# `f(x)=2/(6-x) ,c=-2` 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`

or

`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)=2/(6-x) ` centered at `c=-2` , we may apply Law of Exponent: `1/x^n...

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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`

or

`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)=2/(6-x) ` centered at `c=-2` , 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/(6-x)`

`=2(6-x)^(-1)`

Let `u =6-x` then `(du)/(dx) = -1`

`d/(dx) c*(6-x)^n = c *d/(dx) (6-x)^n`

`= c *(n* (6-x)^(n-1)*(-1)`

` = -cn(6-x)^(n-1)`

`f'(x) =d/(dx)2(6-x)^(-1)`

`=-2*(-1)(6-x)^(-1-1)`

` =2(6-x)^(-2) or 2/(6-x)^2`

`f^2(x) =d/(dx) 2(6-x)^(-2)`

`=-2(-2)(6-x)^(-2-1)`

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

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

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

`=12(6-x)^(-4) or 12/(6-x)^4`

`f^4(x) =d/(dx)12(6-x)^(-4)`

`=-12(-4)(6-x)^(-4-1)`

`=48(6-x)^(-5) or 48/(6-x)^5`

Plug-in `x=-2` for each `f^n(x)` , we get:

`f(-2)=2/(6-(-2))`

`=2/ 8`

`=1/4`

`f'(-2)=2/(6-(-2))^2`

`=2/8^2`

`= 1/32`

`f^2(-2)=4/(6-(-2))^3 `

`=4/8^3`

`=1/128`

`f^3(-2)=12/(6-(-2))^4 `

`=12/8^4`

`= 3/1024`

`f^4(-2)=48/(6-(-2))^5 `

`=48/8^5`

`= 3/2048`

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

`2/(6-x) = sum_(n=0)^oo (f^n(-2))/(n!) (x-(-2))^n`

` = sum_(n=0)^oo (f^n(-2))/(n!) (x+2)^n`

` =1/4+1/32(x+2) +(1/128)/(2!)(x+2)^2 +(3/1024)/(3!)(x+2)^3 +(3/2048)/(4!)(x+2)^4 +...`

` =1/4+1/32(x+2) +(1/128)/2(x+2)^2 +(3/1024)/6(x+2)^3 +(3/2048)/24(x+2)^4 +...`

` =1/4+1/32(x+2) + 1/256(x+2)^2 +1/2048(x+2)^3 + 1/16384(x+2)^4 +...`

` =1/2^2+1/2^5(x+2) + 1/2^8(x+2)^2 +1/2^11(x+2)^3 + 1/2^14(x+2)^4 +...`

` =sum_(n=1)^oo (x+2)^(n-1)/2^(3n-1)`

` =sum_(n=1)^oo( (x+2)^n*(x+2)^(-1))/(2^(3n)2^(-1))`

` =sum_(n=1)^oo (2(x+2)^n)/(2^(3n)(x+2))`

` =sum_(n=1)^oo (2/(x+2))((x+2)/2^3)^n`

` =sum_(n=1)^oo (2/(x+2))((x+2)/8)^n`

Note: Exponents of 2 as 2,5,8,11,14,... follows arithmetic sequence `a_n=a_0+(n-1)d.`

`a_n = 2 +(n-1)3`

`=2+3n-3`

` =3n-1`

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=1)^oo (2/(x+2))((x+2)/8)^n` with  `sum_(n=0)^oo a*r^n` , we determine: `r = (x+2)/8.`

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

`|((x+2)/8)|lt1`

`-1lt(x+2)/8lt1`

Multiply each sides by 8:

`-1*8lt(x+2)/8*8lt1*8`

`-8ltx+2lt8`

Subtract `2` from each sides:

`-8-2ltx+2-2lt8-2`

`-10ltxlt6`

Thus, the power series  of the function `f(x) = 2/(6-x) ` centered at `c=-2` is `sum_(n=1)^oo (x+2)^(n-1)/2^(3n-1)`  with an interval of convergence: `-10ltxlt6` .

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