5.6.E: Problems on Tayior's Theorem
- Page ID
- 24085
Complete the proofs of Theorems \(1,1^{\prime},\) and 2.
Verify Note 1 and Examples (b) and \(\left(\mathrm{b}^{\prime \prime}\right)\).
Taking \(g(t)=(a-t)^{s}, s>0,\) in \((6),\) find
\[
R_{n}=\frac{f^{(n+1)}(q)}{n ! s}(x-p)^{s}(x-q)^{n+1-s} \quad(\text { Schloemilch-Roche remainder }).
\]
Obtain \(\left(5^{\prime}\right)\) and \(\left(5^{\prime \prime}\right)\) from it.
Prove that \(P_{n}\) (as defined) is the only polynomial of degree \(n\) such that
\[
f^{(k)}(p)=P_{n}^{(k)}(p), \quad k=0,1, \ldots, n .
\]
[Hint: Differentiate \(P_{n} n\) times to verify that it satisfies this property.
For uniqueness, suppose this also holds for
\[
P(x)=\sum_{k=0}^{n} a_{k}(x-p)^{k} .
\]
Differentiate \(P n\) times to show that
\[
P^{(k)}(p)=f^{(k)}(p)=a_{k} k ! ,
\]
\(\left.\text { so } P=P_{n} . \text { (Why?) }\right]\)
With \(P_{n}\) as defined, prove that if \(f\) is \(n\) times differentiable at \(p,\) then
\[
f(x)-P_{n}(x)=o\left((x-p)^{n}\right) \text { as } x \rightarrow p
\]
(Taylor's theorem with Peano remainder term).
[Hint: Let \(R(x)=f(x)-P_{n}(x)\) and
\[
\delta(x)=\frac{R(x)}{(x-p)^{n}} \text { with } \delta(p)=0 .
\]
Using the "simplified" L'Hôpital rule (Problem 3 in \(§3\) ) repeatedly \(n\) times, prove \(\left.\text { that } \lim _{x \rightarrow p} \delta(x)=0 .\right]\)
Use Theorem \(1^{\prime}\) with \(p=0\) to verify the following expansions, and prove that \(\lim _{n \rightarrow \infty} R_{n}=0\).
(a) \(\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots-\frac{(-1)^{m} x^{2 m-1}}{(2 m-1) !}+\frac{(-1)^{m} x^{2 m+1}}{(2 m+1) !} \cos \theta_{m} x\)
for all \(x \in E^{1}\) ;
(b) \(\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\cdots+\frac{(-1)^{m} x^{2 m}}{(2 m) !}-\frac{(-1)^{m} x^{2 m+2}}{(2 m+2) !} \sin \theta_{m} x\) for
all \(x \in E^{1} .\)
[Hints: Let \(f(x)=\sin x\) and \(g(x)=\cos x .\) Induction shows that
\[
f^{(n)}(x)=\sin \left(x+\frac{n \pi}{2}\right) \text { and } g^{(n)}(x)=\cos \left(x+\frac{n \pi}{2}\right) .
\]
Using formula \(\left(5^{\prime}\right),\) prove that
\[
\left|R_{n}(x)\right| \leq\left|\frac{x^{n+1}}{(n+1) !}\right| \rightarrow 0 .
\]
Indeed, \(x^{n} / n !\) is the general term of a convergent series
\[
\left.\sum \frac{x^{n}}{n !} \quad \text { (see Chapter } 4, §13, \text { Example }(\mathrm{d})\right).
\]
\(\left.\text { Thus } x^{n} / n ! \rightarrow 0 \text { by Theorem } 4 \text { of the same section. }\right]\)
For any \(s \in E^{1}\) and \(n \in N,\) define
\[
\left(\begin{array}{l}{s} \\ {n}\end{array}\right)=\frac{s(s-1) \cdots(s-n+1)}{n !} \text { with }\left(\begin{array}{l}{s} \\ {0}\end{array}\right)=1 .
\]
Then prove the following.
(i) \(\lim _{n \rightarrow \infty} n\left(\begin{array}{l}{s} \\ {n}\end{array}\right)=0\) if \(s>0\),
(ii) \(\lim _{n \rightarrow \infty}\left(\begin{array}{l}{s} \\ {n}\end{array}\right)=0\) if \(s>-1\),
(iii) For any fixed \(s \in E^{1}\) and \(x \in(-1,1)\).
\[
\lim _{n \rightarrow \infty}\left(\begin{array}{l}{s} \\ {n}\end{array}\right) n x^{n}=0 ;
\]
hence
\[
\lim _{n \rightarrow \infty}\left(\begin{array}{c}{s} \\ {n}\end{array}\right) x^{n}=0 .
\]
\(\left[\text { Hints: }(\mathrm{i}) \text { Let } a_{n}=\left|n\left(\begin{array}{l}{s} \\ {n}\end{array}\right)\right| . \text { Verify that }\right.\)
\[
a_{n}=|s|\left|1-\frac{s}{1}\right|\left|1-\frac{s}{2}\right| \cdots\left|1-\frac{s}{n-1}\right| .
\]
If \(s>0,\left\{a_{n}\right\} \downarrow\) for \(n>s+1,\) so we may put \(L=\lim a_{n}=\lim a_{2 n} \geq 0 .\) (Explain!)
Prove that
\[
\frac{a_{2 n}}{a_{n}}<\left|1-\frac{s}{2 n}\right|^{n} \rightarrow e^{-\frac{1}{2} s} \text { as } n \rightarrow \infty ,
\]
so for large \(n\),
\[
\frac{a_{2 n}}{a_{n}}<e^{-\frac{1}{2} s}+\varepsilon ; \text { i.e., } a_{2 n}<\left(e^{-\frac{1}{2} s}+\varepsilon\right) a_{n} .
\]
With \(\varepsilon\) fixed, let \(n \rightarrow \infty\) to get \(L \leq\left(e^{-\frac{1}{2} s}+\varepsilon\right) L .\) Then with \(\varepsilon \rightarrow 0,\) obtain \(L e^{\frac{1}{2} s} \leq L\).
\(\left.\text { As } e^{\frac{1}{2} s}>1 \text { (for } s>0\right),\) this implies \(L=0,\) as claimed.
(ii) For \(s>-1, s+1>0,\) so by \((i)\),
\[
(n+1)\left(\begin{array}{c}{s+1} \\ {n+1}\end{array}\right) \rightarrow 0 ; \text { i.e., }(s+1)\left(\begin{array}{c}{s} \\ {n}\end{array}\right) \rightarrow 0 . \quad(\text { Why? })
\]
(iii) Use the ratio test to show that the series \(\sum\left(\begin{array}{l}{s} \\ {n}\end{array}\right) n x^{n}\) converges when \(|x|<1\).
\(\text { Then apply Theorem } 4 \text { of Chapter } 4, §13 .]\)
Continuing Problems 6 and \(7,\) prove that
\[
(1+x)^{s}=\sum_{k=0}^{n}\left(\begin{array}{l}{s} \\ {k}\end{array}\right) x^{k}+R_{n}(x) ,
\]
where \(R_{n}(x) \rightarrow 0\) if either \(|x|<1,\) or \(x=1\) and \(s>-1,\) or \(x=-1\) and \(s>0 .\)
[Hints: (a) If \(0 \leq x \leq 1,\) use \(\left(5^{\prime}\right)\) for
\[
R_{n-1}(x)=\left(\begin{array}{c}{s} \\ {n}\end{array}\right) x^{n}\left(1+\theta_{n} x\right)^{s-n}, \quad 0<\theta_{n}<1 . \text { (Verify!) }
\]
Deduce that \(\left|R_{n-1}(x)\right| \leq\left|\left(\begin{array}{c}{s} \\ {n}\end{array}\right) x^{n}\right| \rightarrow 0 .\) Use Problem 7\((\text { iii) if }|x|<1 \text { or Problem } 7(\text { ii })\) if \(x=1\).
(b) If \(-1 \leq x<0,\) write \(\left(5^{\prime \prime}\right)\) as
\[
R_{n-1}(x)=\left(\begin{array}{c}{s} \\ {n}\end{array}\right) n x^{n}\left(1+\theta_{n}^{\prime} x\right) s^{-1}\left(\frac{1-\theta_{n}^{\prime}}{1+\theta_{n}^{\prime} x}\right)^{n-1} . \text { (Check!) }
\]
As \(-1 \leq x<0,\) the last fraction is \(\leq 1 .\) (Why?) Also,
\[
\left(1+\theta_{n}^{\prime} x\right)^{s-1} \leq 1 \text { if } s>1, \text { and } \leq(1+x)^{s-1} \text { if } s \leq 1 .
\]
Thus, with \(x\) fixed, these expressions are bounded, while \(\left(\begin{array}{c}{s} \\ {n}\end{array}\right) n x^{n} \rightarrow 0\) by Problem 7\((\mathrm{i})\) \(\left.\text { or }(\text { iii }) . \text { Deduce that } R_{n-1} \rightarrow 0, \text { hence } R_{n} \rightarrow 0 .\right]\)
Prove that
\[
\ln (1+x)=\sum_{k=1}^{n}(-1)^{k+1} \frac{x^{k}}{k}+R_{n}(x) ,
\]
where \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) if \(-1<x \leq 1\).
[Hints: If \(0 \leq x \leq 1,\) use formula \(\left(5^{\prime}\right)\).
If \(-1<x<0,\) use formula \((6)\) with \(g(t)=\ln (1+t)\) to obtain
\[
R_{n}(x)=\frac{\ln (1+x)}{(-1)^{n}}\left(\frac{1-\theta_{n}}{1+\theta_{n} x} \cdot x\right)^{n} .
\]
\(\text { Proceed as in Problem } 8 .]\)
Prove that if \(f : E^{1} \rightarrow E^{*}\) is of class \(\mathrm{CD}^{1}\) on \([a, b]\) and if \(-\infty<f^{\prime \prime}<0\) on \((a, b),\) then for each \(x_{0} \in(a, b)\),
\[
f\left(x_{0}\right)>\frac{f(b)-f(a)}{b-a}\left(x_{0}-a\right)+f(a) ;
\]
i.e., the curve \(y=f(x)\) lies above the secant through \((a, f(a))\) and \((b, f(b)) .\)
[Hint: This formula is equivalent to
\[
\frac{f\left(x_{0}\right)-f(a)}{x_{0}-a}>\frac{f(b)-f(a)}{b-a} ,
\]
i.e., the average of \(f^{\prime}\) on \(\left[a, x_{0}\right]\) is strictly greater than the average of \(f^{\prime}\) on \([a, b],\) \(\left.\text { which follows because } f^{\prime} \text { decreases on }(a, b) .(\text { Explain! })\right]\)
Prove that if \(a, b, r,\) and \(s\) are positive reals and \(r+s=1,\) then
\[
a^{r} b^{s} \leq r a+s b .
\]
(This inequality is important for the theory of so-called \(L^{p}\) -spaces.)
[Hints: If \(a=b\), all is trivial.
Therefore, assume \(a<b\). Then
\[
a=(r+s) a<r a+s b<b .
\]
Use Problem 10 with \(x_{0}=r a+s b \in(a, b)\) and
\[
f(x)=\ln x, f^{\prime \prime}(x)=-\frac{1}{x^{2}}<0 .
\]
Verify that
\[
x_{0}-a=x_{0}-(r+s) a=s(b-a)
\]
and \(r \cdot \ln a=(1-s) \ln a ;\) hence deduce that
\[
r \cdot \ln a+s \cdot \ln b-\ln a=s(\ln b-\ln a)=s(f(b)-f(a)) .
\]
After substitutions, obtain
\[
\left.f\left(x_{0}\right)=\ln (r a+s b)>r \cdot \ln a+s \cdot \ln b=\ln \left(a^{r} b^{s}\right) .\right]
\]
Use Taylor's theorem (Theorem 1') to prove the following inequalities:
(a) \(\sqrt[3]{1+x}<1+\frac{x}{3}\) if \(x>-1, x \neq 0\).
(b) \(\cos x>1-\frac{1}{2} x^{2}\) if \(x \neq 0\).
(c) \(\frac{x}{1+x^{2}}<\arctan x<x\) if \(x>0\).
(d) \(x>\sin x>x-\frac{1}{6} x^{3}\) if \(x>0\).