1. cn+2=cn; y(x)=c0∑n=0∞x2n+c1∑n=0∞x2n+1=c0+c1x1−x2; ρ=1
2. cn+2=−12cn; ρ=2; y(x)=c0∑n=0∞(−1)nx2n2n+c1∑n=0∞(−1)nx2n+12n
3. (n+2)cn+2=−cn; y(x)=c0∑n=0∞(−1)nx2nn!2n+c1∑n=0∞(−1)nx2n+1(2n+1)!!; ρ=+∞
4. (n+2)cn+2=−(n+4)cn; ρ=1; c0∑n=0∞(−1)n(n+1)x2n+13c1∑n=0∞(−1)n(2n+3)x2n+1
5. 3(n+2)cn+2=ncn; ρ=3–√; y(x)=c0+c1∑n=0∞x2n+1(2n+1)3n
6. (n+1)(n+2)cn+2=(n−3)(n−4)cn; ρ=∞; y(x)=c0(1+6x2+x4)+c1(x+x3)
7. 3(n+1)(n+2)cn+2=−(n−4)2cn; y(x)=c0(1−8x23+8x427)+c1(x−x32+x5120+9∑n=3∞(−1)n[(2n−5)!!]2x2n+1(2n+1)!3n)
8. 2(n+1)(n+2)cn+2=(n−4)(n+4)cn; y(x)=c0(1−4x2+2x4)+c1(x−5x34+7x532+∑n=3∞(2n−5)!!(2n+3)!!x2n+1(2n+1)!2n)
9. (n+1)(n+2)cn+2=(n+3)(n+4)cn; ρ=1; y(x)=c0∑n=0∞(n+1)(2n+1)x2n+c13∑n=0∞(n+1)(2n+3)x2n+1
10. 3(n+1)(n+2)cn+2=−(n−4)cn; y(x)=c0(1+2x23+x427)+c1(x+x36+x5360+3∑n=3∞(−1)n(2n−5)!!x2n+1(2n+1)!3n)
11. 5(n+1)(n+2)cn+2=2(n−5)cn; y(x)=c1(x−4x315+4x5375)+c0(1−x2+x410+x6750+15∑n=4∞(2n−7)!!2nx2n(2n)!5n)
12. c2=0; (n+2)cn+3=cn; y(x)=c0(1+∑n=1∞x3n2⋅5⋯(3n−1))+c1∑n=0∞x3n+1n!3n
13. c2=0; (n+3)cn+3=−cn; y(x)=c0∑n=0∞(−1)nx3nn!3n+c1∑n=0∞(−1)nx3n+11⋅4⋯(3n+1)
14. c2=0; (n+2)(n+3)cn+3=−cn; y(x)=c0(1+∑n=1∞(−1)nx3n3n⋅n!⋅2⋅5⋯(3n−1))+c1∑n=0∞(−1)nx3n+13n⋅n!⋅1⋅4⋯(3n+1)
15. c2=c3=0; (n+3)(n+4)cn+4=−cn; y(x)=c0(1+∑n=1∞(−1)nx4n4n⋅n!⋅3⋅7⋯(4n−1))+c1∑n=0∞(−1)nx4n+14n⋅n!⋅5⋅9⋯(4n+1)
16. y(x)=x
17. y(x)=1+x2
18. y(x)=2∑n=0∞(−1)n(x−1)2nn!2n; converges for all x
19. y(x)=13∑n=0∞(2n+3)(x−1)2n+1; converges if 0<x<2
20. y(x)=2−6(x−3)2; converges for all x
21. y(x)=1+4(x+2)2; converges for all x
22. y(x)=2x+6
23. 2c2+c0=0; (n+1)(n+2)cn+2+cn+cn−1=0 for n≧1; y1(x)=1−x22−x36+⋯; y2(x)=x−x36−x412+⋯
24. y1(x)=1+x33+x55+x645+⋯; y2(x)=x+x33+x46+x55+⋯
25. c2=c3=0, (n+3)(n+4)cn+4+(n+1)cn+1+cn=0 for n≧0; y1(x)=1−x412+x7126+⋯; y2(x)=x−x412−x520+⋯
26. y(x)=c0(1−x630+x972+⋯)+c1(x−x742+x1090+⋯)
27. y(x)=1−x−x22+x33−x424+x530+29x6720+13x7630−143x840320+⋯; y(0.5)≈0.4156
28. y(x)=c0(1−x22+x36+⋯)+c1(x−x36+x412+⋯)
29. y1(x)=1−12x2+1720x6+⋯; y2(x)=x−16x3−160x5+⋯
30. y(x)=c0(1−x22+x36+⋯)+c1(x−x22+x418+⋯)
33. The following figure shows the interlaced zeros of the 4th and 5th Hermite polynomials.
34. The figure below results when we use n=40 terms in each summation. But with n=50 we get the same picture as Fig. 8.2.3 in the text.