Question

id=490136 Use partial fractions to find a power series representation, centered at 0 of 2 f(x) = x2 – 3x + 2 and then find the interval of convergence. Paragraph ВІ E I

Answer 1

We know that x" for all [x]<1. 1-X P=0 2 1 1 2. 2 2 Here given f (x)= x2-3x +2 = 2 2 (x-2)(x-1) x 2x -x +2 x-2 x-1 x-2 x-1 00 2 2 Now Then - -Σ for all <1=[x]<2 x-2 2-X x 21 1- 1 n=0 1 2 2 Now 2 1-1 x" for all [x]<1 X-1 n=0 2n+1 -1 skon Sisi FÅTE O (247) 32 ]" () 14 2" Now let us find the radius of convergence of 2n+1 2" n=0

20+1 21 Let un x". Now lim =lim 2n+2 -1 2+1 r+1 2 *** |x] =[x]. Then the Un+1 – 1).x" x" is absolutely convergent if lim “n+1 <1 then |x|<1 2n+2 -1 =lim 12/2+1 F12 (21+-1 series 22+1 2" nu n=0 Then the radius of convergence of (***)** sp{:{**")convergeet convergent = 1 n=0 2n+1 -1 Then the series is Σ x" convergent for (x/<1>-1<x<1= xe(-1,1). n=0 Now we will check the convergence at the end points of (-1,1). Now for x=-1 the series 2n+1 -1 21 xbecomes 20+1 -1 2" |(-1)" = { 2x(+1)*- (-1)" . Now here we 2 Mao n=0 10 P=0 1 see that 2x(-1)" is divergent series and is convergent then 2n+1 -1 2" (-1)" must be divergent then 29 n=0 N=0 n=0

00 for x =-1 the series 2n+1 -1 2" x" is divergent. n=0 2n+1-1 1 Now for x=1 the series becomes =X2- then by same argument as previous for x=-1 we 2" nal n=0 2" n=0 see that 2n+1 -1 2" x" is divergent for x=1. n=0 2 Then the interval of convergence of f(x) = r - 3x +2 x" is (-1,1). 2

Another method for finding interval of convergence of f(x) :

72 00 1 x 2 Another Method | We have for all = <1=[x]<2 and ---23 2 1-X x" for all x 2 N=0 2 1-1 nal 00 2 [x]<1 then f (x) = x2 – 3x + 2 2+1 -1 x" will happen if for all [x]<2 and []<1. = n=0 2" That is, (x ER:[x]<2}{xe R:\x{<1}={x€ R:[x]<1}=(-1,1) 2 2+1 -1 Then f (x)=72-3x+2 -Σ x" for all re(-1,1). n=0 2"

Happy Chegging :)