Question

Find the Taylor series of the function f (x) = 1/(x + 3) centered at a = 1.Upload a file with your answer and the complete procedure.

Answer 1

a Given:

`f(x)=(1)/(x+3)`

The Taylor series is represented as,

`f(x)=\sum ^(\infty)_(k\mathop=0)(f^k(a))/(k!)(x-a)^k`

To find the desired polynomial, we need to calculate the derivative and evaluate them at given point.

`\begin{gathered} f(x)=(1)/(x+3),f(1)=(1)/(4) \\ f^(\prime)(x)=(d)/(dx)((1)/(x+3))=-(1)/((x+3)^2),f^(\prime)(1)=-(1)/(16) \\ f^(\doubleprime)(x)=(d^2)/(dx^2)((1)/(x+3))=(2)/((x+3)^2),f^(\doubleprime)(1)=(1)/(32) \\ f^(\doubleprime\prime)(x)=\frac{d^3^{}}{dx^3}((1)/(x+3))=-(6)/((x+3)^4),f^(\doubleprime\prime)(1)=-(3)/(128) \\ \ldots\ldots.\ldots\ldots\ldots\ldots\ldots\ldots\text{ So on} \end{gathered}`

Now, use these values to get a polynomial,

`\begin{gathered} f(x)=(f^0(1))/(0!)(x-1)^0+(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)=(1)/(4)-(1)/(16)(x-1)+(1)/(32)((x-1)^2)/(2!)-(3)/(128)((x-1)^3)/(3!) \\ f(x)=(1)/(4)-(1)/(16)(x-1)+(1)/(64)(x-1)^2-(1)/(256)(x-1)^3+.\ldots\ldots\ldots.. \\ f(x)=(1)/(4)-(1)/(4^2)(x-1)+(1)/(4^3)(x-1)^2-(1)/(4^4)(x-1)^3+.\ldots\ldots\ldots.. \end{gathered}`

Answer:

`f(x)=(1)/(4)-(1)/(4^2)(x-1)+(1)/(4^3)(x-1)^2-(1)/(4^4)(x-1)^3+.\ldots\ldots\ldots..`