Math  /  Calculus

QuestionFind the Taylor series for f(x)f(x) centered at the given value of aa. [Assume that ff has a power series expansion. Do not show that Rn(x)0R_{n}(x) \rightarrow 0.] f(x)=sin(x),a=πf(x)=n=0(1)nx2n(2n)!\begin{array}{c} f(x)=\sin (x), \quad a=\pi \\ f(x)=\sum_{n=0}^{\infty}(-1)^{n} \cdot \frac{x^{2 n}}{(2 n)!} \end{array}

Studdy Solution
The Taylor series for f(x)=sin(x)f(x) = \sin(x) centered at a=πa = \pi is:
n=0(1)n+1(2n+1)!(xπ)2n+1\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!} \cdot (x-\pi)^{2n+1}

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