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# 数学代写|数学分析作业代写Mathematical Analysis代考|MTH131LR The Weierstrass Approximation Theorem

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## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|The Weierstrass Approximation Theorem

A frequent problem in Mathematical Analysis is the approximation of a given function $f(x)$ by functions of a special kind.

When $f(x)$ is a continuous function on a closed and bounded interval $[a, b]$, a classical theorem due to Weierstrass provides an exhaustive answer: the approximating functions are polynomials, as long as we interpret the word approximation in the sense of uniform convergence.

Weierstrass Approximation Theorem Any continuous function $f(x)$ on $[a, b]$ is a uniform limit of a sequence of polynomials.

Proof The variable change $x \rightarrow(x-a) /(b-a)$ maps polynomials to polynomials, and reduces us to consider the interval $[0,1]$.

We may also assume $f(0)=f(1)=0$; in fact, once the theorem is proved in that case, setting
$$g(x)=f(x)-f(0)-x[f(1)-f(0)], \quad x \in[0,1],$$
we have $g(0)=g(1)=0$, and the function $f(x)-g(x)$ is a polynomial. So if $g(x)$ is a uniform limit of polynomials, the same is true for $f(x)$.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Abel’s Theorem on Power Series

Consider a power series
$$a_0+a_1 x+a_2 x^2+\ldots+a_k x^k+\ldots$$
whose convergence radius $\varrho$ is a positive real number, and suppose the series also converges at $x=\varrho$ (a similar reasoning holds at $x=-\varrho$ ). Abel’s theorem implies that the sum $f(x)$ of $(1.88)$ is continuous not just on $(-\varrho, \varrho)$, but at $x=\varrho$ as well. Hence the series’ sum at $x=\varrho$ can be computed using the values of $f$ in the limit $x \rightarrow \varrho^{-}$.

We shall discuss two versions of Abel’s theorem. They differ in the proof and in the conclusion. The thesis of Theorem 2 regards the series’ uniform convergence, which implies the sum’s continuity of Theorem 1.
To simplify the notation let us consider the case $\varrho=1$.
This does not affect the result’s generality. In fact, if (1.88) converges at $x=\varrho$, the series of general term $b_k=a_k \varrho^k$ clearly converges.
Since
$$\sum_{k=0}^{\infty} b_k \in \mathbb{R}$$

Abel’s theorem for the case $\varrho=1$ talks about the properties of the power series
$$\sum_{k=0}^{\infty} b_k x^k, \quad-1<x \leq 1,$$
i.e.
$$\sum_{k=0}^{\infty} a_k \varrho^k x^k, \quad-1<x \leq 1$$

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|The Weierstrass Approximation Theorem

Weierstrass Approximation Theorem 任意连续函数 $f(x)$ 在 $[a, b]$ 是洦项式序列的一致极限。

$$g(x)=f(x)-f(0)-x[f(1)-f(0)], \quad x \in[0,1]$$

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Abel’s Theorem on Power Series

$$a_0+a_1 x+a_2 x^2+\ldots+a_k x^k+\ldots$$

$$\sum_{k=0}^{\infty} b_k \in \mathbb{R}$$
㝔例的阿贝尔定理 $\varrho=1$ 谈谈帛级数的性质
$$\sum_{k=0}^{\infty} b_k x^k, \quad-1<x \leq 1$$
IE
$$\sum_{k=0}^{\infty} a_k \varrho^k x^k, \quad-1<x \leq 1$$

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