Respuesta :
a Given:
[tex]f(x)=\frac{1}{x+3}[/tex]The Taylor series is represented as,
[tex]f(x)=\sum ^{\infty}_{k\mathop=0}\frac{f^k(a)}{k!}(x-a)^k[/tex]To find the desired polynomial, we need to calculate the derivative and evaluate them at given point.
[tex]\begin{gathered} f(x)=\frac{1}{x+3},f(1)=\frac{1}{4} \\ f^{\prime}(x)=\frac{d}{dx}(\frac{1}{x+3})=-\frac{1}{(x+3)^2},f^{\prime}(1)=-\frac{1}{16} \\ f^{\doubleprime}(x)=\frac{d^2}{dx^2}(\frac{1}{x+3})=\frac{2}{(x+3)^2},f^{\doubleprime}(1)=\frac{1}{32} \\ f^{\doubleprime\prime}(x)=\frac{d^3^{}}{dx^3}(\frac{1}{x+3})=-\frac{6}{(x+3)^4},f^{\doubleprime\prime}(1)=-\frac{3}{128} \\ \ldots\ldots.\ldots\ldots\ldots\ldots\ldots\ldots\text{ So on} \end{gathered}[/tex]Now, use these values to get a polynomial,
[tex]\begin{gathered} f(x)=\frac{f^0(1)}{0!}(x-1)^0+\frac{f^{\prime}(1)}{1!}(x-1)^1+\frac{f^{^{\doubleprime}}(1)}{2!}(x-1)^2+\frac{f^{^{\doubleprime}^{\prime}}(1)}{3!}(x-1)^3+\text{.}\ldots\ldots\text{...} \\ f(x)=\frac{1}{4}-\frac{1}{16}(x-1)+\frac{1}{32}\frac{(x-1)^2}{2!}-\frac{3}{128}\frac{(x-1)^3}{3!} \\ f(x)=\frac{1}{4}-\frac{1}{16}(x-1)+\frac{1}{64}(x-1)^2-\frac{1}{256}(x-1)^3+.\ldots\ldots\ldots.. \\ f(x)=\frac{1}{4}-\frac{1}{4^2}(x-1)+\frac{1}{4^3}(x-1)^2-\frac{1}{4^4}(x-1)^3+.\ldots\ldots\ldots.. \end{gathered}[/tex]Answer:
[tex]f(x)=\frac{1}{4}-\frac{1}{4^2}(x-1)+\frac{1}{4^3}(x-1)^2-\frac{1}{4^4}(x-1)^3+.\ldots\ldots\ldots..[/tex]