Oral CCP 2015

pZ/p2Z
n2n2+ 13n+ 20
P=
0−1−1
−1 0 −1
1 1 2
R3P
A∈ M3,2(R)B∈ M2,3(R)P=AB BA =I2
E n f ∈ L(E)fn−1̸= 0 fn= 0
aB=a, f(a), f2(a), . . . , fn−1(a)E
f
g E f P
g=P(f)g(a)B
F G E
f∈ L(E) Ker f=FIm f=G
E n f ∈ L(E) Φ L(E)−→ L(E)g7−→ g◦f
Φf
L1L2L(E)E n
∀(u, v)∈L1×L2u◦v+v◦u= 0
p1∈L1p2∈L2Id = p1+p2n=
rg(p1) + rg(p2)
u∈L1x∈Im p2u(x) = 0 x∈Ker p2u(x)∈Ker p2
dim(L1)6n−rg(p2)2dim(L2)
rg(p1) = 0 rg(p1) = n L1={0}L2={0}
A B Mn(C)A B3= 0 AB =BA
A+B
M=A B
C D∈ M2n(C)A∈ Mn(C)
rg M=n D =CA−1B M
A= (aij )∈ Mn(C) (i, j)aij a i =j b i < j c i > j
b̸=c J Mn(C)
x7−→ det(A+xJ)xdet A
det A b =c
ERf∈ L(E)\ {0}f3+f= 0
x∈E x =y+z y ∈Ker f z ∈Ker(f2+ IdE)y=x+f(x)
z=−f2(x)
E= Ker f⊕Ker(f2+ IdE)
∈Ker(f2+ IdE)\ {0}x, f(x)Ker(f2+ IdE)
det(−IdE) dim Ker(f2+ IdE)=2
BE f 
0 0 0
0 0 −1
0 1 0
▹ ◃

R
φC[X]φ(P)(X) = XP (X)
C∞(R,R)
A= (aij )∈ Mn(C)an,1=−1a1,n = 1
rg A A
Ann∈N
E n >2φ E
a∈E\ {0}f E −→ E x 7−→ x+φ(x)a
f E 1f
Ker(f−IdE)
f
f P ∈R3[X]
X2P X4−1
fR3[X]
(x−a)(x−b)y′−nxy =ky a b k ∈R
fCn[X]f(P)(X) = (X−a)(X−b)P′(x)−nXP (X)f
Cn[X]
fdet f
HCE f ∈ L(E)
f(H)⊂H λ ∈CIm(f−λIdE)⊂H
f∈ L(C3)
3 1 2
1 1 0
−112
u E u3=u u
E1, . . . , Epu F E u
Ei
A∈ Mn(R)A3=A+InAMn(C)
X3−X−1
det A > 0
A∈ Mn(R)A2+t A=In
A
A A −InA
n= 3 tr A̸= 0
A B C Mn(R)C=A+B C2= 2A+ 3B C3= 5A+ 6B
C A B
A∈ Mn(C)P∈C[X]P(A)
A
f E P
f P (0) = 0 P′(0) ̸= 0 E= Ker f⊕Im f
M∈ Mn(C)M M2
M
EKn f ∈ L(E)n
g∈ L(E)g◦f=f◦g
f g f g
P∈Kn−1[X]g=P(f)
▹ ◃

A∈ M3(R)A2=
1 0 0
1 2 0
1 2 3
E f ∈ L(E) rg f= 1 f
tr f̸= 0
A B Mn(C)
A B X
Y U =XtY AU =UB U ̸= 0
U∈ Mn(C)AU =UB
AkU=UBkk∈NA B
n+ 1 a0, . . . , anE=Rn[X]
(P|Q) = n
k=0 P(ak)Q(ak)
F=P∈E|n
k=0 P(ak) = 0XnF
u E
v= IdE−uKer v= (Im v)⊥
x∈E n ∈Nfn(x) = 1
n+ 1
n
k=0
uk(x)fn(x)
xKer v
E n u ∈ L(E)
n
k=1ek|u(ek)(e1, . . . , en)
n
j=1 n k=1ej|u(fk)2(e1, . . . , en)
(f1, . . . , fn)
a b E
u u(x) = (a|x)a+ (b|x)b E
Ker u u
A∈ Mn(R) (tr A)26(rg A) tr(A2)
A∈ Mn(R) Φ M∈ Mn(R)7−→ AM −MA
(X1, . . . , Xn)Mn,1(R)
A
(i, j)∈[[1, n]]2Mij =XitXjMij
Φ
λ1, . . . , λpA q1, . . . , qp
rg Φ qi
A∈ Mn(R)t A=−A
XtXAX = 0
B∈ Mn(R)A+B
A∈ Mn(R)
t
AA A = 0
u E
∥u(x)∥6∥x∥x∈E
x∈Ker(u−IdE)∩Im(u−IdE)y u(y)−y=x k ∈N∗uk(y)
x y
E= Ker(u−IdE)⊕Im(u−IdE)
E∥ ∥∞F
F E
▹ ◃

n∈Nx−e−x=n
Run
n∈Nun∈[n, n + 1] (un)
un−n
n∈N∗x+x2+···+xn= 1
R+xn
(xn)ℓ
un−ℓ
un=
n
k=1
1
n+kn+∞
vn=
n
k=1
ln1 + 1
n+kn+∞
>1
1
n(n+ 1)
(a, b)∈R2n∈N∗un= ln n+aln(n+ 2) + bln(n+ 3) (a, b)
n>1un
un= ln2n+ (−1)n−ln(2n)
+∞
k=1
(−1)n
n=−ln 2
+∞
k=11
2k−1−1 2k+∞
k=11
2k+ 1 −1 2k
(a, b, c)∈R31
4X3−X=a X+b 2X−1+c 2X+ 1
+∞
k=1
1
4k3−k
+∞
1
dt
4t3−t
cosπn2lnn+ 1
n
a∈0,π
2b∈R∗
+tana+b
nn
a b
g[0,1] RH x ∈[0,1] 7−→ 1
0|x−t|g(t)dt
H C2[0,1] H′′
(a, b)∈R2f x 7−→ H(x) + ax +b f′′ = 2g
f(0) = f(1) = 0
a∈RE C1[0,1] Rf(0) = 0
f(1) = ainf1
0f′(t)2dt;f∈Ef∈E f(x) = x 0f′(t)dt
fR+−→ Ca b 0< a < b
lim
x→0+b
1
f(t)
tdt ℓ = lim
x→0+bx
ax
f(t)
tdt
+∞
1
f(t)
tdt lim
x→0++∞
x
f(at)−f(bt)
tdt =ℓ
1
0
t−1
ln tdt = ln 2
▹ ◃

I=+∞
0
sin5x
x2dx
∀x∈Rsin5x=sin(5x)−5 sin(3x) + 10 sin x
16
∀A > 0+∞
A
sin5x
x2dx =1 1610 5A
A
sin x
x2dx −15 5A
3A
sin x
x2dx
I
n∈NIn=1
0(1 −x)ne−2xdx
(In)
n∈NInIn+1 (nIn)
(a, b, c)∈R3In=a+b
n+c n2+o1 n2+∞
n∈N∗In=+∞
0
ln(1 + x/n)
x(1 + x2)dx
(In) (nIn)n+∞
shx= 1
n∈NIn=ln(1+√2)
0(sht)ndt (In)
n>2nIn+ (n−1)In−2=√2
In
n∈NBn=+∞
1
dt
1 + t+t2+··· +tn
n Bn
(Bn)
(−1)nBnBn
un= (−1)nπ/2
0cosnt dt
f x 7−→ 1
0
tx
t+ 1 dt
D f
x∈D f(x) + f(x+ 1) f
g x 7−→ +∞
0
e−xt
t+ 1 dt R∗
+
g
g+∞+∞
0
e−t2dt =√π
2x > 0F(x) = +∞
0
1−e−xt2
t2dt
f∈ C∞(R,R)g(x) = f(x)−f(0)
xx̸= 0 g(0) = f′(0)
∀x∈Rg(x) = 1
0f′(xt)dt g ∈ C∞(R,R)
n∈Nt∈[0, π/2] fn(t) = cosntsin t
n>0fn[0, π/2]
(fn)
f x 7−→ +∞
n=0 ln(1 + e−nx)D f
f D D
▹ ◃