An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.


 Marcus Henry
 3 years ago
 Views:
Transcription
1 Improper Inegrls To his poin we hve only considered inegrls f(x) wih he is of inegrion nd b finie nd he inegrnd f(x) bounded (nd in fc coninuous excep possibly for finiely mny jump disconinuiies) An inegrl hving eiher n infinie i of inegrion or n unbounded inegrnd is clled improper. Here re wo exmples +x 2 The firs hs n infinie domin of inegrion nd he inegrnd of he second ends o s x pproches he lef end of he domin of inegrion. We ll sr wih n exmple h illusres he rps h you cn fll ino if you re such inegrls sloppily. Then we ll see how o re hem crefully. Exmple The compuion x = x2 x = = 2 is wrong. In fc, he nswer is ridiculous. The inegrnd >, so he inegrl hs o be x 2 posiive. The flw in he rgumen is h he fundmenl heorem of clculus, which sys h if F (x) = f(x) hen f(x) = F(b) F(), is pplicble only when F (x) exiss nd equls f(x) for ll x b. In his cse F (x) = does no exis for x =. The given x 2 inegrl is improper. We ll see ler h he correc nswer is +. Exmple The creful wy o re n inegrl like h hs bounded inegrnd bu hs +x 2 domin of inegrion h exends o + is o pproxime he inegrl by one wih bounded domin of inegrion, like, nd hen ke he i R. Here is he +x 2 y = f(x) R x forml definiion nd some vriions. We ll work hrough some exmples in deil shorly. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25
2 Definiion 2. () If he inegrl f(x) exiss for ll R >, hen f(x) f(x) when he i exiss (nd is finie). (b) If he inegrl r f(x) exiss for ll r < b, hen when he i exiss (nd is finie). (c) If he inegrl r f(x) f(x) r r f(x) exiss for ll r < R, hen f(x) r c r f(x) + f(x) c when boh is exis (nd re finie). Any c cn be used. When he i(s) exis, he inegrl is sid o be convergen. Oherwise i is sid o be divergen. The creful wy o re n inegrl like h hs finie domin of inegrion bu x whose inegrnd is unbounded ner one i of inegrion is o pproxime he inegrl by one wih domin of inegrion on which he inegrnd is bounded, like, wih >, x nd hen ke he i +. Here is he forml definiion nd some vriions. y y = x x c Joel Feldmn. 25. All righs reserved. 2 Mrch 6, 25
3 Definiion 3. () If he inegrl f(x) exiss for ll < < b, hen when he i exiss (nd is finie). (b) If he inegrl T f(x) f(x) + f(x) exiss for ll < T < b, hen when he i exiss (nd is finie). T f(x) f(x) T b (c) Le < c < b. If he inegrls T f(x) nd f(x) exis for ll < T < c nd c < < b, hen T f(x) f(x) + f(x) T c c+ when boh i exis (nd re finie). When he i(s) exis, he inegrl is sid o be convergen. Oherwise i is sid o be divergen. Ifninegrlhsmorehnone sourceofimpropriey, forexmple ninfiniedominof inegrion nd n inegrnd wih n unbounded inegrnd or muliple infinie disconinuiies, hen you spli i up ino sum of inegrls wih single source of impropriey in ech. For he inegrl, s whole, o converge every erm in h sum hs o converge. For exmple, he inegrl hs domin of inegrion h exends o boh + nd nd, (x 2)x 2 in ddiion, hs n inegrnd which is singulr (i.e. becomes infinie) x = 2 nd x =. So we would wrie (x 2)x 2 = (x 2)x 2 + (x 2)x 2 + (x 2)x (x 2)x (x 2)x (x 2)x 2 (Here he brek poin could be replced by ny number sricly less hn ; could be replced by ny number sricly beween nd 2; nd 3 could be replced by ny number sricly bigger hn 2.) On he righ hnd side he firs inegrl hs domin of inegrion exending o c Joel Feldmn. 25. All righs reserved. 3 Mrch 6, 25
4 he second inegrl hs n inegrnd h becomes unbounded s x, he hird inegrl hs n inegrnd h becomes unbounded s x +, he fourh inegrl hs n inegrnd h becomes unbounded s x 2, he fifh inegrl hs n inegrnd h becomes unbounded s x 2+, nd he ls inegrl hs domin of inegrion exending o +. We re now redy for some (imporn) exmples. Exmple 4 ( x p ) Fix ny p >. The domin of he inegrl exends o + nd he inegrnd is x p x p coninuous nd bounded on he whole domin. So he definiion of his inegrl is R x p x p When p, one niderivive of x p is x p+, nd when p = nd x >, one niderivive p+ of x p is lnx. So x p p [ ] R p if p lnr if p = As R, lnr ends o nd R p ends o when p > (i.e. he exponen is posiive) nd ends o when p < (i.e. he exponen is negive). So { divergen if p x = p if p > p Exmple 4 Exmple 5 ( x p ) Agin fix ny p >. The domin of inegrion of he inegrl is finie, bu he inegrnd becomes unbounded s x pproches he lef end,, of he domin of inegrion. x p x p So he definiion of his inegrl is x p + x p When p, one niderivive of x p is x p+, nd when p = nd x >, one niderivive p+ of x p is lnx. So ] [ p if p x p p + ln if p = c Joel Feldmn. 25. All righs reserved. 4 Mrch 6, 25
5 As +, ln ends o nd p ends o when p > (i.e. he exponen is posiive) nd ends o when p < (i.e. he exponen is negive). So { x = if p < p p divergen if p Exmple 5 Exmple 6 ( x p ) Ye gin fix p >. This ime he domin of inegrion of he inegrl exends o x p +, nd in ddiion he inegrnd becomes unbounded s x pproches he lef end,, x p of he domin of inegrion. So we hve o spli i up. x p = x p + We sw, in Exmple 5, h he firs inegrl diverged whenever p, nd we lso sw, in Exmple 4, h he second inegrl diverged whenever p. So he inegrl diverges x p for ll vlues of p. Exmple 6 x p Exmple 7 ( ) x This is prey suble exmple. I looks from he skech below h he signed re o he y y = x x lef of he y xis should excly cncel he re o he righ of he y xis mking he vlue of he inegrl excly zero. Bu boh of he inegrls x [ ] lnx = + x + x T T x + x T [ ln x ] T + ln ln T = T c Joel Feldmn. 25. All righs reserved. 5 Mrch 6, 25
6 diverge so diverges. x Don mke he miske of hinking h =. I is undefined nd for good reson. For exmple, we hve jus seen h he re o he righ of he y xis is + x = + nd h he re o he lef of he y xis is (subsiue 7 for T bove) + 7 x = If =, he following i should be. [ 7 ] + x + [ ln ] [ x + +ln 7 ln ] + +ln(7) ln7 + = ln7 This ppers o give = ln7. Of course he number 7 ws picked rndom. You cn mke be ny number ll, by mking suible replcemen for 7. Exmple 7 Exmple 8 (Exmple revisied) The creful compuion of he inegrl of Exmple is x2 T [ T T = + x + x2 + ] T x 2 + [ ] + x Exmple 8 Since Exmple 9 ( ) +x 2 r r [ ] R +x rcn x 2 +x 2 r [ rcn x The inegrl converges nd kes he vlue π. +x 2 ] r rcnr = π 2 r rcnr = π 2 Exmple 9 c Joel Feldmn. 25. All righs reserved. 6 Mrch 6, 25
7 Exmple For wh vlues of p does e converge? x(lnx) p Soluion. For x e, he denominor x(lnx) p is never zero. So he inegrnd is bounded on he enire domin of inegrion nd his inegrl is improper only becuse he domin of inegrion exends o + nd we proceed s usul. We hve e x(lnx) p e x(lnx) p lnr du wih u = lnx, du = u p x [ ] (lnr) p if p p ln(lnr) if p = = { divergen if p if p > p jus s in Exmple 4, bu wih R replced by lnr. Exmple Exmple (he gmm funcion) The gmm funcion Γ(x) is defined by he improper inegrl Γ() = x e x We shll now compue Γ(n) for ll nurl numbers n. To ge sred, we ll compue [ ] R Γ() = e x e x e x = nd Γ(2) = xe x [ xe x R + xe x ] e x (by inegrion by prs wih u = x, dv = e x, v = e x, du = ) [ ] R xe x e x = c Joel Feldmn. 25. All righs reserved. 7 Mrch 6, 25
8 For he ls equliy, we used h x xe x =. Now we move on o generl n, using he sme ype of compuion s we jus used o evlue Γ(2). For ny nurl number n, Γ(n+) = x n e x [ x n e x R + x n e x ] nx n e x (by inegrion by prs wih u = x n, dv = e x, v = e x, du = nx n ) n x n e x = nγ(n) To ge o he hird row, we used h x x n e x =. Now h we know Γ(2) = nd Γ(n+) = nγ(n), for ll n N, we cn compue ll of he Γ(n) s. Γ(2) = Γ(3) = Γ(2+)= 2Γ(2) = 2 Γ(4) = Γ(3+)= 3Γ(3) = 3 2 Γ(5) = Γ(4+)= 4Γ(4) = Γ(n) = (n) (n 2) = (n)! Exmple Convergence Tess for Improper Inegrls I is very common o encouner inegrls h re oo compliced o evlue explicily. Numericl pproximion schemes, evlued by compuer, re ofen used insed. You wn o be sure h les he inegrl converges before feeding i ino compuer. Forunely i is usully possible o deermine wheher or no n improper inegrl converges even when you cnno evlue i explicily. For pedgogicl purposes, we re going o concenre on he problem of deermining wheher or no n inegrl f(x) converges, when f(x) hs no singulriies for x. Recll h he firs sep in nlyzing ny improper inegrl is o wrie i s sum of inegrls ech of hs only single source of impropriey eiher domin of inegrion h exends o +, or domin of inegrion h exends o, or n inegrnd which is singulr one end of he domin of inegrion. So we re now going o consider only he firs of hese hree possibiliies. Bu he echniques h we re bou o see hve obvious nlogues for he oher wo possibiliies. c Joel Feldmn. 25. All righs reserved. 8 Mrch 6, 25
9 Now le s sr. Imgine h we hve n improper inegrl f(x), h f(x) hs no singulriies for x nd h f(x) is compliced enough h we cnno evlue he inegrl explicily. The ide is find noher improper inegrl g(x) wih g(x) simple enough h we cn evlue he inegrl g(x) explicily, or les deermine esily wheher or no g(x) converges, nd wih g(x) behving enough like f(x) for lrge x h he inegrl if nd only if g(x) converges. f(x) converges So fr, his is prey vgue sregy. Here is heorem which srs o mke i more precise. Theorem 2 (Comprison). Le be rel number. Le f nd g be funcions h re defined nd coninuous for ll x nd ssume h g(x) for ll x. () If f(x) g(x) for ll x nd if g(x) converges hen f(x) converges. (b) If f(x) g(x) for ll x nd if g(x) diverges hen f(x) diverges. We will no prove his heorem, bu, hopefully, he following supporing rgumens should les pper resonble o you. If g(x) converges, hen he re of { (x,y) } x, y g(x) is finie. When f(x) g(x), he region { (x,y) } x, y f(x) is conined inside { } (x,y) x, y g(x) nd so lso hs finie re. Consequenly he res of boh he regions { (x,y) } { } x, y f(x) nd (x,y) x, f(x) y re finie oo. (Look grph of f(x) nd grph of f(x).) The inegrl f(x) represens he signed re of he union of { (x,y) } x, y f(x) nd { } (x,y) x, f(x) y. If g(x) diverges, hen he re of { (x,y) x, y g(x) } is infinie. When f(x) g(x), he region { (x,y) x, y f(x) } conins he region { (x,y) x, y g(x) } nd so lso hs infinie re. Exmple 3 ( e x2 ) We cnno evlue he inegrl e x2 explicily. Bu we would sill like o ell if i is finie or no. So we wn o find noher inegrl h we cn compue nd h we cn compre o e x2. To do so we pick n inegrnd h looks like e x2, bu whose indefinie inegrl we know like e x. When x, we hve x 2 x nd hence e x2 e x. The inegrl [ ] R e x e x e x ] [e e R = e c Joel Feldmn. 25. All righs reserved. 9 Mrch 6, 25
10 converges. So, by Theorem 2, wih =, f(x) = e x2 nd g(x) = e x, he inegrl e x2 converges oo. Exmple 3 Exmple 4 ( /2 e x2 ) Of course he inegrl is quie similr o he inegrl e /2 e x2 x2 of Exmple 3. Bu we cnno jus repe he rgumen of Exmple 3 becuse i is no rue h e x2 e x when < x <. In fc, for < x <, x 2 < x so h e x2 > e x. However he difference beween nd e /2 e x2 x2 is excly which is clerly well defined /2 e x2 finie number. So we would expec h should be he sum of he proper inegrl /2 e x2 inegrl nd he convergen inegrl /2 e x2 e x2 nd so should be convergen inegrl. This is indeed he cse. The Theorem below provides he jusificion. Exmple 4 Theorem 5. Le < c nd le he funcion f(x) be coninuous for ll x. Then he improper inegrl f(x) converges if nd only if he improper inegrl f(x) converges. c Proof. By definiion he improper inegrl f(x) converges if nd only if he i [ c f(x) = c f(x) + f(x) + c c ] f(x) f(x) exiss nd is finie. (Remember h, in compuing he i, c f(x) is finie consn independen of R nd so cn be pulled ou of he i.) Bu h is he cse if nd only if he i f(x) exiss nd is finie, which in urn is he cse if nd only if he c inegrl f(x) converges. c Exmple 6 Does he inegrl x x 2 +x converge or diverge? Soluion. Our firs sk is o idenify he poenil sources of impropriey for his inegrl. Cerinly he domin of inegrion exends o +. Bu we mus lso check o see if he inegrnd conins ny singulriies. On he domin of inegrion x so he denominor is never zero nd he inegrnd is coninuous. So he only problem is +. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25
11 Our second sk is o develop some inuiion. As he only problem is h he domin of inegrion exends o infiniy, wheher or no he inegrl converges will be deermined by he behvior of he inegrnd for very lrge x. When x is very lrge x x 2 so h he denominor x 2 +x x 2 nd he inegrnd x x =. By Exmple 4, wih p = x 2 +x x 2 x 3 /2, 3/2 x x 2 +x he inegrl converges. So we would expec h converges oo. x 3/2 Our finl sk is o verify h our inuiion is correc. To do so, we wn o pply pr () of Theorem 2 wih f(x) = x nd g(x) being, or possibly some consn imes x 2 +x x 3/2. We will be ble o pply Theorem 2. if we cn show h x is smller hn some x 3/2 x 2 +x consn imes on he domin of inegrion, x. Here goes. x 3/2 x = x 2 +x > x 2 = x 2 +x < x x x = 2 x 2 +x < x = 2 x 3/2 So Theorem 2. nd Exmple 4, wih p = 3 /2 do indeed show h he inegrl converges. Exmple 6 x x 2 +x There is vrin of he comprison Theorem 2 h is ofen esier o pply hn Theorem 2 iself nd h lso fis well wih he sor of inuiion h we developed in Exmple 6. Le s briefly review wh hppened In Exmple 6. We firs idenified he problem x s s he lrge x s. (The inegrl ws improper only becuse he domin of inegrion exended o +.) Our inegrnd, f(x) = x, ws relively compliced bu x 2 +x we found h, for lrge x, f(x) behved like he simple funcion g(x) =. We knew x 3/2 h he inegrl of g(x), over he domin of inegrion of ineres, converged. We hen used Theorem 2 o show h he inegrl of f(x) converged oo. A good wy o formlize f(x) behves like g(x) for lrge x is o require h he i f(x) x g(x) exiss nd is nonzero number. Suppose h his is he cse nd cll he i L. Then mus pproch L s x ends o + nd, in priculr, here mus be number c such f(x) g(x) h f(x) lies beween L nd 2L, h is f(x) lies beween L g(x) nd 2Lg(x), for ll x c. g(x) 2 2 Consequenly, he inegrl of f(x) converges if nd only if he inegrl of g(x) converges, by Theorems 2 nd 5. These considerions led o he following vrin of Theorem 2. c Joel Feldmn. 25. All righs reserved. Mrch 6, 25
12 Theorem 7 ( Limiing comprison). Le < <. Le f nd g be funcions h re defined nd coninuous for ll x nd ssume h g(x) for ll x. () If g(x) converges nd he i f(x) x g(x) exiss (in his cse i is llowed o be ny rel number, including ), hen f(x) converges. (b) If g(x) diverges nd he i exiss nd is nonzero, hen f(x) x g(x) f(x) diverges. There re obvious nlogs of his heorem for he oher ypes of improper inegrls oo. Here is n exmple of how Theorem 7 is used. Exmple 8 Does he inegrl x+sin x converge or diverge? e x +x 2 Soluion. Our firs sk is o idenify he poenil sources of impropriey for his inegrl. The domin of inegrion exends o +. On he domin of inegrion he denominor is never zero so he inegrnd is coninuous. Thus he only problem is +. Our second sk is o develop some inuiion bou he behvior of he inegrnd for very lrge x. When x is very lrge e x x 2, so h he denominor e x +x 2 x 2 nd sinx x, so h he numeror x+sinx x nd he inegrnd x+sinx e x +x 2 x x 2 = x. Since diverges, we would expec x+sin x o diverge oo. x e x +x 2 Our finl sk is o verify h our inuiion is correc. To do so, we se f(x) = x+sinx e x +x 2 g(x) = x c Joel Feldmn. 25. All righs reserved. 2 Mrch 6, 25
13 nd compue f(x) x g(x) x+sinx x e x +x 2 x x (+sinx/x)x (e x /x 2 +)x 2 x x +sinx/x e x /x 2 + = Since g(x) = diverges, by Exmple 4 wih p =, Theorem 7.b now ells us x h f(x) = x+sin x diverges oo. e x +x 2 Exmple 8 c Joel Feldmn. 25. All righs reserved. 3 Mrch 6, 25
4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informationf(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2
Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe
More informatione t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls  hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
More information0 for t < 0 1 for t > 0
8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside
More informationENGR 1990 Engineering Mathematics The Integral of a Function as a Function
ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under
More information5.1The InitialValue Problems For Ordinary Differential Equations
5.The IniilVlue Problems For Ordinry Differenil Equions Consider solving iniilvlue problems for ordinry differenil equions: (*) y f, y, b, y. If we know he generl soluion y of he ordinry differenil
More information( ) ( ) ( ) ( ) ( ) ( y )
8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll
More informationf t f a f x dx By Lin McMullin f x dx= f b f a. 2
Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes
More informationMathematics 805 Final Examination Answers
. 5 poins Se he Weiersrss Mes. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se
More informationThe solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.
[~ o o : o o ill] i 1. Mrices, Vecors, nd GussJordn Eliminion 1 x y = =  z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries
More informationINTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).
INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples
More informationMath 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:
Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial
More informationSeptember 20 Homework Solutions
College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum
More informationREAL ANALYSIS I HOMEWORK 3. Chapter 1
REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs
More informationMotion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Moion Accelerion Pr : Consn Accelerion Accelerion Accelerion Accelerion is he re of chnge of velociy. = v  vo = Δv Δ ccelerion = = v  vo chnge of velociy elpsed ime Accelerion is vecor, lhough in onedimensionl
More informationMTH 146 Class 11 Notes
8. Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he
More informationMAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017
MAT 66 Clculus for Engineers II Noes on Chper 6 Professor: John Quigg Semeser: spring 7 Secion 6.: Inegrion by prs The Produc Rule is d d f()g() = f()g () + f ()g() Tking indefinie inegrls gives [f()g
More informationContraction Mapping Principle Approach to Differential Equations
epl Journl of Science echnology 0 (009) 4953 Conrcion pping Principle pproch o Differenil Equions Bishnu P. Dhungn Deprmen of hemics, hendr Rn Cmpus ribhuvn Universiy, Khmu epl bsrc Using n eension of
More informationProperties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)
Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =
More informationHow to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.
How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels:96377679,735 Emil:hf@scsne.org Commens: 3 ges SubjClss: Funcionl nlsis, comle
More informationChapter Direct Method of Interpolation
Chper 5. Direc Mehod of Inerpolion Afer reding his chper, you should be ble o:. pply he direc mehod of inerpolion,. sole problems using he direc mehod of inerpolion, nd. use he direc mehod inerpolns o
More information3. Renewal Limit Theorems
Virul Lborories > 14. Renewl Processes > 1 2 3 3. Renewl Limi Theorems In he inroducion o renewl processes, we noed h he rrivl ime process nd he couning process re inverses, in sens The rrivl ime process
More informationMATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence
MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,
More information1.0 Electrical Systems
. Elecricl Sysems The ypes of dynmicl sysems we will e sudying cn e modeled in erms of lgeric equions, differenil equions, or inegrl equions. We will egin y looking fmilir mhemicl models of idel resisors,
More informationSome basic notation and terminology. Deterministic Finite Automata. COMP218: Decision, Computation and Language Note 1
COMP28: Decision, Compuion nd Lnguge Noe These noes re inended minly s supplemen o he lecures nd exooks; hey will e useful for reminders ou noion nd erminology. Some sic noion nd erminology An lphe is
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationHow to prove the Riemann Hypothesis
Scholrs Journl of Phsics, Mhemics nd Sisics Sch. J. Phs. Mh. S. 5; (B:56 Scholrs Acdemic nd Scienific Publishers (SAS Publishers (An Inernionl Publisher for Acdemic nd Scienific Resources *Corresonding
More informationA Kalman filtering simulation
A Klmn filering simulion The performnce of Klmn filering hs been esed on he bsis of wo differen dynmicl models, ssuming eiher moion wih consn elociy or wih consn ccelerion. The former is epeced o beer
More information14. The fundamental theorem of the calculus
4. The funmenl heorem of he clculus V 20 00 80 60 40 20 0 0 0.2 0.4 0.6 0.8 v 400 200 0 0 0.2 0.5 0.8 200 400 Figure : () Venriculr volume for subjecs wih cpciies C = 24 ml, C = 20 ml, C = 2 ml n (b) he
More informationSolutions to Problems from Chapter 2
Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5
More information1 jordan.mcd Eigenvalueeigenvector approach to solving first order ODEs.  Jordan normal (canonical) form. Instructor: Nam Sun Wang
jordnmcd Eigenvlueeigenvecor pproch o solving firs order ODEs  ordn norml (cnonicl) form Insrucor: Nm Sun Wng Consider he following se of coupled firs order ODEs d d x x 5 x x d d x d d x x x 5 x x
More informationConvergence of Singular Integral Operators in Weighted Lebesgue Spaces
EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 2, 2017, 335347 ISSN 13075543 www.ejpm.com Published by New York Business Globl Convergence of Singulr Inegrl Operors in Weighed Lebesgue
More informationon the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?
Calculus II MAT 46 Improper Inegrals A mahemaician asked a fence painer o complee he unique ask of paining one side of a fence whose face could be described by he funcion y f (x on he inerval (x + x
More informationQuestion Details Int Vocab 1 [ ] Question Details Int Vocab 2 [ ]
/3/5 Assignmen Previewer 3 Bsic: Definie Inegrls (67795) Due: Wed Apr 5 5 9: AM MDT Quesion 3 5 6 7 8 9 3 5 6 7 8 9 3 5 6 Insrucions Red ody's Noes nd Lerning Gols. Quesion Deils In Vocb [37897] The chnge
More informationPhysics 2A HW #3 Solutions
Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 33 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen
More informationMATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion
More informationMATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)
MATH 14 AND 15 FINAL EXAM REVIEW PACKET (Revised spring 8) The following quesions cn be used s review for Mh 14/ 15 These quesions re no cul smples of quesions h will pper on he finl em, bu hey will provide
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationMagnetostatics Bar Magnet. Magnetostatics Oersted s Experiment
Mgneosics Br Mgne As fr bck s 4500 yers go, he Chinese discovered h cerin ypes of iron ore could rc ech oher nd cerin mels. Iron filings "mp" of br mgne s field Crefully suspended slivers of his mel were
More informationgraph of unit step function t
.5 Piecewie coninuou forcing funcion...e.g. urning he forcing on nd off. The following Lplce rnform meril i ueful in yem where we urn forcing funcion on nd off, nd when we hve righ hnd ide "forcing funcion"
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationSome Inequalities variations on a common theme Lecture I, UL 2007
Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel
More informationPHYSICS 1210 Exam 1 University of Wyoming 14 February points
PHYSICS 1210 Em 1 Uniersiy of Wyoming 14 Februry 2013 150 poins This es is opennoe nd closedbook. Clculors re permied bu compuers re no. No collborion, consulion, or communicion wih oher people (oher
More informationWeek #13  Integration by Parts & Numerical Integration Section 7.2
Week #3  Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by HughesHalle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission
More informationChapter 2. Motion along a straight line. 9/9/2015 Physics 218
Chper Moion long srigh line 9/9/05 Physics 8 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy
More informationChapter 2: Evaluative Feedback
Chper 2: Evluive Feedbck Evluing cions vs. insrucing by giving correc cions Pure evluive feedbck depends olly on he cion ken. Pure insrucive feedbck depends no ll on he cion ken. Supervised lerning is
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationP441 Analytical Mechanics  I. Coupled Oscillators. c Alex R. Dzierba
Lecure 3 Mondy  Deceber 5, 005 Wrien or ls upded: Deceber 3, 005 P44 Anlyicl Mechnics  I oupled Oscillors c Alex R. Dzierb oupled oscillors  rix echnique In Figure we show n exple of wo coupled oscillors,
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationLAPLACE TRANSFORMS. 1. Basic transforms
LAPLACE TRANSFORMS. Bic rnform In hi coure, Lplce Trnform will be inroduced nd heir properie exmined; ble of common rnform will be buil up; nd rnform will be ued o olve ome dierenil equion by rnforming
More informationChallenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k
Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationChapter 2. First Order Scalar Equations
Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.
More informationA LIMITPOINT CRITERION FOR A SECONDORDER LINEAR DIFFERENTIAL OPERATOR IAN KNOWLES
A LIMITPOINT CRITERION FOR A SECONDORDER LINEAR DIFFERENTIAL OPERATOR j IAN KNOWLES 1. Inroducion Consider he forml differenil operor T defined by el, (1) where he funcion q{) is relvlued nd loclly
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationHonours Introductory Maths Course 2011 Integration, Differential and Difference Equations
Honours Inroducory Mhs Course 0 Inegrion, Differenil nd Difference Equions Reding: Ching Chper 4 Noe: These noes do no fully cover he meril in Ching, u re men o supplemen your reding in Ching. Thus fr
More informationPredator  Prey Model Trajectories and the nonlinear conservation law
Predaor  Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories
More informationdy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page
Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,
More informationA 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m
PHYS : Soluions o Chper 3 Home Work. SSM REASONING The displcemen is ecor drwn from he iniil posiion o he finl posiion. The mgniude of he displcemen is he shores disnce beween he posiions. Noe h i is onl
More informationLet us start with a two dimensional case. We consider a vector ( x,
Roaion marices We consider now roaion marices in wo and hree dimensions. We sar wih wo dimensions since wo dimensions are easier han hree o undersand, and one dimension is a lile oo simple. However, our
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information2.7. Some common engineering functions. Introduction. Prerequisites. Learning Outcomes
Some common engineering funcions 2.7 Inroducion This secion provides a caalogue of some common funcions ofen used in Science and Engineering. These include polynomials, raional funcions, he modulus funcion
More informationAverage & instantaneous velocity and acceleration Motion with constant acceleration
Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission
More informationInventory Analysis and Management. MultiPeriod Stochastic Models: Optimality of (s, S) Policy for KConvex Objective Functions
MuliPeriod Sochasic Models: Opimali of (s, S) Polic for Convex Objecive Funcions Consider a seing similar o he Nsage newsvendor problem excep ha now here is a fixed reordering cos (> 0) for each (re)order.
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationSection 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients
Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous
More informationEchocardiography Project and Finite Fourier Series
Echocardiography Projec and Finie Fourier Series 1 U M An echocardiagram is a plo of how a porion of he hear moves as he funcion of ime over he one or more hearbea cycles If he hearbea repeas iself every
More informationSOLUTIONS TO ECE 3084
SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no
More informationFinal Spring 2007
.615 Final Spring 7 Overview The purpose of he final exam is o calculae he MHD β limi in a highbea oroidal okamak agains he dangerous n = 1 exernal ballooningkink mode. Effecively, his corresponds o
More information1. Find a basis for the row space of each of the following matrices. Your basis should consist of rows of the original matrix.
Mh 7 Exm  Prcice Prolem Solions. Find sis for he row spce of ech of he following mrices. Yor sis shold consis of rows of he originl mrix. 4 () 7 7 8 () Since we wn sis for he row spce consising of rows
More informationu(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x
. 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih
More informationSection 5: Chain Rule
Chaper The Derivaive Applie Calculus 11 Secion 5: Chain Rule There is one more ype of complicae funcion ha we will wan o know how o iffereniae: composiion. The Chain Rule will le us fin he erivaive of
More informationGuest Lectures for Dr. MacFarlane s EE3350 Part Deux
Gues Lecures for Dr. MacFarlane s EE3350 Par Deux Michael Plane Mon., 08302010 Wrie name in corner. Poin ou his is a review, so I will go faser. Remind hem o go lisen o online lecure abou geing an A
More informationGreen s Functions and Comparison Theorems for Differential Equations on Measure Chains
Green s Funcions nd Comprison Theorems for Differenil Equions on Mesure Chins Lynn Erbe nd Alln Peerson Deprmen of Mhemics nd Sisics, Universiy of NebrskLincoln Lincoln,NE 685880323 lerbe@@mh.unl.edu
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationSolutions to Assignment 1
MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More information1. Consider a PSA initially at rest in the beginning of the lefthand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.
In Eercise 1, use sndrd recngulr Cresin coordine sysem. Le ime be represened long he horizonl is. Assume ll ccelerions nd decelerions re consn. 1. Consider PSA iniilly res in he beginning of he lefhnd
More informationdt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.
Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies
More informationChapter 7: Solving Trig Equations
Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationHamilton J acobi Equation: Weak S olution We continue the study of the HamiltonJacobi equation:
M ah 5 7 Fall 9 L ecure O c. 4, 9 ) Hamilon J acobi Equaion: Weak S oluion We coninue he sudy of he HamilonJacobi equaion: We have shown ha u + H D u) = R n, ) ; u = g R n { = }. ). In general we canno
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationBiol. 356 Lab 8. Mortality, Recruitment, and Migration Rates
Biol. 356 Lab 8. Moraliy, Recruimen, and Migraion Raes (modified from Cox, 00, General Ecology Lab Manual, McGraw Hill) Las week we esimaed populaion size hrough several mehods. One assumpion of all hese
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationUnit Root Time Series. Univariate random walk
Uni Roo ime Series Univariae random walk Consider he regression y y where ~ iid N 0, he leas squares esimae of is: ˆ yy y y yy Now wha if = If y y hen le y 0 =0 so ha y j j If ~ iid N 0, hen y ~ N 0, he
More information22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 9: The High Beta Tokamak
.65, MHD Theory of Fusion Sysems Prof. Freidberg Lecure 9: The High e Tokmk Summry of he Properies of n Ohmic Tokmk. Advnges:. good euilibrium (smll shif) b. good sbiliy ( ) c. good confinemen ( τ nr )
More informationFinish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details!
MAT 257, Handou 6: Ocober 72, 20. I. Assignmen. Finish reading Chaper 2 of Spiva, rereading earlier secions as necessary. handou and fill in some missing deails! II. Higher derivaives. Also, read his
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationEXERCISES FOR SECTION 1.5
1.5 Exisence and Uniqueness of Soluions 43 20. 1 v c 21. 1 v c 1 2 4 6 8 10 1 2 2 4 6 8 10 Graph of approximae soluion obained using Euler s mehod wih = 0.1. Graph of approximae soluion obained using Euler
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More information