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Redemitz4s

Answered question

2021-11-16

Approximate f by a Taylor polynomial with degree n at the number a.

f (x) = \ln (1 + 2 x), a = 1, n = 3, 0.5 \leq x \leq 1.5

Fesion

Beginner2021-11-17Added 24 answers

Taylor polynomial with degree n at the number a is given by

f (x) \approx T_{n} = \sum_{r = 0}^{n} \frac{f^{r} (a) {(x - a)}^{r}}{r!}

f (x) = \ln (1 + 2 x) \Rightarrow f (1) = \ln 3

f^{'} (x) = \frac{2}{(1 + 2 x)} \Rightarrow f^{'} (1) = \frac{2}{3}

f^{″} (x) = - \frac{4}{{(1 + 2)}^{2}} \Rightarrow f^{″} (1) = - \frac{4}{9}

f^{‴} (x) = \frac{16}{{(1 + 2 x)}^{3}} \Rightarrow f^{‴} (1) = \frac{16}{27}

Taylor polynomial with degree n=3 at the number a=1 is given by

f (x) \approx T_{3} = \sum_{r = 0}^{3} \frac{f^{r} (1) {(x - 1)}^{r}}{r!}

f (x) \approx T_{3} = f (1) + f^{'} (1) (x - 1) + \frac{f^{″} (1) {(x - 1)}^{2}}{2} + \frac{f^{‴} (1) {(x - 1)}^{3}}{6}

f (x) \approx T_{3} = \ln 3 + (\frac{2}{3}) (x - 1) + (- \frac{4}{9}) \frac{{(x - 1)}^{2}}{2} + (\frac{16}{27}) \frac{{(x - 1)}^{3}}{6}

f (x) \approx T_{3} = \ln 3 + (\frac{2}{3}) (x - 1) - (\frac{2}{9}) {(x - 1)}^{2} + (\frac{8}{81}) {(x - 1)}^{3}

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