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Find a Taylor Series Expansion For f(x)=sinx\mathrm{f}(\mathrm{x})=\sin \mathrm{x} With a=π\mathrm{a}=\pi
A) n=0(xπ)2n+1(2n+1)!=(xπ)+(xπ)33!+(xπ)55!+(xπ)77!+\sum_{n=0}^{\infty} \frac{(x-\pi)^{2 n+1}}{(2 n+1) !}=(x-\pi)+\frac{(x-\pi)^{3}}{3 !}+\frac{(x-\pi)^{5}}{5 !}+\frac{(x-\pi)^{7}}{7 !}+\ldots

Question 15

Multiple Choice

Find a Taylor series expansion for f(x) =sinx\mathrm{f}(\mathrm{x}) =\sin \mathrm{x} with a=π\mathrm{a}=\pi .


A) n=0(xπ) 2n+1(2n+1) !=(xπ) +(xπ) 33!+(xπ) 55!+(xπ) 77!+\sum_{n=0}^{\infty} \frac{(x-\pi) ^{2 n+1}}{(2 n+1) !}=(x-\pi) +\frac{(x-\pi) ^{3}}{3 !}+\frac{(x-\pi) ^{5}}{5 !}+\frac{(x-\pi) ^{7}}{7 !}+\ldots
B) n=0(1) n(xπ) 2n+1(2n) !=1(xπ) 22!+(xπ) 44!(xπ) 66!+\sum_{n=0}^{\infty} \frac{(-1) ^{n}(\mathrm{x}-\pi) ^{2 \mathrm{n}+1}}{(2 \mathrm{n}) !}=1-\frac{(\mathrm{x}-\pi) ^{2}}{2 !}+\frac{(\mathrm{x}-\pi) ^{4}}{4 !}-\frac{(\mathrm{x}-\pi) ^{6}}{6 !}+\ldots
C) n=0(1) n(xπ) 2n+1(2n+1) !=(xπ) (xπ) 33!+(xπ) 55!(xπ) 77!+\sum_{n=0}^{\infty} \frac{(-1) ^{n}(\mathrm{x}-\pi) ^{2 \mathrm{n}+1}}{(2 \mathrm{n}+1) !}=(\mathrm{x}-\pi) -\frac{(\mathrm{x}-\pi) ^{3}}{3 !}+\frac{(\mathrm{x}-\pi) ^{5}}{5 !}-\frac{(\mathrm{x}-\pi) ^{7}}{7 !}+\ldots
D) n=0(1) n+1(xπ) 2n+1(2n+1) !=(xπ) +(xπ) 33!(xπ) 55!+(xπ) 77!\sum_{n=0}^{\infty} \frac{(-1) ^{n+1}(x-\pi) ^{2 n+1}}{(2 n+1) !}=-(x-\pi) +\frac{(x-\pi) ^{3}}{3 !}-\frac{(x-\pi) ^{5}}{5 !}+\frac{(x-\pi) ^{7}}{7 !}-\ldots

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