Corrigé du DM 2
Exercice 1 1. (a) ker(f+idR3) =!
x2R3, f(x) =−x"
on résout 8<
:
3x1−3x2+ 3x3= 0
−3x1+ 4x2−4x3= 0
−3x1+ 4x2−4x3= 0 ,
&
3x1−3x2+ 3x3= 0
−3x1+ 4x2−4x3= 0 ,
&
x2−x3=x1
x2−x3= 3/4x1 ,
&
x2−x3= 0 x1= 0
donc ker(f+idR3) =vect(e2+e3) ker(f−2idR3) =!
x2R3, f(x) = 2x"
on résout 8<
:
−3x2+ 3x3= 0
−3x1+x2−4x3= 0
−3x1+ 4x2−7x3= 0
, x2=x3
x1=−x3
donc ker(f+idR3) =vect(−e1+e2+e3)
(b) On doit avoirf(u) =−uetf(w) = 2wdoncu=e2+e3etw=−e1+e2+e3conviennent. Posons v=xe1+ye2+ze3,on doit résoudref(v) =u−vdonc
8<
:
2x−3y+ 3z=−x
−3x+ 3y−4z= 1−y
−3x+ 4y−5z= 1−z ,
&
x−y+z= 0
−3x+ 4y−4z= 1 ,
&
y−z=x
y−z= 1/4 + 3/4x ,
&
y−z=x x= 1 On peut donc choisir par exemplex= 1, y= 1, z= 0.ATTENTION il faut vérifier que la famille(u, v, w)
est une base deR3:On a en effetdet(u, v, w) = '' '' ''
0 1 −1 1 1 1 1 0 1
'' ''
''=−16= 0
(c) Posons P = 0
@ 0 1 −1 1 1 1 1 0 1
1
A,on a doncA0=P−1AP
On peut d’ailleurs le vérifier : en effet
P−1= 0
@ 0 1 −1 1 1 1 1 0 1
1 A
−1
= 0
@ 1 −1 2
0 1 −1
−1 1 −1 1 A 0
@ 0 1 −1 1 1 1 1 0 1
1 A
−10
@ 2 −3 3
−3 3 −4
−3 4 −5 1 A
0
@ 0 1 −1 1 1 1 1 0 1
1 A=
0
@ −1 1 0 0 −1 0
0 0 2
1 A
(d)
, −1 1 0 −1
-n
= (−I2+J)navecJ = , 0 1
0 0 -
donc puisque ces deux matrices commutent et que J2= 0on a d’après la formule du binôme de Newton:
(−I2+J)n= Xn k=0
,n k -
(−1)kJk= (−1)nI2+ (−1)n−1J=
, (−1)n n(−1)n−1 0 (−1)n
-
On remarque que la matriceA0est constituée de deux blocs carrés diagonauxA0=
, −I2+J 0
0 2
- donc d’après les propriétés des produits blocs:
A0n=
, (−I2+J)n 0
0 2n
-
= 0
@ (−1)n n(−1)n−1 0 0 (−1)n 0
0 0 2n
1 A ( on peut le prouver aussi par récurrence. )
page 2 MP Dessaignes 2015
On peut en déduire que
An = (P A0P−1)n=P A0nP−1 An =
0
@ 0 1 −1 1 1 1 1 0 1
1 A
0
@ (−1)n n(−1)n−1 0 0 (−1)n 0
0 0 2n
1 A
0
@ 1 −1 2
0 1 −1
−1 1 −1 1 A
An = 0
@
2n (−1)n−2n 2n−(−1)n (−1)n−2n (−1)n−1n+ 2n (−1)n−(−1)n−1n−2n (−1)n−2n (−1)n−1n−(−1)n+ 2n 2 (−1)n−(−1)n−1n−2n
1 A 2.
(a) 8x2R,limn!1Pn k=0 1
k!xk=ex (b) On a
exp(A0) = lim
n!1
Xn k=0
1 k!
0
@ (−1)k k(−1)k−1 0 0 (−1)k 0
0 0 2k
1 A= lim
n!1
0
@ Pn
k=0 1
k!(−1)k Pn k=0 1
k!k(−1)k−1 0
0 Pn
k=0 1
k!(−1)k 0
0 0 Pn
k=0 1 k!2k
1 A
Or
nlim!1
Xn k=0
1
k!(−1)k=e−1, lim
n!1
Xn k=0
1
k!k(−1)k−1= lim
n!1
Xn k=1
1
(k−1)!(−1)k−1=e−1; lim
n!1
Xn k=0
1 k!2k=e2 donc
exp(A0) = 0
@ e−1 e−1 0 0 e−1 0
0 0 e2
1 A
On a de plus
expA= lim
n!1
Xn k=0
1
k!P A0nP−1= lim
n!1P( Xn k=0
1
k!A0n)P−1=P(expA0)P−1 donc
expA = 0
@ 0 1 −1
1 1 1
1 0 1
1 A
0
@ e−1 e−1 0 0 e−1 0
0 0 e2
1 A
0
@ 1 −1 2
0 1 −1
−1 1 −1 1 A
expA = 0
@ e2 e−1−e2 e2−e−1 e−1−e2 e−1+e2 −e2 e−1−e2 e2 e−1−e2
1 A
3. reprenons la formule deA0n: 0
@ (−1)n n(−1)n−1 0 0 (−1)n 0
0 0 2n
1
Asi l’on remplace brutalementnpar12on obtient
A012 = 0
@
(−1)12 12(−1)12−1 0 0 (−1)12 0
0 0 212
1 A
en posant(−1)12 =ion obtient
”A012” = 0
@ i −2i 0
0 i 0
0 0 p
2 1 A
page 3 MP Dessaignes 2015
On vérifie bien que
(”A012”)2= 0
@ i −2i 0
0 i 0
0 0 p
2 1 A
2
= 0
@ −1 1 0 0 −1 0
0 0 2
1 A=A0
On proposera donc pour matriceBtelle queB2=Ala matriceB=P A012P−1 en effet on peut le vérifier puisque
B2=P A012P−1P A012P−1=P A0P−1=A la valeur exacte deBest donnée par le calcul:
B =
0
@ 0 1 −1 1 1 1 1 0 1
1 A
0
@ i −2i 0
0 i 0
0 0 p
2 1 A
0
@ 1 −1 2
0 1 −1
−1 1 −1 1 A
B =
0
@
p2 i−p
2 p
2−i i−p
2 p
2−12i 32i−p 2 i−p
2 p
2−32i 52i−p 2
1 A
Exercice 2 1. (a) J= 0 BB BB BB
@
0 1 0 0
0 0 1
0 . .. ... ... 0 . .. 0 1
0 0 0
1 CC CC CC A
.On arg(J) =n−1.
D’autre part puisquef2(e1) = 0, f2(e2) = 0, f2(ei) =ei−2pouri2[[3, n]]on a doncJ2= 0 BB BB BB B@
0 0 1 0
0 0 0 . ..
0 . .. ... ... 1 . .. 0 0
0 0 0
1 CC CC CC CA .
De même on trouve pourfkavec16k6n−1quefk(ei) =fk−1(ei−1) =fk−2(ei−2)....=fk−p(ei−p) ainsi lorsquek≥ion trouvefk(ei) =fk−i+1(e1) = 0et pourk < i, fk(ei) =f(ei−k+1) =ei−k
fk(ei) = 0pour touti6ketfk(ei) =ei−kpour touti2[[k+ 1, n]].
Enfin
Jn−1= 0 BB BB BB B@
0 0 0 1
0 0 0 . .. 0 0 . .. ... ...
. .. 0 0
0 0 0
1 CC CC CC CA
etJn= 0
(b) la matriceJkest nulle à partir dek = net donc lorsque l’on applique la formule de la sérieexp(J) = limp!1(Pp
k=0 1
k!Jk)on obtient pour un entier naturelp>n( attention icinest une constante égale à la taille de la matriceJet c’estpqui varie)
Xp k=0
1 k!Jk=
nX−1 k=0
1 k!Jk+
Xp k=n
1 k!Jk=
nX−1 k=0
1 k!Jk exp(J) = lim
p!1( Xp k=0
1
k!Jk) = lim
p!1(
n−1
X
k=0
1 k!Jk) =
n−1
X
k=0
1 k!Jk
page 4 MP Dessaignes 2015
1t2`+B+2 R
ě /Mb mM Z/nZ BMiĕ;`2 Un T`2KB2`V- QM `ûbQmi H2b û[miBQMb 2M 6*hP_AaLh 2i 2M miBHBbMi [mǶmM T`Q/mBi /2 7+i2m`b 2bi MmH bbB HǶmM m KQBMb /2b 7+i2m`b 2bi MmH
ě /Mb mMZ/nZMQM BMiĕ;`2 UnMQM T`2KB2`V- QM T2mi miBHBb2` H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb UpQB` THmb #bV Qm #B2M i2bi2` iQmi2b H2b TQbbB#BHBiûbX .Mb +2 +b- TQm` `ûbQm/`2x4= 1QM T2mi M2 i2bi2` [m2 H KQBiBû /2b +Hbb2b +` ( x)4 =x4XXX G2 MQK#`2 /ǶQTû`iBQMb 2bi HBKBiû 2i QM MǶQm#HB2 Tb /2 `û/mB`2 KQ/mHQ R8 TQm` ûpBi2` H2b KmHiBTHB+iBQMb /2 i`QT ;`M/b MQK#`2bX S` 2t2KTH2- TQm` +H+mH2`84KQ/mHQ R8- QM +QKK2M+2 T` +H+mH2`82= 64 = 1 KQ/mHQ R8 TmBb84= (82)2= ( 1)2= 1KQ/mHQ R8XXX
RX RN 2bi T`2KB2`- /QM+Z/19Z2bi mM +Q`Tb /QM+ BMiĕ;`2X BMbB QM T2mi 7+iQ`Bb2` 2M( ˙x ˙1)·( ˙x2+ ˙x+ ˙1) = ˙0 2i }MH2K2Mi-x˙ = ˙1Qm #B2Mx˙2+ ˙x+ ˙1 = 0X PM +?2`+?2aet˙ b˙ i2Hb [m2a˙+ ˙b= 12ia˙b˙ = 1X PM i`Qmp2 ˙7 2i11X˙
6BMH2K2Mi- H2b bQHmiBQMb bQMi R-d- 2i RRX
kX S` +QMi`2-15MǶ2bi Tb T`2KB2`X S` H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb- bB 2bi HǶBbQKQ`T?BbK2 2Mi`2Z/3Z 2iZ/5Z- QM (1mod15) = (1mod3,1mod5)X
PM `ûbQmi HQ`bx43= 1mod3/MbZ/3Z2i x45= 1mod5 /MbZ/5ZX PM i`Qmp2x32{1mod3,2mod3}2i x52{1mod5,2mod5,3mod5,4mod5}X SQm` +?[m2 +QmTH2(x3, x5)- QM `2i`Qmp2 H +Hbb2 KQ/mHQ R8 +Q``2b@
TQM/Mi2 2M +QMbB/û`Mi 1(x3, x5)X PM i`Qmp2 1(1mod3,1mod5) = 1mod15- 1(1mod3,2mod5) = 7mod15- 1(1mod3,3mod5) = 13mod15- 1(1mod3,4mod5) = 4mod15- 1(2mod3,1mod5) = 11mod15-
1(2mod3,2mod5) = 2mod15- 1(2mod3,3mod5) = 8mod152i 1(2mod3,4mod5) = 14mod15X 6BMH2K2Mi- HǶ2Mb2K#H2 /2 bQHmiBQMb 2bi{˙1,˙2,˙4,˙7,˙8,11,˙ 13,˙ 14˙ }X
jX /27 `+BM2*m#B[m2U-MV,
`2bmHii4()
7Q` F BM `M;2UMV, O H /2`MBĐ`2 pH2m` i2biö2 2bi UM@RVXXX B7 UF jV $W M 44 $W M , O Qm #B2M B7 UF j @ V$W M 44 y,
`2bmHiiXTT2M/UFV
`2im`MU`2bmHiiV
9X *Ƕ2bi /m +Qm`b ,x2i nbQMi T`2KB2`b 2Mi`2 2mtsietseulementsi x 2bi mM ûHûK2Mi BMp2`bB#H2 /2Z/nZbB 2i b2mH2K2Mi bBx2(Z/nZ)⇤X
HQ`bxCard(Z/nZ)⇤ = 12iCard(Z/nZ)⇤='(n)X
8X x2bi ;ûMû`i2m` /2G sietseulementsiH2 bQmb ;`QmT2 2M;2M/`û T`x2bi :sietseulementsiiQmi ûHûK2Mi /2G2bi mM2 TmBbbM+2 U2MiBĕ`2 `2HiBp2V /2 x sietseulementsi G={xk|k2Z}X
G2 MQK#`2 /2 ;ûMû`i2m`b /2U20162bi û;H ¨ '(2016)X G¨ 2M+Q`2- Tb #2bQBM /2 +H+mH2ii2 , 2016 = 2⇥1008 = 22⇥504 = 23⇥252 = 24⇥126 = 25⇥63 = 25⇥32⇥7X
_TT2HQMb 2MbmBi2 [m2'(pk) =pk pk 1TQm`pT`2KB2` 2i [m2 bBn2imbQMi T`2KB2`b 2Mi`2 2mt- /ǶT`ĕb H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb-'(nm) ='(n)⇥'(m)X
6BMH2K2Mi-'(2016) = (25 24)⇥(32 3)⇥(71 70) = 16⇥6⇥6 = 576
S`Q#HĕK2 R
S`iB2 R
RX
kX _a bm7 [mǶBH 2bi BMmiBH2 /2 bmTTQb2` [m2 H2 #mi 2bi mM 2bT+2 p2+iQ`B2H /2 /BK2MbBQM }MB2X a2mH2 H bQm`+2 /QBi HǶāi`2X
X G T`2mp2 b2 7Bi /B`2+i2K2Mi aLa `û+m``2M+2X .ǶBHH2m`b- +2mt [mB QMi +`m 7B`2 mM2 `û+m``2M+2
`2K`[m2`QMi [mǶBHb MǶQMi Tb miBHBbû H2m` ?vTQi?ĕb2 /2 `û+m``2M+2 /Mb HǶ?û`û/BiûXXX
aQBix2NkX HQ`bfk(x) = 0E 2i 2M +QKTQbMi p2+f HBMûB`2-fk+1(x) = 0E- /QM+x2Nk+1X aQBi 2MbmBi2 y 2 Ik+1X AH 2tBbi2 HQ`b x 2 E i2H [m2 fk+1(x) = yX SQbQMb z = f(x)X HQ`b T`
+QMbi`m+iBQM-fk(f(x)) =yX .QM+y2IkX
#X PM pQBi bb2x bQmp2Mi H }M /2 H T`2mp2 , A ={p|Np+1 =Np} 2bi mM2 T`iB2 MQM pB/2 /2N /QM+
/K2i mM THmb T2iBi ûHûK2MiX
S` +QMi`2- BH 2bi ``2 /2 pQB` mM2 T`2mp2 +Q``2+i2 /2A6=;X
PM T2mi TQm` +2H `2;`/2` H2b /BK2MbBQMb , H bmBi2(/BK(Nk))2bi mM2 bmBi2 +`QBbbMi2 Um b2Mb H`;2V /ǶT`ĕb H [m2biBQM T`û+û/2Mi2X *Ƕ2bi mbbB mM2 bmBi2 /Ƕ2MiB2`b2i KDQ`û2 T` /BK(E)+`Nk ⇢EX
R
*2ii2 bmBi2 2bi /QM+ ahhAPLLA_1 5 .QM+ BH 2tBbi2p2Ni2H [m2 /BK(Np) =/BK(Np+1)X *QKK2 /2 THmbNp⇢Np+1-Np=Np+1 2iAMǶ2bi /QM+ Tb pB/2XXX
+X G `ûTQMb2 2bi MQM 5 S` 2t2KTH2 HǶ2M/QKQ`T?BbK2 /2 H T`iB2 j 2bi mM +QMi`2 2t2KTH2- KBb HǶ2M/QKQ`T?BbK2 /2R[X] f :P 7!P0 +QMpB2Mi û;H2K2Mi , TQm` H2b /2mt-Nk =Rk[X]2bi mM2 bmBi2 bi`B+i2K2Mi +`QBbbMi2 /2 MQvmt 5
/X GǶBM+HmbBQMNp0 ⇢Np0+k pB2Mi /2 H +`QBbbM+2 /2 H bmBi2 /2b MQvmtX
_û+BT`Q[m2K2Mi- KQMi`QMb [m2 bB Np0 = Np0+1- HQ`b Np0+1 = Np0+2X aQBi x 2 Np0+2X HQ`b fp0+1(f(x)) = 02i f(x)2Np0+1=Np0X .QM+fp0(x) = 0 2ifp0+1(x) = 0X 6BMH2K2Mi x2Np0+1X S` `û+m``2M+2- QM KQMi`2 HQ`b [m2Np0 =Np0+k TQm` iQmikX
jX X GǶûMQM+û 2bi bQmp2Mi KH +QKT`Bb , BH 7mi KQMi`2` B+B [m2 (Ik) 2bi biiBQMMB`2 2i [m2 H2 `M;
+Q``2bTQM/Mi 2bi H2 KāK2 [m2 TQm` H bmBi2 (Nk)X
PM bBi [m2 H bmBi2 (Ik) 2bi +`QBbbMi2X AH 2M 2bi /2 KāK2 TQm` H bmBi2 /2b /BK2MbBQMb (/BK(Ik))X
*2ii2 bmBi2 2bi KDQ`û2 T` /BK(E)2i ¨ pH2m`b 2MiBĕ`2b- /QM+ biiBQMM2 ¨ T`iB` /ǶmM `M; MQiûq0X S` i?ûQ`ĕK2 /m `M; TTHB[mû ¨ fk U[mB 2bi mM 2M/QKQ`T?BbK2 T` +QKTQbû2 /Ƕ2M/QKQ`T?BbK2bV /BK(Nk) +/BK(Ik) =/BK(E)X
.QM+ /BK(Ik) =/BK(E) /BK(Nk)2i H bmBi2(/BK(Ik))biiBQMM2 2t+i2K2Mi m KāK2 `M; [m2 H bmBi2 /BK(Nk)X
*QKK2Ik⇢Ik+1- H bmBi2(Ik)biiBQMM2 mbbB 2t+i2K2Mi ¨ T`iB` /2 +2 `M; 5
#X S` mM `;mK2Mi /2 /BK2MbBQM Ui?ûQ`ĕK2 /m `M; 2M+Q`2V- BH bm{i /2 KQMi`2` [m2 H bQKK2 2bi /B`2+i2- /QM+ [m2 HǶBMi2`b2+iBQM 2bi `û/mBi2 ¨{0}X
aQBix2Np0\Ip0X HQ`bfp0(x) = 02ix=fp0(y)X .QM+fp0(fp0(y)) =f2p0(y) = 0X .QM+y2N2p0 = Np0X 6BMH2K2Mi-x=fp0(y) = 0X
9X Ip0 2bi /2 /BK2MbBQM }MB2X G #BD2+iBpBiû /ǶmM 2M/QKQ`T?BbK2 2bi /QM+ û[mBpH2Mi2 ¨ b bm`D2+iBpBiû Qm
#B2M bQM BMD2+iBpBiûX A+B- H bm`D2+iBpBiû 2bi THmb bBKTH2 +` T` ?vTQi?ĕb2- Ip0+1 = Ip0- +2 [mB bB;MB}2 2t+i2K2Mif(Ip0)) =Ip0XXX
8X JQMi`QMb [m2 H `2bi`B+iBQM /2 f ¨ Np0 2bi mM 2M/QKQ`T?BbK2 , bQBi x2 Np0 HQ`b fp0(x) = 0X .QM+
fx2Np0+1=Np0X
G `2bi`B+iBQM /2f ¨Np0 2bi MBHTQi2Mi2 /ǶQ`/`2 BM7û`B2m` Qm û;H ¨p0+` TQm` iQmix2Np0-fp0(x) = 0X .2 THmb- HǶQ`/`2 2bi bmTû`B2m` ¨p0+` T` biimi /2p0- BH 2tBbi2x2Ei2H [m2fp0 1(x)6= 02ifp0(x) = 0X
*2 x 2bi mM ûHûK2Mi /2 Np0 i2H [m2fp0 1(x) 6= 0 /QM+ H `2bi`B+iBQM /2 f ¨ Np0 MǶ2bi Tb MBHTQi2Mi2 /ǶQ`/`2 BM7û`B2m` ¨p0 1XXX
eX .ûKQMi`2` [m2 H bmBi2(/BK(Nk+1) /BK(Nk))k2N 2bi /û+`QBbbMi2X
*Ƕ2bi H [m2biBQM /B{+BH2 /2 HǶ2t2`+B+2X AH bǶ;Bi /2 KQMi`2` [m2 HǶû+`i 2Mi`2 /2mt /BK2MbBQMb bm++2bbBp2b /2 MQvmt Biû`ûb p 2M /BKBMmMiX _2K`[mQMb T` H2 i?ûQ`ĕK2 /m `M; [m2 +2i û+`i 2bi H2 KāK2 [m2 +2HmB 2Mi`2 /2mt /BK2MbBQMb /ǶBK;2b Biû`û2b bm++2bbBp2b , /BK(Nk+1) /BK(Nk) =/BK(Ik) /BK(Ik+1)X _2;`/QMb HQ`b +QKK2Mi +2 +QKTQ`i2 +2i û+`iek =/BK(Nk+1) /BK(Nk)X
AMimBiBp2K2Mi- /2Ik+1=f(Ik)⇢Ik- QM T2`/ek /BK2MbBQMbX *Ƕ2bi ¨ /B`2 [mǶBH 2tBbi2 mM 2bT+2 p2+iQ`B2Hk i2H [m2Ik Fk=Ik+1 p2+dim(Fk) =ekX Zm2 b2 Tbb2@i@BH /QM+ 2Mi`2Ik+12i Ik+2\ PM Ik+1=f(Ik) 2iIk+2=f(Ik+1)2i 2M T`iMi /2Ik Fk=Ik+1- QM Q#iB2Mi 2M +QKTQbMi T`f ,Ik+1+f(Fk) =Ik+2X ii2MiBQM , H bQKK2 MǶ2bi THmb 7Q`+ûK2Mi /B`2+i2 /Mb H /2`MBĕ`2 û;HBiûX .QM+ T` H 7Q`KmH2 /2 :`bbKM , ek+1 = dim(Ik+2) dim(Ik+1) dim(f(Fk)) dim(Fk) = ek +` f 2bi mM2 TTHB+iBQM HBMûB`2 U/QM+ [mB #Bbb2 H2b /BK2MbBQMbVX
G bmBi2(ek)2bi /QM+ #B2M /û+`QBbbMi2X
dX aBg2bi mM 2M/QKQ`T?BbK2 MBHTQi2Mi- H bmBi2(Nk)/2b MQvmt Biû`ûb /2g2bi mM2 bmBi2 ah_A*h1J1Lh +`QBbbMi2 Dmb[mǶ¨ mM `M;p0 ¨ T`iB` /m[m2H 2HH2 2bi biiBQMMB`2 ¨Np0 =E +`g2bi MBHTQi2MiX .QM+ H bmBi2 /2b /BK2MbBQMb(/BK(Nk))2bi mbbB mM2 bmBi2 /Ƕ2MiB2`b bi`B+i2K2Mi +`QBbbMi2 /2 y ¨ /BK(E) TmBb biiBQMMB`2 ¨ T`iB` /m `M;p0X .QM+p0/BKE 2igp0 =g/BKE= 0L(E)X
GǶQ`/`2 /2 MBHTQi2M+2 /2g 2bi /QM+ BM7û`B2m` Qm û;H ¨ /BK(E)X
S`iB2 j , mM mi`2 2t2KTH2
aQBiE=R[X]2i P2EX PM /û}MBi (P) =P(X+ 1) P(X)X
RX 2bi HBMûB`2 UiQmi H2 KQM/2 HǶ KQMi`ûX *Ƕ2bi mM 2M/QKQ`T?BbK2 +` (P)2bi mM2 bQKK2 /2 +QKTQbû2 /2 TQHvMƬK2b- /QM+ mM TQHvMƬK2X
AMmiBH2 /2 T`H2` /2 /2;`û /Mb +2ii2 [m2biBQM 5 k
kX F2` = {P 2 K[X]|P(X + 1) P(X) = 0}X lM i2H TQHvMƬK2 2bi /QM+ R@Tû`BQ/B[m2 2i TQm` mM i2H TQHvMƬK2- Q=P P(0) 2bi mM TQHvMƬK2 /K2iiMi mM2 BM}MBiû /2 `+BM2b Um KQBMb iQmb H2b 2MiB2`bV- /QM+ MmH 5U+2i `;mK2Mi `2pB2Mi +QMbiKK2Mi /Mb H2b 2t2`+B+2b 555V
6BMH2K2Mi-P =P(0) 2bi mM TQHvMƬK2 +QMbiMiX
_û+BT`Q[m2K2Mi- iQmi TQHvMƬK2 +QMbiMi +QMpB2Mi /QM+ F2`( ) =R0[X]X
SHmb /B{+BH2 , F2`( k) = Rk 1[X]X SQm` Pk = Xk- QM KQMi`2 T` `û+m``2M+2 [m2 /2;( (Xk)) = k 1X .QM+ T` HBMû`Biû /2 - 2(P) = 0sietseulementsideg(P) 1 2i T` `û+m``2M+2- k(P) = 0sietseulementsideg(P)k 1X 6BMH2K2Mi- F2`( k) =Rk 1[X]X
GǶ2M/QKQ`T?BbK2 MǶ2bi Tb MBHTQi2Mi +` TQm` iQmin2N- F2`( k)6=R[X]XXX
jX m +b Qɍ QM MǶm`Bi Tb 2m HǶB/û2 /Mb H [m2biBQM T`û+û/2Mi2- bBPk =Xk- HQ`b deg( (P)) =k 1X hQmDQm`b T` HBMû`Biû- (Rn[X])⇢Rn 1[X]X
X S` i?ûQ`ĕK2 /m `M;-dim(Im( )) = (n+ 1) dim(ker( )) =n kX P` k(Rn[X])⇢Rn k[X]X S`
BM+HmbBQM /BK2MbBQM- k(Rn[X]) =Rn k[X]X
HQ`bIm( ) =R[X]+` TQm` iQmi TQHvMƬK2P /2 /2;`ûn- BH 2tBbi2 iQmDQm`b m KQBMb mM TQHvMƬK2 Q/2 /2;`ûn+ 1 i2H [m2 (Q) =PX
#X .ǶT`ĕb +2 [mB T`û+ĕ/2- n+1 = 0X .QM+ HǶQ`/`2 /2 MBHTQi2M+2 /2 2bi BM7û`B2m` Qm û;H ¨n+ 1X .2 THmb- n(Xn)6= 0X .QM+ HǶQ`/`2 /2 MBHTQi2M+2 2bi û;H ¨n+ 1X
9X X ii2MiBQM , +Q[mBHH2b /Mb H2 bmD2i , HB`2 ǴJQMi`2` [m2 TQm` iQmi 2MiB2` Mim`2Hp- p(S) = Xp
k=0
( 1)p k
✓T k
◆ P(X+ k).ǴX *2mt [mB QMi #Q`/û H [m2biBQM QMi ;ûMû`H2K2Mi +Q``B;û
S` `û+m``2M+2 bm` pX C2 M2 7Bb [m2 HǶ?û`û/Biû ,
p+1(S) = ( p(S)) = Xp
k=0
( 1)p k
✓p k
◆
P(X+k)
!
=
= Xp
k=0
( 1)p k
✓p k
◆
P(X+k+ 1) Xp
k=0
( 1)p k
✓p k
◆
P(X+k)
= Xp
k=0
( 1)p k
✓p k
◆
P(X+k+ 1) + Xp
k=0
( 1)p k+1
✓p k
◆
P(X+k)
=
p+1X
k0=1
( 1)p k0+1
✓ p k0 1
◆
P(X+k0) + Xp
k=0
( 1)p k+1
✓p k
◆
P(X+k)
= ( 1)p+1
✓p 0
◆
P(X) + Xp
k=1
( 1)p k+1P(X+k)
✓ p k 1
◆ +
✓p k
◆ !
+ ( 1)0
✓p p
◆
P(X+p+ 1)
= ( 1)p+1
✓p 0
◆
P(X) + Xp
k=1
( 1)p k+1P(X+k)
✓p+ 1 k
◆!
+ ( 1)0
✓p+ 1 p+ 1
◆
P(X+p+ 1)
=
p+1X
k=0
( 1)p+1 k
✓p+ 1 k
◆
P(X+k)
_û+m``2M+2 ûi#HB2 UiQmi2 `2bb2K#HM+2 p2+ H2 #BMƬK2 /2 L2riQM M2 b2`Bi 7Q`imBi2XXXV
#X PM bBi [m2 n+1= 0X .QM+ TQm` iQmiP- n+1(P) = 0X
.Mb +2ii2 7Q`KmH2- 2M BbQHMi H2 i2`K2k= 0- QM Q#iB2Mi , ( 1)n+1P(X) =Pn+1
k=1( 1)n+1 k n+1k P(X+ k)X
PM TQb2 k= ( 1)k+1 n+1k XXX
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