# 微积分2|Calculus 2 MAST10006代写

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This subject will extend knowledge of calculus from school. Students are introduced to hyperbolic functions and their inverses, the complex exponential and functions of two variables. Techniques of differentiation and integration will be extended to these cases. Students will be exposed to a wider class of differential equation models, both first and second order, to describe systems such as population models, electrical circuits and mechanical oscillators. The subject also introduces sequences and series including the concepts of convergence and divergence.

Find the mass of a ball 9 of radius $a$ whose density is numerically equal to the distance from a fixed diametral plane.

Let the ball be the inside of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$, and let the fixed diametral plane be $z=0$. Then $M=\iiint|z| d V$. Use the upper hemisphere and double the result. In spherical coordinates.
\begin{aligned} M &=2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{a} z \cdot \rho^{2} \sin \phi d \rho d \phi d \theta=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \int_{0}^{a} \rho \cos \phi \cdot \rho^{2} \sin \phi d \rho d \phi d \theta \ &\left.=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \frac{1}{4} \rho^{4} \cos \phi \sin \phi\right]{0}^{a} d \phi d \theta=\frac{1}{2} a^{4} \int{0}^{2 \pi} \frac{1}{2} \sin ^{2} \phi \int_{0}^{\pi / 2} d \theta=\frac{1}{4} a^{4} \int_{0}^{2 \pi} d \theta=\frac{1}{4} a^{4} \cdot 2 \pi=\frac{1}{2} \pi a^{4} \end{aligned}

## MAST10006 COURSE NOTES ：

Find the surface area $S$ of the part of the sphere $x^{2}+y^{2}+z^{2}=4 z$ inside the paraboloid $z=x^{2}+y^{2}$.
$\square$ The region $\mathscr{R}$ under the spherical cap (Fig. 44-33) is obtained by finding the intersection of $x^{2}+y^{2}+z^{2}=4 z$ and $z=x^{2}+y^{2}$. This gives $z(z-3)=0$. Hence, the paraboloid cuts the sphere when $z=3$, and $\mathscr{B}$ is the disk $x^{2}+y^{2} \leq 3 . \quad S=\iint_{a} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} d A . \quad 2 x+2 z \frac{\partial z}{\partial x}=4 \frac{\partial z}{\partial x}, \quad \frac{\partial z}{\partial x}=-\frac{x}{z-2} . \quad$ Similarly, $\frac{\partial z}{\partial y}=-\frac{y}{z-2}$. Hence,
$$1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=1+\frac{x^{2}}{(z-2)^{2}}+\frac{y^{2}}{(z-2)^{2}}=\frac{(z-2)^{2}+x^{2}+y^{2}}{(z-2)^{2}}=\frac{\left(x^{2}+y^{2}+z^{2}\right)-4 z+4}{(z-2)^{2}}=\frac{4}{(z-2)^{2}}$$
Therefore,
$$\left.S=\iint_{\pi} \frac{2}{z-2} d A=\int_{0}^{2 w} \int_{0}^{\sqrt{3}} \frac{2}{\sqrt{4-r^{2}}} r d r d \theta=-\int_{0}^{2 \theta} 2 \sqrt{4-r^{2}}\right]{0}^{\sqrt{3}} d \theta=-2 \int{0}^{2 \pi}(1-2) d \theta=2 \cdot 2 \pi=4 \pi$$

# 物理1|Physics 1 PHYC10003代写

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This subject is designed for students with a sound background in physics, and aims to provide a strong understanding of a broad range of physics principles.

An atomic clock is placed in a jet airplane. The clock measures a time interval of $3600 \mathrm{~s}$ when the jet moves with speed $400 \mathrm{~m} / \mathrm{s}$.

• How much larger a time interval does an identical clock held by an observer at rest on the ground measure?

We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be the rest frame of the atomic clock. It follows from that
and from that
$$\gamma \simeq 1+\beta^{2} / 2$$
It follows that $\delta t=3.2 \mathrm{~ns}$ when $v=400 \mathrm{~m} / \mathrm{s}$ and $\Delta t^{\prime}=3600 \mathrm{~s}$.

## PHYC10003 COURSE NOTES ：

Two spaceships approach each other, each moving with the same speed as measured by a stationary observer on the Earth. Their relative speed is $0.70 c$,

• Determine the velocities of each spaceship as measured by the stationary observer on Earth.
Solution
Text Eq. (1.32) gives the Lorentz velocity transformation:
$$u_{x}^{\prime}=\frac{u_{x}-v}{1-u_{x} v / c^{2}}$$
where $u_{x}$ is the velocity of an object measured in the $S$ frame, $u_{x}^{\prime}$ is the velocity of the object measured in the $S^{\prime}$ frame and $v$ is the velocity of the $S^{\prime}$ frame along the $x$ axis of $S$.

We take the $S$ frame to be attached to the Earth and the $S^{\prime}$ frame to be attached to the spaceship moving to the right with velocity $v$. The other spaceship has velocity $u_{x}=-v$ in $S$ and velocity $u_{x}^{\prime}=-0.70 c$ in $S^{\prime}$.
It follows fromthat
$$0.70=\frac{2 \beta}{1+\beta^{2}}$$
solving which yields $\beta=0.41$. As measured by the stationary observer on Earth, the spaceships are moving with velocities $\pm 0.41 c$.

# 量子、统计和复合物理|PHYS3034/PHYS3935 Quantum, Statistical and Comp Phys代写 sydney代写

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You will learn how to use computer-aided design (CAD) software to design 3D printable molecular models that convey intermolecular interactions like hydrogen-bonding. The models will incorporate movable (i.e., rotation, flexion) and magnetic design elements to represent the formation of dynamic/weak bonds, which will serve as a useful visualisation and communication tool for complex molecular structures. The designs will be printed remotely and students will be able to keep their printed designs.

is also a nonnegative-definite Hermitian matrix, and we obtain the matrix inequality
$$\sum_{j=1}^{m}\left(x_{j}^{2}+y_{j}^{2}\right) M_{j} \geq X^{2}+Y^{2} \pm i[X, Y]$$
Since $S=S^{*}>0$ from the assumption, the above formula results in
$$\sum_{j=1}^{m}\left(x_{j}^{2}+y_{j}^{2}\right) S^{\frac{1}{2}} M_{j} S^{\frac{1}{2}} \geq S^{\frac{1}{2}}\left(X^{2}+Y^{2}\right) S^{\frac{1}{2}} \pm i S^{\frac{1}{2}}[X, Y] S^{\frac{1}{2}}$$

If $[X, Y]=0$, then $\Gamma(X, Y ; S)=\operatorname{Tr}\left[S\left(X^{2}+Y^{2}\right)\right]$ holds. In this case, if we define $M=\left{M_{j} ; x_{j} ; y_{j}\right}$ by (4)-(6), then (1) and (2) obviously hold and
$$\sum_{j=1}^{m} x_{j}^{2} M_{j}=X^{2}, \quad \sum_{j=1}^{m} y_{j}^{2} M_{j}=Y^{2}$$
hold, so that we have $\Delta(M ; S)=\operatorname{Tr}\left[S\left(X^{2}+Y^{2}\right)\right]$. Therefore, it gives the minimum of $\Delta$ and $\Delta_{*}(X, Y ; S)=\Gamma(X, Y ; S)$ holds.

## PHYS3034/PHYS3935 COURSE NOTES ：

$$\frac{1}{2 \pi \mu} \int(x+i y) E_{x, y} d x d y=X+i Y,$$
which can be also written as
$$\frac{1}{2 \pi \mu} \int x E_{x, y} d x d y=X, \quad \frac{1}{2 \pi \mu} \int y E_{x, y} d x d y=Y .$$
Moreover, if we multiply $X-i Y$ from the right in both sides, from $[X, Y]=i \mu I$, we have
$$\frac{1}{2 \pi \mu} \int\left(x^{2}+y^{2}\right) E_{x, y} d x d y=X^{2}+Y^{2}+\mu I .$$

# 电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

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The students are cultivated as high-quality innovative professionals. They are well developed in the
aspects of morality, intelligence, physique, and aesthetic. They possess the basic theoretical knowledge in
the fields of Materials Science and Physics，and are well trained in the applied research, technological
development, and engineering. They can research physical properties and laws of materials at the level of
molecule, atom and electron，and apply to develop new materials preparative technology，advanced
function materials and equipments. The graduates are expected to work in various industries, universities,
and research institutes in the fields of over function materials and the related (such as energy engineering,
electric power, etc.), engaging in product design, technological development, scientific research, and
management, and playing important and leading roles in the fields with international competitiveness and
innovation.

$$\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u-u_{\mathrm{e}}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}-2 u u_{\mathrm{e}}\right)\right),$$
and for the lower section of windings, just above (1),
$$\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u+u_{e}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}+2 u u_{e}\right)\right) \text {. }$$

For the upper section of the coil containing $N$ wires , combiningand yields an excess of positive charges:
$$\Delta Q_{+}^{\prime}=N \Delta q_{+}^{\prime}=\frac{N q u_{\mathrm{e}} u}{c^{2}} .$$
For the lower section of windings (1), the combination of yields an excess of negative charges:
$$\Delta Q_{-}^{\prime}=N \Delta q_{-}^{\prime}=-\frac{N q u_{\mathrm{e}} u}{c^{2}}$$

## PHYS3035/PHYS3935 COURSE NOTES ：

Derivation: Each end of the bar (or coil) produces a flux density $B_{t}=$ $\frac{\Phi}{4 \pi R^{2}}$ at the point of observation according. Only the difference of the two values is important, so that in the first principal orientation
$$B=\frac{\phi}{4 \pi}\left(\frac{1}{(R-l / 2)^{2}}-\frac{1}{(R+1 / 2)^{2}}\right)$$
When the distance $R$ is sufficiently large compared to the length $I$ of the bar or coil, we can neglect $P^{2}$ relative to $R^{2}$, and for the magnitude of $B$, we then obtain
$$B=\frac{1}{2 \pi} \frac{\Phi l}{R^{3}}=\frac{\mu_{0}}{2 \pi} \frac{m}{R^{3}}$$
Correspondingly, for the second principal orientation, we find
$$B=\frac{\mu_{0}}{4 \pi} \frac{m^{3}}{R^{3}}$$

# 电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

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The students are cultivated as high-quality innovative professionals. They are well developed in the
aspects of morality, intelligence, physique, and aesthetic. They possess the basic theoretical knowledge in
the fields of Materials Science and Physics，and are well trained in the applied research, technological
development, and engineering.

$$G(q, \omega)=\frac{1}{q^{2}-q_{0}^{2}}\left(U-\frac{q q}{q_{0}^{2}}\right)$$
which sometimes is useful. Like the Huygens propagator, the dyadic Green function is also singular for $q=q_{0}$.

The plane-wave representation of the dyadic Green function for the magnetic field is obtained by inserting the Fourier integral transformation
$$\boldsymbol{G}{\mathrm{M}}(\boldsymbol{R} ; \omega)=(2 \pi)^{-3} \int{-\infty}^{\infty} \boldsymbol{G}_{\mathrm{M}}(q, \omega) \mathrm{e}^{i \boldsymbol{q} \cdot \boldsymbol{R}} \mathrm{d}^{3} q$$

Calculations analogous to those carried out to determine $\boldsymbol{G}(\boldsymbol{R} ; \omega)$ result in
$$\boldsymbol{G}{\mathrm{M}}(\boldsymbol{q}, \omega)=\frac{q}{q{0}} \frac{1}{q^{2}-q_{0}^{2}} \boldsymbol{U} \times \hat{\boldsymbol{q}}$$
an expression which also is singular for $q=q_{0}$. The folding theorem in $\boldsymbol{r}$-space gives when applied to
$$\boldsymbol{B}(q, \omega)=\frac{i \mu_{0} \omega}{c_{0}} \boldsymbol{G}{\mathrm{M}}(q, \omega) \cdot J(q, \omega)$$ A current density parallel to the $q$-direction does not give rise to a magnetic field, and the magnetic field generated by a current density perpendicular to $\hat{q}$ always lies in a plane perpendicular to $\hat{q}$. To prove the above-mentioned claims we expand the unit dyad after a triple set of orthogonal unit vectors $\hat{q}{\perp}^{(1)}, \hat{q}{\perp}^{(2)}$, and $\hat{q}$, with $\hat{\boldsymbol{q}}{1}^{(1)} \times \hat{\boldsymbol{q}}{2}^{(2)}=\hat{\boldsymbol{q}}$, i.e., $$\boldsymbol{U}=\hat{\boldsymbol{q}}{\perp}^{(1)} \hat{q}{\perp}^{(1)}+\hat{\boldsymbol{q}}{\perp}^{(2)} \hat{\boldsymbol{q}}_{\perp}^{(2)}+\hat{q} \hat{q}$$

## PHYS3035/PHYS3935 COURSE NOTES ：

In the mixed representation, the electric field, $\boldsymbol{E}(z ; q |, \omega)$, from a current density distribution, $\boldsymbol{J}(z: q |, \omega)$, is outside the distribution given by
$$E\left(z ; q_{|}, \omega\right)=i \mu_{0} \omega \int_{-\infty}^{\infty} G\left(z-z^{\prime} ; q_{1}, \omega\right) \cdot J\left(z^{\prime} ; q_{|}, \omega\right) \mathrm{d} z^{\prime}$$
where
$$G\left(Z ; q_{|}, \omega\right)=\Gamma\left(s g n Z ; q_{1}, \omega\right) \mathrm{e}^{i x_{\perp}|Z|}$$
with
$$\Gamma\left(\operatorname{sgn} Z: q_{|}, \omega\right)=\frac{i}{2 q_{0}^{2} \kappa \perp}\left[q_{0}^{2} U-q_{|} q_{|}-\kappa_{\perp}^{2} \hat{z} \bar{z}-\left(q_{|}^{\hat{z}}+\hat{z} q_{|}\right) \kappa_{\perp} \operatorname{sgn} Z\right]$$
Let us now assume that the source current density is nonvanishing only on a plane sheet located at $z^{\prime}=z_{0}$. Thus,
$$J\left(z^{\prime} ; q_{|}, \omega\right)=J_{0}\left(q_{|}, \omega\right) \delta\left(z^{\prime}-z_{0}\right)$$

# 物理学|PHYS1002 Physics 1 (Fundamentals)代写 monash代写

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This course is a real-time, online course where the instructor and students meet via web conferencing tools, at scheduled days and times. Instructors and students share information, ideas and learning experiences in a virtual course environment. Participation in synchronous courses requires students to have reliable, high-speed internet access, a computer (ideally with a webcam), and a headset with a microphone.

Let $Q_{B} Q_{R}, T_{R}$ and $W$, respectively, denote, the heat drawn by the refrigerator from the body at temperature $T$, out of which $Q_{R}$ is rejected in the room at temperature $T_{R}$ and a work $W$ is performed on the refrigerator. Now,
$$Q_{R}=Q_{B}+W \text { and } Q_{B}=Q_{R}-W$$
And the Coefficient of performance of the refrigerator, $(\mathrm{COF})=\frac{W}{Q_{R}}=\frac{\left(T_{R}-T\right)}{T_{R}}$ or
$$Q_{R}=\frac{W T_{R}}{\left(T_{R}-T\right)}$$

Substituting this value of $Q_{R}$ in Eq. S-7.10.1 gives:
\begin{aligned} &Q_{B}=Q_{R}-W=\frac{W T_{R}}{\left(T_{R}-T\right)}-W=W\left(\frac{T}{\left(T_{R}-T\right)}\right) \ &W=\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B} \end{aligned}
shows that in order to draw a quantity of heat $Q_{B}$ from the body at temperature $T$ the amount of work required to be done is $W$ whichis equal to $\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B}$.

## PHS2061 COURSE NOTES ：

The thermal capacity (at constant volume) is defined as $C_{V} \equiv\left(\frac{\partial U}{\partial T}\right){V}$, which in the present case is $K T^{3}$. Using the relation $\left(\frac{\partial S}{\partial T}\right){V}=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right){V}=\frac{K T^{3}}{T}=K T^{2}$, and hence, $$\Delta S=\int{T_{R}}^{0} K T^{2} d T=-\frac{K}{3} T_{R}^{3}$$
Change in internal energy of the body may be calculated using the relation
$$d U=C d T \quad \text { or } \quad \Delta U=\int_{T_{R}}^{0} K T^{3} d T=-\frac{K}{4} T_{R}^{4}$$

# 量子和热物理学|PHS2061 Quantum and thermal physics代写 monash代写

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The equation of state for Van der Waals gas is,
$$P=\frac{R T}{(v-b)}-\frac{a}{v^{2}}$$
Therefore, $\left(\frac{\partial P}{\partial T}\right){v}=\frac{R}{(v-b)}$ and $\left(\frac{\partial P}{\partial v}\right){T}=\frac{2 a(v-b)^{2}-R T v^{3}}{(v-b)^{2} v^{3}}$

$c_{P}-c_{v}=-T \frac{\left[\left(\frac{\partial P}{\partial T}\right){v}\right]^{2}}{\left(\frac{\partial P}{\partial v}\right){T}}=\frac{R}{1-2 a(v-b)^{2} / R T v^{3}}$
Also,
$$\kappa_{T}=-\frac{1}{v}\left(\frac{\partial v}{\partial P}\right){T}=-\frac{1}{v}\left(\frac{\partial P}{\partial v}\right){T}^{-1}=\frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}}$$
But $\kappa_{s}=\frac{\kappa_{T}}{\gamma}$ and $\gamma=1.66$ for a monatomic gas
Hence,
Similarly,
\begin{aligned} &\kappa_{s}=0.6 \kappa_{T}=0.6 \frac{(v-b)^{2} v^{2}}{R T v^{3}-2 a(v-b)^{2}} \ &\beta=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{v}=\frac{R v^{2}(v-b)}{R T v^{3}-2 a(v-b)^{2}} \end{aligned}

## PHS2061 COURSE NOTES ：

Since
$$d F=d U-T d s-S d T$$
The difference in the value of $F$ of two nearby equilibrium states of an open system may be given by
$$d F=-P d V-S d T+\mu d \mathbb{N}$$
$7.61$
Also
$$d F=\left(\frac{\partial F}{\partial V}\right){T, \mathbb{N}} d V+\left(\frac{\partial F}{\partial T}\right){V, \mathbb{N}} d T+\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T} d \mathbb{N}$$ Comparison of $$\mu=\left(\frac{\partial F}{\partial \mathbb{N}}\right){V, T}$$

# 工程物理学|PHS1002 Physics for engineering代写 monash代写

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An exact first-degree first-order ODE is one of the form
$$A(x, y) d x+B(x, y) d y=0 \quad \text { and for which } \quad \frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$$
In this case $A(x, y) d x+B(x, y) d y$ is an exact differential, $d U(x, y)$ . In other words
$$A d x+B d y=d U=\frac{\partial U}{\partial x} d x+\frac{\partial U}{\partial y} d y$$

from which we obtain
\begin{aligned} &A(x, y)=\frac{\partial U}{\partial x} \ &B(x, y)=\frac{\partial U}{\partial y} \end{aligned}
Since $\partial^{2} U / \partial x \partial y=\partial^{2} U / \partial y \partial x$ we therefore require
$$\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$$

## PHS1002 COURSE NOTES ：

Substituting $y=v x$ we obtain
$$v+x \frac{d v}{d x}=v+\tan v .$$
Cancelling $v$ on both sides, rearranging and integrating gives
$$\int \cot v d v=\int \frac{d x}{x}=\ln x+c_{1}$$
But
$$\int \cot v d v=\int \frac{\cos v}{\sin v} d v=\ln (\sin v)+c_{2}$$
so the solution to the $\mathrm{ODE}$ is $y=x \sin ^{-1} A x$, where $A$ is a constant.

# 基础物理学|PHS1001 Foundation physics代写 monash代写

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$$\int_{a V} v_{i} n_{j} \mathrm{~d} A=\int_{V} \frac{\partial v_{i}}{\partial x_{j}} \mathrm{~d} V .$$
GrEEN’s identities also follow from adequate application of the Divergence Theorem. The first GREEN’s identity is obtained by choosing in $f=\phi v$. This yields
\begin{aligned} \int_{\partial V} \phi v \cdot n \mathrm{~d} A &=\int_{V} \operatorname{div}(\phi v) \mathrm{d} V \ &=\int_{V}{\phi \operatorname{div} v+v \operatorname{grad} \phi} \mathrm{d} V . \end{aligned}

If we substitute here $v=\operatorname{grad} \psi$ and observe that div $v=\operatorname{div} \operatorname{grad} \psi=\Delta \psi$, where $\Delta$ is the LAPLACE operator, then
$$\int_{V}{\operatorname{grad} \phi \cdot \operatorname{grad} \psi} \mathrm{d} V=-\int_{V} \phi \Delta \psi \mathrm{d} V+\int_{\partial V} \phi \frac{\partial \psi}{\partial n} \mathrm{~d} A$$
which is GREEN’s first identity.
If in the roles of $\phi$ and $\psi$ are interchanged, we obtain
$$\int_{V}{g r a d \psi \cdot \operatorname{grad} \phi} \mathrm{d} V=-\int_{V} \psi \Delta \phi \mathrm{d} V+\int_{\partial V} \psi \frac{\partial \phi}{\partial n} \mathrm{~d} A$$

## PHS1001 COURSE NOTES ：

which, when combined, yields
$$\int_{\mathcal{C}} \phi \mathrm{d} x=-\int_{A_{C}}(\operatorname{grad} \phi) \times n \mathrm{~d} A$$
If $\phi$ in is the ith component of a vector field, then it also implies
$$\int_{\mathcal{C}} v \otimes \mathrm{d} x=-\int_{A_{\mathcal{C}}}(\operatorname{grad} v) \times \boldsymbol{n} \mathrm{d} A$$
A further interesting formula is obtained by selecting
$$e_{i} \times \int_{\mathcal{C}} v_{i} \mathrm{~d} x=\oint_{\mathcal{C}} v \times \mathrm{d} x=-\int_{A_{C}} \underbrace{e_{i} \times \operatorname{grad} v_{i}}{-\text {curl } v} \times n \mathrm{~d} A$$ so that $$\oint{\mathcal{C}} v \times \mathrm{d} x=\int_{A_{\mathcal{C}}}(\operatorname{curl} v) \times \boldsymbol{d} A$$

# 数学建模|MTH2040 Mathematical modelling代写 monash代写

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If $|\alpha|=\nu$, then $\lambda_{1}=\lambda_{2}=-\alpha$ and the corresponding generalized eigenspace is spanned by
$$z^{1}=\left[\begin{array}{c} 1 \ -\alpha \end{array}\right], \quad z^{2}=\left[\begin{array}{l} 0 \ 1 \end{array}\right]$$
where $z^{1}$ is a real eigenvector and $z^{2}$ is a generalized eigenvector of second order. By (34) the corresponding (generalized) real eigenmotions are
$$x^{1}(t)=e^{-\alpha t}\left[\begin{array}{c} 1 \ -\alpha \end{array}\right], \quad x^{2}(t)=e^{-\alpha t}\left[\begin{array}{l} 0 \ 1 \end{array}\right]+t e^{-\alpha t}\left[\begin{array}{c} 1 \ -\alpha \end{array}\right]$$

If $|\alpha|<v$ (hence $z^{2}=\overline{z^{1}}$ ) then $\lambda_{1,2}=-\alpha \pm x \sqrt{\nu^{2}-\alpha^{2}}$ and the corresponding real modes are by $(44),(45)$ and $(50)$
$$x^{1}(t)=e^{-\alpha t}\left(\cos \left(\sqrt{u^{2}-\alpha^{2}} t\right)\left[\begin{array}{c} 1 \ -\alpha \end{array}\right]-\sin \left(\sqrt{u^{2}-\alpha^{2}} t\right)[\sqrt{0}]\right)$$

## MTH2040 COURSE NOTES ：

is called the Fibonacci sequence and plays a role in various fields of mathematics. The eigenvalues of the matrix in (59) are $\lambda_{1,2}=(1 \pm \sqrt{5}) / 2$ and the corresponding real modes are
$$z^{1}(t)=\left(\frac{1+\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c} (1+\sqrt{5}) / 2 \ 1 \end{array}\right], \quad z^{2}(t)=\left(\frac{1-\sqrt{5}}{2}\right)^{t}\left[\begin{array}{c} (1-\sqrt{5}) / 2 \ 1 \end{array}\right]$$
The solution of the initial value problem is of the form $x(t)=\alpha_{1} z^{1}(t)+\alpha_{2} z^{2}(t)$ for some $\left(\alpha_{1}, \alpha_{2}\right) \in \mathbb{R}^{2}$. Since $x(0)=[1,0]^{\top}$ we obtain $\alpha_{1}=1 / \sqrt{5}, \alpha_{2}=-1 / \sqrt{5}$ and hence the following analytic expression for the Fibonacci sequence
$$\xi(t)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}\right]$$