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# 5.6.E: Problemas en el teorema de Tayior

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## Ejercicio$$\PageIndex{1}$$

Completar las pruebas de Teoremas$$1,1^{\prime},$$ y 2.

## Ejercicio$$\PageIndex{2}$$

Verificar Nota 1 y Ejemplos (b) y$$\left(\mathrm{b}^{\prime \prime}\right)$$.

## Ejercicio$$\PageIndex{3}$$

Tomando$$g(t)=(a-t)^{s}, s>0,$$ en$$(6),$$ hallazgo
\ [
R_ {n} =\ frac {f^ {(n+1)} (q)} {n! s} (x-p) ^ {s} (x-q) ^ {n+1-s}\ quad (\ texto {resto Schloemilch-Roche}).
\]
Obtener$$\left(5^{\prime}\right)$$ y$$\left(5^{\prime \prime}\right)$$ de ella.

## Ejercicio$$\PageIndex{4}$$

Demostrar que$$P_{n}$$ (como se define) es el único polinomio de grado$$n$$ tal que
\ [
f^ {(k)} (p) =P_ {n} ^ {(k)} (p),\ quad k=0,1,\ ldots, n.
\]
[Pista: Diferenciar$$P_{n} n$$ tiempos para verificar que satisface esta propiedad.
Por singularidad, supongamos que esto también se mantiene para
\ [
P (x) =\ sum_ {k=0} ^ {n} a_ {k} (x-p) ^ {k}.
\]
Diferenciar$$P n$$ tiempos para mostrar que
\ [
P^ {(k)} (p) =f^ {(k)} (p) =a_ {k} k! ,
\]
$$\left.\text { so } P=P_{n} . \text { (Why?) }\right]$$

## Ejercicio$$\PageIndex{5}$$

Con$$P_{n}$$ como se define, probar que si$$f$$ es$$n$$ tiempos diferenciables en$$p,$$ entonces
\ [
f (x) -P_ {n} (x) =o\ left ((x-p) ^ {n}\ right)\ text {as} x\ rightarrow p
\]
(teorema de Taylor con término de resto Peano).
[Pista: Dejar$$R(x)=f(x)-P_{n}(x)$$ y
\ [
\ delta (x) =\ frac {R (x)} {(x-p) ^ {n}}\ texto {con}\ delta (p) =0.
\]
Usando la regla “simplificada” L'Hôpital (Problema 3 in$$§3$$) repetidamente$$n$$ veces, probar$$\left.\text { that } \lim _{x \rightarrow p} \delta(x)=0 .\right]$$

## Ejercicio$$\PageIndex{6}$$

Utilice el Teorema$$1^{\prime}$$ con$$p=0$$ para verificar las siguientes expansiones, y probarlo$$\lim _{n \rightarrow \infty} R_{n}=0$$.
(a)$$\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots-\frac{(-1)^{m} x^{2 m-1}}{(2 m-1) !}+\frac{(-1)^{m} x^{2 m+1}}{(2 m+1) !} \cos \theta_{m} x$$
para todos$$x \in E^{1}$$;
(b)$$\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\cdots+\frac{(-1)^{m} x^{2 m}}{(2 m) !}-\frac{(-1)^{m} x^{2 m+2}}{(2 m+2) !} \sin \theta_{m} x$$ para
todos$$x \in E^{1} .$$
[Consejos: Let$$f(x)=\sin x$$ e$$g(x)=\cos x .$$ Inducción muestra que
\ [
f^ {(n)} (x) =\ sin\ left (x+\ frac {n\ pi} {2}\ right)\ text {y} g^ {(n)} (x) =\ cos\ izquierda (x+\ frac {n\ pi} {2}\ derecha).
\]
Usando fórmula$$\left(5^{\prime}\right),$$ demostrar que
\ [
\ izquierda|R_ {n} (x)\ derecha|\ leq\ izquierda|\ frac {x^ {n+1}} {(n+1)!} \ derecha|\ fila derecha 0.
\]
En efecto,$$x^{n} / n !$$ es el término general de una serie convergente
\ [
\ left. \ suma\ frac {x^ {n}} {n!} \ quad\ text {(ver Capítulo} 4, §13,\ text {Ejemplo} (\ mathrm {d})\ right).
\]
$$\left.\text { Thus } x^{n} / n ! \rightarrow 0 \text { by Theorem } 4 \text { of the same section. }\right]$$

## Ejercicio$$\PageIndex{7}$$

Para cualquiera$$s \in E^{1}$$ y$$n \in N,$$ definir
\ [
\ left (\ begin {array} {l} {s}\\ {n}\ end {array}\ right) =\ frac {s (s-1)\ cdots (s-n+1)} {n!} \ text {con}\ left (\ begin {array} {l} {s}\\ {0}\ end {array}\ right) =1.
\]
Entonces prueba lo siguiente.
(i)$$\lim _{n \rightarrow \infty} n\left(\begin{array}{l}{s} \\ {n}\end{array}\right)=0$$ si$$s>0$$,
(ii)$$\lim _{n \rightarrow \infty}\left(\begin{array}{l}{s} \\ {n}\end{array}\right)=0$$ si$$s>-1$$,
(iii) Para cualquier fijo$$s \in E^{1}$$ y$$x \in(-1,1)$$.
\ [
\ lim _ {n\ fila derecha\ infty}\ left (\ begin {array} {l} {s}\\ {n}\ end {array}\ right) n x^ {n} =0;
\]
por lo tanto
\ [
\ lim _ {n\ fila derecha\ infty}\ left (\ begin {array} {c} {s}\\ {n}\ end {array}\ derecha) x^ {n} =0.
\]
$$\left[\text { Hints: }(\mathrm{i}) \text { Let } a_{n}=\left|n\left(\begin{array}{l}{s} \\ {n}\end{array}\right)\right| . \text { Verify that }\right.$$
\ [
a_ {n} =|s|\ izquierda|1-\ frac {s} {1}\ derecha|\ izquierda|1-\ frac {s} {2}\ derecha|\ cdots\ izquierda|1-\ frac {s} {n-1}\ derecha|.
\]
Si$$s>0,\left\{a_{n}\right\} \downarrow$$ por$$n>s+1,$$ eso podemos poner$$L=\lim a_{n}=\lim a_{2 n} \geq 0 .$$ (¡Explique!)
Demostrar que

\ [\ frac {a_ {2 n}} {a_ {n}} <\ izquierda|1-\ frac {s} {2 n}\ derecha|^ {n}\ fila derecha e^ {-\ frac {1} {2} s}\ texto {as} n\ fila derecha\ infty,
\]
así para grande$$n$$,
\ [
\ frac {a_ {2 n}} {a_ {n}} <e^ {- \ frac {1} {2} s} +\ varepsilon;\ text {i.e.,} a_ {2 n} <\ left (e^ {-\ frac {1} {2} s} +\ varepsilon\ derecha) a_ {n}.
\]
Con$$\varepsilon$$ fijo, vamos$$n \rightarrow \infty$$ a obtener$$L \leq\left(e^{-\frac{1}{2} s}+\varepsilon\right) L .$$ Luego con$$\varepsilon \rightarrow 0,$$ obtener$$L e^{\frac{1}{2} s} \leq L$$.
$$\left.\text { As } e^{\frac{1}{2} s}>1 \text { (for } s>0\right),$$esto implica$$L=0,$$ como se reclama.
(ii) Para$$s>-1, s+1>0,$$ así por$$(i)$$,
\ [
(n+1)\ left (\ begin {array} {c} {s+1}\\ {n+1}\ end {array}\ right)\ rightarrow 0;\ text {i.e.,} (s+1)\ left (\ begin {array} {c} {s}\\ {n}\ end {array}\ right)\ right tarrow 0. \ quad (\ text {¿Por qué? })
\]
(iii) Usa la prueba de ratio para mostrar que la serie$$\sum\left(\begin{array}{l}{s} \\ {n}\end{array}\right) n x^{n}$$ converge cuando$$|x|<1$$.
$$\text { Then apply Theorem } 4 \text { of Chapter } 4, §13 .]$$

## Ejercicio$$\PageIndex{8}$$

Continuar Problemas 6 y$$7,$$ demostrar que
\ [
(1+x) ^ {s} =\ sum_ {k=0} ^ {n}\ left (\ begin {array} {l} {s}\\ {k}\ end {array}\ right) x^ {k} +R_ {n} (x),
\]
donde$$R_{n}(x) \rightarrow 0$$ si cualquiera$$|x|<1,$$ o$$x=1$$ y$$s>-1,$$ o$$x=-1$$ y$$s>0 .$$
[Consejos: (a) Si se$$0 \leq x \leq 1,$$ usa$$\left(5^{\prime}\right)$$ para
\ [
R_ {n-1} (x) =\ left (\ begin {array} {c} {s}\\ {n}\ end {array}\ right) x^ {n}\ left (1+\ theta_ {n} x\ right) ^ {s-n},\ quad 0<\ theta_ {n} <1. \ text {(¡Verifica!) }
\]
Deducir que$$\left|R_{n-1}(x)\right| \leq\left|\left(\begin{array}{c}{s} \\ {n}\end{array}\right) x^{n}\right| \rightarrow 0 .$$ Use Problema 7$$(\text { iii) if }|x|<1 \text { or Problem } 7(\text { ii })$$ si$$x=1$$.
(b) Si$$-1 \leq x<0,$$ escribe$$\left(5^{\prime \prime}\right)$$ como
\ [
R_ {n-1} (x) =\ left (\ begin {array} {c} {s}\\ {n}\ end {array}\ right) n x^ {n}\ left (1+\ theta_ {n} ^ {\ prime} x\ right) s^ {-1}\ left (\ frac {1-\ theta_ {n} ^ {\ prime}} {1+\ theta_ {n} ^ {\ prime} x}\ derecha) ^ {n-1}. \ text {(¡Cheque!) }
\]
Como es$$-1 \leq x<0,$$ la última fracción$$\leq 1 .$$ (¿Por qué?) Además,
\ [
\ izquierda (1+\ theta_ {n} ^ {\ prime} x\ derecha) ^ {s-1}\ leq 1\ texto {si} s>1,\ texto {y}\ leq (1+x) ^ {s-1}\ texto {si} s\ leq 1.
\]
Así, con$$x$$ fijo, estas expresiones están delimitadas, mientras que$$\left(\begin{array}{c}{s} \\ {n}\end{array}\right) n x^{n} \rightarrow 0$$ por Problema 7$$(\mathrm{i})$$$$\left.\text { or }(\text { iii }) . \text { Deduce that } R_{n-1} \rightarrow 0, \text { hence } R_{n} \rightarrow 0 .\right]$$

## Ejercicio$$\PageIndex{9}$$

Demostrar que

\ [\ ln (1+x) =\ suma_ {k=1} ^ {n} (-1) ^ {k+1}\ frac {x^ {k}} {k} +R_ {n} (x),
\]
donde$$\lim _{n \rightarrow \infty} R_{n}(x)=0$$ si$$-1<x \leq 1$$.
[Consejos: Si$$0 \leq x \leq 1,$$ usa fórmula$$\left(5^{\prime}\right)$$.
Si$$-1<x<0,$$ usa fórmula$$(6)$$ con$$g(t)=\ln (1+t)$$ para obtener
\ [
R_ {n} (x) =\ frac {\ ln (1+x)} {(-1) ^ {n}}\ left (\ frac {1-\ theta_ {n}} {1+\ theta_ {n} x}\ cdot x\ derecha) ^ {n}.
\]
$$\text { Proceed as in Problem } 8 .]$$

## Ejercicio$$\PageIndex{10}$$

Demostrar que si$$f : E^{1} \rightarrow E^{*}$$ es de clase$$\mathrm{CD}^{1}$$ on$$[a, b]$$ y si$$-\infty<f^{\prime \prime}<0$$ on$$(a, b),$$ entonces para cada$$x_{0} \in(a, b)$$,
\ [
f\ left (x_ {0}\ right) >\ frac {f (b) -f (a)} {b-a}\ left (x_ {0} -a\ right) +f (a);
\]
es decir, la curva$$y=f(x)$$ se encuentra por encima de la secante through$$(a, f(a))$$ y$$(b, f(b)) .$$
[Pista: Esta fórmula es equivalente a
\ [
\ frac {f\ left (x_ {0}\ right) -f (a)} {x_ {0} -a} >\ frac {f (b) -f (a)} {b-a},
\]
es decir, el promedio de$$f^{\prime}$$ on$$\left[a, x_{0}\right]$$ es estrictamente mayor que el promedio de$$f^{\prime}$$ en$$[a, b],$$$$\left.\text { which follows because } f^{\prime} \text { decreases on }(a, b) .(\text { Explain! })\right]$$

## Ejercicio$$\PageIndex{11}$$

Demostrar que si$$a, b, r,$$ y$$s$$ son reales positivos y$$r+s=1,$$ luego
\ [
a^ {r} b^ {s}\ leq r a+s b.
\]
(Esta desigualdad es importante para la teoría de los llamados$$L^{p}$$ -espacios.)
[Consejos: Si$$a=b$$, todo es trivial.
Por lo tanto, asuma$$a<b$$. Entonces
\ [
a =( r+s) a<r a+s b<b.
\]
Usa el Problema 10 con$$x_{0}=r a+s b \in(a, b)$$ y
\ [
f (x) =\ ln x, f^ {\ prime\ prime} (x) =-\ frac {1} {x^ {2}} <0.
\]
Verifica que
\ [
x_ {0} -a=x_ {0} - (r+s) a=s (b-a)
\]
y de$$r \cdot \ln a=(1-s) \ln a ;$$ ahí deducir que
\ [
r\ cdot\ ln a+s\ cdot\ ln b-\ ln a=s (\ ln b-\ ln a) =s (f (b) -f (a)).
\]
Después de las sustituciones, obtener

\ [\ izquierda.f\ izquierda (x_ {0}\ derecha) =\ ln (r a+s b) >r\ cdot\ ln a+s\ cdot\ ln b=\ ln\ izquierda (a^ {r} b^ {s}\ derecha). \ derecho]
\]

## Ejercicio$$\PageIndex{12}$$

Utilice el teorema de Taylor (Teorema 1') para probar las siguientes desigualdades:
(a)$$\sqrt[3]{1+x}<1+\frac{x}{3}$$ si$$x>-1, x \neq 0$$.
b)$$\cos x>1-\frac{1}{2} x^{2}$$ si$$x \neq 0$$.
c)$$\frac{x}{1+x^{2}}<\arctan x<x$$ si$$x>0$$.
d)$$x>\sin x>x-\frac{1}{6} x^{3}$$ si$$x>0$$.

5.6.E: Problemas en el teorema de Tayior is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.