Tag Archives: 数学方法代写

数学方法|PH30025 Mathematical methods代写

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这是一份bath巴斯大学PH30025作业代写的成功案

数学方法|PH30025 Mathematical methods代写

Suppose the complex numbers $z_{1}$ and $z_{2}$ are represented inby the points $P_{1}$ and $P_{2}$ respectively. Also, let $A$ be the point $z=1$. Then, with $O P_{2}$ as base corresponding to $\mathrm{OA}$, the triangle $\mathrm{OP}{2} \mathrm{P}$ is constructed to be similar to triangle $\mathrm{OAP}{1}$. Now,
$$
\begin{gathered}
\mathrm{OP} / r_{2}=\mathrm{OP} / \mathrm{OP}{2}=\mathrm{OP}{1} / \mathrm{OA}=r_{1} / 1 \
\mathrm{OP}=r_{1} r_{2} .
\end{gathered}
$$
Also
$$
\angle \mathrm{POP}{2}=\angle \mathrm{P}{1} \mathrm{OA}=\theta_{1}
$$
and so
$$
\angle \mathrm{POA}=\theta_{1}+\theta_{2} \text {. }
$$


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PH30025 COURSE NOTES ：

$\operatorname{Cosh}^{-1} x$ may be expressed in terms of logarithms as follows:
If
$$
y=\cosh ^{-1} x
$$
then
$x=\cosh y=1 / 2\left(\mathrm{e}^{y}+\mathrm{e}^{-y}\right)$
and so
$$
e^{2 y}-2 x e^{y}+1=0 \text {. }
$$
Regarding this as a quadratic equation in $\mathrm{e}^{y}$,
$$
e^{y}=x \pm\left(x^{2}-1\right)^{1 / 2}
$$



数学方法|MA30059 Mathematical methods 2代写

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ToTopics will be chosen from the following: Elliptic equations in two independent variables: Harmonic functions. Mean value property. Maximum principle (several proofs). Dirichlet and Neumann problems. Representation of solutions in terms of Green’s functions.

这是一份Bath巴斯大学MA30059 作业代写的成功案

数学方法|MA30059 Mathematical methods 2代写

$$
|\Gamma(z)| \leq \Gamma(x), \quad \operatorname{Re}(z)=x>0
$$
If $f(z)$ is a complex function of $z$, then writing $f(z)$ as
$$
f(z)=u(x, y)+i v(x, y),
$$
it follows that the Cauchy-Riemann equations [7] must be satisfied, i.e.,
$$
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
$$
For $f(z)=\Gamma(z)$, we have, using the relation,
$$
\begin{aligned}
t^{z} &=e^{z \ln t}=e^{x \ln t+i y \ln t} \
&=e^{x \ln t}[\cos (y \ln t)+i \sin (y \ln t)]
\end{aligned}
$$

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MA30059 COURSE NOTES ：

Therefore, we conclude that
$$
R(m, n) \equiv \int_{0}^{1} x^{m}(\ln x)^{n} d x=\frac{(-1)^{n} n !}{(m+1)^{n+1}}
$$
Note that for $n=0$, we obtain
$$
R(m, n)=\int_{0}^{1} x^{m} d x=\frac{1}{m+1}
$$
and for $m=0$,
$$
R(0, n)=\int_{0}^{1}(\ln x)^{n} d x=(-1)^{n} n !
$$




数学方法 Mathematical Methods MTH1002/MTH1002-JD

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这是一份exeter埃克塞特大学MTH1002作业代写的成功案例

数学结构 Mathematical Structures MTH1001
问题 1.

Now we determine the curves for which $L^{2}=4 \pi A$. To do this we find conditions for which $A$ is equal to the upper bound we obtained for it above. First note that
$$
\sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)=\sum_{n=1}^{\infty} n^{2}\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$

证明 .

implies that all the coefficients except $a_{0}, c_{0}, a_{1}, b_{1}, c_{1}$ and $d_{1}$ are zero. The constraint,
$$
\pi \sum_{n=1}^{\infty} n\left(a_{n} d_{n}-b_{n} c_{n}\right)=\frac{\pi}{2} \sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$
then becomes
$$
a_{1} d_{1}-b_{1} c_{1}=a_{1}^{2}+b_{1}^{2}+c_{1}^{2}+d_{1}^{2}
$$

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MTH1002/MTH1002-JD COURSE NOTES ：

The Fourier sine series has the form
$$
x(1-x)=\sum_{n=1}^{\infty} a_{n} \sin (n \pi x)
$$
The norm of the eigenfunctions is
$$
\int_{0}^{1} \sin ^{2}(n \pi x) d x=\frac{1}{2}
$$
The coefficients in the expansion are
$$
\begin{aligned}
a_{n} &=2 \int_{0}^{1} x(1-x) \sin (n \pi x) \mathrm{d} x \
&=\frac{2}{\pi^{3} n^{3}}(2-2 \cos (n \pi)-n \pi \sin (n \pi)) \
&=\frac{4}{\pi^{3} n^{3}}\left(1-(-1)^{n}\right)
\end{aligned}
$$