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数学分析代写 Mathematical Analysis代考

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数学分析代写Mathematical Analysis

这些理论通常是在实数和复数及函数的背景下研究的。分析是由微积分演变而来，它涉及到分析的基本概念和技术。分析可以区别于几何学；然而，它可以应用于任何有近似性定义的数学对象空间（拓扑空间）或对象之间的特定距离（公制空间）。

数学分析包含几个不同的主题，列举如下：

度量空间Metric spaces代写

在数学中，公制空间是一个集合，其中集合中的元素之间的距离概念（称为公制）被定义。

许多分析都发生在一些公制空间中；最常用的是实线、复平面、欧几里得空间、其他矢量空间和整数。没有公制的分析的例子包括度量理论（描述大小而不是距离）和函数分析（研究不需要有任何距离感的拓扑向量空间）。

序列和限制Sequences and limits代写

参数推断检验是在遵循某些参数的数据上进行的：数据将是正常的（即分布与钟形曲线平行）；数字可以加、减、乘、除；在比较两组或多组时，变异是相等的。

其他相关科目课程代写：

  • Real analysis实分析
  • Complex analysis复分析

数学分析的历史

数学分析正式发展于17世纪科学革命期间，但它的许多想法可以追溯到早期的数学家。分析学的早期成果隐含在古希腊数学的早期。例如，芝诺的二分法悖论中就隐含了一个无限的几何和。(严格来说，这个悖论的意义在于否认无限之和的存在）。后来，希腊数学家如Eudoxus和Archimedes在使用穷举法计算区域和实体的面积和体积时，更明确但非正式地使用了极限和收敛的概念。对无限小数的明确使用出现在阿基米德的《机械定理的方法》中，这部作品在20世纪被重新发现。在亚洲，中国数学家刘徽在公元3世纪用穷举法求出了圆的面积。从耆那教文献中可以看出，印度人早在公元前4世纪就已经掌握了算术和几何数列之和的公式。 在印度数学中，早在公元前2000年的吠陀文献中就发现了算术数列的特殊例子，并隐含在其中。

数学分析代写 Mathematical Analysis代考

Mathematical analysis was formally developed during the scientific revolution of the 17th century, but many of its ideas can be traced back to early mathematicians. The earliest results of analysis are implied in early ancient Greek mathematics. For example, an infinite geometric sum is implied in Zeno’s dichotomous paradox. (Strictly speaking, the paradox aims to deny the existence of infinite sums.) Later, Greek mathematicians such as Eudoxus and Archimedes used the concepts of limit and convergence more explicitly but informally when they used the exhaustive method to calculate the area and volume of regions and entities. The explicit use of infinitesimals appears in Archimedes’ Method of Mechanical Theorems, a work rediscovered in the 20th century. In Asia, Chinese mathematician Liu Hui used the exhaustive method to find the area of a circle in the 3rd century CE. Jain literature shows that Indians had learned formulas for summing arithmetic and geometric series as early as the 4th century BCE. In Indian mathematics, particular examples of arithmetic series are found implicitly in the Vedic literature as early as 2000 BCE.

数学分析相关课后作业代写

问题 1.

Parameters with Order Restrictions. Let $X_1, \ldots, X_n$ be indpendent random variables with $$ X_i \sim P_{\theta_i}, \text { for } i=1, \ldots, n $$ (a). For $P_\theta=N(\theta, 1)$, determine the maximum likelihood estimate of $$ \left(\theta_1, \ldots, \theta_n\right) $$ when there are no restrictions on the $\theta_i$.

证明 .

The likelihood function for the independent normal distribution is given by

$L\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right)=\prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}\left(x_i-\theta_i\right)^2}$

The log-likelihood function is then

$\ell\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right)=-\frac{n}{2} \log (2 \pi)-\sum_{i=1}^n \frac{1}{2}\left(x_i-\theta_i\right)^2$

To find the maximum likelihood estimates (MLEs), we differentiate the log-likelihood function with respect to each parameter and set the resulting equations equal to zero. Specifically,

$\frac{\partial \ell}{\partial \theta_i}=\sum_{j=1}^n\left(x_j-\theta_j\right) \frac{\partial \theta_j}{\partial \theta_i}=\theta_i-x_i=0 \quad$ for $i=1, \ldots, n$

Therefore, the MLE for $\theta_i$ is simply $\hat{\theta}_i = x_i$ for $i=1,\ldots,n$. This is intuitive since the normal distribution is symmetric and the maximum likelihood estimator for the mean is the sample mean.

Note that there are no restrictions on the $\theta_i$, so we don’t need to worry about any constraints on the estimates.


数学分析课后作业代写的应用代写

数学分析学是数学的一个分支，涉及连续函数、极限和相关理论，如微分、积分、度量、无限序列、数列和分析函数。

英国论文代写Viking Essay为您提供数学分析作业代写Mathematical Analysis代考服务

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数学分析|MATH2023/MATH2923 Analysis代写 Sydney代写

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这是一份Sydney悉尼大学MATH2023/MATH2923的成功案例

数学分析|MATH2023/MATH2923 Analysis代写 Sydney代写


We have
$$
\Phi_{p}(f)=|\Omega|^{-\frac{1}{p}}|f|_{L^{p}(\Omega)} .
$$
By corollary $\Phi_{p}(f)$ viewed as a function of $p$ is increasing with
$$
\Phi_{p}(f) \leq|f|_{L^{\infty}(\Omega)}
$$ that
$$
|f|_{L^{\infty}(\Omega)} \leq \lim {p \rightarrow \infty} \Phi{p}(f)
$$
For $K \in \mathbf{R}$ let
$$
A_{K}:={x \in \Omega|| f(x) \mid \geq K} .
$$
The set $A_{k}$ is measureable since $f$ is and $\left|A_{K}\right|>0$ if $K<|f|_{L^{\infty}(\Omega)}$. Moreover,
$$
\Phi_{p}(f) \geq|\Omega|^{-\frac{1}{p}}\left(\int_{A_{K}}|f(x)|^{p} d x\right)^{1 / p} \geq|\Omega|^{-\frac{1}{p}}\left|A_{K}\right|^{\frac{1}{p}} K .
$$
Passing to the limit $p \rightarrow \infty$ we obtain
$$
\lim {p \rightarrow \infty} \Phi{p}(f) \geq K
$$
Because this holds for all $K<|f|_{L^{\infty}(\Omega)}$ we conclude
$$
\lim {p \rightarrow \infty} \Phi{p}(f) \geq|f|_{L^{\infty}(\Omega)}
$$



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MATH2023/MATH2923 COURSE NOTES ：

Let $\left(f_{k}\right){k \in \mathrm{N}} \subset L^{P}(\Omega)$ be a Cauchy sequence. It suffices to show that $\left(f{k}\right)$ has a convergent subsequence. For every $i \in \mathrm{N}$ there is an integer $N_{i}$ so that
$$
\left|f_{n}-f_{m}\right|_{L P(\Omega)} \leq 2^{-i} \text { whenever } n, m \geq N_{i} .
$$
We construct a subsequence $\left(f_{k_{i}}\right) \subset\left(f_{k}\right)$ so that
$$
\left|f_{k_{i+1}}-f_{k_{i}}\right|_{L P(\Omega)} \leq 2^{-i}
$$
by setting $k_{i}:=\max \left{i, N_{i}\right}$. In order to simplify notation we will from now on assume that
$$
\left|f_{k+1}-f_{k}\right|_{L F(\Omega)} \leq 2^{-k}
$$
so that
$$
M:=\sum_{k \in \mathbf{N}}\left|f_{k+1}-f_{k}\right|_{L^{p}(\Omega)}<\infty
$$















数学分析 Mathematical Analysis  MA140-10/MA152-15 

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这是一份warwick华威大学MA140-10/MA152-15的成功案例

数学分析 Mathematical Analysis  MA140-10/MA152-15 


Proof As is shown in linear algebra, the matrix $A$ that represents $T$ is a product of elementary matrices
$$
A=E_{1} \cdots E_{k} .
$$
Each elementary $2 \times 2$ matrix is one of the following types:
$$
\left[\begin{array}{ll}
\lambda & 0 \
0 & 1
\end{array}\right] \quad\left[\begin{array}{ll}
1 & 0 \
0 & \lambda
\end{array}\right] \quad\left[\begin{array}{ll}
0 & 1 \
1 & 0
\end{array}\right] \quad\left[\begin{array}{ll}
1 & \sigma \
0 & 1
\end{array}\right]
$$
where $\lambda>0$. The first three matrices represent isomorphisms whose effect on $I^{2}$ is obvious: $I^{2}$ is converted to the rectangles $\lambda I \times I, I \times \lambda I, I^{2}$. In each case, the area agrees with the magnitude of the determinant. The fourth isomorphism converts $I^{2}$ to the parallelogram$\Pi$ is Riemann measurable since its boundary is a zero set. By Fubini’s Theorem, we get
$$
|\Pi|=\int \chi_{\Pi}=\int_{0}^{1}\left[\int_{x=\sigma x}^{x=1+\sigma y} 1 d x\right] d y=1=\operatorname{det} E .
$$

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MA140-10/MA152-15 COURSE NOTES ：

$$
d x_{I}: \varphi \mapsto \int_{l^{k}} \frac{\partial \varphi_{I}}{\partial u} d u
$$
where this integral notation is shorthand for
$$
\int_{0}^{1} \ldots \int_{0}^{1} \frac{\partial\left(\varphi_{i_{1}}, \ldots, \varphi_{i_{k}}\right)}{\partial\left(u_{1}, \ldots, u_{k}\right)} d u_{1 \ldots} . d v_{k}
$$
If $f$ is a smooth function on $\mathbb{R}^{n}$ then $f d x_{l}$ is the functional
$$
f d x_{I}: \varphi \mapsto \int_{I^{k}} f(\varphi(u)) \frac{\partial \varphi_{I}}{\partial u} d u
$$










数学分析 Analysis MATH41220-WE01/MATH1051-WE01/MATH3011-WE01

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这是一份durham杜伦大学MATH41220-WE01/MATH1051-WE01/MATH3011-WE01作业代写的成功案例

数学分析 Analysis MATH41220-WE01/MATH1051-WE01/MATH3011-WE01
问题 1.

Thus $\left(d\left(p_{n}, q_{n}\right)\right)$ is a Cauchy sequence in $\mathbb{R}$, and because $\mathrm{R}$ is complete,
$$
L=\lim {n \rightarrow \infty} d\left(p{n}, q_{n}\right)
$$
exists. Let $\left(p_{n}^{\prime}\right)$ and $\left(q_{n}^{\prime}\right)$ be sequences that are co-Cauchy with $\left(p_{n}\right)$ and $\left(q_{n}\right)$, and let
$$
L^{\prime}=\lim {n \rightarrow \infty} d\left(p{n}^{\prime}, q_{n}^{\prime}\right) .
$$


证明 .

Then
$$
\left|L-L^{\prime}\right| \leq\left|L-d\left(p_{n}, q_{n}\right)\right|+\left|d\left(p_{n}, q_{n}\right)-d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)\right|+\left|d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)-L^{\prime}\right| .
$$
As $n \rightarrow \infty$, the first and third terms tend to 0 . the middle term is
$$
\left|d\left(p_{n}, q_{n}\right)-d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)\right| \leq d\left(p_{n}, p_{n}^{\prime}\right)+d\left(q_{n}, q_{n}^{\prime}\right) .
$$

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MATH41220-WE01/MATH1051-WE01/MATH3011-WE01 COURSE NOTES ：

$$
|n|{p}=\frac{1}{p^{k}} $$ where $p^{k}$ is the largest power of $p$ that divides $n$. (The norm of 0 is by definition 0 .) The more factors of $p$, the smaller the $p$-norm. Similarly, if $x=a / b$ is a fraction, we factor $x$ as $$ x=p^{k} \cdot \frac{r}{s} $$ where $p$ divides neither $r$ nor $s$, and we set $$ |x|{p}=\frac{1}{p^{k}} .
$$
The $p$-adic metric on $Q$ is
$$
d_{p}(x, y)=|x-y|_{p} .
$$








量化方法|MA10214 Quantitative methods代写

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The laws of Probability, Bayes’ Theorem, Decision Trees. Binomial, Poisson and Normal distributions and their applications; the relationship between these distributions. Different types of sample; sampling distributions of means, standard deviations and proportions. Confidence intervals and hypothesis testing; types of error, significance levels and P values. Power. Quality control: Acceptance sampling and Shewhart charts.

这是一份Bath巴斯大学学院MA10214作业代写的成功案

量化方法|MA10214 Quantitative methods代写

Let the exercise price be $E$ and the barrier be at $B$ where $B<E$.
Since the Black-Scholes partial differential equation governs the price of the option we can, as before, look for solutions of the form:
$$
c(S, E){d o}=A{1} S^{m_{1}}+A_{2} S^{m_{2}}
$$
subject to the boundary conditions: (i) $c_{d b}(B, E)=0$ and (ii) $c(\infty, E){d o}=S$, see the previous section. From (i) we have: $$ c{d b}(B, E)=A_{1} B^{m_{1}}+A_{2} B^{m_{2}}=0, \text { so } \quad A_{1}=-A_{2} B^{m_{2}-m_{n}}
$$
Therefore
$$
c_{d m}(S, E)=-A_{2} B^{m_{2}-m_{n}} S^{m_{1}}+A_{2} S^{m_{2}}
$$
From (ii), as $S \rightarrow \infty$ :
$$
c_{d m}(S, E)=-A_{2} B^{m_{2}-m_{n}} S^{m_{1}}+A_{2} S^{m_{2}}=S
$$
However, since $m_{2}<0$, we have $A_{2} S^{m_{2}} \rightarrow 0$, as $S \rightarrow \infty$, giving
$$
c_{d o}(S, E)=-A_{2} B^{m_{2}-m_{1}} S^{m_{n}}=S
$$

英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MA10214 COURSE NOTES ：

The Black approximation, $C_{B L}$, can be expressed more concisely in terms of our previously defined notation as:
$$
C_{B L}(S, E, \tau)=\max \left(v_{1}, v_{2}\right)
$$
where $v_{1}$ and $v_{2}$ are the following European calls
$$
v_{1}=c\left(S_{D}, E, \tau\right) \quad \text { and } \quad v_{2}=c\left(S_{D}^{+}, E, \tau_{1}\right), \quad \tau=T-t \quad \tau_{1}=T-t_{n}
$$
and
$$
S_{D}=S-\sum_{i=1}^{n} D_{i} \quad \text { and } \quad S_{D}^{+}=S-\sum_{i=1}^{n-1} D_{i}
$$



数学分析|MA20219 Analysis 2B代写

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If we adapt the ideas seen in previous analysis units, and in particular the idea of a derivative, to complex variables, then we find a remarkably beautiful theory. In particular, a lot of the complications coming from limited differentiability or from various notions of convergence fall away. Of course, this comes at a price, which is that complex differentiability is much more restrictive than the real counterpart. So we will have a very powerful theory, but it will apply only to a relatively small set of functions (which still includes rational, exponential, and trigonometric functions and much more).

这是一份Bath巴斯大学学院MA20219作业代写的成功案

数学分析|MA20219 Analysis 2B代写

Let $U=\mathbb{C} \backslash{1-i,-1-i, \log 2+\pi i}$. Consider the function $f: U \rightarrow \mathbb{C}$ given by
$$
f(z)=\frac{e^{z}+2}{(z-\log 2-\pi i)\left(z^{2}+2 i z-2\right)}, \quad z \in U .
$$
Let $\gamma:[-\pi / 2, \pi / 2+20] \rightarrow \mathbb{C}$ such that
$$
\gamma(t)= \begin{cases}10 e^{i t} & \text { if }-\pi / 2 \leq t \leq \pi / 2 \ (10+\pi / 2-t) i & \text { if } \pi / 2<t \leq \pi / 2+20\end{cases}
$$
Set $\Gamma=\gamma([-\pi / 2, \pi / 2+20])$.
You may use any result from the lectures and exercises to answer the following questions. If you need to determine any winding numbers, you may use geometric intuition.
(a) Show that $f$ has poles of order 1 at $1-i$ and $-1-i$.
(b) Sketch the curve $\Gamma$, including its position relative to the points $1-i$ and $-1-i$. Indicate the orientation of $\gamma$.
(c) The singularity of $f$ at $\log 2+\pi i$ is removable. Using this information (for which no proof is required), or otherwise, find
$$
\int_{\gamma} f(z) d z .
$$

英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MA20219 COURSE NOTES ：

i. The singularities of the given function are 0 and 2 . The distance from the centre to the nearest singularity (apart from the centre itself) is therefore 2. So by a result from the lecture notes, the punctured disc of convergence has radius 2 and therefore is $B_{2}(0) \backslash{0}$.
ii. We have (using the geometric series) for $z \in B_{2}(0)$
$$
\frac{1}{z-2}=\frac{-1}{2} \frac{1}{1-\frac{z}{2}}=\frac{-1}{2} \sum_{k=0}^{\infty}\left(\frac{z}{2}\right)^{k}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k}
$$
so that for $z \in B_{2}(0) \backslash{0}$
$$
\frac{1}{z(z-2)}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k-1}=\sum_{n=-1}^{\infty} \frac{-1}{2^{n+2}} z^{n}
$$



代数 Algebra 1B MA10210

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这是一份BATH巴斯大学MA10210作业代写的成功案例

数学分析 Analysis 1 MA10207
问题 1.

An equivalent system is, therefore,
$$
\begin{aligned}
&x+y-z=0 \
&-4 y+2 z=0
\end{aligned}
$$

证明 .

This has infinitely many solutions: $2 y=z, x=z-y=2 y-y=y$. Hence, $\operatorname{ker} T={(y, y, 2 y): y \in \mathbb{R}}$
$$
={y(1,1,2): y \in \mathbb{R}}=\langle(1,1,2)\rangle,
$$
and nullity $T=1$. Now $T((1,0,0))=(1,3,5), T((0,1,0))=(1,-1,1)$, $T((0,0,1))=(-1,-1,-3)$. But $(1,0,0),(0,1,0),(0,0,1)$ is a basis for $\mathbb{R}^{3}$, so $(1,3,5),(1,-1,1),(-1,-1,-3)$ is a spanning sequence for im $T$, by Theorem 6.4.3. These vectors must be linearly dependent, since, by the dimension theorem,c

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MA10210 COURSE NOTES ：

to above is $T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ given by
$$
T((x, y, z))=(x-y+2 z, 2 x+y-z, x-4 y+7 z)
$$
Now $T((1,0,0))=(1,2,1), \quad T((0,1,0))=(-1,1,-4), \quad T((0,0,1))=(2,-1,7)$ span im $T$, by Theorem $6.4 .3$. However,
$$
(1,2,1)=5(-1,1,-4)+3(2,-1,7),
$$
so these three vectors are linearly dependent. Since $(1,2,1),(-1,1,-4)$ are linearly independent, these two vectors form a basis for im $T$. Hence $\operatorname{rank} A=\operatorname{rank} T=2$.








数论 5E: Number Theory/4H: Number Theory MATHS5074_1/MATHS4108_1

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这是一份GLA格拉斯哥大学MATHS5074_1作业代写的成功案例

数论 5E: Number Theory/4H: Number Theory MATHS5074_1/MATHS4108_1 4
问题 1.

If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then the function $(f \circ \phi) \phi^{\prime}:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$
\int_{\phi(\alpha)}^{\phi(\beta)} f(x) d x=\int_{\alpha}^{\beta} f(\phi(t)) \phi^{\prime}(t) d t
$$

证明 .

If $f:[a, b] \rightarrow \mathbb{R}$ is integrable and $\phi^{\prime}(t) \neq 0$ for every $t \in(\alpha, \beta)$, then the function $(f \circ \phi)\left|\phi^{\prime}\right|:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$
\int^{b} f(x) d x=\int^{\beta} f(\phi(t))\left|\phi^{\prime}(t)\right| d t
$$

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MATHS5074_1/MATHS4108_1 COURSE NOTES ：

Now take $\lambda \in \operatorname{Sp}(T)$. Put
$$
\lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) .
$$
Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain
$$
g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) .
$$
Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus,
$$
f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T))
$$
i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$








数学分析 Analysis 1 MA10207

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这是一份BATH巴斯大学MA10207作业代写的成功案例

数学分析 Analysis 1 MA10207
问题 1.

Let $X$ be a (nonzero) complex Banach space and let $T$ be a bounded endomorphism of $X$; i.e., $T \in B(X)$. For $h \in \mathscr{H}(\mathrm{Sp}(T))$, the contour integral with kernel the resolvent $R(T, \cdot)$ of $T$ is denoted by
$$
\mathscr{R}_{T} h:=\frac{1}{2 \pi i} \oint h(z) R(T, z) d z
$$

证明 .

and called the Riesz-Dunford integral (of the germ $h$ ). If $f$ is a tunction holomorphic in a neighborhood about $\operatorname{Sp}(T)$ then put $f(T):=\mathscr{R}{T} f:=\mathscr{R}{T} \bar{f}$. We also use more suggestive designations like
$$
f(T)=\frac{1}{2 \pi i} \oint \frac{f(z)}{z-T} d z
$$

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MA10207 COURSE NOTES ：

Now take $\lambda \in \operatorname{Sp}(T)$. Put
$$
\lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) .
$$
Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain
$$
g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) .
$$
Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus,
$$
f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T))
$$
i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$








数学分析|Mathematical Analysis代写 MATH0048

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这是一份ucl伦敦大学学院 math0048作业代写的成功案

数学分析|Mathematical Analysis代写 MATH0048
问题 1.

According to $5.6 .9$, for $|\lambda|>|T|$ there is an operator $(1-T / \lambda)^{-1}$ presenting the sum of the Neumann series; i.e.,
$$
R(T, \lambda)=\frac{1}{\lambda}\left(1-{ }^{T} / \lambda\right)^{-1}=\frac{1}{\lambda} \sum_{k=0}^{\infty} \frac{T^{k}}{\lambda^{k}}
$$


证明 .

It is clear that
$$
|R(T, \lambda)| \leq \frac{1}{|\lambda|} \cdot \frac{1}{1-|T | /| \lambda \mid} . \triangleright
$$

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MATH0048 COURSE NOTES ：

This entails sesquilinearity:
$$
\begin{aligned}
(y, x)=&(y, x){\mathbb{R}}-i(i y, x){\mathbb{R}}=(x, y){\mathbb{R}}-i(x, i y){\mathbb{R}} \
&=(x, y){\mathbb{R}}+i(i x, y){\mathbb{R}}=(x, y)^{*}
\end{aligned}
$$
since
$(x, i y){\mathbf{R}}=1 / 4\left(|x+i y|^{2}-|x-i y|^{2}\right)$ ${ }^{1}{ }^{1} / 4\left(|i||y-i x|^{2}-|-i||i x+y|^{2}\right)=-(i x, y){\mathbb{R}}$.