Review : Taylor Series — Question 9

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Question 9

Let T4(x)=1−x2/2+x4/24T_4(x)=1-x^2/2+x^4/24, the fourth-degree Maclaurin polynomial of cos⁡x\cos x. Consider a rational competitor Q(x)=1+ax21+bx2,a,b∈ℝ.Q(x)=\frac{1+ax^2}{1+bx^2},\qquad a,b\in\mathbb R. A student believes that replacing a Taylor polynomial by a rational expression with the same local coefficients must improve accuracy.

Tasks

  1. Determine a,ba,b so that QQ and cos⁡x\cos x agree through degree four at 00, and identify any real poles of QQ.

  2. Find the leading nonzero terms of cos⁡x−T4(x)\cos x-T_4(x) and cos⁡x−Q(x)\cos x-Q(x). Compute the limiting ratio of their absolute errors as x→0x\to 0, x≠0x\ne 0.

  3. Use an alternating Taylor enclosure for cos⁡1\cos 1 to prove which approximation is closer at x=1x=1, without relying on a calculator value of cos⁡1\cos 1.

  4. Evaluate the student’s claim and explain why matching coefficients alone cannot rank all approximations on a whole interval.

Original worksheet page 1: question and worked solution for 6-2-009
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Question 9 – Solution

Strategy. Match coefficients, then examine the first coefficient on which the competing approximations disagree.

Step 1: Solve the matching equations. Near 00, Q=1+(a−b)x2+(b2−ab)x4+⋯Q=1+(a-b)x^2+(b^2-ab)x^4+\cdots. Thus a−b=−1/2a-b=-1/2 and −b(a−b)=1/24-b(a-b)=1/24, giving b=112,a=−512,Q(x)=12−5x212+x2.\boxed{b=\frac 1{12},\quad a=-\frac 5{12},\quad Q(x)=\frac{12-5x^2}{12+x^2}.} Since 12+x2>012+x^2>0, there are no real poles.

Step 2: Compare the leading errors. The rational geometric expansion has x6x^6 coefficient −1/288-1/288. Hence cos⁡x−T4(x)=−x6720+O(x8),cos⁡x−Q(x)=x6480+O(x8).\cos x-T_4(x)=-\frac{x^6}{720}+O(x^8),\qquad \cos x-Q(x)=\frac{x^6}{480}+O(x^8). Therefore lim⁡x→0|cos⁡x−Q(x)||cos⁡x−T4(x)|=3/2\boxed{\lim_{x\to 0}\frac{|\cos x-Q(x)|}{|\cos x-T_4(x)|}=3/2}. In particular, the rational approximation is worse for all sufficiently small nonzero xx.

Step 3: Make an exact comparison at one. Alternating bounds give T6(1)=389720<cos⁡1<T4(1)=1324,Q(1)=713.T_6(1)=\frac{389}{720}<\cos 1<T_4(1)=\frac{13}{24},\qquad Q(1)=\frac 7{13}. Here Q(1)<389/720Q(1)<389/720, and the midpoint of the two approximations is 12(7/13+13/24)=337/624<389/720\tfrac 12(7/13+13/24)=337/624<389/720. Thus cos⁡1\cos 1 lies above their midpoint and between them: T4(1)T_4(1) is strictly closer than Q(1)Q(1).

Step 4: State the proper conclusion. The claim fails both locally and at x=1x=1. Matching four derivatives fixes contact at the center, not the next error coefficient or behavior far away. For 0<|x|≤10<|x|\le 1, alternating bounds give cos⁡x−T4(x)<0\cos x-T_4(x)<0 and cos⁡x−Q(x)≥x6(1312−1720)>0,\cos x-Q(x)\ge x^6\left(\frac 1{312}-\frac 1{720}\right)>0, since T4−Q=x6/[288(1+x2/12)]T_4-Q=x^6/[288(1+x^2/12)]. These signs explain the plotted errors.

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Original worksheet page 2: question and worked solution for 6-2-009

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