mathrm{K}{e 2}<\ldots \mathrm{K}{\mathrm{en}}\right)
    $$
  4. The Implications
    According to Gordon’s uncertainty hypothesis, rational, risk averse investors adopt a “bird in the hand” philosophy to compensate for the non-payment of future dividends.

They prefer dividends now, rather than later, even if retentions are more profitable than distributions (i.e. $r>\mathrm{K}{\mathrm{e}}$ ). They prefer high dividends to low dividends period by period. (i.e. $\mathrm{D}{1}>\mathrm{D}{2}$ ). . Near dividends and higher payouts are discounted at a lower rate ( $\mathrm{K}{\mathrm{ct}}$ now dated) ,
Thus, investors require a higher overall average return on equity $\left(\mathrm{K}_{c}\right.$ ) from firms that retain a higher proportion of earnings with obvious implications for share price. It will fall.




















金融学基础|FINM1001 Foundations of Finance代写

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This course is designed to familiarise students with the components of the financial system as well as to introduce them to the three basic ideas underpinning finance, namely the time value of money, diversification and arbitrage. In doing so, the course provides students with introductory exposure to financial transactions, institutions and markets including money markets, stock markets, foreign exchange and derivative markets and the instruments traded therein. It also provides students with a solid foundation for later studies in finance.

这是一份anu澳大利亚国立大学FINM1000的成功案例

金融学基础|FINM1001 Foundations of Finance代写


问题 1.

If you wish to provide 520,000 for your newborn’s University education, how much should you imvest now, given the interest rate that will accrue on the investment is $10 \%$ p.a compounded monthly?

证明 .

In order to determine how much you should invest now, calculate the present value of $\$ 20,000$ received 18 years from now, bearing in mind that interest is compounded monthly.
$$
\begin{aligned}
P V &=\frac{\$ 20,000}{\left(1+\frac{0.10}{12}\right)^{12_{x} 13}} \
&=\$ 3,330.73
\end{aligned}
$$






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FINM1001 COURSE NOTES :

A company needs $\$ 10,000$ in 5 years to replace a piece of equipment. How much must be invested each year at $8 \%$ p.a compounded semi-annualy in order to provide for this replacement?

To determine the amount the company must invest annually, simply use the future value of an annuity formula, bearing in mind that the interest rate is compounded semi-annually. Therefore, as investments will be made on an annual basis, we must calculate an annual effective interest rate to use in the annuity calculation.
$$
\begin{aligned}
r &=\left(1+\frac{0.08}{2}\right)^{2}-1 \
&=0.0816 \
&=8.16 \%
\end{aligned}
$$
$$
\begin{aligned}
\$ 10,000 &=F\left[\frac{(1.0816)^{3}-1}{0.0816}\right] \
F &=\frac{\$ 10,000}{\left[\frac{(1.0816)^{5}-1}{0.0816}\right]} \
&=\$ 1,699.14
\end{aligned}
$$




















科学计算 Scientific Computing MATH3511/MATH6111

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这是一份anu西澳大学MATH3511/MATH6111的成功案例

科学计算 Scientific Computing MATH3511/MATH6111


To be more specific, suppose that
$$
F(h)=a_{0}+a_{1} h^{p}+\mathcal{O}\left(h^{r}\right)
$$
as $h \rightarrow 0$ for some $p$ and $r$, with $r>p$. We assume that we know the values of $p$ and $r$, but not $a_{0}$ or $a_{1}$. Indeed, $F(0)=a_{0}$ is the quantity we seek. Suppose that we have computed $F$ for two stepsizes, say, $h$ and $q h$ for some $q>1$. Then we have
$$
F(h)=a_{0}+a_{1} h^{p}+\mathcal{O}\left(h^{r}\right)
$$
and
$$
F(q h)=a_{0}+a_{1}(q h)^{p}+\mathcal{O}\left(h^{r}\right) .
$$
This system of two linear equations in the two unknowns $a_{0}$ and $a_{1}$ is easily solved to obtain
$$
a_{0}=F(h)+\frac{F(h)-F(q h)}{q^{p}-1}+\mathcal{O}\left(h^{r}\right) .
$$
Thus, the accuracy of the improved value, $a_{0}$, is $\mathcal{O}\left(h^{r}\right)$ rather than only $\mathcal{O}\left(h^{p}\right)$.

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MATH3511/MATH6111 COURSE NOTES :

For example, applying Euler’s method to this equation using a fixed stepsize $h$, we have
$$
y_{k+1}=y_{k}+\lambda y_{k} h=(1+\lambda h) y_{k},
$$
which means that
$$
y_{k}=(1+\lambda h)^{k} y_{0} .
$$
Provided $\lambda<0$, the exact solution decays to zero as $t$ increases, as will the computed solution if $|1+\lambda h|<1$. This result agrees with our earlier stability analysis because $J=\lambda$ for this ODE. We also note that the growth factor $1+\lambda h$ agrees with the series expansion
$$
e^{\lambda h}=1+\lambda h+\frac{(\lambda h)^{2}}{2}+\frac{(\lambda h)^{3}}{6}+\cdots
$$
through terms of first order in $h$, and hence Euler’s method is first-order accurate. Especially for more complicated numerical methods, a linear ODE is easier to work with than a general ODE, and it produces essentially the same stability result if we equate $\lambda$ with the Jacobian $J$ at a given point. An important caveat, however, is that $\lambda$ is constant, whereas the Jacobian $J$ varies for a nonlinear equation, and hence the stability can potentially change.











数学物理学的高级课题 Advanced Topics in Mathematical Physics MATH3351/MATH6211

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这是一份anu澳大利亚国立大学MATH3351/MATH6211的成功案例

数学物理学的高级课题 Advanced Topics in Mathematical Physics MATH3351/MATH6211


$$
\sigma_{y}=\sigma F-(T F) \sigma
$$
where
$$
F=u \sigma+v+w\left[T^{-1}(\sigma)\right]^{-1} \text {. }
$$
One can check the by direct substitution of the operator $\sigma$ and by use of the equation for $\phi$.
Remark 2.33. Theorem $2.32$ is evidently valid for the spectral problem
$$
\lambda \psi=u T \psi+v \psi+w T^{-1} \psi
$$
with the only correction being that the last term for the transform $v[1]$ is absent. The equation goes to the “Riccati equation” analog for the function $\sigma$ :
$$
\mu=u \sigma+v+w\left[T^{-1}(\sigma)\right]^{-1}
$$

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MATH3351/MATH6211 COURSE NOTES :

$$
\xi_{t}=[v / 2+(u+\beta I) T] \xi=Z \xi
$$
which is solved by
$$
\xi=\exp (Z t) \xi_{0} .
$$
Plugging $\Phi$ into (2.135) yields the spectral problem for the difference shift operators:
$$
\mu \Phi(x)=\xi^{-1}[u \xi \Phi(x+1)+v \xi \Phi+w \xi \Phi(x-1)]
$$
Separating variables again, a class of particular solutions is built as
$$
\Phi=\eta \exp (\Sigma x)
$$
hence, we arrive at the matrix spectral problem for $\eta$ :
$$
\mu \eta=\xi^{-1}[u \xi \eta \exp (\Sigma)+v \xi \eta+w \xi \eta \exp (-\Sigma)]
$$
with the operator on the right-hand side and, therefore, spectral parameter $\mu$ parameterized by $t$. Finally, the matrix $\sigma$ is composed as
$$
\sigma=\xi(t) \eta \exp (\Sigma) \eta^{-1} \xi^{-1}(t)
$$











高级代数2:场扩展和伽罗瓦理论 Advanced Algebra 2: Field extensions and Galois Theory MATH3345/MATH6215

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这是一份anu澳大利亚国立大学MATH3345/MATH6215的成功案例

高级代数2:场扩展和伽罗瓦理论 Advanced Algebra 2: Field extensions and Galois Theory MATH3345/MATH6215


Thus, assume that each $a_{i}$ is $n$-powerless. Observe that in this case it suffices to prove the theorem for $n_{1}=\cdots=n_{t}=n$ as clearly
$$
\left(\mathbf{Q}\left(\sqrt[\pi t]{a_{1}}, \ldots, \sqrt[n_{t}]{a_{t}}\right) / \mathbf{Q}\right) \leq n_{1} \cdots n_{t}
$$
and
$$
\left(\mathbf{Q}\left(\sqrt[n]{a_{1}}, \ldots, \sqrt[n]{a_{t}}\right) / \mathbf{Q}\left(\sqrt[n]{a_{1}}, \ldots, \sqrt[n_{t}]{a_{t}}\right)\right) \leq\left(n / n_{1}\right) \cdots\left(n / n_{t}\right)
$$
So
$$
\left(\mathbf{Q}\left(\sqrt[n]{a_{1}}, \ldots, \sqrt[n]{a_{t}}\right) / \mathbf{Q}\right)=n^{t}
$$
forces both inequalities to be equalities.

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MATH3345/MATH6215 COURSE NOTES :

In the above situation, let
$$
\theta=\alpha_{1} Y_{1}+\cdots+\alpha_{n} Y_{n} \in \tilde{\mathbf{E}}
$$
and let
$$
F(Z)=\prod_{\tau^{\prime} \in S_{n}}\left(Z-\tau^{\prime}(\theta)\right) \in \tilde{\mathbf{E}}[Z] .
$$
The polynomial $F(Z)$ is a symmetric function of its roots, so by its coefficients are functions of the elementary symmetric functions of its roots and hence of $Y_{1}, \ldots, Y_{n}$ and the coefficients of $f(X)$. Thus, $F(Z) \in$ $\tilde{\mathbf{F}}[Z]$. Now we may factor $F(Z)$ into a product of irreducibles in $\tilde{\mathbf{F}}[Z]$,
$$
F(Z)=F_{1}(Z) \cdots F_{t}(Z)
$$
One of these is divisible by $Z-\theta$ in $\tilde{\mathbf{E}}(Z)$. Renumbering if necessary we may assume it is $F_{1}(Z)$. Since $F(Z)$ is invariant under $S_{n}$, the action of $S_{n}$ permutes these factors.











高级分析2:勒贝斯格积分和希尔伯特空间 Advanced Analysis 2: Lebesgue Integration and Hilbert Spaces MATH3320/MATH6212

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这是一份anu澳大利亚国立大学MATH3320/MATH6212的成功案例

高级分析2:勒贝斯格积分和希尔伯特空间 Advanced Analysis 2: Lebesgue Integration and Hilbert Spaces MATH3320/MATH6212


Suppose that a function $f(x)$ is bounded on the interval $[A, B]$, where $A, B \in \mathbb{R}$ and $A<B$.
DEFINITION. The real number
$$
I^{-}(f, A, B)=\sup {\Delta} s(f, \Delta), $$ where the supremum is taken over all dissections $\Delta$ of $[A, B]$, is called the lower integral of $f(x)$ over $[A, B]$. DEFINITION. The real number $$ I^{+}(f, A, B)=\inf {\Delta} S(f, \Delta),
$$
where the infimum is taken over all dissections $\Delta$ of $[A, B]$, is called the upper integral of $f(x)$ over $[A, B]$.

REMARK. Since $f(x)$ is bounded on $[A, B]$, it follows that $s(f, \Delta)$ and $S(f, \Delta)$ are bounded above and below. This guarantees the existence of $I^{-}(f, A, B)$ and $I^{+}(f, A, B)$.

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MATH3320/MATH6212 COURSE NOTES :

$$
S(f, \Delta)-s(f, \Delta)<\frac{\epsilon}{c} $$ It is easy to see that $$ S(c f, \Delta)=c S(f, \Delta) \quad \text { and } \quad s(c f, \Delta)=c s(f, \Delta) $$ Hence $$ S(c f, \Delta)-s(c f, \Delta)<\epsilon . $$ It follows from Theorem $2 \mathrm{D}$ that $c f \in \mathcal{R}([A, B])$. Also, (13) clearly implies $I^{+}(c f, A, B)=c I^{+}(f, A, B)$. Suppose next that $c<0$. Since $f \in \mathcal{R}([A, B])$, it follows from Theorem $2 \mathrm{D}$ that for every $\epsilon>0$, there exists a dissection $\Delta$ of $[A, B]$ such that
$$
S(f, \Delta)-s(f, \Delta)<-\frac{\epsilon}{c}
$$
It is easy to see that
$$
S(c f, \Delta)=c s(f, \Delta) \quad \text { and } \quad s(c f, \Delta)=c S(f, \Delta)
$$
Hence
$$
S(c f, \Delta)-s(c f, \Delta)<\epsilon .
$$
It follows from Theorem $2 \mathrm{D}$ that $c f \in \mathcal{R}([A, B])$. Also, (14) clearly implies $I^{+}(c f, A, B)=c I^{-}(f, A, B)$.
(c) Note simply that
$$
\int_{A}^{B} f(x) \mathrm{d} x \geq(B-A) \inf _{x \in[A, B]} f(x)
$$
where the right hand side is the lower sum corresponding to the trivial dissection.
(d) Note that $g-f \in \mathcal{R}([A, B])$ in view of (a) and (b). We now apply (c) to the function $g-f$.
Next, we investigate the question of breaking up the interval $[A, B]$ of integration.











分形几何与混沌 Fractal Geometry and Chaotic MATH3062/MATH6116

0

这是一份anu澳大利亚国立大学MATH3062/MATH6116的成功案例

分形几何与混沌 Fractal Geometry and Chaotic MATH3062/MATH6116


$$
N_{\delta / n}(F) \leqslant N_{\delta}^{\prime}(F) .
$$
If $\delta \sqrt{n}<1$ then
$$
\frac{\log N_{\delta / n}(F)}{-\log (\delta \sqrt{n})} \leqslant \frac{\log N_{\delta}^{\prime}(F)}{-\log \sqrt{n}-\log \delta}
$$
so taking limits as $\delta \rightarrow 0$
$$
\operatorname{dim}{\mathrm{B}} F \leqslant \lim {\delta \rightarrow 0} \frac{\log N_{\delta}^{\prime}(F)}{-\log \delta}
$$
and
$$
\overline{\operatorname{dim}}{\mathrm{B}} F \leqslant \varlimsup{\delta \rightarrow 0} \frac{\log N_{\delta}^{\prime}(F)}{-\log \delta} .
$$
On the other hand, any set of diameter at most $\delta$ is contained in $3^{n}$ mesh cubes of side $\delta$ (by choosing a cube containing some point of the set together with its neighbouring cubes). Thus
$$
N_{\delta}^{\prime}(F) \leqslant 3^{n} N_{\delta}(F)
$$

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MATH3029/MATH6109 COURSE NOTES :

(Note that $\mathcal{H}^{s, 0}$ is just a minor variant of $s$-dimensional Hausdorff measure where only rectangles are allowed in the $\delta$-covers.) The dimension print, print $F$, of $F$ is defined to be the set of non-negative pairs $(s, t)$ for which $\mathcal{H}^{s, t}(F)>0$.
Using standard properties of measures, it is easy to see that we have monotonicity
$$
\text { print } F_{1} \subset \text { print } F_{2} \text { if } F_{1} \subset F_{2}
$$
and countable stability
$$
\operatorname{print}\left(\bigcup_{i=1}^{\infty} F_{i}\right)=\bigcup_{i=1}^{\infty} \text { print } F_{i} \text {. }
$$
Moreover, if $(s, t)$ is a point in print $F$ and $\left(s^{\prime}, t^{\prime}\right)$ satisfies
$$
\begin{gathered}
s^{\prime}+t^{\prime} \leqslant s+t \
t^{\prime} \leqslant t
\end{gathered}
$$
then $\left(s^{\prime}, t^{\prime}\right)$ is also in print $F$.











概率论与应用 Probability Theory with Applications MATH3029/MATH6109

0

这是一份anu澳大利亚国立大学MATH3029/MATH6109的成功案例

概率论与应用 Probability Theory with Applications MATH3029/MATH6109


$$
p(A \mid D I)=\frac{p(A \mid I)}{p(D \mid I)}\left[p\left(D \mid A H_{0} I\right) p\left(H_{0} I\right)+\sum_{i=1}^{n} p\left(D \mid H_{i} I\right) p\left(H_{i} \mid I\right)\right]
$$
From this we see that if the different ways of being broken do not in themselves tell him different things about the data:
$$
p\left(D \mid H_{i} I\right)=p\left(D \mid H_{1} I\right), \quad 1 \leq i \leq n
$$
then enumer ation of the $n$ different ways of being broken is unnecessary: the calculation re duces to finding the likelihoo $\mathrm{d}$
$$
L \equiv p\left(D \mid A H_{0} I\right) p\left(H_{0} \mid I\right)+p\left(D \mid H_{1} I\right)\left[1-p\left(H_{0} \mid I\right)\right]
$$
and only the total probability of being broken:
$$
p\left(\bar{H}{0} \mid I\right)=\sum{i}^{n} p\left(H_{i} \mid I\right)=1-p\left(H_{0} \mid I\right)
$$

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MATH3029/MATH6109 COURSE NOTES :

$$
p(\theta \mid D, I)=p(\theta \mid I) \frac{p(d \mid \theta I)}{p(D \mid I)}
$$
for some parameter $\theta$, suppose we decide to introduce some additional evidence $E$. Then another application of Bayes’ theorem updates that conclusion to
$$
p(\theta \mid E, D, I)=p(\theta \mid D, I) \frac{p(E \mid \theta, D, I)}{p(E \mid D, I)}
$$
so the necessary and sufficient condition that the new information will change our conclusions is that, on some region of the par ameter space of positive measure the likelihood ratio in (8-73) differs from unity:
$$
p(E \mid \theta, D, I) \neq p(E \mid D, I) .
$$
But if the evidence $E$ was something already implied by the data and prior information, then
$$
p(E \mid \theta, D, I)=p(E \mid D, I)=1
$$