Taylor Polynomial

Theorem. Suppose that $f$ is a function for which
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all exists. Let
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and define
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Then
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Proof. Define
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and $g(x)=(x-a)^n$. We prove that
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Notice that
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Thus
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and
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We may therefore apply l'H\opital's Rule $n-1$ times to obtain
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Since $Q$ is a polynomial of degree $n-1$, its $(n-1)$st derivative is constant; in fact, MATH. Thus
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Therefore
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Definition. The polynomial $P_{n,a}(x)$ is called the Taylor polynomial of degree $n$ for $f$ at $a$.

Order of Equality

Two functions $f$ and $g$ are said to be equal up to order $n$ at $a$ if
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The above theorem says that the Taylor polynomial equals $f$ up to order $n$ near $a$.

Uniqueness

Theorem. Let $P$ and $Q$ be two polynomials in $(x-a)$, of degree $\leq n$, and suppose that $P$ and $Q$ are equal up to order $n$ neat $a$. Then $P=Q.$

Proof. Let $R=P-Q$. Then $R$ is a polynomial of degree $\leq n$. Write
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The given condition
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implies that
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For $j=0$, we have
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Thus $b_0=0$ and
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Therefore
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For $j=1$, we have
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Thus $b_1=0$ and
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Continuing this way we see that
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Corollary. Let $f$ be $n$-times differentiable at $a$, and suppose that $P$ is a polynomial in $(x-a)$ of degree $\leq n$, which equals $f$ up to order $n$ at $a$. Then $P\equiv P_{n,a}$, the Taylor polynomial of degree $n$ for $f $ near a.

Derivation of the Taylor Expansion from Integrations by Parts

Write
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Adding up these identities, we have
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Taylor's Theorem

Theorem. Suppose that $f^{\prime }$, MATH are defined on $[a,x]$ and $R_{n,a}(x)$ is defined by
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Then

(1) there exists some $t$ in $(a,x)$ such that
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(2) there exists some $t$ in $(a,x)$ such that
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Moreover, if $f^{(n+1)}$ is integrable on $[a,x]$ then

(3) MATH

Proof. Fix $a$ and $x$. Define
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for $t$ in $[a,x]$. Then
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Applying the Mean Value Theorem to $\ S$ on the interval $[a,x]:$there exists some $\ t$ in $(a,x)$ such that
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Note that
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Therefore
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This is called the Cauchy form of remainder.

Set $g(t)=(x-t)^{n+1}$. Now apply the Cauchy Mean value Theorem to $S$ and $g $ there exists some $t$ in $(a,x)$ such that
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Therefore
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or
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This is called the Lagrange form of the remainder.

If $f^{(n+1)}$ is integrable on $[a,x]$, it follows from the Fundamental Theorem of Calculus that
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Therefore
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This is called the integral form of the remainder.

Some Consequences


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From the equation
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we have
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for all $x>-1.$

From the equation
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we have
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