`h(x)=1/(1-5x) ,c=0` Find a power series for the function, centered at c and determine the interval of convergence.

Expert Answers
marizi eNotes educator| Certified Educator

A power series centered at `c=0` is follows the formula:

`sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...`

The given function` h(x)= 1/(1-5x)` resembles the power series centered at `c=0` :

`(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n`

or

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

To evaluate the given function `h(x) =1/(1-5x)` centered at `c=0` , we may apply Law of exponents: `1/x^n = x^(-n)` .

`h(x)= (1-5x) ^(-1)`

Apply the aforementioned formula for power series on  `(1-5x) ^(-1) or (1+(-5x))^(-1)` , we may replace "`x` " with "`-5x`  " and "`k` " with "`-1` ". We let:

`(1+(-5x))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-5x) ^n `

`=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-5)^nx ^n`

`=1+(-1)(-5)^1x +(-1(-2))/(2!)(-5)^2x ^2+(-1(-2)(-3))/(3!)(-5)^3x ^3+(-1(-2)(-3)(-4))/(4!)(-5)^4x ^4+...`

`=1+5x +2/2*25*x ^2+(-6)/6(-125)x ^3+24/24*625*x ^4+...`

`=1+5x +25x ^2+125x ^3+625x ^4+...`

`= sum_(n=0)^oo (5x)^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 (5x)^n`  with `sum_(n=0)^oo a*r^n` , we determine: `r = 5x` .

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

`|5x|lt1`

`-1 lt5xlt1`

Divide each part by `5` :

`(-1)/5 lt(5x)/5lt1/5`

`-1/5ltxlt1/5`

Thus, the power series of the  function ` h(x)=1/(1-5x)`  centered at `c=0` is  `sum_(n=0)^oo(5x)^n` with an interval of convergence: `-1/5 ltxlt1/5` .