`f(x) = 1/(3-x) , c=1` Find a power series for the function, centered at c and determine the interval of convergence.

To determine the power series centered at c, we may apply the formula for Taylor series:

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

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

To list the `f^n(x)` for the given function `f(x)=1/(3-x)` centered at `c=1` , 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) =1/(3-x)`

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

Let `u =3-x` then `(du)/(dx) = -1`

`d/(dx) c*(3-x)^n = c *d/(dx) (3-x)^n`

` = c *(n* (3-x)^(n-1)*(-1)`

` = -cn(3-x)^(n-1)`

`f'(x) =d/(dx) (3-x)^(-1)`

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

`=(3-x)^(-2) or 1/(3-x)^2`

`f^2(x) =d/(dx) (3-x)^(-2)`

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

`=2(3-x)^(-3) or 2/(3-x)^3`

`f^3(x) =d/(dx)2(3-x)^(-3)`

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

`=6(3-x)^(-4) or 6/(3-x)^4`

`f^4(x) =d/(dx)6(3-x)^(-4)`

`=-6(-4)(3-x)^(-4-1)`

`=24(3-x)^(-5) or 24/(3-x)^5`

Plug-in `x=1` for each `f^n(x)` , we get:

`f(1)=1/(3-1) =1/2`

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

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

`f^3(1)=6/(3-1)^4 = 3/8`

`f^4(1)=24/(3-1)^5 = 3/4`

Plug-in the values on the formula for Taylor series, we get:

`1/(3-x) = sum_(n=0)^oo (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 +(f^4(1))/(4!)(x-1)^4 +...`

` =1/2+1/4(x-1) +(1/4)/(2!)(x-1)^2 +(3/8)/(3!)(x-1)^3 +(3/4)/(4!)(x-1)^4 +... `

` =1/2+1/4(x-1) +(1/4)/2(x-1)^2 +(3/8)/6(x-1)^3 +(3/4)/24(x-1)^4 +...`

` =1/2+1/4(x-1) + 1/8(x-1)^2 +1/16(x-1)^3 +1/32(x-1)^4 +...`

` =sum_(n=1)^oo ((x-1)/2)^n`

To determine the interval of convergence, we may apply geometric series test wherein the series `sum_(n=0)^oo a*r^n` is convergent if `|r|lt1 or -1 ltrlt 1` . If `|r|gt=1` then the geometric series diverges.

By comparing `sum_(n=0)^oo ((x-1)/2)^n ` or `sum_(n=0)^oo 1*((x-1)/2)^n ` with `sum_(n=0)^oo a*r^n` , we determine: `r = (x-1)/2` .

Apply the condition for convergence of geometric series: `|r|lt1` .

`|(x-1)/2|lt1`

`-1lt(x-1)/2lt1`

Multiply each sides by 2:

`-1*2lt(x-1)/2*2lt1*2`

`-2ltx-1lt2`

Add 1 on each sides:

`-2+1ltx-1+1lt2+1`

`-1ltxlt3`

Thus, the power series of the function `f(x) = 1/(3-x) ` centered at `c=1 ` is `sum_(n=1)^oo ((x-1)/2)^n` with an interval of convergence: `-1ltxlt3` .

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