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Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 17/latex

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\setcounter{section}{17}






\subtitle {The Taylor formula}






\image{ \begin{center}
\includegraphics[width=5.5cm]{\imageinclude {Taylor_Brook_Goupy_NPG.jpg} }
\end{center}
\imagetext {Brook Taylor (1685-1731)} }

\imagelicense { Taylor Brook Goupy NPG.jpg } {Louis Goupy} {Astrochemist} {Commons} {PD} {}

So far, we have only considered power series of the form \mathl{\sum_{k=0}^\infty c_k x^k}{.} Now we allow that the variable $x$ may be replaced by a \quotationshort{shifted variable}{} $x-a$, in order to study the local behavior in the \keyword {expansion point} {} $a$. Convergence means, in this case, that some
\mathrelationchain
{\relationchain
{ \epsilon }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} exists, such that for
\mathrelationchaindisplay
{\relationchain
{x }
{ \in} {] a- \epsilon, a+ \epsilon[ }
{ } { }
{ } { }
{ } { }
} {}{}{,} the series converges. In this situation, the function, presented by the power series, is again differentiable, and its derivative is given as in Theorem 16.1 . For a convergent power series
\mathrelationchaindisplay
{\relationchain
{ f(x) }
{ \defeq} { \sum _{ k= 0}^\infty c_k (x-a)^{ k } }
{ } { }
{ } { }
{ } { }
} {}{}{,} the polynomials \mathl{\sum_{k=0}^n c_k (x-a)^k}{} yield polynomial approximations for the function $f$ in the point $a$. Moreover, the function $f$ is arbitrarily often differentiable in $a$, and the higher derivatives in the point $a$ can be read of from the power series directly, namely
\mathrelationchaindisplay
{\relationchain
{ f^{(n)} (a) }
{ =} { n ! c_n }
{ } { }
{ } { }
{ } { }
} {}{}{.} We consider now the question whether we can find, starting with a differentiable function of sufficiently high order, approximating polynomials \extrabracket {or a power series} {} {.} This is the content of the \keyword {Taylor expansion} {.}


\inputdefinition
{ }
{

Let
\mathrelationchain
{\relationchain
{I }
{ \subseteq }{ \R }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} denote an interval,
\mathdisp {f \colon I \longrightarrow \R} { }
an $n$-times differentiable function, and
\mathrelationchain
{\relationchain
{a }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.} Then
\mathrelationchaindisplay
{\relationchain
{ T_{ a,n } ( f) ( x ) }
{ \defeq} { \sum_{ k = 0}^{ n } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k } }
{ } { }
{ } { }
{ } { }
} {}{}{}

is called the \definitionword {Taylor polynomial of degree}{} $n$ for $f$ in the point $a$.

}

So
\mathrelationchaindisplay
{\relationchain
{ T_{ a,0 } ( f) ( x ) }
{ \defeq} { f(a) }
{ } { }
{ } { }
{ } { }
} {}{}{} is the constant approximation,
\mathrelationchaindisplay
{\relationchain
{ T_{ a,1 } ( f) ( x ) }
{ \defeq} { f(a) + f'(a)(x-a) }
{ } { }
{ } { }
{ } { }
} {}{}{} is the linear approximation,
\mathrelationchaindisplay
{\relationchain
{ T_{ a,2 } ( f) ( x ) }
{ \defeq} { f(a) + f'(a)(x-a) + { \frac{ f^{\prime \prime} (a) }{ 2 } } (x-a)^2 }
{ } { }
{ } { }
{ } { }
} {}{}{} is the quadratic approximation,
\mathrelationchaindisplay
{\relationchain
{ T_{ a,3 } ( f) ( x ) }
{ \defeq} { f(a) + f'(a)(x-a) + { \frac{ f^{\prime \prime} (a) }{ 2 } } (x-a)^2+ { \frac{ f^{\prime \prime \prime} (a) }{ 6 } } (x-a)^3 }
{ } { }
{ } { }
{ } { }
} {}{}{} is the approximation of degree $3$, etc. The Taylor polynomial of degree $n$ is the \extrabracket {uniquely determined} {} {} polynomial of degree $\leq n$ with the property that its derivatives and the derivatives of $f$ at $a$ coincide up to order $n$.




\inputfaktbeweisnichtvorgefuehrt
{Real function/Taylor formula/(n+1)-times continuously differentiable/Lagrange/Fact}
{Theorem}
{}
{

\factsituation {Let $I$ denote a real interval,
\mathdisp {f \colon I \longrightarrow \R} { }
an \mathl{(n+1)}{-}times differentiable function, and
\mathrelationchain
{\relationchain
{a }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} an inner point of the interval.}
\factconclusion {Then for every point
\mathrelationchain
{\relationchain
{x }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} there exists some
\mathrelationchain
{\relationchain
{c }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} such that
\mathdisp {f( x) = \sum_{ k = 0}^{ n } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k } + \frac{ f^{ (n+1) } ( c )}{ (n+1)! } (x-a)^{ n+1 }} { . }
}
\factextra {Here, $c$ may be chosen between \mathcor {} {a} {and} {x} {.}}
}
{Real function/Taylor formula/(n+1)-times continuously differentiable/Lagrange/Fact/Proof

}







\image{ \begin{center}
\includegraphics[width=5.5cm]{\imageinclude {Sintay.svg} }
\end{center}
\imagetext {The real sine function, together with several approximating Taylor polynomials (of odd degree).} }

\imagelicense { Sintay.svg } {} {Qualc1} {Commons} {CC-by-sa 3.0} {}




\inputfactproof
{Real function/Taylor formula/(n+1)-times continuously differentiable/Estimate for error/Fact}
{Corollary}
{}
{

\factsituation {Suppose that $I$ is a bounded closed interval,
\mathdisp {f \colon I \longrightarrow \R} { }
is an \mathl{(n+1)}{-}times continuously differentiable function,
\mathrelationchain
{\relationchain
{a }
{ \in }{ I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} an inner point, and
\mathrelationchain
{\relationchain
{B }
{ \defeq }{ {\max { \left( \betrag { f^{(n+1)}(c) } , c \in I \right) } } }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.}}
\factconclusion {Then, between \mathl{f(x)}{} and the $n$-th Taylor polynomial, we have the estimate
\mathrelationchaindisplay
{\relationchain
{ \betrag { f(x) - \sum_{ k = 0}^{ n } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k } } }
{ \leq} { \frac{B}{ (n+1)!}\betrag { x-a }^{n+1} }
{ } { }
{ } { }
{ } { }
} {}{}{.}}
\factextra {}
}
{

The number $B$ exists due to Theorem 11.13 , since the \mathl{(n+1)}{-}th derivative \mathl{f^{(n+1)}}{} is continuous on the compact interval $I$. The statement follows, therefore, directly from Theorem 17.2 .

}






\subtitle {Criteria for extrema}

We have seen in the 15th lecture that it is a necessary condition for a differentiable function to have a local extremum at a point, that its derivative equals $0$ at this point. We give now an important sufficient criterion, which relies on higher derivatives.




\inputfactproof
{Real function/Extrema/Higher derivatives/Fact}
{Theorem}
{}
{

\factsituation {Let $I$ denote a real interval,
\mathdisp {f \colon I \longrightarrow \R} { }
an \mathl{(n+1)}{-}times continuously differentiable function, and
\mathrelationchain
{\relationchain
{a }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} an inner point of the interval.}
\factcondition {Suppose that
\mathdisp {f'(a)= f^{\prime \prime}(a) = \ldots = f^{(n)}(a)=0 \text{ and } f^{(n+1)}(a) \neq 0} { }
is fulfilled.}
\factsegue {Then the following statements hold.}
\factconclusion {\enumerationthree {If $n$ is even, then $f$ does not have a local extremum in $a$. } {Suppose that $n$ is odd. In case
\mathrelationchain
{\relationchain
{ f^{(n+1)}(a) }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} the function $f$ has an isolated local minimum in $a$. } {Suppose that $n$ is odd. In case
\mathrelationchain
{\relationchain
{ f^{(n+1)}(a) }
{ < }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} the function $f$ has an isolated local maximum in $a$. }}
\factextra {}
}
{

Under the given conditions, the Taylor formula becomes
\mathrelationchaindisplay
{\relationchain
{ f(x)-f(a) }
{ =} { \frac{ f^{ (n +1) } ( c )}{ (n +1)! } (x-a)^{ n +1 } }
{ } { }
{ } { }
{ } { }
} {}{}{,} with some $c$ \extrabracket {depending on $x$} {} {} between \mathcor {} {a} {and} {x} {.} Depending on whether \mathcor {} {f^{(n+1)}(a)>0} {or} {f^{(n+1)}(a) < 0} {} holds, we have \extrabracket {due to the continuity of the $(n+1)$-th derivative} {} {} \mathcor {} {f^{(n+1)}(x)>0} {or} {f^{(n+1)}(x) < 0} {} for
\mathrelationchain
{\relationchain
{x }
{ \in }{ [a-\epsilon,a+\epsilon] }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} for a suitable
\mathrelationchain
{\relationchain
{\epsilon }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.} For these $x$, we have
\mathrelationchain
{\relationchain
{c }
{ \in }{ [a-\epsilon,a+\epsilon] }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} so that the sign of \mathl{f^{(n+1)}(c)}{} depends on the sign of \mathl{f^{(n+1)}(a)}{.}
For $n$ even, \mathl{n+1}{} is odd and therefore the sign of \mathl{(x-a)^{n+1}}{} changes at
\mathrelationchain
{\relationchain
{x }
{ = }{a }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} \extrabracket {for
\mathrelationchain
{\relationchain
{x }
{ < }{a }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} the sign is negative, and for
\mathrelationchain
{\relationchain
{x }
{ > }{a }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} the sign is positive} {} {.} Since the sign of \mathl{f^{(n+1)}(c)}{} does not change, the sign of \mathl{f(x)-f(a)}{} is changing. This means that there can not be an extremum.
Suppose now that $n$ is odd. Then \mathl{n+1}{} is even, hence
\mathrelationchain
{\relationchain
{ (x-a)^{n+1} }
{ > }{ 0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} for all
\mathrelationchain
{\relationchain
{x }
{ \neq }{a }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} in the neighborhood. This means, in the neighborhood, in case
\mathrelationchain
{\relationchain
{ f^{(n+1)}(a) }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} that
\mathrelationchain
{\relationchain
{ f(x) }
{ > }{ f(a) }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} holds, and we have an isolated minimum in $a$. If
\mathrelationchain
{\relationchain
{ f^{(n+1)}(a) }
{ < }{ 0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} then
\mathrelationchain
{\relationchain
{ f(x) }
{ < }{ f(a) }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} holds, and we have an isolated maximum in $a$.

}


A special case of this is that in case
\mathrelationchain
{\relationchain
{ f'(a) }
{ = }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} and
\mathrelationchain
{\relationchain
{ f^{\prime \prime}(a) }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} then we have an isolated minimum, and in case
\mathrelationchain
{\relationchain
{ f'(a) }
{ = }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} and
\mathrelationchain
{\relationchain
{ f^{\prime \prime}(a) }
{ < }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} we have an isolated maximum.






\subtitle {The Taylor series}




\inputdefinition
{ }
{

Let
\mathrelationchain
{\relationchain
{I }
{ \subseteq }{\R }
{ }{ }
{ }{ }
{ }{ }
} {}{}{} denote an interval,
\mathdisp {f \colon I \longrightarrow \R} { }
an infinitely often differentiable function, and
\mathrelationchain
{\relationchain
{a }
{ \in }{I }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.} Then
\mathdisp {\sum_{ k = 0}^{ \infty } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k }} { }

is called the \definitionword {Taylor series}{} of $f$ in the point $a$.

}




\inputfactproof
{Convergent power series/R/Taylor series/Coincidence/Fact}
{Theorem}
{}
{

\factsituation {Let \mathl{\sum _{ n= 0}^\infty c_n x^{ n }}{} denote a power series which converges on the interval \mathl{]-r, r[}{,} and let
\mathdisp {f \colon ]-r,r[ \longrightarrow \R} { }
denote the function defined via Theorem 12.2 .}
\factconclusion {Then $f$ is infinitely often differentiable, and the Taylor series of $f$ in $0$ coincides with the given power series.}
\factextra {}
}
{

That $f$ is infinitely often differentiable, follows directly from Theorem 16.1 by induction. Therefore, the Taylor series exists in particular in the point $0$. Hence, we only have to show that the $n$-th derivative has \mathl{c_n n!}{} as its value. But this follows also from Theorem 16.1 .

}





\inputexample{}
{

We consider the function
\mathdisp {f \colon \R \longrightarrow \R , x \longmapsto f(x)} { , }
given by
\mathrelationchaindisplay
{\relationchain
{ f(x) }
{ \defeq} { \begin{cases} 0,\, \text{ if } x \leq 0\, , \\ e^{- \frac{1}{x} },\, \text{ if } x > 0 \, . \end{cases} }
{ } { }
{ } { }
{ } { }
} {}{}{} We claim that this function is infinitely often differentiable, which is only in $0$ not directly clear. We first show, by induction, that all derivatives of \mathl{e^{- \frac{1}{x} }}{} have the form \mathl{p { \left( \frac{1}{x} \right) } e^{- \frac{1}{x} }}{} with certain polynomials
\mathrelationchain
{\relationchain
{p }
{ \in }{\R [Z] }
{ }{ }
{ }{ }
{ }{ }
} {}{}{,} and that therefore the limit for \mathl{x \rightarrow 0,\, x >0}{} equals $0$ \extrabracket {see Exercise 17.16 and Exercise 17.17 } {} {.} Therefore, the limit exists for all derivatives and is $0$. So all derivatives in $0$ have value $0$, and therefore the Taylor series in $0$ is just the zero series. However, the Function $f$ is in no neighborhood of $0$ the zero function, since
\mathrelationchain
{\relationchain
{ e^{- \frac{1}{x} } }
{ > }{0 }
{ }{ }
{ }{ }
{ }{ }
} {}{}{.}

}






\subtitle {Power series ansatz}

Taylor series/R/Power series/Section