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4.5: Teorema de Taylor
https://espanol.libretexts.org/Matematicas/Analisis/Introducci%C3%B3n_al_An%C3%A1lisis_Matem%C3%A1tico_I_(Lafferriere%2C_Lafferriere_y_Nguyen)/04%3A_Diferenciaci%C3%B3n/4.05%3A_Teorema_de_Taylor
Consideremos la función $g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .$ Entonces\[g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x}...Consideremos la función $g(t)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(t)}{k !}(x-t)^{k}-\frac{\lambda}{(n+1) !}(x-t)^{n+1} .$ Entonces $g(\bar{x})=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(\bar{x})}{k !}(x-\bar{x})^{k}-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=f(x)-P_{n}(x)-\frac{\lambda}{(n+1) !}(x-\bar{x})^{n+1}=0 .$ y $g(x)=f(x)-\sum_{k=0}^{n} \frac{f^{(k)}(x)}{k !}(x-x)^{k}-\frac{\lambda}{(n+1) !}(x-x)^{n+1}=f(x)-f(x)=0 .$ Por el teorema de Rolle, existe $c$ en el medio $\bar{x}$ y $x$ tal que\(g^{\…

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