→ ybite:p(x) = 5為一多項式,且其微分過後為0 02/11 16:01
→ ybite:我另有想法:如果以{1,x,x^2,x^3}為Basis,微分應該可表為 02/11 16:02
→ ybite:一個矩陣, [ 0 1 0 0 ] 02/11 16:03
→ ybite: [ 0 0 2 0 ] 02/11 16:03
→ ybite: [ 0 0 0 3 ],其nullity=1,因此ker(D)!=0(歪了XD 02/11 16:03
→ ybite: [ 0 0 0 0 ] 02/11 16:04
推 BenLinus:同意樓上! 02/11 16:44
→ peropero1:由定義:ker(D)={x|D(x)=0,x屬於P3}, 02/11 17:35
→ peropero1:所以x需為常數才成立 所以ker(D)!=0 02/11 17:35
→ SkullMaster:我了解了,謝謝各位板指教:) 02/11 17:43