a Express in Power Series developed around X00 fx1x316x4 b W

Solution

Let\'s write f(x) as g(x)*h(x), where
g(x) = 1+x³, and h(x) = 1/(16+x ⁴)

We want to put h(x) in the form a/(1-r), the sum of a geometric series, so that we can derive its power series.

Divide h(x) by 16 in the numerator and denominator (this keeps the value of h(x) unchanged.

Then, h(x) = 1/16/(1+x ⁴/16) or h(x) =  1/16/(1 - (-x⁴/16))

h(x) is now in the form a/(1-r), with a = 1/16 and r = -x⁴/16

Then, we may express h(x) as a + ar + ar² + a³ ..., or  1/16 +  1/16(-x⁴/16) ... =

1/16 - 1/256x⁴ ..., and g(x)h(x) =

(1+x³)(1/16 - 1/256x⁴ ...) = 1/16 + 1/16x³ - 1/256x⁴ - 1/256x⁷ ...

We will consider the n\'th term as 1/16(1+x³)(-x⁴/16)ⁿ for n=0, 1, 2, 3, ...

Then, the ratio is 1/16(1+x³)(-x⁴/16)ⁿ⁺¹ /(1/16(1+x³)(-x⁴/16)ⁿ ) =

1/16(1+x³)(-x⁴/16)ⁿ (-x⁴/16)/(1/16(1+x³)(-x⁴/16)ⁿ ) =

(-x⁴/16)

Then, the radius of convergence can be found by |r| = 1, or |-x⁴/16| = 1, or

x⁴/16 = 1, or

x⁴ = 16, or |x| = 2

d) Note that the ratio is fixed in this problem, so the radius becomes an inequality.

-2 < x < 2 (when x = ±2, all terms are equal in magnitude and alternating, as the ratio is -x⁴/16 = -1, so there is no limit to the sum)