Final Review of Advanced Calculus（1）

Chasse_neige

1. Taylor's Theorem

1.1 Taylor's Theorem with Peano's Form of Remainder

When $f(x)$ is n times differentiable on an open interval $I$ containing the point $a$

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n+o((x-a{)}^n)$$

Specially, when $a=0$, the polnomial

$$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\cdots +\frac{f^{(n)}(0)}{n!}{x}^n+o(x^n)$$

is called Maclaurin's polynomial.

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1.2 Taylor's Theorem with Lagrange's Form of Remainder

If a function $f(x)$ is $(n+1)$ times differentiable on an open interval $I$ containing the point $a$, then for any $x$ in $I$, there exists a point $c$ between $a$ and $x$ such that:

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n+R_n(x)$$

where $R_n(x)$ is the remainder term given by:

$$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a{)}^{n+1}$$

Here, $c$ is some point between $a$ and $x$. The remainder term $R_n(x)$ represents the error in approximating $f(x)$ by the $n$-th degree Taylor polynomial centered at $a$. The form of the remainder provided by Lagrange gives a specific way to bound this error, depending on the $(n+1)$-th derivative of $f$ at some point $c$ in the interval.

Proof.

Given that $f(x)$ is $(n+1)$ times differentiable, then we can use the Cauchy's Mean Value Theorem on the functions $F(t)=f(a+t\mathrm{\Delta }x)-P_n(t\mathrm{\Delta }x)$ and $G(t)=(t\mathrm{\Delta }x{)}^{n+1}$, where $t$ is an arbitrary constant between $0$ and $1$

$$\frac{F(t)-F(0)}{G(t)-G(0)}=\frac{(f^{\prime }(a+t_1\mathrm{\Delta }x)-P_n^{\prime }(t_1\mathrm{\Delta }x))\mathrm{\Delta }x}{(n+1)(t_1\mathrm{\Delta }x{)}^n\mathrm{\Delta }x}=\dots =\frac{f^{(n)}(a+t_{n+1}\mathrm{\Delta }x)}{(n+1)!}$$

then let $t=1$ and we can get the Lagrange's form.

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1.3 Taylor's Theorem with Cauchy's Form of Remainder

Cauchy's Form of the Remainder is another way to express the remainder term $R_n(x)$. It states that:

$$R_n(x)=\frac{f^{(n+1)}(c)}{n!}(x-c{)}^n(x-a)$$

where $c$ is some number between $a$ and $x$ .

Proof.

Given that $f(x)$ is $(n+1)$ times differentiable, then we can use the Cauchy's Mean Value Theorem on the functions $F(t)=f(x)-P_n(x-t\mathrm{\Delta }x)$ and $G(t)=x-t\mathrm{\Delta }x$, where $t$ is an arbitrary constant between $0$ and $1$

$$\frac{F(t)-F(0)}{G(0)-G(t)}=\frac{f^{(n+1)}(x-t_1\mathrm{\Delta }x)(t_1\mathrm{\Delta }x{)}^n\mathrm{\Delta }x}{n!\mathrm{\Delta }x}$$

then let $t=1$ and we can get the Cauchy's form.

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Applications:

Using the Taylor's theorem with Lagrange's form of remainder to estimate the Taylor series's error.

Example 1

Given that$f\in C^3[a-h,a+h]$, $h>0$ and $|f(x)|\le M$, prove that

$$|\frac{f(a+h)+f(a-h)-2f(a)}{h^2}-f^{″}(a)|\le \frac{Mh}{3}$$

Proof.

First use $f$'s Taylor Series about a:

$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2+\frac{f^{‴}(a+t(x-a))}{6}(x-a{)}^3$$

so it is trivial that

$$f(a+h)=f(a)+f^{\prime }(a)h+\frac{f^{″}(a)}{2}{h}^2+\frac{f^{‴}(a+th)}{6}{h}^3$$$$f(a-h)=f(a)-f^{\prime }(a)h+\frac{f^{″}(a)}{2}{h}^2+\frac{f^{‴}(a-th)}{6}{h}^3$$

so

$$|f(a+h)+f(a-h)-2f(a)|=|f^{″}(a)h^2+\frac{f^{‴}(a+th)+f^{‴}(a-th)}{6}{h}^3|\le |f^{″}(a)h^2+\frac{1}{3}M{h}^3|$$

Example 2 Linear Interpolation

Suppose that the function $f:[a,b]\to \mathbb{R}$ between two points $(a,f(a))$ and $(b,f(b))$. The linear function used to estimate the points between $a$ and $b$ is

$$l(x)=\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)$$

and $l$ is called the linear interpolation on the interval. Given that on the interval $[a,b]$, $\underset{x\in [a,b]}{sup}|f^{″}|=M$ and $f^{″}$ is continuous on the interval, prove that the error of the linear interpolation $|f(x)-l(x)|\le \frac{1}{8}(b-a{)}^{2}M$

Proof.

Given that $f^{″}$ is continuous on the interval $[a,b]$, then $f$ can be estimated by the Taylor Series

$$f(a)=f(x)+f^{\prime }(x)(a-x)+\frac{1}{2}(a-x{)}^2{f}^{″}(\xi ),a<\xi <b$$$$f(b)=f(x)+f^{\prime }(x)(b-x)+\frac{1}{2}(b-x{)}^2{f}^{″}(\eta ),a<\eta <b$$

so the error equals to

$$|f(x)-l(x)|=|\frac{(b-x)(a-x)}{2}(\frac{x-a}{b-a}{f}^{″}(\xi )+\frac{b-x}{b-a}{f}^{″}(\eta ))|$$

Notice that $\frac{x-a}{b-a}{f}^{″}(\xi )+\frac{b-x}{b-a}{f}^{″}(\eta )$ is actually between the two values $f^{″}(\xi )$ and $f^{″}(\eta )$ and $f^{″}$ is continuous on the closed interval $[a,b]$, according to the Intermediate Value Theorem, there exists a $\zeta \in [a,b]$ so that

$$|f^{″}(\zeta )|=|\frac{x-a}{b-a}{f}^{″}(\xi )+\frac{b-x}{b-a}{f}^{″}(\eta )|\le M$$

while the Quadratic Function $(b-x)(x-a)$ has a maximum $\frac{1}{4}(b-a{)}^2$ on the interval $[a,b]$, then the error $|f(x)-l(x)|\le \frac{1}{8}(b-a{)}^{2}M$ is trivial.

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2. Definite Integral (Riemann)

2.1 Definition

If a function $f$ is defined on the interval $[a,b]$ and for any partition $\pi$ on the interval, the limit

$$|I-\sum _{i=1}^{n}f({\xi }_i)\mathrm{\Delta }{x}_i|<ϵ$$

exists, where ${\xi }_i$ is an arbitrary constant in $[x_{i-1},x_i](1\le i\le n)$, we can term $I$ the Riemann Integral, which is the limit of the Riemann Sum $\sum _{i=1}^{n}f({\xi }_i)\mathrm{\Delta }{x}_i$ and the function $f$ is Riemann Integrable on the interval. The Riemann Integral can be denoted as

$${\int }_a^{b}f(x)dx=\underset{||\pi ||\to 0}{lim}\sum _{i=1}^{n}f({\xi }_i)\mathrm{\Delta }{x}_i$$

2.2 Properties of Riemann Integrable Functions

2.2.1 The Newton-Leibniz Formula

If the function $f$ is defined on the interval $I$ and has a continuous anti derivative $F(x)$ on it, then the Riemann integral can be denoted as

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

which is easy to prove consider the Lagrange Mean Value Theorem.

2.2.2 Integrable-Caused Properties

1. If a function $f$ is Riemann integrable on the interval $I$, it is bounded on the interval.
2. Integral Mean Value Theorem: If the functions $f$ and $g$ are continuous on the interval $I$ and $g$ does not change its sign on the interval, then there exists a $\xi \in [a,b]$ so that

$${\int }_a^{b}f(x)g(x)dx=f(\xi ){\int }_a^{b}g(x)dx$$

Specially, let $g=1$, the formula would become

$${\int }_a^{b}f(x)dx=f(\xi )(b-a)$$

Applications:

Integral Inequalities:

1. Cauchy-Schwarz Inequality:

$${({\int }_a^{b}f(x)g(x)dx)}^2\le {\int }_a^b{f}^2(x)dx{\int }_a^b{g}^2(x)dx$$

Proof.

$${\int }_a^b(f(x)+\lambda g(x){)}^{2}dx\ge 0$$

where $\lambda$ is an arbitrary constant.

$${\lambda }^2{\int }_a^b{g}^2(x)dx+2\lambda {\int }_a^{b}f(x)g(x)dx+{\int }_a^b{f}^2(x)dx\ge 0$$

Using the criterion $\mathrm{\Delta }\le 0$ can get the inequality.

2. Hölder Inequality Let $f$ and $g$ be measurable functions, and let $p$ and $q$ be positive real numbers such that $p>1$ and $q>1$, with the relationship:

$$\frac{1}{p}+\frac{1}{q}=1$$

Then, Hölder Inequality states that:

$${\int }_a^b|f(x)g(x)|dx\le {({\int }_a^b|f(x){|}^{p}dx)}^{1/p}{({\int }_a^b|g(x){|}^{q}dx)}^{1/q}$$

Proof. Notice that $e^x$ is a convex on $\mathbb{R}$. Use the Jensen Inequality,

$$e^{\frac{1}{p}\ln a^p+\frac{1}{q}\ln b^q}\le \frac{1}{p}{e}^{\ln a^p}+\frac{1}{q}{e}^{\ln b^q}$$

we can get the Yong inequality

$$\frac{1}{p}{a}^p+\frac{1}{q}{b}^q\ge ab$$

Then let $a=\frac{|f|}{{({\int }_a^b|f(x){|}^{p}dx)}^{1/p}}$ and $b=\frac{|g|}{{({\int }_a^b|g(x){|}^{q}dx)}^{1/q}}$ and we can get the inequality.

3. Minkowski Inequality Let $f$ and $g$ be measurable functions on $\mathbb{R}$, and let $1\le p<\mathrm{\infty }$. Minkowski's Inequality states that:

$${({\int }_a^b|f(x)+g(x){|}^{p}dx)}^{1/p}\le {({\int }_a^b|f(x){|}^{p}dx)}^{1/p}+{({\int }_a^b|g(x){|}^{p}dx)}^{1/p}$$

Proof.

$${\int }_a^b|f(x)+g(x){|}^{p}dx\le {\int }_a^b|f(x)+g(x){|}^{p-1}|f|dx+{\int }_a^b|f(x)+g(x){|}^{p-1}|g|dx\le {({\int }_a^b|f(x)+g(x){|}^{p}dx)}^{\frac{p-1}{p}}{({\int }_a^b|f(x){|}^{p}dx)}^{1/p}+{({\int }_a^b|f(x)+g(x){|}^{p}dx)}^{\frac{p-1}{p}}{({\int }_a^b|g(x){|}^{p}dx)}^{1/p}$$

where the second $\le$ is based on the Hölder Inequality.

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2.2.3 Riemann Integrability Theorem

If a function $f$ is continuous on the interval $I$, then the following three properties are tantamount:

1. $$f\in R[a,b]$$
2. $$\underset{||\pi ||\to 0}{lim}\sum _{i=1}^n{\omega }_i\mathrm{\Delta }{x}_i=0({\omega }_i=\underset{x\in [x_{i-1},x_i]}{sup}f-\underset{x\in [x_{i-1},x_i]}{inf}f)$$
3. $$\underset{―}{I}=\bar{I}$$

So it is not hard to prove that any monotonous or continuous function on a closed interval is Riemann integrable. (To prove the latter, just consider the Cantor Theorem to make the solution trivial).

Lebesgue Theorem

Lebesgue's Theorem on Riemann Integrability provides a criterion for determining whether a bounded function is Riemann integrable. The theorem states that a bounded function $f$ defined on a closed interval $[a,b]$ is Riemann integrable if and only if the set of points where $f$ is discontinuous has measure zero.

• Measure Zero: A set has measure zero if, for any $ϵ>0$, it can be covered by a countable collection of intervals whose total length is less than $ϵ$. Examples of sets with measure zero include finite sets, countable sets, and certain dense sets like the rational numbers in an interval.
• Discontinuity Set: The set of discontinuities of a function is the set of points where the function is not continuous. If this set has measure zero, the function is considered "almost everywhere" continuous.
• Integrability: This theorem implies that a function can have many discontinuities and still be Riemann integrable, as long as the discontinuities are "sparse" enough to form a set of measure zero.

Applications:

Integral Approximation

Example 1

If $f\in R[a,b]$, prove that for any $ϵ>0$, there always exists an step fuction/ piecewise linear function/ continuous and differentiable function that guarantees

$${\int }_a^b|f(x)-g(x)|dx<ϵ$$

Proof.

$$\because f\in R[a,b]$$$$\therefore \mathrm{\exists }\pi ,s.t.\mathrm{\forall }ϵ>0,\sum _{i=1}^n{\omega }_i\mathrm{\Delta }{x}_i<ϵ$$

1. For step functions: Choose $$ g(x) = \sum_{k=1}^{n} f(x_{k-1}) I_{[x_{k-1}, x_{k}]} $$
2. For piecewise linear functions: Choose $$ g(x) = \sum_{k=1}^{n} \left( f(x_{k-1}) + \frac{f(x_{k}) - f(x_{k-1})}{\Delta x_{k}} (x - x_{k-1}) \right) I_{[x_{k-1}, x_{k}]} $$
3. For continuous and differentiable functions: According to the Weierstrass First Approximation Theorem, we can choose a polnomial $P_n(x)$, so that

$$\underset{x\in [a,b]}{sup}|P_n(x)-f(x)|<ϵ$$

For example, the Bernstein polynomial for $f\in [a,b]\to f^{\prime }\in [0,1]$

$$B_n(f^{\prime })(x)=\sum _{k=0}^n{f}^{\prime }(\frac{k}{n})C_n^k{x}^k(1-x{)}^{n-k}$$

can be a template.

Example 2 Riemann-Lebesgue Lemma

If $f$ is a Riemann integrable function on the interval $[a,b]$ and $g$ is a periodic function with period $T$ that is integrable on $[0,T]$

$$\underset{p\to \mathrm{\infty }}{lim}{\int }_a^{b}f(x)g(px)dx=\frac{1}{T}{\int }_0^{T}g(x)dx{\int }_a^{b}f(x)dx$$

Proof.

According to the additivity principle,

$${\int }_a^{b}f(x)g(px)dx=\sum _{k=1}^n{\int }_{x_{k-1}}^{x_k}f(x)g(px)dx$$

Using the Integral Mean Value Theorem, we can deduce that for every interval $[x_{k-1},x_k]$, there exists a ${\xi }_k\in [x_{k-1},x_k]$ so that

$${\int }_{x_{k-1}}^{x_k}f(x)g(px)dx=f({\xi }_k){\int }_{x_{k-1}}^{x_k}g(px)dx$$

Choose a partition $\pi :a=x_0<a+\frac{T}{p}=x_1<\cdots <x_n=b$, so

$${\int }_{x_{k-1}}^{x_k}g(px)dx=\frac{1}{p}{\int }_0^{T}g(t)dt$$

Therefore, we can add the parts and use the definition of the Riemann Integral to get the lemma:

$$\underset{p\to \mathrm{\infty }}{lim}{\int }_a^{b}f(x)g(px)dx=\underset{p\to \mathrm{\infty }}{lim}\frac{1}{p}\sum _{k=1}^{n}f({\xi }_k){\int }_0^{T}g(t)dt=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}f({\xi }_k)\frac{b-a}{nT}{\int }_0^{T}g(t)dt=\frac{1}{T}{\int }_0^{T}g(x)dx{\int }_a^{b}f(x)dx$$

Example 3

If $f$ is a differentiable funcition on the interval $[a,b]$, given the Integral Mean Value Theorem

$${\int }_a^{x}f(x)dx=f(\xi (x))(x-a)$$

Prove that

$$\underset{x\to a+}{lim}\frac{\xi (x)-a}{x-a}=\frac{1}{2}$$

Proof.

Given that $f$ is differentiable, we can write its Taylor series

$$f(\xi (x))=f(a)+f^{\prime }(a)(\xi -a)+o(\xi -a)$$$$\therefore {\int }_a^{x}f(x)dx=(f(a)+f^{\prime }(a)(\xi -a)+o(\xi -a))(x-a)$$$$\therefore \underset{x\to a+}{lim}\frac{{\int }_a^{x}f(x)dx}{(x-a{)}^2}=\frac{1}{2}{f}^{\prime }(a)=f^{\prime }(a)\frac{\xi -a}{x-a}$$$$\therefore \underset{x\to a+}{lim}\frac{\xi (x)-a}{x-a}=\frac{1}{2}$$