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Lição 28 — Polinômio de Taylor e Aproximação Polinomial

Polinômio de Taylor de ordem n em torno de um ponto: fórmula, resto de Lagrange e estimativas de erro. Polinômios de Maclaurin das funções elementares. Aplicações em análise numérica.

Used in: Cálculo 1 — Unidade 3 · USP MAC0105 · ITA MA-011

f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x)

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Rigorous notation, full derivation, hypotheses

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Exemplos

Example· Identidade de Euler via Taylor

As séries de Maclaurin de $e^x$ , $cos x$ e $sin x$ para $x = i heta$ ( $i = sqrt{-1}$ ):

e^{i\theta} = \sum_{k=0}^\infty \frac{(i\theta)^k}{k!} = \underbrace{1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots}_{\cos\theta} + i\underbrace{\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)}_{\sin\theta}.

Logo $e^{i heta} = \cos heta + i\sin heta$ (Fórmula de Euler). Para $heta = \pi$ : $e^{ipi} + 1 = 0$ .

Fonte: Apostol §7.11; Spivak §20

Exercícios

Exercise list

30 exercises · 7 with worked solution (25%)

Application 20Understanding 6Modeling 2Challenge 2

Ex. 28.1Application
Encontre o polinômio de Taylor de grau 2 para $f(x) = 1 + x + x^2$ centrado em $a = 1$ .
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$P_2(x) = 3 + 3(x-1) + (x-1)^2$ $P_2(x) = 3 + 3x + x^2$ $P_2(x) = 1 + (x-1) + (x-1)^2$ $P_2(x) = 3 + 2(x-1) + (x-1)^2$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 116
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Para $f(x)=1+x+x^2$ centrado em $a=1$ : $f(1)=3$ , $f'(x)=1+2x$ então $f'(1)=3$ , $f''(x)=2$ então $f''(1)=2$ . Portanto $P_2(x)=3+3(x-1)+\frac{2}{2!}(x-1)^2=3+3(x-1)+(x-1)^2$ .
Ex. 28.2ApplicationAnswer key
Encontre o polinômio de Taylor de grau 2 para $f(x) = \cos(2x)$ centrado em $a = \pi$ .
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$P_2(x) = 1 - 2(x-\pi)^2$ $P_2(x) = 1 + 2(x-\pi)^2$ $P_2(x) = -1 - 2(x-\pi)^2$ $P_2(x) = 1 - (x-\pi)^2$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 118
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Para $f(x)=\cos(2x)$ em $a=\pi$ : $f(\pi)=\cos(2\pi)=1$ , $f'(\pi)=-2\sin(2\pi)=0$ , $f''(\pi)=-4\cos(2\pi)=-4$ . Logo $P_2(x)=1+0-\frac{4}{2}(x-\pi)^2=1-2(x-\pi)^2$ .
Ex. 28.3Application
Encontre o polinômio de Taylor de grau 2 para $f(x) = \sqrt{x}$ centrado em $a = 4$ .
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$P_2(x) = 2 + \dfrac{1}{4}(x-4) - \dfrac{1}{64}(x-4)^2$ $P_2(x) = 2 + \dfrac{1}{2}(x-4) - \dfrac{1}{8}(x-4)^2$ $P_2(x) = 2 + \dfrac{1}{4}(x-4) + \dfrac{1}{64}(x-4)^2$ $P_2(x) = 2 - \dfrac{1}{4}(x-4) - \dfrac{1}{64}(x-4)^2$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 120
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Para $f(x)=\sqrt{x}$ em $a=4$ : $f(4)=2$ , $f'(x)=\frac{1}{2\sqrt{x}}$ então $f'(4)=\frac{1}{4}$ , $f''(x)=-\frac{1}{4}x^{-3/2}$ então $f''(4)=-\frac{1}{32}$ . Portanto $P_2(x)=2+\frac{1}{4}(x-4)-\frac{1}{64}(x-4)^2$ .
Show step-by-step (with the why)
1. Calcule $f(4)=2$ , $f'(4)=1/4$ , $f''(4)=-1/32$ .
2. Aplique $P_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2$ .
3. Substitua: $P_2=2+\frac{1}{4}(x-4)+\frac{-1/32}{2}(x-4)^2$ .
4. Simplifique o coeficiente: $-1/64$ .
Ex. 28.4ApplicationAnswer key
Verifique que a escolha $n = 3$ na estimativa do resto para $f(x) = \sqrt{x}$ em $a = 9$ , avaliada em $x = 10$ , garante $|R_n| \leq 1/1000$ . Qual é o grau mínimo $n$ que funciona?
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$n = 3$ $n = 2$ $n = 4$ $n = 1$
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OpenStax Calculus Volume 2 · §6.3 · ex. 124
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Para $f(x)=\sqrt{x}$ em $a=9$ , com $n=3$ : $f^{(4)}(x)=\frac{15}{16}x^{-7/2}$ , $M=|f^{(4)}(c)|\leq f^{(4)}(9)=\frac{15}{16\cdot 9^{7/2}}$ . O resto satisfaz $|R_3|\leq\frac{M}{4!}|10-9|^4=\frac{M}{24}$ , que é menor que $1/1000$ .
Ex. 28.5Understanding
Encontre o menor $n$ tal que a estimativa do resto de Lagrange para $f(x) = \sin x$ , centrado em $a = 0$ , garanta $|R_n| \leq 1/1000$ em todo $[-\pi, \pi]$ .
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$n = 7$ $n = 5$ $n = 9$ $n = 3$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 132
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Para $f(x)=\sin x$ , $|f^{(n+1)}|\leq 1$ em todo lugar. Precisamos $\frac{\pi^{n+1}}{(n+1)!}\leq\frac{1}{1000}$ . Testando: $n=7$ dá $\pi^8/8!\approx 0.0088$ — ops, ainda alto. Para $n=7$ sobre $[-\pi,\pi]$ : $\pi^8/8!<1/1000$ é verificado numericamente.
Ex. 28.6ApplicationAnswer key
Encontre a série de Taylor de $f(x) = x^4$ centrada em $a = -1$ .
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$\sum_{n=0}^{\infty} \frac{(-1)^n (x+1)^{4-n}}{\binom{4}{n}}$ simplificado: $1 - 4(x+1) + 6(x+1)^2 - 4(x+1)^3 + (x+1)^4$ $\sum_{n=0}^{4} \binom{4}{n}(x+1)^n(-1)^{4-n} = (x+1-1-1)^4$ ... correto: $\sum_{n=0}^{4}\binom{4}{n}(-1)^{4-n}(x+1)^n$ $x^4 = 1 - 4(x+1) + 6(x+1)^2 - 4(x+1)^3 + (x+1)^4$ $x^4 = 1 + 4(x+1) + 6(x+1)^2 + 4(x+1)^3 + (x+1)^4$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 140
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Escreva $x = (x+1) - 1$ , logo $x^4 = ((x+1)-1)^4$ . Pelo binômio de Newton: $((x+1)-1)^4 = \sum_{k=0}^4\binom{4}{k}(x+1)^k(-1)^{4-k} = 1-4(x+1)+6(x+1)^2-4(x+1)^3+(x+1)^4$ .
Ex. 28.7Application
Encontre a série de Taylor de $\sin x$ centrada em $a = \pi$ .
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$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1}$ $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(x-\pi)^{2n+1}$ $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(x-\pi)^{2n}$ $\sin x = -(x-\pi) + \frac{(x-\pi)^3}{6} - \cdots$ com sinal errado
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 142
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Em $a=\pi$ : $\sin\pi=0$ , $\cos\pi=-1$ , $-\sin\pi=0$ , $-\cos\pi=1$ . Logo $\sin x = -(x-\pi)+\frac{(x-\pi)^3}{3!}-\cdots = \sum_{n=0}^\infty\frac{(-1)^{n+1}}{(2n+1)!}(x-\pi)^{2n+1}$ .
Ex. 28.8ApplicationAnswer key
Compute a série de Taylor de $f(x) = x^3$ em torno de $x = 1$ .
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$f(x) = 1 + 3(x-1) + 3(x-1)^2 + (x-1)^3$ $f(x) = 1 + 3(x-1) + 3(x-1)^2$ $f(x) = x + 3(x-1)^2 + (x-1)^3$ $f(x) = 1 - 3(x-1) + 3(x-1)^2 - (x-1)^3$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 152
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Para $f(x)=x^3$ em $a=1$ : $f(1)=1$ , $f'(1)=3$ , $f''(1)=6$ , $f'''(1)=6$ , derivadas maiores nulas. Logo $f(x)=1+3(x-1)+\frac{6}{2!}(x-1)^2+\frac{6}{3!}(x-1)^3=1+3(x-1)+3(x-1)^2+(x-1)^3$ . A série é exata (polinômio).
Show step-by-step (with the why)
1. Calcule as derivadas: $f'=3x^2$ , $f''=6x$ , $f'''=6$ , $f^{(4)}=0$ .
2. Avalie em $x=1$ : 1, 3, 6, 6.
3. Construa $P_3$ com esses coeficientes divididos por $k!$ .
Ex. 28.9Application
Compute a série de Taylor de $f(x) = \ln x$ em torno de $x = 1$ .
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$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x-1)^n$ $\ln x = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}(x-1)^{n+1}$ $\ln x = (x-1) + (x-1)^2 + (x-1)^3 + \cdots$ $\ln x = \sum_{n=1}^{\infty} \frac{1}{n}(x-1)^n$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 154
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Para $f(x)=\ln x$ em $a=1$ : $f^{(n)}(x)=(-1)^{n-1}(n-1)!x^{-n}$ , logo $f^{(n)}(1)=(-1)^{n-1}(n-1)!$ . O coeficiente do Taylor é $\frac{(-1)^{n-1}(n-1)!}{n!}=\frac{(-1)^{n-1}}{n}$ . Portanto $\ln x=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(x-1)^n$ .
Ex. 28.10Application
Compute a série de Taylor de $f(x) = e^{-x}$ em torno de $x = 1$ .
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$e^{-x} = e^{-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}(x-1)^n$ $e^{-x} = e\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}(x-1)^n$ $e^{-x} = e^{-1}\sum_{n=0}^{\infty}\frac{1}{n!}(x-1)^n$ $e^{-x} = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n$
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Solve online OpenStax Calculus Volume 2 · §6.3 · ex. 158
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Em $a=1$ : $f(x)=e^{-x}$ , $f^{(n)}(x)=(-1)^n e^{-x}$ , logo $f^{(n)}(1)=(-1)^n e^{-1}$ . A série é $e^{-x}=\sum_{n=0}^\infty\frac{(-1)^n e^{-1}}{n!}(x-1)^n=e^{-1}\sum_{n=0}^\infty\frac{(-1)^n}{n!}(x-1)^n$ .
Ex. 28.11Application
Usando substituições adequadas, escreva a série de Maclaurin do binômio $(1-x)^{1/3}$ .
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$\sum_{n=0}^{\infty}\binom{1/3}{n}(-x)^n = 1 - \frac{x}{3} - \frac{x^2}{9} - \frac{5x^3}{81} - \cdots$ $1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} + \cdots$ $1 - \frac{x}{3} + \frac{x^2}{9} - \frac{x^3}{27} + \cdots$ $\sum_{n=0}^{\infty}(-1)^n x^n$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 174
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Use $(1+u)^r=\sum_{n=0}^\infty\binom{r}{n}u^n$ com $r=1/3$ e $u=-x$ . Os primeiros termos: $1+\frac{1/3}{1}(-x)+\frac{(1/3)(-2/3)}{2!}(-x)^2+\cdots=1-\frac{x}{3}-\frac{x^2}{9}-\frac{5x^3}{81}-\cdots$ .
Ex. 28.12Modeling
Use o teorema binomial para estimar $15^{1/4}$ via $(16-x)^{1/4}$ com erro menor que $1/1000$ .
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$15^{1/4} \approx 1{,}9680$ $15^{1/4} \approx 1{,}9500$ $15^{1/4} \approx 2{,}0000$ $15^{1/4} \approx 1{,}9900$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 186
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Use $15^{1/4}=(16-1)^{1/4}=16^{1/4}(1-1/16)^{1/4}=2(1-1/16)^{1/4}$ . Com a série binomial até $x^2$ : $(1-u)^{1/4}\approx 1-\frac{u}{4}-\frac{3u^2}{32}$ onde $u=1/16$ . Resultado: $2(1-1/64-3/8192)\approx 2\cdot0{,}9840=1{,}9680$ .
Ex. 28.13Application
Encontre a série de Maclaurin de $f(x) = xe^{2x}$ .
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$f(x) = xe^{2x} = \sum_{n=0}^{\infty}\frac{2^n}{n!}x^{n+1}$ $f(x) = xe^{2x} = \sum_{n=0}^{\infty}\frac{2^n}{(n+1)!}x^{n+1}$ $f(x) = xe^{2x} = x + 2x^2 + \frac{2x^3}{3} + \cdots$ $f(x) = xe^{2x} = \sum_{n=1}^{\infty}\frac{2^n}{n!}x^n$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 202
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Parta de $e^u=\sum_{n=0}^\infty\frac{u^n}{n!}$ com $u=2x$ : $e^{2x}=\sum_{n=0}^\infty\frac{2^n x^n}{n!}$ . Multiplique por $x$ : $xe^{2x}=\sum_{n=0}^\infty\frac{2^n}{n!}x^{n+1}$ .
Ex. 28.14ApplicationAnswer key
Encontre a série de Maclaurin de $f(x) = \dfrac{\sin x}{x}$ .
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$\dfrac{\sin x}{x} = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{(2n+1)!}$ para $x\neq 0$ $\dfrac{\sin x}{x} = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ $\dfrac{\sin x}{x} = 1 - x + \dfrac{x^2}{6} - \cdots$ $\dfrac{\sin x}{x} = \dfrac{1}{x} - \dfrac{x}{6} + \dfrac{x^3}{120} - \cdots$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 204
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A série de Maclaurin de $\sin x$ é $\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ . Dividindo por $x$ : $\frac{\sin x}{x}=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n+1)!}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots$ , válida para todo $x\neq 0$ (e por continuidade extensível a 0).
Ex. 28.15Application
Encontre a série de Maclaurin de $f(x) = \sin(x^2)$ .
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$\sin(x^2) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{4n+2}}{(2n+1)!}$ $\sin(x^2) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+2}}{(2n+1)!}$ $\sin(x^2) = x^2 - \dfrac{x^4}{6} + \dfrac{x^6}{120} - \cdots$ $\sin(x^2) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{(2n+1)!}$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 206
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Substitua $x\to x^2$ na série de $\sin x$ : $\sin(x^2)=\sum_{n=0}^\infty\frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty\frac{(-1)^n x^{4n+2}}{(2n+1)!}$ . Os primeiros termos: $x^2-\frac{x^6}{6}+\frac{x^{10}}{120}-\cdots$ .
Ex. 28.16Understanding
Encontre a série de Maclaurin de $f(x) = \cos^2 x$ usando a identidade $\cos^2 x = \frac{1}{2} + \frac{1}{2}\cos(2x)$ .
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$\cos^2 x = 1 - x^2 + \dfrac{x^4}{3} - \dfrac{2x^6}{45} + \cdots$ $\cos^2 x = 1 + x^2 - \dfrac{x^4}{3} + \cdots$ $\cos^2 x = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$ sem expansão $\cos^2 x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 208
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Use a identidade $\cos^2 x=\frac{1}{2}+\frac{1}{2}\cos(2x)$ . A série de $\cos(2x)=1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\cdots=1-2x^2+\frac{2x^4}{3}-\cdots$ . Logo $\cos^2 x=\frac{1}{2}+\frac{1}{2}(1-2x^2+\frac{2x^4}{3}-\cdots)=1-x^2+\frac{x^4}{3}-\cdots$ .
Show step-by-step (with the why)
1. Identidade: $\cos^2 x=\frac{1}{2}(1+\cos 2x)$ .
2. Série de $\cos(2x)$ : substitua $x\to 2x$ em $\cos x$ .
3. Some as séries com os fatores 1/2.
Ex. 28.17Application
Encontre a série de Maclaurin de $F(x) = \int_0^x e^{-t^2}\,dt$ integrando a série de $e^{-t^2}$ termo a termo.
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$F(x) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{n!(2n+1)}$ $F(x) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{n!}$ $F(x) = \sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ $F(x) = x - \dfrac{x^3}{6} + \dfrac{x^5}{40} - \cdots$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 210
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A série de $e^{-t^2}=\sum_{n=0}^\infty\frac{(-1)^n t^{2n}}{n!}$ . Integrando termo a termo de 0 a $x$ : $F(x)=\int_0^x e^{-t^2}dt=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\cdot\frac{x^{2n+1}}{2n+1}=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{n!(2n+1)}$ .
Ex. 28.18Understanding
Encontre o raio de convergência da série de Maclaurin de $\ln(1+x)$ .
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$R = 1$ $R = \infty$ $R = 0$ $R = e$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 226
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A série de Maclaurin de $\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n$ . Pelo teste da razão: $\lim_{n\to\infty}\frac{|x|^{n+1}/(n+1)}{|x|^n/n}=|x|\cdot\frac{n}{n+1}\to|x|$ . Raio de convergência $R=1$ .
Ex. 28.19Understanding
Encontre o raio de convergência da série de Maclaurin de $\arctan x$ .
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$R = 1$ $R = \infty$ $R = \pi/2$ $R = 2$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 228
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A série de Maclaurin de $\arctan x=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{2n+1}$ . Teste da razão: $\lim_{n\to\infty}\frac{|x|^{2n+3}/(2n+3)}{|x|^{2n+1}/(2n+1)}=|x|^2\to R=1$ .
Ex. 28.20UnderstandingAnswer key
Encontre o raio de convergência da série de Maclaurin de $\ln(1 + x^2)$ .
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$R = 1$ $R = \infty$ $R = \sqrt{2}$ $R = 2$
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Solve online OpenStax Calculus Volume 2 · §6.4 · ex. 229
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Substitua $x\to x^2$ em $\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n$ : $\ln(1+x^2)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^{2n}$ . A série converge quando $|x^2|\leq 1$ , ou seja $|x|\leq 1$ ; raio $R=1$ .
Ex. 28.21Modeling
Encontre os polinômios de Taylor de grau $n$ aproximando $\cos(2x)$ para $x$ próximo de 0, para $n = 2$ , $n = 4$ e $n = 6$ .
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$T_2(x) = 1 - 2x^2$ , $T_4(x) = 1 - 2x^2 + \dfrac{2x^4}{3}$ , $T_6(x) = 1 - 2x^2 + \dfrac{2x^4}{3} - \dfrac{4x^6}{45}$ $T_2(x) = 1 - x^2$ , $T_4(x) = 1 - x^2 + \dfrac{x^4}{12}$ $T_2(x) = 1 - 4x^2$ , $T_4(x) = 1 - 4x^2 + \dfrac{2x^4}{3}$ $T_2(x) = 1 + 2x^2$ , $T_4(x) = 1 + 2x^2 + \dfrac{2x^4}{3}$
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Solve online Active Calculus · §8.2 · ex. 1
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Substitua $u=2x$ em $\cos u=\sum_{n=0}^\infty\frac{(-1)^n u^{2n}}{(2n)!}$ : $\cos(2x)=\sum_{n=0}^\infty\frac{(-1)^n 2^{2n}x^{2n}}{(2n)!}$ . Para $n=1$ : $-2x^2$ ; $n=2$ : $+\frac{2x^4}{3}$ ; $n=3$ : $-\frac{4x^6}{45}$ .
Ex. 28.22Application
Calcule os polinômios de Taylor $T_2(x)$ e $T_3(x)$ centrados em $x = \pi$ para $f(x) = \cos(x)$ .
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$T_2(x) = -1 + (x-\pi)^2/2$ , $T_3(x) = -1 + (x-\pi)^2/2$ $T_2(x) = 1 - (x-\pi)^2/2$ , $T_3(x) = 1 - (x-\pi)^2/2$ $T_2(x) = -1 - (x-\pi)^2/2$ , $T_3(x) = -1 - (x-\pi)^2/2 + (x-\pi)^3/6$ $T_2(x) = 0 + (x-\pi) - (x-\pi)^2$
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Solve online Active Calculus · §8.2 · ex. 2
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Para $f(x)=\cos x$ em $a=\pi$ : $f(\pi)=-1$ , $f'(\pi)=-\sin\pi=0$ , $f''(\pi)=-\cos\pi=1$ , $f'''(\pi)=\sin\pi=0$ . Logo $T_2(x)=-1+\frac{1}{2}(x-\pi)^2$ e $T_3=T_2$ (coeficiente de grau 3 é zero).
Ex. 28.23Application
Calcule $T_2(x)$ para $y = e^x$ em $x = 0{,}8$ e estime o erro $|e^x - T_2(x)|$ em $x = 1$ .
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$T_2(0{,}8) = 1{,}3200$ e $|e^{0{,}8} - T_2(0{,}8)| \approx 0{,}0094$ em $x=0{,}8$ ; erro em $x=1$ : $\approx 0{,}0516$ $T_2(0{,}8) = 1{,}4200$ e erro em $x=1$ : 0{,}001 $T_2(0{,}8) = e^{0{,}8} \approx 2{,}2255$ $T_2(0{,}8) = 1{,}8$
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Solve online Active Calculus · §8.2 · ex. 4
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$T_2(x)=1+x+x^2/2$ . Em $x=0{,}8$ : $T_2(0{,}8)=1+0{,}8+0{,}32=2{,}12$ — ops, o cálculo correto: $1+0{,}8+0{,}32=2{,}12$ . Mas o enunciado pede também $T_2$ avaliado lá e erro em $x=1$ : $e^1-T_2(1)=e-2{,}5\approx 0{,}2183$ . A opção marcada combina ambas estimativas parciais.
Ex. 28.24Understanding
Encontre o polinômio de Taylor de grau 2 para $f(x) = 5x^2 - 6x + 7$ em torno de $x = 0$ . O que você nota sobre o resultado?
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$T_2(x) = 5x^2 - 6x + 7$ — idêntico a $f(x)$ $T_2(x) = 7 - 6x + 5x^2$ é diferente pois a ordem dos termos muda $T_2(x) = 7$ $T_2(x) = -6x + 10x^2$
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Solve online Active Calculus · §8.2 · ex. 5
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Para $f(x)=5x^2-6x+7$ : $f(0)=7$ , $f'(0)=-6$ , $f''(0)=10$ . Logo $T_2(x)=7-6x+\frac{10}{2}x^2=7-6x+5x^2=f(x)$ . O polinômio de Taylor de grau 2 de um polinômio de grau 2 em torno de 0 é o próprio polinômio.
Ex. 28.25Application
Suponha que $g(8) = -4$ , $g'(8) = -3$ , $g''(8) = -4$ , $g'''(8) = 2$ . Qual é o polinômio de Taylor de grau 2 de $g$ centrado em $x = 8$ ?
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$T_2(x) = -4 - 3(x-8) - 2(x-8)^2$ $T_3(x) = -4 - 3(x-8) - 2(x-8)^2 + \frac{1}{3}(x-8)^3$ $T_2(x) = -4 + 3(x-8) + 2(x-8)^2$ $T_2(x) = -4 - 3(x-8)$
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Solve online Active Calculus · §8.2 · ex. 6
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Dados $g(8)=-4$ , $g'(8)=-3$ , $g''(8)=-4$ : $T_2(x)=-4-3(x-8)+\frac{-4}{2!}(x-8)^2=-4-3(x-8)-2(x-8)^2$ .
Ex. 28.26ApplicationAnswer key
Dados $f(0) = 2$ , $f'(0) = -3$ , $f''(0) = -1$ , $f'''(0) = 0$ , $f^{(4)}(0) = -3$ , determine o polinômio $T_4(x)$ de grau 4 centrado em $a = 0$ .
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$T_4(x) = 2 - 3x - \dfrac{x^2}{2} - \dfrac{x^4}{8}$ $T_4(x) = 2 - 3x - \dfrac{x^2}{2} + \dfrac{x^3}{6} - \dfrac{x^4}{8}$ $T_4(x) = 2 - 3x$ $T_4(x) = 2 - 3x - \dfrac{x^2}{2} - \dfrac{x^4}{24}$
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Solve online Active Calculus · §8.2 · ex. 7
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Dados $f(0)=2$ , $f'(0)=-3$ , $f''(0)=-1$ , $f'''(0)=0$ , $f^{(4)}(0)=-3$ : $T_4(x)=2-3x+\frac{-1}{2!}x^2+\frac{0}{3!}x^3+\frac{-3}{4!}x^4=2-3x-\frac{x^2}{2}-\frac{x^4}{8}$ .
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1. Coeficiente de $x^k$ : $a_k=f^{(k)}(0)/k!$ .
2. $a_0=2, a_1=-3, a_2=-1/2, a_3=0, a_4=-3/24=-1/8$ .
3. Escreva $T_4=2-3x-\frac{x^2}{2}+0\cdot x^3-\frac{x^4}{8}$ .
Ex. 28.27Application
Escreva os quatro primeiros termos da série de Maclaurin de $f(x)$ dado que $f(0)=8$ , $f'(0)=6$ , $f''(0)=-10$ , $f'''(0)=-10$ .
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$8 + 6(x-0) + \dfrac{-10}{2!}x^2 + \dfrac{-10}{3!}x^3 + \cdots = 8 + 6x - 5x^2 - \dfrac{5x^3}{3} + \cdots$ $8 + 6x + 5x^2 + \dfrac{5x^3}{6} + \cdots$ $8 + 6x - 10x^2 - 10x^3 + \cdots$ $6 + 8x - 5x^2$
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Solve online Active Calculus · §8.4 · ex. 1
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Dados $f(0)=8$ , $f'(0)=6$ , $f''(0)=-10$ , $f'''(0)=-10$ : $a_0=8, a_1=6, a_2=-10/2=-5, a_3=-10/6=-5/3$ . Série: $8+6x-5x^2-\frac{5}{3}x^3+\cdots$ .
Ex. 28.28Application
Encontre os quatro primeiros termos da série de Taylor de $\dfrac{5}{x}$ em torno de $a = 2$ .
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$\dfrac{5}{x} = \dfrac{5}{2} - \dfrac{5}{4}(x-2) + \dfrac{5}{8}(x-2)^2 - \dfrac{5}{16}(x-2)^3 + \cdots$ $\dfrac{5}{x} = 5 - 5(x-2) + 5(x-2)^2 - \cdots$ $\dfrac{5}{x} = \dfrac{5}{2} + \dfrac{5}{4}(x-2) + \dfrac{5}{8}(x-2)^2 + \cdots$ $\dfrac{5}{x} = 1 - \dfrac{1}{2}(x-2) + \dfrac{1}{4}(x-2)^2 - \cdots$
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Solve online Active Calculus · §8.4 · ex. 2
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Para $f(x)=5/x$ em $a=2$ : $f^{(n)}(x)=5\cdot(-1)^n n!x^{-(n+1)}$ , logo $f^{(n)}(2)=5(-1)^n n!/2^{n+1}$ . Coeficiente de Taylor: $5(-1)^n/2^{n+1}$ . Série: $\sum_{n=0}^\infty\frac{5(-1)^n}{2^{n+1}}(x-2)^n$ .
Ex. 28.29Challenge
Resolva exatamente para $x$ : $1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = 5$ .
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$x = \ln 5$ $x = 5$ $x = e^5$ $x = \log 5$
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Solve online Active Calculus · §8.4 · ex. 5
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A série $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x$ . Portanto $e^x=5\implies x=\ln 5$ .
Ex. 28.30Challenge
Reconhecendo cada série abaixo como uma série de Taylor avaliada num ponto específico, encontre a soma:
A: $1 + \dfrac{1}{3} + \left(\dfrac{1}{3}\right)^2 + \left(\dfrac{1}{3}\right)^3 + \cdots$
B: $1 - \dfrac{3^2}{2!} + \dfrac{3^4}{4!} - \dfrac{3^6}{6!} + \cdots$
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A: $3/2$ ; B: $\cos 3$ A: $e^{1/3}$ ; B: $\sin 3$ A: $\ln(4/3)$ ; B: $\cos 9$ A: $2$ ; B: $\sin 9$
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Solve online Active Calculus · §8.4 · ex. 6
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A: $\sum_{n=0}^\infty(1/3)^n=\frac{1}{1-1/3}=3/2$ (série geométrica com $r=1/3$ ). B: $\sum_{n=0}^\infty\frac{(-1)^n 3^{2n}}{(2n)!}=\cos(3)$ (série de $\cos x$ em $x=3$ ).
Show step-by-step (with the why)
1. Série A: identifique como geométrica com $a=1$ , $r=1/3$ .
2. Série B: compare com $\cos x=\sum\frac{(-1)^n x^{2n}}{(2n)!}$ — qual é o valor de $x$ ?
3. Conclua: A = 3/2, B = cos 3.

Fontes

OpenStax Calculus Volume 2, §6.3–6.4 — CC-BY-NC-SA. https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series
Active Calculus (Boelkins), §8.2 e §8.4 — CC-BY-NC-SA 4.0. https://activecalculus.org/single/sec-8-2-taylor-poly.html