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Valentin [98]
3 years ago
13

Find the taylor polynomial t3(x) for the function f centered at the number

Mathematics
1 answer:
inysia [295]3 years ago
7 0

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

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

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

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