1、Taylors Theorem And Its Applications2Overview0()U x variable x in 0 x,where is a fixed point,can be calculated simplyThis approximation is simple and is easily used,but it has low precision,since and the approximation is,and also thisIn Chapter 2,we had seen that the value of a smooth function for s
2、omeby approximation 000()()()()f xf xfxxx.()f xthe difference between the accurate value of 0 xx just a infinitesimal of higher order with respect to is very small.0|xx approximation can only be used in the case of In this lecture,we will form a new technique to approximate the value of a function w
3、ith higher order of computation error.3Taylors Theorem000()()()f xfxxx 2010200()()()()nnnP xaa xxaxxaxxL()nP xis using a line,()f x,to approximate a curve.It can be(1)n n find a suitable polynomials of degreefto approximate a given curve of function,such that the approximate error is.That isIf we ca
4、n do so,what are the coefficients of,and how can we obtain000()()()()f xf xfxxx It is easy to see that the approximation,in fact,()f ximaged that if we use a suitable curve to approximate curve,the applicablewill be wider and the precision will be improved.It seems nature that we chosethe polynomial
5、s as the suitable curve.0()nxx an infinitesimal of higher order with respect to 0()()()nnf xP xo xx.them?Then the question becomes to,we try to4Taylors TheoremTaylor PolynomialPeano remainderTaylor coefficientsTaylor,Brook(1685-1731)English mathematician 200000()000()0000()()()()()()1!2!()()()!()()(
6、)!nnnknknkfxfxf xf xxxxxfxxxo xxnfxxxo xxk L L(Taylors theorem with Peano remainder)Suppose that the function f is differentiable of order n at the point x0.Then5Taylors TheoremTheorem(Taylors theorem with Lagrange remainder)Suppose that the function f is differentiable of order 1n in a interval I,0
7、 xI.Then for any xI,there exists at least one point ,which lies between x and 0 x,such that Taylor,Brook(1685-1731)English mathematician Lagrange remainderLagrange Formula200000()(1)1000()(1)10000()()()()()()1!2!()()()()!(1)!()()()()!(1)!nnnnknnknkfxfxf xf xxxxxfxfxxxxnnfxfxxxxkn L6Taylors Theorem0M
8、 Compare to the Peano remainder,the Lagrange remainder can be used to estimate the error more precisely.such that(1)|()|nfxM ,xa b ,In fact,if the function f is differentiable of order n+1 on a,b,and there is a constant then0nR()nP xWe can see from the inequality that n as.This means that to approxi
9、mate a differentiable function,the error may be make arbitrarilyif we utilize the polynomial ,a bof any order in the whole interval small(1)110|()|()|()(1)!(1)!nnnnfMRxxxbann .n is taken large enough.while7Taylors TheoremColin Maclaurin(1698-1746)Scottish mathematician If we let 00 x ,then the Lagra
10、nge formula becomes()(1)21()(1)10(0)(0)(0)()()(0)1!2!(1)!(0)(),01.!(1)!nnnnknnknkffffxf xfxxxxnnffxxxkn Land this formula is called the Maclaurin formula.8Maclaurin Formulae for Some Elementary Functions2112!(1)!nxxnxxeexxnn L(,)x ()xf xe Maclaurin formula for the exponential function()()kxfxe Since
11、 we havewhere 01 and.()(0)1kf and,9Maclaurin Formulae for Some Elementary Functions()()sin,0,1,2,2kfxxkkn L()10,2(0),1,2,(1),21kmkmfmkm L35721121cossin(1)(1)3!5!7!(21)!(21)!mmmmxxxxxxxxmm L(,)x ()sinf xx()cosf xx Maclaurin formula for and()sinf xx Let,since or()(0)1kf,thenwhere and 01.we have 10Macl
12、aurin Formulae for Some Elementary FunctionsSimilarly,we have242122coscos1(1)(1)2!4!(2)!(22)!mmmmxxxxxxmm L()sinf xx()cosf xx Maclaurin formula for and(,)x where and 01.11Maclaurin Formulae for Some Elementary Functions()1(1)!()(1),1,2,(1)kkkkfxknx L()1(0)(1)(1)!kkfk 234111ln(1)(1)(1)234(1)(1)nnnnnx
13、xxxxxxnnx L()ln(1)f xx Maclaurin formula for We had known thatsoThus(1,)x (0,1)Where and.12Maclaurin Formulae for Some Elementary Functions()()(1)(1)(1),0,1,2,kkfxkxkn LL()(0)(1)(1)kfk L211(1)(1)(1)(1)12!(1)()(1)!(1)nnnnxxxxnnxnx LLL()(1)()f xxR Maclaurin formula for We had known thatsoThus(0,1)(1,)
14、x Where and.13Some Applications of Taylors Theoremxy xysin(1)Approximate calculations14Some Applications of Taylors Theoremxy xysin!33xxy o(1)Approximate calculations15Some Applications of Taylors Theoremxy xysin!33xxy o!5!353xxxy (1)Approximate calculations16Some Applications of Taylors Theoremxy x
15、ysin!33xxy !5!353xxxy !7!5!3753xxxxy o(1)Approximate calculations17Some Applications of Taylors Theoremxysin!11!9!7!5!3119753xxxxxxy o(1)Approximate calculations18Some Applications of Taylors Theorem1111(1),(0,1)2!neRn L3(1)(1)!(1)!neRnn(1)Approximate calculations,we have3ee since,soIt may be seen t
16、hat(1)0nRn ,as e,which means that is approximated and estimation of the error.eApproximate calculation of the value of Solutionxe1x in the maclaurin formulae for Let by the approximation formula:11122!3!enLThe error can be made arbitrarily small as long if Finish.n is taken large enough.19Some Appli
17、cations of Taylors TheoremExample The Taylor approximation of degree 2 for the function()cosf xx is 2212xP.For what value of x is the error in this approximation not greater that 0.1?(1)Approximate calculations(4)422()()()()4!fRxf xP xx|cos|1 4|0.124x Since the error of approximation of Taylor is wh
18、en lies between the points x0 and(4)()cosf.Since and for all,we know that the error in the approximation is no more than 4|24x.,we only need 1.241.24x.In order to have Solution:20Some Applications of Taylors Theorem(1)Approximate calculations()xf 2n 5320 xx Example Find an approximate value of a rea
19、l root for the equation on the parameter;we denote this root by.This function can be seen is differentiable to any order.Thus,by Taylor,we obtain where is a very small parameter.It is easy to prove that this equation has a real root and it is dependentby the given function.as an implicit function de
20、terminedIt can be prove that()xf the implicit function formula and taking for instance,2(0)(0)(0)2!fxfff .Solution212(0)(0)20)!(fxfff Some Applications of Taylors Theorem(1)Approximate calculations5320 xx Find an approximate value of a real root for the equation where is a very small parameter.0 2x
21、450dxdxxxdd Solution(continued),we have,that is We take different to each side of the given,we haveWe will determine the coefficients in the last equation.Take equation with respect to 0(0)dxfd Differentiating both sides of the given equation again,we haveso that(0)2f.2(0)(00)2!)fffxf 405xx 1.40 22S
22、ome Applications of Taylors TheoremSo that 22432(5)2020d xdxdxxxddd(1)Approximate calculations5320 xx Find an approximate value of a real root for the equation where is a very small parameter.Solution (continued)220(0)d xfd therefore,the second degree approximate of the real root is 2(0)(0)(0)2!xfff
23、f 22403200 x.23402025dxdxxddx 11600 Finish.23Some Applications of Taylors Theoremsin x30sinlimxxxx(2)Finding limitsSolution,we haveFinish.30sinlim.xxxx Find By Maclaurin formula with Peano remainder to 3430()3!limxxxo xxx 4301()lim3!xo xx16 24Some Applications of Taylors TheoremExample Suppose that(
24、)0fx and()f x is an equivalent infinitesimal with x as 0 x.Prove that()f xx,if 0 x .(3)Proving inequalities(0)0f ProofBy the Maclaurin formula we have2()()(0)(0)2!ff xffxx ,x0lies between and .On the other hand,by the assumption we know2()(0)(0)()2!fffxxxo x .Thus,()()f xxo x.Then,we have(0)1f and .
25、25Some Applications of Taylors TheoremExample Suppose that()0fx and()f x is an equivalent infinitesimal with x as 0 x.Prove that()f xx,if 0 x .(3)Proving inequalities 2()()2!ff xxx Proof(continued)()0fx Since,we have()0fx ()f xx 0 x Similarly,we can prove that if then,while.So that()f xx 0 x as.Finish.
