A.11.2: Section 11.2 Answers
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- Jun 8, 2024
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( \newcommand{\kernel}{\mathrm{null}\,}\)
2. F(x)=2+2π∑∞n=1(−1)nnsinnπx;F(x)={2,x=−1,2−x,−1<x<1,2,x=1
3. F(x)=−π2−12∑∞n=1(−1)nn2cosnx−4∑∞n=1(−1)nnsinnx;F(x)={−3π2,x=−π,2x−3x2,−π<x<π,3π2,x=π
4. F(x)=−12π2∑∞n=1(−1)ncosnπxn2;F(x)=1−3x2,−1≤x≤1
5. F(x)=2π−4π∑∞n=114n2−1cos2nx;F(x)=|sinx|,−π≤x≤π
6. F(x)=−12sinx+2∑∞n=2(−1)nnn2−1sinnx;F(x)=xcosx,−π≤x≤π
7. F(x)=−2π+π2cosx−4π∑∞n=14n2+1(4n2−1)2cos2nx;F(x)=|x|cosx,−π≤x≤π
8. F(x)=1−12cosx−2∑∞n=2(−1)nn2−1cosnx;F(x)=xsinx,−π≤x≤π
9. F(x)=π2sinx−16π∑∞n=1n(4n2−1)2sin2nx;F(x)=|x|sinx,−π≤x≤π
10. F(x)=1π+12cosπx−2π∑∞n=1(−1)n4n2−1cos2nπx;F(x)=f(x),−1≤x≤1
11. F(x)=14πsinπx−8π2∑∞n=1(−1)nn(4n2−1)2sin2nπx;−14π∑∞n=1(−1)nn(n+1)sin(2n+1)πxF(x)=f(x),−1≤x≤1
12. F(x)=12sinπx−4π∑∞n=1(−1)nn4n2−1sin2nπx;F(x)={0,−1≤x<12,−12,x=−12,sinπx,−12<x<12,12,x=12,0,12<x≤1
13. F(x)=1π+1πcosπx−2π∑∞n=21n2−1(1−nsinnπ2)cosnπx;F(x)={0,−1≤x<12,12,x=−1,|sinπx|,−12<x<12,12,x=1,0,12<x≤1
14. F(x)=1π2+14πcosπx+2π2∑∞n=1(−1)n4n2+1(4n2−1)2cos2nπx+14π∑∞n=1(−1)n2n+1n(n+1)cos(2n+1)πx;F(x)={0,−1≤x<12,14,x=−12,xsinπx,−12<x<12,14,x=12,0,12<x≤1,
15. F(x)=1−8π2∑∞n=11(2n+1)2cos(2n+1)πx4−4π∑∞n=1(−1)nnsinnπx4;F(x)={2,x=−4,0,−4<x<0,x,0≤x<4,2,x=4
16. F(x)=12+1π∑∞n=11nsin2nπx+8π3∑∞n=01(2n+1)3sin(2n+1)πx;F(x)={12,x=−1,x2,−1<x<0,12,x=0,1−x2,0<x<1,12,x=1
17. F(x)=34+1π∑∞n=11nsinnπ2cosnπx2+3π∑∞n=11n(cosnπ−cosnπ2)sinnπx2
18. F(x)=52+3π∑∞n=11nsin2nπ3cosnπx3+1π∑∞n=11n(cosnπ−cos2nπ3)sinnπx3
20. F(x)=sinhππ(1+2∑∞n=1(−1)nn2+1cosnx−2∑∞n=1(−1)nnn2+1sinhnx)
21. F(x)=−πcosx−12sinx+2∑∞n=2(−1)nnn2−1sinnx
22. F(x)=1−12cosx−πsinx−2∑∞n=2(−1)nn2−1cosnx
23. F(x)=−2sinkππ∑∞n=1(−1)nnn2−k2sinnx
24. F(x)=sinkππ[1k−2k∑∞n=1(−1)nn2−k2cosnx]
