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# 数学代写|实分析代写Real Analysis代考|A Glance at Our Example

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## 数学代写|实分析代写Real Analysis代考|A Glance at Our Example

For the series given in equation (5.35) on page 192 , we see that
$$E_k(x, 0)=0-\frac{f_k(x)-0}{x-0}=\frac{-x^2}{\left(1+k x^2\right)\left(1+(k-1) x^2\right)} .$$

This should look familiar. It is precisely the summand that we saw in the last section where we showed that
$$\mathcal{E}(x, 0)=\sum_{k=1}^{\infty} E_k(x, 0)=\sum_{k=1}^{\infty} \frac{-x^2}{\left(1+k x^2\right)\left(1+(k-1) x^2\right)}=-1,$$
(since $x$ is not 0 ). No matter how close we take $x$ to $0, \mathcal{E}(x, 0)$ will remain -1 . It cannot be made arbitrarily small. This confirms what we already knew; we cannot differentiate this series at $x=0$ by differentiating each summand.

The Solution
Uniform convergence of $\sum_{k=1}^{\infty} f_k(x)$ is not enough to guarantee that term-by-term differentiation can be used. Uniform convergence of the series of derivatives, $\sum_{k=1}^{\infty} f_k^{\prime}(x)$, is sufficient.

Theorem 5.7 (Term-by-term Differentiation). Let $f_1+f_2+f_3+\cdots$ be a series of functions that converges at $x=a$ and for which the series of derivatives, $f_1^{\prime}+f_2^{\prime}+$ $f_3^{\prime}+\cdots$, converges uniformly over an open interval I that contains a. It follows that

1. $F=f_1+f_2+f_3+\cdots$ converges uniformly over the interval $I$,
2. $F$ is differentiable at $x=a$, and
3. for all $x \in I, F^{\prime}(x)=\sum_{k=1}^{\infty} f_k^{\prime}(x)$.
Proof: The key to this proof is defining the function
$$g_k(x)=\frac{f_k(x)-f_k(a)}{x-a}, \quad x \neq a$$

## 数学代写|实分析代写Real Analysis代考|Integration

In his derivation of the formula for the coefficients of a Fourier series, Joseph Fourier assumed that the integral of a series is the sum of the integrals. This is a questionable procedure that will sometimes fail. It is correct, however, when the series in question converges uniformly over the interval of integration.

Theorem 5.8 (Term-by-term Integration). Let $f_1+f_2+f_3+\cdots$ be uniformly convergent over the interval $[a, b]$, converging to $F$. If each $f_k$ is integrable over $[a, b]$, then so is $F$ and
$$\int_a^b F(x) d x=\sum_{k=1}^{\infty} \int_a^b f_k(x) d x$$
Before proceeding with the proof, we need to face one major obstacle: we have not yet defined integration. The reason for this is that defining integration is not easy. It requires a very profound understanding of the nature of the real number line. In fact, it will not be until the sequel to this book, A Radical Approach to Lebesgue’s Theory of Integration, that we do justice to the question of integration. The modern definition was not determined until the 20 th century.

In the meantime, you will have to rely on whatever definition of integration you prefer. Fortunately, to prove this theorem we only need two properties of the integral:
\begin{aligned} \int_a^b(f(x)+g(x)) d x & =\int_a^b f(x) d x+\int_a^b g(x) d x, \ \left|\int_a^b f(x) d x\right| & \leq \int_a^b|f(x)| d x \end{aligned}

## 数学代写|实分析代写Real Analysis代考|A Glance at Our Example

$$E_k(x, 0)=0-\frac{f_k(x)-0}{x-0}=\frac{-x^2}{\left(1+k x^2\right)\left(1+(k-1) x^2\right)}$$

$$\mathcal{E}(x, 0)=\sum_{k=1}^{\infty} E_k(x, 0)=\sum_{k=1}^{\infty} \frac{-x^2}{\left(1+k x^2\right)\left(1+(k-1) x^2\right)}=-1,$$
(自从 $x$ 不为 0 ) 。无论我们走多近 $x$ 到 $0, \mathcal{E}(x, 0)$ 将保持 -1 。它不能任意变小。这证实了我们已经 知道的；我们无法区分这个系列 $x=0$ 通过区分每个被加数。

1. $F=f_1+f_2+f_3+\cdots$ 在区间内均匀收敛 $I$,
2. $F$ 可微于 $x=a$ ，和
3. 对全部 $x \in I, F^{\prime}(x)=\sum_{k=1}^{\infty} f_k^{\prime}(x)$.
证明: 这个证明的关键是定义函数
$$g_k(x)=\frac{f_k(x)-f_k(a)}{x-a}, \quad x \neq a$$

## 数学代写|实分析代写Real Analysis代考|Integration’

$$\int_a^b F(x) d x=\sum_{k=1}^{\infty} \int_a^b f_k(x) d x$$

$$\int_a^b(f(x)+g(x)) d x=\int_a^b f(x) d x+\int_a^b g(x) d x,\left|\int_a^b f(x) d x\right| \leq \int_a^b|f(x)| d x$$

## MATLAB代写

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