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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)= \sqrt{25-x^2} $ in $ x_0=2.5$.

Function and value of the function

$ f(x)= \sqrt{25-x^2} $,      $ f(2.5)= 4.330127018922194 $

Derivatives, derivatives at $2.5$ and coefficients of Taylor polynomial

$ i$$f^{(i)}(x)$$f^{(i)}(x_0)$$\frac{f^{(i)}(x_0)}{i!}$
1$ -{{x }\over{\sqrt{25-x^2}}} $$ -0.57735026918963 $$ -0.57735026918963 $
2$ -{{ 25{}\sqrt{25-x^2}}\over{x^4-50{}x^2+625}} $$ -0.3079201435678 $$ - 0.1539600717839 $
3$ {{75{}x{}\sqrt{25-x^2}}\over{\left(x-5\right)^3{} \left(x+5\right)^3}} $$ -0.12316805742712 $$ -0.020528009571187 $
4$ {{75 {}\left(4{}x^2+25\right)}\over{\left(x-5\right)^3{}\left(x+5\right)^ 3{}\sqrt{25-x^2}}} $$ -0.13137926125559 $$ -0.0054741358856498 $

Polinomios de Taylor

$ T_{4}(x)= {{1 }\over{2}}{}\left(5{}\sqrt{3}\right)-{{1}\over{3}}{}\sqrt{3}{}\left( x-2.5\right)-{{1}\over{45}}{}\left(4{}\sqrt{3}\right){}\left(x-2.5 \right)^2-{{1}\over{675}}{}\left(8{}\sqrt{3}\right){}\left(x-2.5 \right)^3-{{1}\over{10125}}{}\left(32{}\sqrt{3}\right){}\left(x-2.5 \right)^4 $