A.7.1: Section 7.1 Answers
( \newcommand{\kernel}{\mathrm{null}\,}\)
1.
- R=2;I=(−1,3)
- R=1/2;I=(3/2,5/2)
- R=0
- R=16;I=(−14,18)
- R=∞;I=(−∞,∞)
- R=4/3;I=(−25/3,−17/3)
3.
- R=1;I=(0,2)
- R=√2;I=(−2−√2,−2+√2)
- R=∞;I=(−∞,∞)
- R=0
- R=√3;I=(−√3,√3)
- R=1:I=(0,2)
5.
- R=3;I=(0,6)
- R=1;I=(−1,1)
- R=1/√3;I=(3−1/√3,3+1/√3)
- R=∞;I=(−∞,∞)
- R=0
- R=2;I=(−1,3)
11. bn=2(n+2)(n+1)an+2+(n+1)nan+1+(n+3)an
12. b0=2a2−2a0bn=(n+2)(n+1)an+2+[3n(n−1)−2]an+3(n−1)an−1,n≥1
13. bn=(n+2)(n+1)an+2+2(n+1)an+1+(2n2−5n+4)an
14. bn=(n+2)(n+1)an+2+2(n+1)an+1+(n2−2n+3)an
15. bn=(n+2)(n+1)an+2+(3n2−5n+4)an
16. b0=−2a2+2a1+a0,bn=−(n+2)(n+1)an+2+(n+1)(n+2)an+1+(2n+1)an+an−1,n≥2
17. b0=8a2+4a1−6a0,bn=4(n+2)(n+1)an+2+4(n+1)2an+1+(n2+n−6)an−3an−1,n≥1
21. b0=(r+1)(r+2)a0,bn=(n+r+1)(n+r+2)an−(n+r−2)2an−1,n≥1.
22. b0=(r−2)(r+2)a0,bn=(n+r−2)(n+r+2)an+(n+r+2)(n+r−3)an−1,n≥14
23. b0=(r−1)2a0,b1=r2a1+(r+2)(r+3)a0,bn=(n+r−1)2an+(n+r+1)(n+r+2)an−1+(n+r−1)an−2,n≥2
24. b0=r(r+1)a0,b1=(r+1)(r+2)a1+3(r+1)(r+2)a0,bn=(n+r)(n+r+1)an+3(n+r)(n+r+1)an−1+(n+r)an−2,n≥2
25. b0=(r+2)(r+1)a0,b1=(r+3)(r+2)a1,bn=(n+r+2)(n+r+1)an+2(n+r−1)(n+r−3)an−2,n≥2
26. b0=2(r+1)(r+3)a0,b1=2(r+2)(r+4)a1,bn=2(n+r+1)(n+r+3)an+(n+r−3)(n+r)an−2,n≥2