Taylor 泰勒级数

Not in BZB, but try any standard calculus text such as
Calculus by Thomas and Finney. Calculus by Grossman
Topic 8
Higher derivatives — quick review
Consider the function y = (1 + x )e −x . Derivatives are y = −xe −x and y = (x − 1)e −x .
(k ) (k ) (N )
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
Taylor series Solutions to DEs
Topic 8
General Taylor polynomial of degree N
Note that all higher derivatives of P2 (x ) are zero.
Topic 8
Quadratic approximation
From these equations we see that P2 (c ) = a0 P2 (c ) = a1 P2 (c ) = 2a2 Since we want the parabola to be a good approximation to f (x ) near x = c we equate P2 (c ) = f (c ) P2 (c ) = f (c ) P2 (c ) = f (c )
Topic 8
Outline Introduction Taylor polynomials Taylor series Solutions to DEs
Topic 8 — Taylor polynomials and series
Andrew Rechnitzer and Aleks Owczarek
Taylor series Solutions to DEs
= f (c ) + f (c )(x − c )
Topic 8
An example
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
Topic 8
Quadratic approximation
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
Hence we can write 2 P2 (x ) = f (c ) + f (c )(x − c ) + 1 2 f (c )(x − c )
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
We will start by considering the simplest polynomial — a line. The tangent line to a curve gives a pretty good approximation over small distances. The equation of the tangent line to the curve at c is y − f (c ) = f (c )(x − c ) y or
Taylor series Solutions to DEs
Topic 8
Quadratic approximation
f (x) P2 (x) P1 (x)
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
The University of Melbourne
May 9, 2005
Topic 8
Topic outline
Outline Introduction Taylor polynomials Taylor series Solutions to DEs
Introduction Taylor polynomials Taylor series Solutions to DEs
Find the tangent line to y = (x + 1)e −x at x = 2. At x = 2 y = 3e −2 and y = −2e 2 .
Taylor series Solutions to DEs
So the tangent line is (y − 3e −2 ) = −2e −2 (x − 2). Or approximately y = 0.4060 − 0.2707(x − 2). So at x = 2.5 the tangent line gives y = 0.2707. The real function gives y = 3.5e −2.5 = 0.2873 — so not too bad.
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
The N th degree Taylor polynomial is PN (x ) = f (c ) + f (c )(x − c ) + 1 f (c )(x − c )2 2! 1 1 (N ) + f (3) (c )(x − c )3 + · · · + f (c )(x − c )N 3! N!
Taylor series Solutions to DEs
c
P2 (x ) — a parabola that approximates f (x ) near (about) x = c . P2 (x ) = a0 + a1 (x − c ) + a2 (x − c )2 P2 (x ) = 2a2 P2 (x ) = a1 + 2a2 (x − c )
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
Taylor series Solutions to DEs
f (x) f (x0 ) P(x) P(x0 )
c
x0
Topic 8
The tangent line — a linear approximation
We can get another type of approximation of the solution of a DE by using the information encoded in the derivative and the higher derivatives.
Topic 8
1 −2 e (x − 2)2 So P1 (x ) = 3e −2 − 2e −2 (x − 2) + 2
Gives P2 (2.5) = 0.2876 compared with f (2.5) = 0.2873.
Topic 8
General Taylor polynomial of degree N
1 So a0 = f (c ), a1 = f (c ) and a2 = 2 f (c ).
Outline Introduction Taylor polynomials
Definition Linear Quadratic General Example
Taylor series Solutions to DEs
Topic 8
Higher derivatives — quick review
Outline Introduction
Higher derivatives
Taylor polynomials Taylor series Solutions to DEs
y tells us about the rate of change of y . y tells us about the rate of change of y . y (3) tells us about the rate of change of y . and so on. . .
Topic 8
Introduction
Outline Introduction
Higher derivatives
Taylor polynomials Taylor series Solutions to DEs
Taylor polynomials — approximation to a function near some point. Taylor series — exact representation of a function.
We find the polynomial PN (x ) = a0 + a1 (x − c ) + · · · + aN (x − c )N so that PN (c ) = f (c ) PN (c ) = f (c ) PN (c ) = f (c ) . . . PN (c ) = f (N ) (c ) Exercise: Show that PN (c ) = k !ak . Note that PN (c ) = 0 for k > N .