Question

x+1 4x – 7 c) f(x)= ; c= 0 2x2 + 3x – 2 d) f(x)= 2x2 + 5x – 3 ; C = 2

Find a power series for the function, centered at c, and determine the interval of convergence.

Answer 1

We first express both the function into standard form for the power series expansion and then we find the interval of convergence.

It is to be noted that both the power series expansion constant are in the summation form not in product form so to perform ratio (or root) test we separately calculated the intervals and finally we took the intersection of the interval for interval of convergence.

like ratio test) Now we calculate the limit (almost (-1) + 3 2 n+1 ( First passt of series), oh lim no I N12 (-1) n. 3 m 2 121 2 12] (x+2) = lim 2 for ( naa -1) . It will be convergent 1 < 1%, (2 <x<2 CA semes) Again L = lim (1 1/ second part of - 2121 6. It will be convergent for 2121 <1 B²)x 220 ex. <h2 prirloce ) 0 .e. b < x <t2 & from both the interval of convergence will be 24 x < 2 N Now we check end points. Si At x=+1 (second part of the semes) the sames will be E (antbru (31)- E an zu E bn an. gives myun second 2n+2 (+ 5)? lim pant su CV (fails) synstues ont E 2 (+ 22 h Raatio test first part nal 1 1 n e lim 2. m44 Root test gives, lim | 2 + (+1) 17 no +S41 Teos him 2n - 2 1+ h = (fails). <1 finally, interval of convergence - <x<b

f(x) fly) W 90 @ family X+1 ; C = 2x² +5 x-3 2x24 5x-3 f(x) 2x² + 6x-2x-3 (x-1) (x+3) = (2x+1)(2+3) Now X+1 O Powen R+1) 13/12 (1 Tray) + +243 z 3 + Stay) First solving A. 2x-1 x+3 SIS A(x+3)+B(2x-1) On't now wort (2x-1) (x+3) for A B we on li va get, A + 2B = 1 alwinst ano noong sa wohl to A 73 73 ON Mie Code (8 fox) = -(+22) 3(+243) B = (63A-B = 317 2/7 svip soot visu and to Now we put y = x-2 to find the power samies at the center C=2. a Secon + 142 +3 toot to fry) 264+2)-1 Y+2 +3 14th 2 + uc 2 y + 3

2 35 t PO 1+ 2y It 3 5 n=0 i fly) = lomoto power Semes will be sobe)= 2 (o"6333" +ão conclusion [***". 4 + şi loom] yn poo [6%)"+ 3565)"] (x-2)^ providing N=o (x-21 <1. Fingt part, N1 L = lim Nya oh (x-2)+) (-2/310 (x-2)n = 12-21 & Internal of convergence, ģ12-21 <1 i pe-21 < 2 / 2 2-/ < x < 2+ 3 .. t3 <<< 3 - Second part, L= line 13 - 15.01 (x-2) 2+1 12/25 (-4) (x2)n + 12-21 convergent for > 5 12-21 <1 [X-2 1 <5 - 3 <* <]