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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

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