Taylor Series 泰勒级数 英文版

Taylor Serise

Abstract:Taylor series and Taylor formula are very important in approximate calculation.In this paper，I will discuss the application of Taylor series in many ways in the calculuation of duel function.

Key words: Taylor formula Taylor series Application Duel function

Definition:

Let f be a function with derivatives of all orders throughout some interval containing x0 as interior point. Then the Taylor series generated by f at x=x0 is: f(xo)f(n)(x0)2(xx0)(xx0)f(x0)f(x0)(xx0)2!n! The above function expansion is referred to as the Taylor series. In the Taylor formula, take x0 = 0, the series is called Maclaurin series. Taylor’s Theorem If f is differentiable through order n+1 in an open interval I containing x0, the for each x in I, there exists a number  between x and x0 such that

f(n)(x0)f(n1)()f(x0)2(xx0)(xx0)n1f(x)f(x0)f(x0)(xx0)(xx0)n!(n1)!2! Taylor forluma of Duel function: For fx,y，then

fx,y2fx0,y0

xx0yy0fx0,y0xy1xx0yy0fx0,y02!xyApplication:

1. Accurate to 10,find the approximately value of e.

9x2xnee1xxn12!n!(n1)!Solution : （01）

xWhen x1 ，

e39 (n1)!<(n1)!<10 , 取n12

13！6.210

39 then 13!<10

9

so,

e1111123!12!2.718281828

x3sinxx62. Proof that

x3f(x)sinxx6 Let

Then

x2f(x)cosx12 f(x)xsinx

f(x)1cosx

And f(0)f(0)f(0)0， f(x)0,x(0,2)

11f(0)x2f()x323!，x(0,2)

Therefore

f()3x03!

f(x)f(0)f(0)x

x3sinxx6 x(0,2) So

x2(1x)(1x) 3. Find the Maclaurin series of111x224(1x)4(1x) 2(1x)(1x)(1x)Solution : =

111n1()x(1)nxn21x4n04n0

11n1nx(1(1)n)xn2n04n0 11(1)nn(n1)x2n02 x1,1

4. f(x) is second order differentiable in the interval a,b，andf(x)0，proof that:

bab1)f(x)dx.a2ba

f(

f(x)f(abababf()ab2)f()(x)(x)22222

Because f(x)0，

ababab)f()(x)222，  is between x0andx。

f(x)f( abf(x)dxf(abbabbab)dxf()(x)dxaa222

ab1ab)(ba)f()[(bx0)2(ax0)2]222

f(

f(ab)(ba)2

f(ab1b)f(x)dx.a2ba

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