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Infinite Series

9.7 Taylor Polynomials and Approximations

Focus on taylor polynomial construction, approximate a function value, taylor error bound.

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Ideas in this lesson

01Taylor polynomial constructionWorked steps · five questions02Approximate a function valueWorked steps · five questions03Taylor error boundWorked steps · five questions
Idea 1 of 3

Taylor polynomial construction

The coefficient at degree kk is ff⁽k^{k}⁾ak\frac{a}{k}!, multiplying (x−a)k(x - a)^{k}. A degree −n-n polynomial stops after degree nn.

Worked example 1

For f(x)=x3f(x) = x^{3}, find the coefficient of (x−2)(x - 2) in the Taylor polynomial centered at 22.
  1. f′(x)=3x2f'(x) = 3x^{2}; f′′(x)=6xf''(x) = 6x; ff‴(x)=6(x) = 6
  2. c1=3(2)21!=12c_1 = \dfrac{3(2)^{2}}{1!} = 12

Answer: 1212

Worked example 2

For f(x)=x3f(x) = x^{3}, find the coefficient of (x−3)2(x - 3)^{2} in the Taylor polynomial centered at 33.
  1. f′(x)=3x2f'(x) = 3x^{2}; f′′(x)=6xf''(x) = 6x; ff‴(x)=6(x) = 6
  2. c2=6(3)2!=9c_2 = \dfrac{6(3)}{2!} = 9

Answer: 99

Idea 2 of 3

Approximate a function value

Substitute into a finite Taylor polynomial near its center. More terms change the approximation, not the exact function.

Worked example 1

Use the degree −1-1 Maclaurin polynomial of exe^{x} to estimate e0.5e^{0.5}.
  1. ff⁽k^{k}⁾(0)=1(0) = 1; P1(x)=x01+x1P_1(x) = \dfrac{x^{0}}{1} + \dfrac{x}{1}
  2. P1(12)=32≈1.5P_1(\dfrac{1}{2}) = \dfrac{3}{2} \approx 1.5

Answer: 32≈1.5\dfrac{3}{2} \approx 1.5

Worked example 2

Use the degree −2-2 Maclaurin polynomial of exe^{x} to estimate e0.3333333333e^{0.3333333333}.
  1. ff⁽k^{k}⁾(0)=1(0) = 1; P2(x)=x01+x1+x22P_2(x) = \dfrac{x^{0}}{1} + \dfrac{x}{1} + \dfrac{x^{2}}{2}
  2. P2(13)=2518≈1.39P_2(\dfrac{1}{3}) = \dfrac{25}{18} \approx 1.39

Answer: 2518≈1.39\dfrac{25}{18} \approx 1.39

Idea 3 of 3

Taylor error bound

Bound the next derivative on the interval and apply the Lagrange remainder formula.

Worked example 1

For exe^{x} on [0,0.5][0, 0.5], give the Lagrange error bound of its degree −1-1 Maclaurin polynomial.
  1. M=e12M = e^{\dfrac{1}{2}}; n=1n = 1
  2. ∣R1(12)∣≤e12(12)22!=e128≈0.21|R_1(\dfrac{1}{2})| \le \dfrac{e^{\dfrac{1}{2}}(\dfrac{1}{2})^{2}}{2!} = \dfrac{e^{\dfrac{1}{2}}}{8} \approx 0.21

Answer: e128≈0.21\dfrac{e^{\dfrac{1}{2}}}{8} \approx 0.21

Worked example 2

For exe^{x} on [0,0.3333333333][0, 0.3333333333], give the Lagrange error bound of its degree −2-2 Maclaurin polynomial.
  1. M=e13M = e^{\dfrac{1}{3}}; n=2n = 2
  2. ∣R2(13)∣≤e13(13)33!=e13162≈0.00861|R_2(\dfrac{1}{3})| \le \dfrac{e^{\dfrac{1}{3}}(\dfrac{1}{3})^{3}}{3!} = \dfrac{e^{\dfrac{1}{3}}}{162} \approx 0.00861

Answer: e13162≈0.00861\dfrac{e^{\dfrac{1}{3}}}{162} \approx 0.00861

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §9.7,p9.7, p. 636636. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.