微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets

微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets

简单的说,学好微积分(数学分析)是一个毁灭自己的先天直觉然后重新塑造一个后天直觉。

转变思维永远不是简单,但是不转变,贪图一时的捷径只是饮鸩止渴罢了。高中的时候,我一个同学很背单词的时候喜欢用汉字去拼那些单词的发音,还喜欢学各种解题技巧,这个时候我和他的成绩是一样的。

国外的老师较为看重学生homework的完成情况,对于同学们来说,完成一门科目作业并获得不错的成绩是尤为重要的事情。但对于不少同学来说,在自身英语说存在局限的情况下,当数学基础较为薄弱时,微积分作业的难度一下子就提升了,很难独立完成微积分作业。Calculus-do™提供的专业微积分代写能为大家解决所有的学术困扰,我们不仅会帮大家完成作业,还提供相应的数学知识辅导课程,以此来提高同学们学习能力。

我们提供的econ代写服务范围广, 其中包括但不限于:

  • 单变量微积分
  • 多变量微积分
  • 傅里叶级数
  • 黎曼积分
  • ODE
  • 微分学
微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets

微积分作业代写calclulus代考|Larger exceptional sets

As integration theory developed over the centuries since Newton it became clear that the theory required quite large exceptional sets, certainly larger than just a few points. But this also requires us to characterize those sets that can be so neglected and also to describe what we must require of an indefinite integral so that these sets can be ignored.

At a calculus level we can easily go one step further, even if we cannot quite approach the full generality needed. The key is to push Lemma $1.2$ and Lemma $1.3$ much further. We cannot do this with help from the mean-value theorem as before: indeed the proof is deferred to the next chapter where we introduce a new technique needed for the integration theory.

LEMMA 1.5. Suppose that $F$ and $G$ are both continuous functions on an interval $[a, b]$. Suppose that there is a sequence of points $e_{1}, e_{2}, e_{3}, \ldots$ of points from $[a, b]$ and that $F^{\prime}(x)=G^{\prime}(x)$ for all $a<x<b$ except possibly at the points $e_{1}, e_{2}, e_{3}$, … Then
$$
F(b)-F(a)=G(b)-G(a)
$$

微积分作业代写calclulus代考|A version of the Newton integral for elementary calculus

Assuming Lemma $1.5$ for the moment we can introduce a much improved version of the Newton integral.

Definition $1.6$ (Modified Newton Integral). Suppose that $f$ is a function defined on an interval $(a, b)$ except possibly at the points of a sequence $e_{1}, e_{2}, e_{3}$, .. from $[a, b]$. Suppose that we can find a continuous function $F:[a, b] \rightarrow \mathbb{R}$ so that $F^{\prime}(x)=f(x)$ for every $x$ with $a<x<b$ with perhaps the exception of the points $e_{1}, e_{2}, e_{3}, \ldots$. Then we will say that $F$ is an indefinite integral of $f$ on $[a, b]$ and we will write
$$
\int_{a}^{b} f(x) d x=F(b)-F(a)
$$
and call the latter the definite integral of $f$ on $[a, b]$.
The only justification needed would be to use Lemma $1.5$ to check that if $F$ and $G$ both qualify to be indefinite integrals of $f$ on an interval $[a, b]$, then $F$ and $G$ differ by a constant so that $F(b)-F(a)=G(b)-G(a)$. Thus the definite integral is unambiguously defined.

There is one subtle point here that might be missed. Suppose that two functions $F$ and $G$ both qualify to be indefinite integrals of $f$. That means that there is some sequence of points $e_{1}, e_{2}, e_{3}, \ldots$ from $[a, b]$ and that $F^{\prime}(x)=f(x)$ provided $x$ is not one of the points in this sequence. It also means that there is some sequence of points $e_{1}^{\prime}, e_{2}^{\prime}, e_{3}^{\prime}, \ldots$ from $[a, b]$ [not necessarily the same sequence as before] and that $G^{\prime}(x)=f(x)$ provided $x$ is not one of the points in this sequence.

Accordingly, we observe that $F^{\prime}(x)=G^{\prime}(x)$ except possibly at points belonging to the combined sequence
$$
e_{1}, e_{1}^{\prime}, e_{2}, e_{2}^{\prime}, e_{3}, e_{3}^{\prime}, e_{4}, e_{4}^{\prime}, \ldots
$$allowing us to apply Lemma 1.5. From that lemma we deduce that $F$ and $G$ differ by a constant and that $F(b)-F(a)=G(b)-G(a)$. Thus the definite integral is unambiguously defined no matter what indefinite integral we choose to use.

微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets

微积分作业代写calclulus代考|Larger exceptional sets


自牛顿以来的几个世纪以来,随着积分理论的发展,很明显该理论需要相当大的异常
集,当然不仅仅是几个点。但这也要求我们描述那些可以如此忽略的集合,并苗述我们 必须要求的不定积分,以便可以忽略这些集合。
在微积分水平上,我们可以轻松地更进一步,即使我们不能完全接近所需的全部一般 性。关键是推引理1.2和引理1.3更进一步。我们不能像以前那样借助中值定理来做到这 一点: 事实上,证明推迟到下一章我们介绍积分理论所需的新技术。
引理 1.5。假设 $F$ 和 $G$ 都是区间上的连续函数 $[a, b]$. 假设有一个点序列 $e_{1}, e_{2}, e_{3}, \ldots$ 的 点数 $[a, b]$ 然后 $F^{\prime}(x)=G^{\prime}(x)$ 对所有人 $a<x<b$ 除了可能在点 $e_{1}, e_{2}, e_{3} , \ldots$ 然 后
$$
F(b)-F(a)=G(b)-G(a)
$$


微积分作业代写calclulus代考|A version of the Newton integral for elementary calculus


假设引理1.5目前我们可以引入牛顿积分的改进版本。
定义1.6 (修正牛顿积分) 。假设 $f$ 是在区间上定义的函数 $(a, b)$ 除了可能在序列的点 $e_{1}, e_{2}, e_{3} , \ldots$ 从 $[a, b]$. 假设我们可以找到一个连续函数 $F:[a, b] \rightarrow \mathbb{R}$ 以便
$F^{\prime}(x)=f(x)$ 对于每个 $x$ 和 $a<x<b$ 也许除了点 $e_{1}, e_{2}, e_{3}, \ldots$. 然后我们会说 $F$ 是 一个不定积分 $f$ 上 $[a, b]$ 我们会写
$$
\int_{a}^{b} f(x) d x=F(b)-F(a)
$$
并称后者为定积分 $f$ 上 $[a, b]$.
唯一需要的理由是使用引理 $1.5$ 检查是否 $F$ 和 $G$ 两者都有资格成为的不定积分 $f$ 在一个区 间 $[a, b]$ ,然后 $F$ 和 $G$ 相差一个常数,使得 $F(b)-F(a)=G(b)-G(a)$. 因此,定 积分是明确定义的。
这里有一个微妙的地方可能会被遗漏。假设有两个函数 $F$ 和 $G$ 两者都有资格成为的不定 积分 $f$. 这意味着有一些点序列 $e_{1}, e_{2}, e_{3}, \ldots$ 从 $[a, b]$ 然后 $F^{\prime}(x)=f(x)$ 假如 $x$ 不是这 个序列中的点之一。这也意味着有一些点序列 $e_{1}^{\prime}, e_{2}^{\prime}, e_{3}^{\prime}, \ldots$ 从 $[a, b][$ 不一定和以前的 顺序相同 $] G^{\prime}(x)=f(x)$ 假如 $x$ 不是这个序列中的点之一。
因此,我们观察到 $F^{\prime}(x)=G^{\prime}(x)$ 可能在属于组合序列的点除外
$$
e_{1}, e_{1}^{\prime}, e_{2}, e_{2}^{\prime}, e_{3}, e_{3}^{\prime}, e_{4}, e_{4}^{\prime}, \ldots
$$
允许我们应用引理 1.5。从这个引理我们推断出 $F$ 和 $G$ 相差一个常数,并且 $F(b)-F(a)=G(b)-G(a)$. 因此,无论我们选择使用什么不定积分,定积分都 是明确定义的。

微积分网课代修|积分学代写Integral Calculus代考|MA1030C Larger exceptional sets
微积分作业代写calclulus代考

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