Lecture Summry Note This document ws written in the following wy (which might explin its structure): First, I copied Prof. Immoglu s non-exmple notes for the review. Then I merged the in-clss exmples from the review nd our ssistnt s (Tim) review. The lnguge is inconsistent. Tle of Contents Integrtion... 3. Fcts... 3.2 Properties... 3.3 Mittelwertstz der Integrlrechnung... 3.4 Fundmentl theorem of Clculus... 3.5 How do we clculte integrls?... 3.6 Improper Integrls... 4.7 Exmples... 4.7. Exmples BP 203... 4.7.2 Exmple Spring 200... 4.8 Additionl Wisdom... 5 2 Differentil Equtions... 5 2. Liner differentil equtions with constnt coefficients... 5 2.. Finding the homogeneous solution yh of Ly = 0... 5 2..2 How to find the specil solution of Ly = x using the method of Anstz... 5 2.2 Boundry or initil vlue prolems... 6 2.3 Solving DGL y seprtion of vriles... 6 2.4 Ansätze... 6 2.5 Exmples... 6 2.5. Exmple Spring 20... 6 2.5.2 Exmple Summer 203... 7 2.5.3 Exmple Spring 20... 7 3 Differentition in Rn... 7 3. Differentition rules... 8 3.2 Directionl derivtive... 8 3.3 Higher prtil derivtives... 8 3.4 The extrem of function f: Ω. R... 8 3.5 Line integrl... 9 3.6 div, rot,... 9 3.7 Tngentileene usrechnen... 0 3.8 Potentil usrechnen... 0 3.9 Additionl Wisdom... 0 4 Integrtion in Rn... 0 7/30/204 Linus Metzler
4. Sustitution in Rn... 0 4.2 Green s theorem... 4.3 Additionl Wisdom... 7/30/204 Linus Metzler 2
Integrtion Let f: R R e continuous function, P = { = x o < x x < < x n = } prtition of the intervl [, ] nd ξ k [x k, x k+ ] points in ech suintervl. Then the sum S(f, P, ξ) = n k=0 f(ξ k )(x k+ x k ) is clled the Riemnn sum ttched to f nd to P. For I k = [x k, x k+ ], U(f, P) = n k=0 (inf f) (x k+ x k ) nd O(f, P) = I k n k=0 (sup f) (x k+ x k ) re clled the lower nd upper Riemnn sums. Similrly f dx = sup{u(f, p), p P(I)} I k nd f dx = inf{o(f, p), p P(I)} re clled lower nd upper integrls. f is clled Riemnn integrle if f dx =. f dx. Fcts - Ech continuous function is Riemnn integrle - Ech monotonic function is Riemnn integrle.2 Properties - Let f, g Riemnn integrle on I, α, β R. Then. (αf + βg)dx = α f dx + β g dx 2. If f(x) g(x) x [, ] then f dx 3. f dx f(x) dx 4. (inf I 5. f dx f) ( ) f(x) dx = f(x) dx c 6. f dx = f(x) dx 7. f(x) 0 x D f dμ D 0.3 Mittelwertstz der Integrlrechnung g dx (sup f) ( ) I + f(x) dx,, c R c f: [, ] R continonous. Then ξ [, ] such tht f(x) dx = f(ξ)( )..4 Fundmentl theorem of Clculus 7/30/204 Linus Metzler 3 x. Let f: [, ] R continuous. Define F(x) f(t) dt x [, ]. Then F is differentile nd F = f. F is clled primitive (Stmmfunktion) of f 2. If G is n nohte rpmimirve of f then G = F + c for some constnt c 3. Let F e ny pmirmitve of f, then f(x) dx = F() F().5 How do we clculte integrls? Never forget the constnt!. Prtil integrtion: follows product rule for differentition f(x)g (x) dx = f(x)g(x) f (x)g(x) dx f(x)g (x) dx = f(x)g(x) f (x)g(x) dx 2. Sustitution: follows chin rule for differentition f(x) dx = f(φ(y))φ (y) dy f(x) dx = f(φ(y))φ (y) dy 3. Prtil frctions: to integrtion rtionl functions of the form P(x) φ() φ() P(x) = P(x) Q(x) (x 2 +)(x ) 2 (x+2) Anstz: P(x) = Ax+B + C + D Q(x) x 2 + x Bechte Vielfchheiten, C Nullstellen Bsic types of integrtion of rtionl functions Polynomil: n x n x dx = n+ n + E (x ) 2 x+2 n+ + c Q(x), P, Q re polynomils
Inverse powers: dx = { log x x 0, für = (x x 0 ) r r, für 2 r (x x 0 ).6 Improper Integrls The improper integrl of n integrle function f on (, ) which is integrle on ny suintervl [, ]. We define the improper integrl f(x) dx Integrlkriterium Fcts lim lim f(x) dx. - f(x) is defined on [, [ - f(x) 0 x [, [ - f(x) monoton fllend f (x) 0 - n= f(n) konvergiert f(x) dx konvergiert s. s R, > 0, dx = {, s > x s s, s 2. If f is on [, ) continuous nd c nd s > so tht f(x) c/x s x, then f(x) dx converges 3. If f is in [, ) continuous nd c > 0 such tht f(x) c/x, x, then f(x) dx diverges to..7 Exmples.7. Exmples BP 203-2 +ln x x dx - cos x cosh x dx - x2 x+2 x 3 x 2 +x u = + ln x du = dx x 2 + ln x dx x u = cos x u = sin x u I = (cos x) +ln 2 = u /2 du v = cosh x v = sinh x = u3/2 3/2 v cosh x dx = uv vu = (cos x)(cosh x) + (sin x) from to + ln 2 7/30/204 Linus Metzler 4 u v sinh x dx = cos x cosh x + [sin x cosh x cos x cosh x dx] I = [cos x sinh x + sin x cosh x] + C 2 x 3 x 2 + x = x 2 (x ) + (x ) = (x )(x 2 + ) x2 x + 2 x 3 x 2 + x = A Bx + C + x x 2 + A(x 2 + ) + (Bx + C)(x ) = x 2 x + 2 A =, C =, B = 0; x2 x + 2 x 3 x 2 + x = x x 2 + dx = ln x tn x + C.7.2 Exmple Spring 200 Untersuche, o ds untere Integrl x 2 + x dx konvergiert. x 2 + x x 2 x 2 + x x 2 dx converges
or lim dx = lim x(x + ) x x + dx = lim[ln x ln x + ] x = lim ln x + = lim ln + ln 2 = ln 2.8 Additionl iisdom From http://en.wikipedi.org/wiki/differentition_under_the_integrl_sign: ( f(x) dx ) = f(), ( f(x) dx ) = f() φ(α) f(x, α) dx, dφ = f(x, α) dx + f(, α) f(, α) dα (Leiniz) α α α n! (2πn) ( n e )n (Stirling) Solids of revolution when integrting prllel to the xis of revolution: V = π f 2 (x) dx when integrting perpendiculr to the xis of revolution: V = 2π x f(x) dx Don t forget +C when integrting 2 Differentil Equtions 2. Liner differentil equtions with constnt coefficients To solve liner differentil equtions of the form Ly = (x) where L dn + dx n n + + dx n + dx 0, (x) function, i R.. Find homogenous solutiony H. Nmely solution of Ly = 0. 2. Find specil solution y S of Ly = (x) using the method of Anstz vom Typ der rechten Seite. 3. The generl solution is given y y = y H + y S 2.. Finding the homogeneous solution y H of Ly = 0. Find the chrcteristic polynomil of L. Nmely P L (λ) = λ n + n λ n + + λ + 0 2. Fct: if λ, λ r C re the pirwise distinct roots of p(λ) = 0 with ssocited multiplicities m,, m R, then the functions x x k e λ jx, j r, 0 k m j form system of fundmentl solutions of the homogenous eqution Ly = 0. Note: if L hs rel coffeicients, every pir of cimplex conjugte, non-rl roots λ j = μ λ ± iν j of multiplicity m j give fundmentl solution x k e (μ j±iν j )x = x k e μ j(cos ν j x ± i sin ν j x) for 0 k < m j. So one cn s sis tke x k e μ jx cos ν j x nd x k e μ jx sin ν j x insted of x k e (μ j+iν j )x nd x k e (μ j iν j )x. Then the generl homogenous solutions is of the form y H (x) = c jk x k e λ jx j= with constnts c jk. 2..2 How to find the specil solution of Ly = (x) using the method of Anstz Fcts r m j k=0. Let λ C. If λ is not solution of p L (λ) =??, then the inhomongoues DGL Ly = e λx hs prticulr solution y = p L (λ) eλx 2. Let λ C, m its multiplicity s solution of p L(λ) = 0 (m cn e zero which mens λ is not solution of p L (λ) = 0). Let Q(x) polynomil of degree k. Then prticulr solution of Ly(x) = Q(x)e λx is of the form y(x) = R(x)e λx for polynomil R(x) of degree k + m 3. If L hs rel coefficients. Let μ, ν R, m the multiplicity of μ ± iν s solution of p L (λ) = 0 (m = 0 mens μ ± iν is root of p L ). Let Q(x), R(x) e polynomil of degree k. The prticulr solution of the inhomogeneous DGL Ly = Q(x)e μx cos νx + R(x)e μx sin x is of the form y(x) = s(x)e μx cos νx + T(x)e μx sin x for polynomils S, T of degree k + m d n d 7/30/204 Linus Metzler 5
2.2 Boundry or initil vlue prolems y( ) = A y(0) = A y( When we re given DGL Ly = (x) together with either oundry vlues 2 ) = A 2 y or initil vlues (0) = A 2, y( n ) = A n y n (0) = A n we first find the generl solution y = y H + y S. Then we determine the constnts c,, c n in the homogenous solution using the given oundry/initil vlues. 2.3 Solving DGL y seprtion of vriles Fcts - If f: Ω R is differentile in x 0 R, then the prtil derivtives exists nd the differentil df(x 0 ) hs the mtrix representtion ( (x x 0 ) (x x n 0 )) = f the grdient of f. - f diff in x 0 f is continous in x 0 - If ll prtil derivtives of f exists nd continuous, then f is differentile. Using the lst two fcts nd the definition of differentiility, one cn study if given is differentile of not. Recipe - Sei die DG in der Form df(x) = g(x)h(f(x)) dx - Sei nun y = f(x), dnn dy = g(x)h(x) dx - Flls h(y) 0, dnn dy = g(x)dx h(y) dy h(y) - Alterntive Nottion: = g(x) dx - werden nun eide Seiten nch x integriert, dnn - y y dy = ln y 2.4 Ansätze,, c, d R, μ, ν R, n N, X n = Polynomil of degree x Störfunktion q(x) e μx sin νx cos νx e μx sin νx e μx cos νx P n (x)e μx P n (x)e μx sin νx Q n (x)e μx cos νx dy h(y) dx dx = g(x) dx dy = g(x) dx h(y) Anstz für y p (x) 2.5. Exmple Spring 20 ) Bestimme lle Lösungen y = y(x) der DGL y (4) y = 0 welche für x eschränkt leien. Chrcteristic polynomil: x 4 = 0 (x 2 )(x 2 + ) = 0 λ = ± ex, e x λ = ±i cos x, sin x y H (x) = c e x + c x 2 + c 3 cos x + c 4 sin x The solutions tht remin ounded s x re of the form c 3 cos x + c 4 sin x ) Bestimme eine Lösung y = y(x) der DGL y (4) y = e x + x 7/30/204 Linus Metzler 6 e μx c sin νx + d cos νx e μx (c sin νx + d cos νx) R n (x)e μx e μx (R n (x) sin νx + S n (x) cos νx) Bem Liegt eine Linerkomintion der Störfunktionen vor, so ht mn uch ls Anstz eine entsprechende Linerkomintion zu wählen. Bem 2 Flls λ = μ + iν eine m fche Nullstelle des chrkteristischen Polynoms von (H) ist, so muss mn den Anstz für y p (x) mit dem Fktor x m multiplizieren. 2.5 Exmples
y 4 y = e x y p y 4 y = x y p2 Superposition: y p = y p + y p2 y = c e x + c 2 e x + + (x) y p = Cxe x nd y p2 = Dx + E Try y p = Cxe x + Dx + E, put this in y 4 y = e x + x y p (4) (x) = C[ 4e x + xe x ] y p (4) (x) y p (x) = C[ 4e x + xe x ] [Cxe x + Dx + E] = e x + x c =, D = 4 2.5.2 Exmple Summer 203 ) Für welche Werte des Prmters R stret die llgemeine Lösung der DGL y + 2y + y = 0 unhängig von den Anfngsedingungen gegen 0 für x? λ 2 2 ± 4 4 + 2λ + = 0 λ,2 = = ± 2 For < 0: there re 2 complex conjugte roots. Let = 2. Then ± i c e x cos x + c 2 e x sin x 0 s x For = 0: ( ) is doule root. The solution c e x + c 2 e x x 0 independent of the initil conditions. For > 0: then one of the roots will e positive if >. Tht will led to λ = + > 0 which leds to growing solution. We do not wnt > or < 0. If < then λ,2 < 0 ) Finden Sie eine homogene DGL 2. Ordnunug mit konstnten Koeffizienten, deren llgemeine Lösung y(x) = e x + 2xe x ist. Ws sind dnn die Anfngsedingungen ei x = 0? y = e x + 2xe x We re looking for 2 nd DGL. By looking t the eqution, you cn see tht λ = with multiplicity 2 (i. e. doule root of the chr. pol. ). (λ + ) 2 = λ 2 + 2λ + y + 2y + y = 0 + initil vlues c =, c 2 = 2 y GH generl homogenous solution = c e x + c 2 xe x y(0) = e 0 = nd y (0) = e x + 2[e x xe x ] for x = 0 = 2.5.3 Exmple Spring 20 Bestimme die Lösung y = y(x) der DGL y = e x y mit y(0) = 0 y = ex e y dyey = e x dx e y dy = e x dx e y = e x + c y = ln e x + c 0 = y(0) = ln e 0 + c = ln + c c = 0 3 Differentition in R n A function f: Ω R n R is differentile in x 0 if there exists liner mp A: R n R such tht f(x) = f(x x 0 ) + A(x x o ) + R(x, x 0 ) where lim x x0 R(x,x 0 ) x x 0 = 0. In this cse A is clled the differentil of f t x 0 nd it is dented y (df)(x 0 ). Let (A, A 2,, A n ) e mtrix representtion of the liner mp A: R n R. The f differentile t x 0 mens f(x) = f(x 0 ) + A (x x 0 ) + A 2 (x 2 x 0 2 ) + + A n (x n x 0 n ) + R(x, x 0 ). A prtil derivtive is defined s f( ( ) lim,.., i, i +h, i+,, n ) f(,, i,, n ) i h 0 h Fct Let h(s, t) e continuously differentile function of two vriles nd (t) differentile function of one (t) vrile. Define u(t) h(s, t) ds. Then u is diffenetle nd u (t) = h((t), t) t(t) + (t) h t (s, t) d?? 7/30/204 Linus Metzler 7
In prticulr if u(t) is defined s definite integrl of h(s, t) in the vrile s, u(t) h(s, t) ds, then u is differentile nd one cn interchnge the order of differentition nd integrtion. Tht is d dt u(t) = d dt h(s, t) ds. t 3. Differentition rules Let f, g: Ω R differentile in x 0. Then:. d(f ± g)(x 0 ) = df(x 0 ) + dg(x 0 ) 2. d(fg)(x 0 ) = g(x 0 )df(x 0 ) + f(x 0 )dg(x 0 ) 3. If g(x 0 ) 0 then d(f/g)(x 0 ) = g(x 0 )fd(x 0 ) f(x 0 )dg(x 0 ) (g(x 0 )) 2 4. Let h: R R e differentile in g(x 0 ). Then d(h g)(x 0 ) = h (g(x 0 )) dg(x 0 ) h(s, t) ds = 5. Let H: I R Ω R n e differentile in x 0 I nd f: Ω R differentile in H(t 0 ). Then d dt (f H)(t 0 ) = df(h(t 0 )) H (t 0 ) where H(t) = (H (t), H 2 (t),, H n (t)), H (t) = (H (t), H 2 (t),, H n (t)) d 6. Chin rule: (f g)(t dt 0 ) = df(g(t 0 )) g (t 0 ); d (f(x(t), y(t))) = df(x(t), y(t)) (t) dt (x y (t) ) = (x(t), y(t)) x x (t) + (x(t), y(t)) y y (t) 3.2 Directionl derivtive The directionl derivtive of f is in the direction of unit vector e R n {0} is given y d e f(x 0 ) = f(x 0 ) e. 3.3 Higher prtil derivtives One cn similrly define higher order prtil derivtives for functions f C m (Ω). Fct (Schwrz) If f C 2 (Ω) then 2 f = 2 f x i x j x j x independent of the order of differentition. i nd in generl for f C^m(Ω), ll prtil derivtives of order m re Using higher order derivtives one cn nlogous to the -dimensionl cse define Tylor pproximtion of f. Fct Let f C m (Ω), f: Ω R, Ω R, x 0, x Ω. Then f(x ) = f(x 0 ) + f(x 0 )(x x 0 ) + 2 x 0 i )(x j x 0 j ) + R(f, x, x 0 ) where lim x x 0 R(f,x,x 0 ) x x 0 0. 2 i,j= x i x j (x 0 )(x i The nlog of the second derivtive is given y the mtrix of prtil derivtives of order 2. The mtrix is clled the Hesse mtrix of f. Hess f 2 f ( 2 f x i x j) i,j= n 3.4 The extrem of function f: Ω. R Definition A point x Ω is clled criticl point if f(x) = 0 Fct If f is differentile nd x 0 is locl extrem of f, then x 0 is criticl point. Fct Let x 0 e criticl point of f. Then we hve. x 0 is locl minimum if 2 f(x 0 ) is positive definite (det Hess f > 0 tr Hess f > 0) 2. x 0 is locl mximum if 2 f(x 0 ) is negtive definite (det Hess f > 0 tr Hess f < 0) 3. Otherwise x 0 is sddle point (det Hess f < 0) To find extrem of f on region Ω.. Find criticl points f = 0, x 0 is criticl point 2. Check the nture of criticl points y Hess(f)(x 0 ) 3. Check the criticl points tht rise from here 7/30/204 Linus Metzler 8
H f (x) ( 2 f x i x j (x)) i,j=,,n Flls 2D: H f (x) ( 2 f x y (x)) = i,j=,,n (x) (x) x x x x 2 = (x) (x) x 2 x x 2 x 2 (x) ( x n x x x (x) ( y x (x) x x n (x) x 2 x n (x) x n 2 (x) (x) x n x n )n n x y (x) y y (x) ) 2 2 The Jcoi-Mtrix works similr to the Hesse-Mtrix, ut it only uses the first derivtives. Fct Let f: Ω R e continours nd differentile on n open set Ω R n. Let Ω e the oundry of Ω. Then every glol extrem of f is either criticl point of f in Ω or glol extrml point of f x. 3.5 Line integrl Let v: Ω R n e vector field nd curve with prmeteriztion : [, ] Ω, t (t). Then the line integrl of v long is deinfed s v ds v((t)), (t) dt Fcts.. v ds is independent of the prmeteriztion of the pth 2. v ds = + 2 v ds + v ds 2 3. v ds = v ds, where is the sme pth s in opposite direction 4. If v is the grdient vector field ssocited to function f i.e. v = df, then v ds = f(()) f(()), : [, ] Ω 5. Wir können den Begriff des Wegintegrls uf Wege erweitern, die stückweise C sind. Ein stückweise C -Weg ist eine stetige Aildung : [, ] R n mit einer endlichen Unterteilung des Intervlls = c 0 < c < < c n = so dss [ci,c i+ ]: [c i, c i+ ] R n, (i = 0 n ) in C ist. Dnn definiert mn: λ n i=0 λ ( [ci,c i+ ] ) Equivlent once cn write everything in terms of forms: λ = λ dx + λ 2 dx 2 + + λ n dx n then λ λ((t)) (t) dt Fcts λ: Ω L(R n R) continuous forms, then the following re equivlent. f C (Ω) st df = λ 2. For every two continuous C pths, 2 with the sme eginning nd end points: λ = λ 2 3. For every closed curve, λ = 0 Definition A vector field v: Ω R n is clled conservtive if v ds = 0 for ll closed curves. Fct For simply connected region Ω, we hve: v conservtive v = f for some function f. 3.6 div, rot, div K K = K + K 2 + K 3 x y z grd f = f = ( (x x 0 ),, x n (x 0 )), in 3D: ( x y z), Richtungsleitung: f r = 7/30/204 Linus Metzler 9
rot K K = K 3 K 2 y z K K 3 z x K 2 K x ( y ) div(fk) = f K + f div K div(k L) = L rot K K rot L 0 rot(grd f) = ( 0) 0 div(rot K) = 0 div(f rot K) = grd f rot K 3.7 Tngentileene usrechnen Um die Tngentileene m Grph G(t) uszurechnen, git es drei Möglichkeiten. T(x, y) = f(x 0, y 0 ) + ( )(x x x 0 ) + ( )(y y y 0 ) = df(x, y) -. Möglichkeit: T(x, y) usrechnen f(x, y) - 2. Möglichkeit: Tngentilvektoren; Tngentilvektoren sind immer: ( ) - 3. Möglichkeit: Linerkomintion 3.8 Potentil usrechnen Sei F = (6xy + 4z 2, 3x 2 + 3y 2, 8xz) ein Vektorfeld. Bestimme ds Potentil f - - - =y 3 +h(z) = 6xy + x 4z2 (6xy + 4z 2 )dx = 3x 2 y + 4xz + g(y, z) = f(x, y, z) = y 3x2 + 3y 2 = y 3x2 + g g (y, z) = y y 3y2 3y 2 dy = y 3 + h(z) = 8xz = z z z (3x2 y + 4xz + y 3 + h(z)) = 8xz + g (z) - f(x, y, z) = 3x 2 y + 4xz + y 3 + c 3.9 Additionl iisdom = g z g (z) = 0, 0 dz = c d f(x(t), y(t)) = df(x(t), y(t)) (t) dt (x y (t) ) = (x(t), y(t)) x x (t) + (x(t), y(t)) y y (t) (chin rule) Ein geschlossener Weg in einem konservtiven Vektorfeld ist = 0, E ds = 0. 4 Integrtion in R n The Riemnn integrl in R n is constructed in n nlog wy to the cse n = with Riemnn sums over suintervls replced with sums over surectngles, with dx replced with n-ddimensionl volume element dvol n which we denote either y dvol n or dμ(x ). Fct For rectngle Q = [, ] [c, d] R 2 : f dμ Q d c d. c = f dy dx = f dx dy Fuini F(t)dt = J f(x, y) dxdy = J I f(x, y)dydx = I J f(x, y) d(x, y) I J 4. Sustitution in R n Let u, v R n open, Φ: u v ijective with det Φ 0 u. Then for f = v R continuous we hve f(x )dμ(x ) v = f(φ(y)) det(dφ(y)) dμ(y ) u Theorem 9.7 U, V R open, Φ: U V ijective, continuous, differentile, det dφ(y ) 0 y U, f: V R continuous. f(x )dμ(x ) = f(φ(y )) det dφ(y ) dμ. dφ(y ) is the Jcoi mtrix. V Φ(U)=V https://www.youtue.com/wtch?v=tsljeont9y 7/30/204 Linus Metzler 0
4.2 Green s theorem Let Ω R 2 whose oundry Ω hs C prmeteriztion. Let U Ω nd f = Q P where P, Q x y C (U). Then ( Q P ) dμ = Ω P dx + Q dy x y Ω OR Let V = (P, Q) e vector field then v ds = Ω rot v dμ where rot V = Q Ω P nd the line integrl is tken x y round the oundry of Ω in counter-clockwise direction. 4.3 Additionl iisdom - Prmeterintegrl 2 d dx ( (x) (x) f(x, t) dt (x) ) = f(x, (x)) (x) f(, (x)) (x) + f x (x, t) dt - Mehrdimensionle Integrtion Bei der Koordintentrnsformtion ds r (o.ä.) nicht vergessen (x) 2 Differentition under the integrl sign 7/30/204 Linus Metzler