`f(x)=2/x n=3,c=1` Find the n'th Taylor Polynomial centered at c

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Taylor series is an example of infinite series derived from the expansion of f(x) about a single point. It is represented by infinite sum of` f^n(x)` centered at `x=c.` The general formula for Taylor series is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) =f(c)+f'(c)(x-c) +(f^2(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f^4(c))/(4!)(x-c)^4 +...`

To determine the Taylor polynomial of degree `n=3 ` from the given function `f(x)=2/x ` centered at `x=1` , we may apply the definition of Taylor series.

To determine the list `f^n(x)` up to `n=3` , we may apply Law of Exponent: `1/x^n = x^-n`  and  Power rule for derivative: `d/(dx) x^n= n *x^(n-1)` .

`f(x) = 2/x or 2x^(-1)`

`f'(x) = d/(dx) 2/x`

            `= d/(dx) 2x^(-1)`

           `= 2*d/(dx) x^(-1)`

           `=2*(-1 *x^(-1-1))`

           `=-2x^(-2) or -2/x^2`

`f^2(x)= d/(dx) -2x^(-2)`

            `=-2 *d/(dx) x^(-2)`

           `=-2 *(-2x^(-2-1))`

           `=4x^(-3) or 4/x^3`

`f^3(x)= d/(dx) 4x^(-3)`

           `=4 *d/(dx) x^(-3)`

          `=4 *(-3x^(-3-1))`

          `=-12x^(-4) or -12/x^4`

Plug-in `x=1` , we get:

`f(2)=2/1 =2`

`f'(2)=-2/1^2 = -2`

`f^2(2)=4/1^3 =4`

`f^3(2)=-12/1^4 = -12`

Applying the formula for Taylor series, we get:

`sum_(n=0)^3 (f^n(1))/(n!) (x-1)^n`

`=f(1)+f'(1)(x-1) +(f^2(1))/(2!)(x-1)^2 +(f^3(1))/(3!)(x-1)^3`

`=2+(-2)(x-1) +4/(2!)(x-1)^2 +(-12)/(3!)(x-1)^3`

`=2-2(x-1) +4/2(x-1)^2 -12/6(x-1)^3`

`=2-2(x-1) +2(x-1)^2 -2(x-1)^3`

The Taylor polynomial of degree `n=3`  for the given function `f(x)=2/x` centered at `x=1` will be:

`P_3(x)=2-2(x-1) +2(x-1)^2 -2(x-1)^3`

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