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Polinomios de Taylor

$ n$-degree Taylor polynomial for the function $ f(x)$ around $ x=x_0$\[\begin{align*} T_n(x)&=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots\\&+\frac{f^{(i)}(x_0)}{i!}(x-x_0)^i+\cdots \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \end{align*}\]

Problema

Find $ 4 $-degree Taylor polynomial for function $ f(x)= \sin \left(5{}x\right) $ in $ x_0=114.592$.

Function and value of the function

$ f(x)= \sin \left(5{}x\right) $,      $ f(114.592)= 0.92841989327408 $

Derivatives, derivatives at $114.592$ and coefficients of Taylor polynomial

$ i$$f^{(i)}(x)$$f^{(i)}(x_0)$$\frac{f^{(i)}(x_0)}{i!}$
1$ 5{}\cos \left(5{}x\right) $$ 1.857663194533349 $$ 1.857663194533349 $
2$ -25{}\sin \left(5{}x\right) $$ -23.21049733185189 $$ - 11.60524866592595 $
3$ -125{}\cos \left(5{}x\right) $$ - 46.44157986333373 $$ -7.740263310555622 $
4$ 625{}\sin \left(5{}x\right) $$ 580.2624332962973 $$ 24.17760138734572 $

Polinomios de Taylor

$ T_{4}(x)= \sin \left({{14324}\over{ 25}}\right)+5{}\cos \left({{14324}\over{25}}\right){}\left(x-114.592 \right)-{{1}\over{2}}{}\left(25{}\sin \left({{14324}\over{25}} \right)\right){}\left(x-114.592\right)^2-{{1}\over{6}}{}\left(125{} \cos \left({{14324}\over{25}}\right)\right){}\left(x-114.592\right)^ 3+{{1}\over{24}}{}\left(625{}\sin \left({{14324}\over{25}}\right) \right){}\left(x-114.592\right)^4 $