## 数学代写|实分析作业代写Real analysis代考|Taylor Polynomials and Taylor Series

We now consider the problem of representing a function $f$ in terms of a power series in greater detail. Newton derived the power series expansion of many of the elementary functions by algebraic techniques or term-by-term integration. For example, the series expansion of $1 /(1+x)$ can easily be obtained by long division, which upon term-by-term integration gives the power series expansion of $\ln (1+x)$. Maclaurin and Taylor were among the first mathematicians to use Newton’s calculus in determining the coefficients in the power series expansion of a function. Both realized that if a function $f(x)$ had a power series expansion $\sum a_k(x-c)^k$, then the coefficients $a_k$ had to be given by $f^{(k)}(c) / k !$

DEFINITION 8.7.13 Let $f$ be a real-valued function defined on an open interval $I$, and let $c \in I$ and $n \in \mathbb{N}$. Suppose $f^{(n)}(x)$ exists for all $x \in I$. The polynomial
$$T_n(f, c)(x)=\sum_{k=0}^n \frac{f^{(k)}(c)}{k !}(x-c)^k$$
is called the Taylor polynomial of order $n$ of $f$ at the point $c$. If $f$ is infinitely differentiable on $I$, the series
$$\sum_{k=0}^{\infty} \frac{f^{(k)}(c)}{k !}(x-c)^k$$
is called the Taylor series of $f$ at $c$.
For the special case $c=0$, the Taylor series of a function $f$ is often referred to as the Maclaurin series. The first three Taylor polynomials $T_0, T_1, T_2$, are given specifically by
\begin{aligned} & T_0(f, c)(x)=f(c), \ & T_1(f, c)(x)=f(c)+f^{\prime}(c)(x-c), \ & T_2(f, c)(x)=f(c)+f^{\prime}(c)(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^2, \end{aligned}
The Taylor polynomial $T_1(f, c)$ is the linear approximation to $f$ at $c$; that is, the equation of the straight line passing through $(c, f(c))$ with slope $f^{\prime}(c)$.

## 数学代写|实分析作业代写Real analysis代考|Lagrange Form of the Remainder

Our first result, due to Joseph Lagrange (1736-1813), is called the Lagrange form of the remainder. This result, sometimes also referred to as Taylor’s theorem, was previously proved for the special case $n=2$ in Lemma 5.4.3.
THEOREM 8.7.16 Suppose $f$ is a real-valued function on an open interval $I, c \in I$ and $n \in \mathbb{N}$. If $f^{(n+1)}(t)$ exists for every $t \in I$, then for any $x \in I$, there exists a $\zeta$ between $x$ and $c$ such that
$$R_n(x)=R_n(f, c)(x)=\frac{f^{(n+1)}(\zeta)}{(n+1) !}(x-c)^{n+1}$$
Remark. Continuity of $f^{(n+1)}$ is not required.
Proof. Fix $x \in I$, and let $M$ be defined by
$$f(x)=T_n(f, c)(x)+M(x-c)^{n+1}$$
To prove the result we need to show that $(n+1) ! M=f^{(n+1)}(\zeta)$ for some $\zeta$ between $x$ and $c$. To accomplish this, set
\begin{aligned} g(t) & =f(t)-T_n(f, c)(t)-M(t-c)^{n+1} \ & =R_n(t)-M(t-c)^{n+1} \end{aligned}
First, since $T_n$ is a polynomial of degree less than or equal to $n$,
$$g^{(n+1)}(t)=f^{(n+1)}(t)-(n+1) ! M$$
Also, since $T_n^{(k)}(f, c)(c)=f^{(k)}(c), k=0,1, \ldots, n$,
$$g(c)=g^{\prime}(c)=\cdots=g^{(n)}(c)=0$$
For convenience, let’s assume $x>c$. By the choice of $M, g(x)=0$. By the mean value theorem applied to $g$ on the interval $[c, x]$, there exits $x_1$, $c<x_1<x$, such that
$$0=g(x)-g(c)=g^{\prime}\left(x_1\right)(x-c)$$

# 实分析代考

## 数学代写|实分析作业代写Real analysis代考|Taylor Polynomials and Taylor Series

$$T_n(f, c)(x)=\sum_{k=0}^n \frac{f^{(k)}(c)}{k !}(x-c)^k$$

$$\sum_{k=0}^{\infty} \frac{f^{(k)}(c)}{k !}(x-c)^k$$

$$T_0(f, c)(x)=f(c), \quad T_1(f, c)(x)=f(c)+f^{\prime}(c)(x-c), T_2(f, c)(x)=f(c)+f^{\prime}(c)$$

## 数学代写|实分析作业代写Real analysis代考|Lagrange Form of the Remainder

$$R_n(x)=R_n(f, c)(x)=\frac{f^{(n+1)}(\zeta)}{(n+1) !}(x-c)^{n+1}$$

$$f(x)=T_n(f, c)(x)+M(x-c)^{n+1}$$

$$g(t)=f(t)-T_n(f, c)(t)-M(t-c)^{n+1} \quad=R_n(t)-M(t-c)^{n+1}$$

$$g^{(n+1)}(t)=f^{(n+1)}(t)-(n+1) ! M$$

$$g(c)=g^{\prime}(c)=\cdots=g^{(n)}(c)=0$$

$$0=g(x)-g(c)=g^{\prime}\left(x_1\right)(x-c)$$

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