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

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\zwischenueberschrift{The Taylor formula}
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 \anfuehrung{shifted variable} $x-a$, in order to study the local behavior in the \stichwort {expansion point} $a$. Convergence means, in this case, that some \mavergleichskette {\vergleichskette { \epsilon } { > }{0 } { }{ } { }{ } { }{ } } exists, such that for \mavergleichskettedisp {\vergleichskette {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 \mavergleichskettedisp {\vergleichskette { 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 \mavergleichskettedisp {\vergleichskette { 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 \zusatzklammer {or a power series} {.} This is the content of the \stichwort {Taylor expansion} {.} \inputdefinition { } {
Let \mavergleichskette {\vergleichskette {I } { \subseteq }{ \R } { }{ } { }{ } { }{ } } denote an interval, \mathdisp {f \colon I \longrightarrow \R} { } an $n$-times differentiable function, and \mavergleichskette {\vergleichskette {a } { \in }{I } { }{ } { }{ } { }{ } }{.} Then \mavergleichskettedisp {\vergleichskette { T_{ a,n } ( f) ( x ) } { \defeq} { \sum_{ k = 0}^{ n } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k } } { } { } { } { } { } { } }
So \mavergleichskettedisp {\vergleichskette { T_{ a,0 } ( f) ( x ) } { \defeq} { f(a) } { } { } { } { } { } { } } is the constant approximation, \mavergleichskettedisp {\vergleichskette { T_{ a,1 } ( f) ( x ) } { \defeq} { f(a) + f'(a)(x-a) } { } { } { } { } { } { } } is the linear approximation, \mavergleichskettedisp {\vergleichskette { 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, \mavergleichskettedisp {\vergleichskette { 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 \zusatzklammer {uniquely determined} polynomial of degree $\leq n$ with the property that its derivatives and the derivatives of $f$ at $a$ coincide up to order $n$.
\faktsituation {Let $I$ denote a real interval, \mathdisp {f \colon I \longrightarrow \R} { } an \mathl{(n+1)}{-}times differentiable function, and \mavergleichskette {\vergleichskette {a } { \in }{I } { }{ } { }{ } { }{ } } an inner point of the interval.} \faktfolgerung {Then for every point \mavergleichskette {\vergleichskette {x } { \in }{I } { }{ } { }{ } { }{ } }{,} there exists some \mavergleichskette {\vergleichskette {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 }} { . } } \faktzusatz {Here, $c$ may be chosen between \mathkor {a} {and} {x} {.}} } {Real function/Taylor formula/(n+1)-times continuously differentiable/Lagrange/Fact/Proof
\faktsituation {Suppose that $I$ is a bounded closed interval, \mathdisp {f \colon I \longrightarrow \R} { } is an \mathl{(n+1)}{-}times continuously differentiable function, \mavergleichskette {\vergleichskette {a } { \in }{ I } { }{ } { }{ } { }{ } } an inner point, and \mavergleichskette {\vergleichskette {B } { \defeq }{ {\max { \left( \betrag { f^{(n+1)}(c) } , c \in I \right) } } } { }{ } { }{ } { }{ } }{.}} \faktfolgerung {Then, between \mathl{f(x)} and the $n$-th Taylor polynomial, we have the estimate \mavergleichskettedisp {\vergleichskette { \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} } { } { } { } { } { } { } }{.}} \faktzusatz } {
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 .
\zwischenueberschrift{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.
\faktsituation {Let $I$ denote a real interval, \mathdisp {f \colon I \longrightarrow \R} { } an \mathl{(n+1)}{-}times continuously differentiable function, and \mavergleichskette {\vergleichskette {a } { \in }{I } { }{ } { }{ } { }{ } } an inner point of the interval.} \faktvoraussetzung {Suppose that \mathdisp {f'(a)= f^{\prime \prime}(a) = \ldots = f^{(n)}(a)=0 \text{ and } f^{(n+1)}(a) \neq 0} { } is fulfilled.} \faktuebergang {Then the following statements hold.} \faktfolgerung {\aufzaehlungdrei {If $n$ is even, then $f$ does not have a local extremum in $a$. } {Suppose that $n$ is odd. In case \mavergleichskette {\vergleichskette { f^{(n+1)}(a) } { > }{0 } { }{ } { }{ } { }{ } }{,} the function $f$ has an isolated local minimum in $a$. } {Suppose that $n$ is odd. In case \mavergleichskette {\vergleichskette { f^{(n+1)}(a) } { < }{0 } { }{ } { }{ } { }{ } }{,} the function $f$ has an isolated local maximum in $a$. }} \faktzusatz } {
Under the given conditions, the Taylor formula becomes \mavergleichskettedisp {\vergleichskette { f(x)-f(a) } { =} { \frac{ f^{ (n +1) } ( c )}{ (n +1)! } (x-a)^{ n +1 } } { } { } { } { } { } { } }{,} with some $c$ \zusatzklammer {depending on $x$} between \mathkor {a} {and} {x} {.} Depending on whether \mathkor {f^{(n+1)}(a)>0} {or} {f^{(n+1)}(a) < 0} holds, we have \zusatzklammer {due to the continuity of the $(n+1)$-th derivative} \mathkor {f^{(n+1)}(x)>0} {or} {f^{(n+1)}(x) < 0} for \mavergleichskette {\vergleichskette {x } { \in }{ [a-\epsilon,a+\epsilon] } { }{ } { }{ } { }{ } }{,} for a suitable \mavergleichskette {\vergleichskette {\epsilon } { > }{0 } { }{ } { }{ } { }{ } }{.} For these $x$, we have \mavergleichskette {\vergleichskette {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 \mavergleichskette {\vergleichskette {x } { = }{a } { }{ } { }{ } { }{ } } \zusatzklammer {for \mavergleichskette {\vergleichskette {x } { < }{a } { }{ } { }{ } { }{ } }{,} the sign is negative, and for \mavergleichskette {\vergleichskette {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 \mavergleichskette {\vergleichskette { (x-a)^{n+1} } { > }{ 0 } { }{ } { }{ } { }{ } } for all \mavergleichskette {\vergleichskette {x } { \neq }{a } { }{ } { }{ } { }{ } } in the neighborhood. This means, in the neighborhood, in case \mavergleichskette {\vergleichskette { f^{(n+1)}(a) } { > }{0 } { }{ } { }{ } { }{ } }{,} that \mavergleichskette {\vergleichskette { f(x) } { > }{ f(a) } { }{ } { }{ } { }{ } } holds, and we have an isolated minimum in $a$. If \mavergleichskette {\vergleichskette { f^{(n+1)}(a) } { < }{ 0 } { }{ } { }{ } { }{ } }{,} then \mavergleichskette {\vergleichskette { f(x) } { < }{ f(a) } { }{ } { }{ } { }{ } } holds, and we have an isolated maximum in $a$.
A special case of this is that in case \mavergleichskette {\vergleichskette { f'(a) } { = }{0 } { }{ } { }{ } { }{ } } and \mavergleichskette {\vergleichskette { f^{\prime \prime}(a) } { > }{0 } { }{ } { }{ } { }{ } }{,} then we have an isolated minimum, and in case \mavergleichskette {\vergleichskette { f'(a) } { = }{0 } { }{ } { }{ } { }{ } } and \mavergleichskette {\vergleichskette { f^{\prime \prime}(a) } { < }{0 } { }{ } { }{ } { }{ } } we have an isolated maximum.
\zwischenueberschrift{The Taylor series}
\inputdefinition { } {
Let \mavergleichskette {\vergleichskette {I } { \subseteq }{\R } { }{ } { }{ } { }{ } } denote an interval, \mathdisp {f \colon I \longrightarrow \R} { } an infinitely often differentiable function, and \mavergleichskette {\vergleichskette {a } { \in }{I } { }{ } { }{ } { }{ } }{.} Then \mathdisp {\sum_{ k = 0}^{ \infty } \frac{ f^{( k )}(a)}{ k !} (x-a)^{ k }} { }
\faktsituation {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 .} \faktfolgerung {Then $f$ is infinitely often differentiable, and the Taylor series of $f$ in $0$ coincides with the given power series.} \faktzusatz } {
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 .
\inputbeispiel {
We consider the function \mathdisp {f \colon \R \longrightarrow \R , x \longmapsto f(x)} { , } given by \mavergleichskettedisp {\vergleichskette { 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 \mavergleichskette {\vergleichskette {p } { \in }{\R [Z] } { }{ } { }{ } { }{ } }{,} and that therefore the limit for \mathl{x \rightarrow 0,\, x >0} equals $0$ \zusatzklammer {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 \mavergleichskette {\vergleichskette { e^{- \frac{1}{x} } } { > }{0 } { }{ } { }{ } { }{ } }{.}
\zwischenueberschrift{Power series ansatz}
Taylor series/R/Power series/Section