Lecture 23: Taylor's Formula

The role of Taylor's formula: establishing the relationship between higher-order derivatives and functions, using higher-order derivatives to study functions;

1.1 What is Taylor's Formula?

• Popular Science Article: (99+ private messages / 85 messages) How to explain Taylor's formula in simple terms? - Zhihu (zhihu.com)

What is Taylor's Formula

• Essence: This is about using polynomial functions to approximate smooth functions
• Explanation: Complex equations are approximated using a series of simple, computable equations. For example, turning the area of an irregular figure into the area of a series of (many) squares and then adding them together;

1.1.1 Definition of Taylor's Formula

#####Theorem: Taylor's Theorem - with the remainder of Peano

Description: Taylor's formula with Peano remainder: Suppose $f(x)$ is differentiable in $n$ order at $x_0$, then $$f(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+o((x-x_0)^n)$$

Explanation

• Meaning:
• Write a general function as a general polynomial + remainder ;
• Additional items:
• $$P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$
• $f(x)$ n Taylor polynomials at $x_{0}$:
• $$R_n(x)=o((x-x_0)^n)$$
• Disadvantages:
• 1. Only qualitative descriptions of the remaining items can be given, not quantitative analysis;
• 2. It can only be used when x0 is approaching; it cannot be used from a distance;

Additional Information

• First, the relationship between functions and differentials
• If $f(x)$ is differentiable at $x_0$, then $\Delta y\approx dy$
• $f(x)\approx f(x_0)+f^{\prime}(x_0)(x-x_0)$
• At this point, it is approximately equal to, because geometrically, $f^{\prime}(x_0)(x-x_0)$ actually uses a straight line to simulate the value of the curve;
• Therefore, a single differentiation can be performed using higher-order infinitesimals to further approximate the value of the curve, i.e., the value of f(x).
• $f(x)=f(x_0)+f^{\prime}(x_0)(x-x_0)+o(x-x_0)$
• If $f(x)$ is n-th differentiable at $x_0$, does there exist an n-th degree polynomial $P_n(x)=a_0+a_1(x-x_0)+a_2(x-x_0)^2+\cdots+a_n(x-x_0)^n$ such that:
• $f(x)=P_n(x)+o((x-x_0)^n)$
• Therefore, we get $a_0=f(x_0),\quad a_k=\frac{f^{(k)}(x_0)}{k!}\quad k=1,2\cdots n$
• So: $P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$

#####Theorem: Taylor's Formula - Lagrange

Description: Let $f(x)$ be in the interval $I$ and $n+1$ order is differentiable $, x_0\in I, $ then $\forall x\in I, \exists\xi\in I\quad(\xi$ is between $x_0$ and $x$), making $$f(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}\big(x-x_0\big)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}\big(x-x_0\big)^{n+1}$$

Taylor's formula with Lagrangian remainders:

$$R_{n}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_{0})^{n+1}$$

Explanation

• Provides the specific expression for the remaining terms
• Quantitative analysis can be performed under given conditions;
• It guarantees error, with zero as x0 approaches zero;

Corollary 1: Taylor's formula for zero

$\begin{gathered}\text{If }x_0=0,\text{ then}\\f(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n+\frac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1},\quad0<\theta<1.\\\text{This is the Maclaurin formula for }f(x).\end{gathered}$

Corollary 2: Maclaurin formulas for several elementary functions

• $e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}e^{\theta x}\quad\quad x\in(-\infty,+\infty)$
• Because the derivative of e is still itself, so for each order of derivative, use 1 to substitute;
• $\sin x=x-\frac{x^3}{3!}+\cdots+\left(-1\right)^{m-1}\frac{x^{2m-1}}{\left(2m-1\right)!}+\left(-1\right)^m\frac{\cos\theta x}{\left(2m+1\right)!}x^{2m+1}$
• $\begin{aligned}(1+x)^\alpha&=1+cx+\cdots+\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}x^n\\&+\frac{\alpha(\alpha-1)\cdots(\alpha-n)(1+\beta x)^{x-n-1}}{(n+1)!}x^{n+1}\quad x\in(-1,+\infty)\end{aligned}$

1.2 Content Summary

1.2.1 Commonalities

Commonalities

• 1. Use simple polynomial functions to approximate general functions;
• 2. Functions are associated with higher-order derivatives;
• Conclusion:
• 1. When the problem requiresstudying higher-order derivatives, Taylor's formula is often used;
• 2. If the study is a local property -> Taylor's formula for the remaining Peano terms;
• 3. If the study is global properties -> the Taylor formula for the remaining Lagrange

Core: Both Taylor formulas write f(x) asone polynomial + one remainder;

• Peano Remaining Terms
• $R_n(x)=o((x-x_0)^n)$
• Lagrange remainder
• $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$

Essence: Use polynomials to approximate the value of f(x) -> Use known point information to represent unknown points;

• Why? Because solving polynomials is simple

1.2.2 Analysis of the Use of Two Taylor Formulas

Differences

• Other items differ

Peano: Qualitative, local description -> Study the local morphology of functions using the Peano formula

• 1. Study the limits of the function approaching a point;
• 2. Study the extremum of the function;

Lagrange: A quantitative, overall description

• 1. Study maximum/minimum;
• 2. Proving inequalities -> studying an interval;

Supplement: A special case of Taylor's theorem when it comes to Lagrange's mean value theorem;

• The four major mean value theorems:
• The first three establish the relationship between $f(x)$ and the first derivative;
• Taylor's Theorem: Study the relationship between $f(x)$ and higher-order derivatives
• Commonality: Studying the relationship between functions and derivatives, laying the foundation forusing derivatives to study functions;

1.3 Example Problems

Example Question: $\text{ Find the limit }\lim_{x\to0}\frac{\cos x-e^{-\frac{x^2}2}}{x^4}$

• Analysis
• The expression has type $0/0$, but L'Hôpital's rule is not the intended method here.
• So you can consider using Taylor
• Since it is a limit, local Taylor's formula is used
• Since x is to the fourth power, write it as four terms;
• Analysis
• Solution process:
• $cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o\left(x^4\right)$
• Because:
• $\mathbb{e}^x=1+x+\frac{\mathbb{x}}{2_1^2}$
• De: $e^{-\frac{X^{2}}{2}}=1-\frac{X^{2}}{2}+\frac{1}{2!}\left(-\frac{X^{2}}{2}\right)^{2}+o(X^{4})$
• Therefore, we have the original expression =
• $\operatorname*{lim}_{x\rightarrow0}\frac{-\frac{1}{12}x^{4}+0(x^{4})}{x^{4}}=-\frac{1}{12}$