for R: f(x)=a0+∑j=1∞(ajcos(jx)+bjsin(jx)): a0=2π1∫−ππf(x)dx; aj=π1∫−ππf(x)cos(jx)dx; bj=π1∫−ππf(x)sin(jx)dx
for C: fn:=⟨f,un⟩=2π1∫−ππf(x)e−inxdx; a0=2π1f0; an=2π1(fn+f−n); bn=2π−i(fn−f−n)
∣∣f−∑j=−nnfjuj∣∣∞≤πn1∣∣f′∣∣2 .33 make Bsp. .34
10
10.1 Normed VS
Normed spaces examples
(C(Iˉ,R),∣∣⋅∣∣p), ∣∣f∣∣p:=(∫I∣f∣p)p1 is normed space
The space of sequences (lp,∣∣⋅∣∣p) s.t. ∣∣⋅∣∣p:x↦(∑j=1∞∣ξ∣p)p1 & infinite vectors in C, lp:={x=(ξn)n∈N∣∀j∈Nξj∈C,∑j=1∞∣ξj∣p<∞}, bounded seq. in C, similarly for p→∞
∇f(x0)⋅h0 for ∣h0∣ is the Richtungsableitung von f in Richtung h0 an der Stelle x0 .12
To Compute: Dx(h)=limt→0tf(x+th)−f(x), f diff in x, ∀h∈RnDx(h)=⟨∇f(x),h⟩
Höhelinien: Nf(c):={x∈U∣f(x)=c} clear, U is the domain of f:R2→R
Falllinien: for β:R→R2 s.t. β′(t)=∇f(β(t))uhm
divf:Rn→R,x↦divf(x):=∑i=1n∂ifi(x); rotf:R3→R3,x↦(∂2f3(x)−∂3f2(x),∂3f1(x)−∂1f3(x),∂1f2(x)−∂2f1(x))T; Δ:C2(Rn,R)→C(Rn,R),f↦∑i=1n∂i2fi, Δ(f)=div(∇(f)), called Laplace-Operator .14
f:U(⊂Rn)→Rn inj. diff in a s.t. detf′(a)=0 (so that f′(a) is an inv. matrix), for b:=f(a) for f(a) is an inner point of f(U) (or you can’t derivate) def. g:=f−1:f(U)→Rn. then g′(b)=(f′(a))−1, see those as matrices. Maybe it is to say when ⋅−1 and ⋅′ commute .16
Actually you need only detf′(a)=0 .17
Q! dom(g) wtf? .18
f∈C1(U,Rn), inv., f(U) open, f−1∈C1 then f is a Diffeomorphismus on U. .19
11.2 Mittelwertsatz
for a,b∈Rn,t∈(0,1): γab(t):=a+t(b−a); Γab:={x∈Rn∣∃t∈(0,1)x=γab(t)}
Mittelwertsatz: ∃c∈Γabf(b)=f(a)+f′(c)(b−a), Γab is a the Gerade from a to b, for Im(f)⊆R
It’s ok γ:[0,1]→U s.t. γ(0)=a∧γ(1)=b with γ diff. so f(b)=f(a)+f′(γ(θ))γ′(θ), θ∈(0,1)
for f:U→Rn, f′=0⇔f=k for k∈R
∂1f,...,∂nf cont. around a⇒f diff. .23, comp both lim. ∂if(a), for Im(f)⊂Rn
11.3 Höhere Ableitungen
f′′′:U→L(Rn,L(Rn,L(Rn,Rm))) or f′′:U×Rn×Rn→Rm,(x,h,k)↦f′′(x,h,k)=(f′′(x)h)k
In general f(p):U×(Rn)p→Rm,(x,h1,...,hp)↦f(p)(x,h1,...,hp)
also remember: ∂j∂if(x)=f′′(x,ei,ej) .24
Schwarz: if f∈C2(Rn,Rm)∧U open ⇒∀i,j∈{1,...,n}∂i∂jf(x)=∂j∂if(x) .25
check .26
Taylor: f(x+h)=f(x)+f(1)(x,h)+...+p!1f(p)(x,h,...,h)+Rp(x,h), .27 where Rp(x,h)=(p+1)!1f(p+1)(x+θh,h,...,h) for θ∈(0,1); Im(f)⊂R and f(n)=f(x+h)=∑∣α∣=0,α∈N0npα!1∂αf(x)hα+Rp(x,h) where: hα:=h1α1⋅...⋅hnαn same for the others
e.g. f(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)+21fxx(a,b)(x−a)2+21fyy(a,b)(y−b)2+fxy(a,b)(x−a)(y−b)+R2(x,h) for R2(x,y)=61f′′′(x+θh,h,h,h) for θ∈(0,1).
Im(f)⊂R: a critical point of f⇔∇f(a)=0⇔∀h∈Rnf′(a,h)=0
If a extremstelle of f⇒a critical point of f .29
for f∈C2(Rn,R) then Hessf(x):=(∂i∂jf(x))i,j∈{1,...,n} .31, .32, .34
pd(Hessf(a))⇒a minimum; nd(Hessf(a))⇒a maximum; Hessf(a) indef. ⇒a not an extr.
M convex if the line between any two m1,m2 is in M .35
im(f)⊂R op, conv: f conv. ⇔∀x1,x2,x1=x2∀t∈(0,1)f(x1+t(x2−x1))≤f(x1)+t(f(x2)−f(x1))
if only < it is streng convec, also f conv. ⇒f cont. .37
f convex ⇔∀x1,x2∈Uf(x1+t(x2−x1))≤f(x1)+t(f(x2)−f(x1)), with > for streng convex .36
.38 if f diff.
if pd(Hessf)⇒f streng conv.; spd(Hessf)⇒f conv.
11.4 Extrema mit Nebenbedingungen
ext: Berechnung der Extrema von f:Rn→R (n=2 in the Klausur)
∇f(x)=!0 and get the points, Pi
Compute Hessf(x) and insert Pi
pd(Hessf(Pi))⇒Pi is a min.
nd(Hessf(Pi))⇒Pi is a max.
if Hessf(Pi) is indef: Pi is Sattelpunkt
else: we can’t say (for spd and snd)
im(f)⊂R, else there can’t be an exercise but to compute the Jacobi.
Checks:
Check if ∧i=1n(g1=0,...,gm=0) (for m<n) holds for the extrema of f.
Check that M={x∈Rn∣g(x)=0} is compact (necess. to use this method).
Remember M=g−1({0})∧g stet ∧ ({0} abg.) ⇒M abg.
Solve ∑i=1n(∇f(x))i+λi(∇g(x))i=!0∧g1=!0∧...∧gm=!0, and get Pi for some λi
Compute f(Pi) and compare them, they are extrema (not rel.), check if minima or maxima
If we have n−m=1 then h:R→R
notice h=f(ϕ(t),ψ(t)) for some ϕ and ψ then see simply .41 write alg.
Define F:Rn×Rm→R,(x,λ)↦f(x)+∑i=1mλigi(x) and then (λ is Lagrange Moltiplikator) ∇F(x,λ)=(∇f(x)+∑i=1mλigi(x),g1(x),...,gn(x))T set all to 0, .42 is an ez application. .43
spd(HessF(x0,λ0))∨snd(HessF(x0,λ0))⇒x0 is extr. of f∣M for M:={x∈U∣gj(0)∀j∈{1,...,m}}
compute ∇F(x,λ)=0 and get x0 and λ0, then compute HessF(x0,λ0) check if definit.
12
12.1 Weglängen
γ:[a,b]→Rn, if γ stetig diff. and ∀s∈(a,b)γ′(s)=0 then γ is glatt; def. t=s↦∣γ′(s)∣γ′(s)
L(γ):=∫ab∣γ′(s)∣ds; additiv, limn→∞L(γn)=L(γ)
dom(γ)=[a,b]∼dom(γ~)=[c,d] if ex. ϕ:[a,b]→[c,d] s.t. stetig diff. and bij with ∀s∈[c,d]ϕ′(s)>0, γ~(s~)=γ(ϕ(s~)); the equivalence classes on ∼ are a orientierte stückwise glatte Kurve.
Every class has rapresentatvie a γ s.t. ∣γ′(s)∣=1. The Bogenlänge of γ is ∫0s∣γ′(t)∣dt .6 .7
k(s):=∣γ′′(s)∣ is called Krümmung, ⟨γ′(s),γ′′(s)⟩=0 direction of k and t are ⊥ .7
12.2 Kurve in der Ebene und im Raum
for ∣γ′(s)∣=1, (set speed to 1) we have t(s)=(x′(s)y′(s)); n(s)=(−y′(s)x′(s)), now define α,β,δ,ϵ s.t. t′(s)=α(s)t(s)+β(s)n(s); n′(s)=δ(s)t(s)+ϵ(s)n(s) like if they were funcitons. But α=ϵ=0 and β=−δ called Frenetsche Gleichungen
def. τ(s):=⟨n′(s),b(s)⟩=⟨−b′(s),n(s)⟩ called Torsion.
12.3 m-dimensionale Flächen im R
γ:U(⊂Rm)→Rn, m<n so γ′:U→L(Rm,Rn). τ:=γ(u)+γ′(u,h), so τ is aff. subsp. in Rn
γ(u,v)=(f(u)cos(v)f(u)sin(u)g(u))Rotationsfläche, check γ′(u,v) is full rank then γ(u,v) is a 2-Fläche
dom(γ)=U & dom(γ~)=U~ then γ∼γ~⇔∃Φ:U~→U s.t. γ~=Φ(γ) and Φ is a diffeomorph. .12 then the class is called m-Fläche with the gleiche Orientierung if det(Φ′)>0. uhm what?
.13
13
13.1 Measure Theory
A⊆P(X) is a σ-algebra if: (i) ∅∈A, (ii) A∈A→Ac∈A, (iii) (∀n∈NAn∈A)→⋃n∈NAn∈A .1
char. of σ-al., (i) X∈A (ii) B⊆A→A∖B∈A (iii) A∩B∈A (iv) ⋃n∈NAn∈A Blatt 9.4
see blatt other rules & given a fixed X⋂i∈IAi is a σ-algebra, so σ(E):=⋂{A∣E⊂A} is the smallest σ-algebra containing E.
μ:A→[0,∞] a measure on A if (i) μ(∅)=0, (ii) μ(⋃n∈N⋅An)=∑n∈Nμ(An)
μ is said σ-endlich if ex. (An)n∈N⊂A s.t. ⋃n∈NAn=X and ∀n∈Nμ(An)<∞ .3
μ({x∈X∣¬M(x)})=0 : M(x) is true fast überall.
δx(A)=χA(x)=1 if x∈A, 0 if x∈A; ζ(A):=∣A∣ if ∣A∣<ℵ0 , ∞ if ∣A∣≥ℵ0 .5
AE:={A∩E∣A∈A} is a σ-al. μ∣E:=μ∣AE then (E,AE,μ∣E) is a measure space
σ(f):={f−1(B)∣B∈B} for f:X→Y and (Y,B), (X,A) is the σ-al. generated by f.
⇔ 2. ⇒ 3. (3. ⇒ 2. if μ endlich, i.e. μ(x)<∞)
μ is σ-additiv
(∀n∈NAn⊂An+1), so ⋃n∈NAn=:A(∈A): limn→∞μ(An)=μ(A)
(∀n∈NAn+1⊂An), ⋂n∈NAn=∅ and μ(A1)<∞limn→∞μ(An)=0
(Borel-σ-algebra) B(Rn):=σ({(a,b]∣a,b∈Rn∧a<b}), all open and closed sets are B-mess. .9
(i) λ({x})=0 (ii) if ∃j∈{1,..,n}ajbj→λ(∏j=1n[aj,bj]) (iii) λ((a,b))=λ([a,b))=... ez (iv) λ(Q)=0 (v) λ is σ-end., since Rn=⋃N∈N∏j=1n[−N,N].
λ(C)=1−31⋅∑n=0∞(32)n=0 but ∣C∣=∣R∣⇒λ(C)=0
f:X→S, (X,A), (S,S)fA-S-meas. ⇔f−1(S)⊂A i.e. for all B∈S: f−1(B)∈A, A-meas.: (S,S)=(Rn,B(Rn)); Borel-Meas if also (X,A)=(Rm,B(Rm)). Meas. comes from meas.
ext. for f meas. f:Rn→Rn and M∈B(Rn)⇒f−1(M)∈B(Rn)
contant func. are always meas.; f,g meas ⇒g∘f meas.
for (X,A), (S,σ(E)): f:X→SA-σ(E)meas. ⇔f−1(E)⊂A .15
How to find E of B? Check if with the rules of a σ-al. you get (a,b] like in .16.(ii)⇒(i)
For im(f)⊂R: f Borel-meas. ⇔∀a∈R{x∈X∣f(x)>a}∈B(Rn)⇔... .16
f stetig ⇒f Borel-meas.; (fk)k∈N Borel-meas ⇒f Borel meas.; f,g meas. ⇒f⋅g,f+g,max{f,g}, for r>0: ∣f∣r and for k∈N: fk are all meas. .18
13.2 Lebesgue Integral
f=∑i=1kciχAi for χAi(x):=1 if x∈Ai, 0 if x∈Ai, is a Stufenf., B(X,A;R) meas. bound func.
like in 1-dim: (sk)k∈N→f. if f∈B(X,A;R) then a (sk)k∈N conv. glm. so the space of the sk is dense in B(X,A;R) respect to the ∣∣⋅∣∣∞; f+:=max{f,0}, f−:=min{0,f}; f=f+−f− .20
μ translationsinvariant if μ∘Φ−1=μ for all translation of the form Φ:x↦x+c
μ bewegungsinvariant if μ∘Φ−1=μ for all bewegungen of the form Φ:x↦Sx+c for S orth.
μ translat. on (Rn,B(Rn)) s.t. μ(W)<∞ for W an Einheitswürfel then μ=μ(W)λ.
λ is bewegungs.;
T sq. inv. matrix λ(T(B))=∣det(T)∣⋅λ(B) .42
U,V⊂Rn op. Φ:U→V a C1-Diffeo, let f:V→R meas:f int. on V=Φ(U)⇔(f∘Φ)⋅∣detΦ′∣ integr. ⇒∫Φ(U)f(x)dx=∫Uf(Φ(z))⋅∣detΦ′(z)∣dz .43
.44
13.5 Kurven- und Flächenintegrale
Vector Field: v:D(⊂Rn)→Rn; Γ⊂Rn=im(γ) and ω:Γ×Rn→Rn stet. linear in the second argument, then ω is an 1-Form in Γ. Similarly for any U⊂Rn instead of Γ we call ω an 1-Form.
for ω(x,h)=⟨ω(x),h⟩, ω(x)=(ω1(x),...,ωn(x))T for ωi:=ω(x,ei)
ext: example: for any f diff. then (x,h)↦⟨∇f(x),h⟩ is an 1-Form
call ∫Γω:=∫baω(γ(t),γ′(t))dtKurvenintegral; then lin. and for Γ=Γ1+Γ2: ∫Γω=∫Γ1ω+∫Γ2ω
Compute: ∫ΓΩ=∫Γ⟨ω(γ(t)),γ′(t)⟩dt, like a vectorfield ⋅ the directon of γ. For Ω:Γ×R2→R and ω:R2→R2, so ω is like v and Ω must give us something in one dimension so that we are allowed to integrate.
for Γ1,Γ2 have the same starting and ending point, and ∫Γ1ω=∫Γ2ω then ω is wegunabhängig
U sternförmig to p⇔∀x∈Upx⊂U.
Exakt, How to prove:
if you have a f s.t. ω(x)=∇f(x), call f Stammfunktion and ω exakt. .49
ext: e.g. for U sternf. findf s.t. f(x)=∫Γ(x0,x)ω, for γ:t↦x0+t(x−x0)uhm
ext: e.g. for all i find: f=∫ωidti+c(x1,...,xi−1,xi+1,...,xn)uhm
ω has U open and zusamm.: ω exakt ⇔ω wegunabähngig .50
for ωi=∂if: ω exakt ⇒∂jωi=∂j∂if=∂i∂jf=∂iωj, conver. holds if U sternfö. for one p
Then thanks to Schwarz if f∈C2(U,Rn) and U sternf. then f is exakt.
γ Par. von m-Fl. g:u↦γ′(u)Tγ′(u)=(⟨∂iγ(u),∂jγ(u)⟩)i,j=1,...,m .55.i
Δ:C2(Rn,R)→C(Rn,R),f↦∑i=1n∂i2f, Δ(f)=div(∇(f)), called Laplace-Operator 11.14
ext: Dy(h):=limt→0tf(y+th)−f(y),
Mittelwertsatz: ∃c∈Γabf(b)=f(a)+f′(c)(b−a), Γab is a the Gerade from a to b, for Im(f)⊆R
Taylor in Rn: f(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)+21fxx(a,b)(x−a)2+21fyy(a,b)(y−b)2+fxy(a,b)(x−a)(y−b)+R2(x,h) for R2(x,y)=61f′′′(x+θh,h,h,h) for θ∈(0,1). 11.27
f convex ⇔∀x1,x2∈Uf(x1+t(x2−x1))≤f(x1)+t(f(x2)−f(x1)), with < for streng conv. 11.36
Solve ∑i=1n(∇f(x))i+λi(∇g(x))i=!0∧g1=!0∧...∧gm=!0, and get Pi for some λi