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 `f(x)= 2/(1-x^2)` centered at `c=0` resembles the power series:
`(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+...`
Apply Law of exponents: `1/x^n = x^(-n)` , we get: `f(x)= 2(1-x^2) ^(-1)`
Apply...
See This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
the aforementioned formula for power series on ` (1-x^2) ^(-1) or (1+(-x^2)) ^(-1)` , we may replace "`x` " with "`-x^2` " and "`k` " with "`-1` ". We let:
`(1+(-x^2))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-x^2) ^n`
`=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-1)^nx ^(2n)`
`=1+(-1)(-1)^1x ^(2*1) +(-1(-2))/(2!)(-1)^2x ^(2*2)+(-1(-2)(-3))/(3!)(-1)^3x ^(2*3)+(-1(-2)(-3)(-4)/(4!)(-1)^4x ^(2*4)+...`
`=1+(-1)*(-1)x^2 +2/2*1*x ^4 +(-6)/6*(-1)*x^6+24/24*1*x^8+...`
`=1+1x^2 +2/2x ^4 +6/6x^6+24/24x^8+...`
`=1+x^2 +x ^4 +x^6+x^8+...`
Applying `(1+(-x^2))^(-1) =1+x^2 +x ^4 +x^6+x^8+...` , we get:
`2*(1+(-x^2))^(-1) =2*[ 1+x^2 +x ^4 +x^6+x^8+...]`
`=2+2x^2 +2x ^4 +2x^6+2x^8+...`
`= sum_(n=0)^oo 2x^(2n)`
The power series of the function `f(x)=2/(1-x^2)` centered at `c=0` is:
`2/(1-x^2) =sum_(n=0)^oo 2x^(2n)`
or
`2/(1-x^2) =2+2x^2 +2x ^4 +2x^6+2x^8+...`
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.
The series `sum_(n=0)^oo 2x^(2n)` is the same as `sum_(n=0)^oo 2(x^2)^n` .
By comparing `sum_(n=0)^oo2(x^2)^n` with `sum_(n=0)^oo a*r^n` , we determine: `r =x^2` .
Apply the condition for convergence of geometric series: `|r|lt1` .
`|x^2|lt1`
`|x|^2lt1`
For `xgt=0` , we replace `|x|` with `x` .
`x^2lt1`
Applying `(f(x))^2lta` then ` f(x)ltsqrt(a)` and `f(x) gt-sqrt(a)` .
Let `f(x) =x` and `a =1` then `+- sqrt(a)=+- sqrt(1) = +-1` .
The interval will be `xlt1` and `xgt-1` and combine as `-1ltxlt1` .
For `xlt0` , we replace `|x|` with `-x.`
`(-x)^2lt1`
`x^2 lt1`
The interval will also be `-1ltxlt1` .
Combining the ranges from `xlt0` and or `xgt=0` , we get the interval for convergence as: `-1ltxlt1` .
Check the convergence at endpoints that may satisfy `|x^2|=1` .
Let `x=-1` on `sum_(n=0)^oo 2(x^2)^n` , we get:
`sum_(n=0)^oo2((-1)^2)^n=sum_(n=0)^oo 2(1)^n`
Using geometric series test, the `r =1` satisfy `|r| gt=1` . Thus, the series diverges at `x=-1` .
Let `x=1` on `sum_(n=0)^oo 2(x^2)^n` , we get:
`sum_(n=0)^oo2(1^2)^n=sum_(n=0)^oo 2(1)^n`
Using geometric series test, the `r =1` satisfy `|r| gt=1` . Thus, the series diverges at `x=1` .
Thus, the power series `sum_(n=0)^oo 2(x^2)^n` or `sum_(n=0)^oo2x^(2n)` has an interval of convergence: `-1 ltxlt1` .