DIFFERENTIATING UNDER THE INTEGRAL SIGN


 Paulina Cunningham
 4 years ago
 Views:
Transcription
1 DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I hd lerned to do integrls by vrious methods shown in book tht my high school physics techer Mr. Bder hd given me. [It] showed how to differentite prmeters under the integrl sign it s certin opertion. It turns out tht s not tught very much in the universities; they don t emphsize it. But I cught on how to use tht method, nd I used tht one dmn tool gin nd gin. [If] guys t MIT or Princeton hd trouble doing certin integrl, [then] I come long nd try differentiting under the integrl sign, nd often it worked. So I got gret reputtion for doing integrls, only becuse my bo of tools ws different from everybody else s, nd they hd tried ll their tools on it before giving the problem to me. ichrd Feynmn [, pp. 71 7] 1. Introduction The method of differentition under the integrl sign, due originlly to Leibniz, concerns integrls depending on prmeter, such s 1 e t d. Here t is the etr prmeter. (Since is the vrible of integrtion, is not prmeter.) In generl, we might write such n integrl s (1.1) b f(, t) d, where f(, t) is function of two vribles like f(, t) e t. Emple 1.1. Let f(, t) ( + t 3 ). Then 1 f(, t) d 1 ( + t 3 ) d. An ntiderivtive of ( + t 3 ) with respect to is 1 6 ( + t3 ) 3, so 1 ( + t 3 ) d ( + t3 ) ( + t3 ) 3 t t3 + t 6. This nswer is function of t, which mkes sense since the integrnd depends on t. We integrte over nd re left with something tht depends only on t, not. An integrl like b f(, t) d is function of t, so we cn sk bout its tderivtive, ssuming tht f(, t) is nicely behved. The rule is: the tderivtive of the integrl of f(, t) is the integrl of the tderivtive of f(, t): (1.) d dt b f(, t) d 1 b f(, t) d. t
2 KEITH CONAD This is clled differentition under the integrl sign. If you re used to thinking mostly bout functions with one vrible, not two, keep in mind tht (1.) involves integrls nd derivtives with respect to seprte vribles: integrtion with respect to nd differentition with respect to t. Emple 1.. We sw in Emple 1.1 tht 1 ( + t3 ) d 4/3 + t 3 + t 6, whose tderivtive is 6t + 6t 5. According to (1.), we cn lso compute the tderivtive of the integrl like this: The nswers gree. (.1) d dt 1 ( + t 3 ) d t ( + t3 ) d ( + t 3 )(3t ) d (1t + 6t 5 ) d 6t + 6t 5 1 6t + 6t 5.. Euler s fctoril integrl in new light For integers n, Euler s integrl formul for n! is n e d n!, which cn be obtined by repeted integrtion by prts strting from the formul (.) e d 1 when n. Now we re going to derive Euler s formul in nother wy, by repeted differentition fter introducing prmeter t into (.). For ny t >, let tu. Then d t du nd (.) becomes te tu du 1. Dividing by t nd writing u s (why is this not problem?), we get (.3) e t d 1 t. This is prmetric form of (.), where both sides re now functions of t. We need t > in order tht e t is integrble over the region. Now we bring in differentition under the integrl sign. Differentite both sides of (.3) with respect to t, using (1.) to tret the left side. We obtin so (.4) e t d 1 t, e t d 1 t.
3 DIFFEENTIATING UNDE THE INTEGAL SIGN 3 Differentite both sides of (.4) with respect to t, gin using (1.) to hndle the left side. We get Tking out the sign on both sides, (.5) e t d t 3. e t d t 3. If we continue to differentite ech new eqution with respect to t few more times, we obtin nd Do you see the pttern? It is (.6) 3 e t d 6 t 4, 4 e t d 4 t 5, 5 e t d 1 t 6. n e t d n! t n+1. We hve used the presence of the etr vrible t to get these equtions by repetedly pplying d/dt. Now specilize t to 1 in (.6). We obtin n e d n!, which is our old friend (.1). Voilá! The ide tht mde this work is introducing prmeter t, using clculus on t, nd then setting t to prticulr vlue so it disppers from the finl formul. In other words, sometimes to solve problem it is useful to solve more generl problem. Compre (.1) to (.6). 3. A dmped sine integrl We re going to use differentition under the integrl sign to prove t sin e d π rctn t for t >. Cll this integrl F (t) nd set f(, t) e t (sin )/, so (/t)f(, t) e t sin. Then F (t) e t (sin ) d. The integrnd e t sin, s function of, cn be integrted by prts: e ( sin cos ) sin d 1 + e. Applying this with t nd turning the indefinite integrl into definite integrl, F (t) e t (t sin + cos ) (sin ) d 1 + t e t.
4 4 KEITH CONAD As, t sin + cos oscilltes lot, but in bounded wy (since sin nd cos re bounded functions), while the term e t decys eponentilly to since t >. So the vlue t is. Therefore F (t) e t (sin ) d t. We know n eplicit ntiderivtive of 1/(1 + t ), nmely rctn t. Since F (t) hs the sme tderivtive s rctn t, they differ by constnt: for some number C, (3.1) t sin e d rctn t + C for t >. We ve computed the integrl, up to n dditive constnt, without finding n ntiderivtive of e t (sin )/. To compute C in (3.1), let t on both sides. Since (sin )/ 1, the bsolute vlue of the integrl on the left is bounded from bove by e t d 1/t, so the integrl on the left in (3.1) tends to s t. Since rctn t π/ s t, eqution (3.1) s t becomes π + C, so C π/. Feeding this bck into (3.1), t sin (3.) e d π rctn t for t >. If we let t + in (3.), this eqution suggests tht (3.3) sin d π, which is true nd it is importnt in signl processing nd Fourier nlysis. It is delicte mtter to derive (3.3) from (3.) since the integrl in (3.3) is not bsolutely convergent. Detils re provided in n ppendi. (4.1) The improper integrl formul 4. The Gussin integrl e / d π 1 is fundmentl to probbility theory nd Fourier nlysis. The function π e / is clled Gussin, nd (4.1) sys the integrl of the Gussin over the whole rel line is 1. The physicist Lord Kelvin (fter whom the Kelvin temperture scle is nmed) once wrote (4.1) on the bord in clss nd sid A mthemticin is one to whom tht [pointing t the formul] is s obvious s twice two mkes four is to you. We will prove (4.1) using differentition under the integrl sign. The method will not mke (4.1) s obvious s 4. If you tke further courses you my lern more nturl derivtions of (4.1) so tht the result relly does become obvious. For now, just try to follow the rgument here stepbystep. We re going to im not t (4.1), but t n equivlent formul over the rnge : π π (4.) e / d. Cll the integrl on the left I. For t, set F (t) e t (1+ )/ 1 + d.
5 DIFFEENTIATING UNDE THE INTEGAL SIGN 5 Then F () d/(1 + ) π/ nd F ( ). Differentiting under the integrl sign, F (t) te t (1+ )/ d te t / Mke the substitution y t, with dy t d, so F (t) e t / e y / dy Ie t /. e (t) / d. For b >, integrte both sides from to b nd use the Fundmentl Theorem of Clculus: Letting b, (5.1) b F (t) dt I b e t / dt F (b) F () I π I I π I π. b e t / dt. I lerned this from Michel ozmn [6], who modified n ide on mth.stckechnge [9]. 5. Higher moments of the Gussin For every integer n we wnt to compute formul for n e / d. (Integrls of the type n f() d for n, 1,,... re clled the moments of f(), so (5.1) is the nth moment of the Gussin.) When n is odd, (5.1) vnishes since n e / is n odd function. Wht if n,, 4,... is even? The first cse, n, is the Gussin integrl (4.1): (5.) e / d π. To get formuls for (5.1) when n, we follow the sme strtegy s our tretment of the fctoril integrl in Section : stick t into the eponent of e / nd then differentite repetedly with respect to t. For t >, replcing with t in (5.) gives π (5.3) e t / d. t Differentite both sides of (5.3) with respect to t, using differentition under the integrl sign on the left: π / e t d t 3/, so π (5.4) e t / d t 3/. Differentite both sides of (5.4) with respect to t. After removing common fctor of 1/ on both sides, we get (5.5) 4 e t / d 3 π t 5/.
6 6 KEITH CONAD Differentiting both sides of (5.5) with respect to t few more times, we get nd Quite generlly, when n is even 6 e t / d 3 5 π t 7/, 8 e t / d π t 9/, 1 e t / d π t 11/. n e t / d (n 1) π t n/ t, where the numertor is the product of the positive odd integers from 1 to n 1 (understood to be the empty product 1 when n ). In prticulr, tking t 1 we hve computed (5.1): n e / d (n 1) π. As n ppliction of (5.4), we now compute ( 1 )! : 1/ e d, where the nottion ( 1 )! nd its definition re inspired by Euler s integrl formul (.1) for n! when n is nonnegtive integer. Using the substitution u 1/ in 1/ e d, we hve ( ) 1! 1/ e d We re going to compute by looking t its tderivtive: ue u (u) du u e u du u e u du π 3/ by (5.4) t t π. 6. A cosine trnsform of the Gussin F (t) (6.1) F (t) cos(t)e / d sin(t)e / d.
7 DIFFEENTIATING UNDE THE INTEGAL SIGN 7 This is good from the viewpoint of integrtion by prts since e / is the derivtive of e /. So we pply integrtion by prts to (6.1): nd Then u sin(t), dv e d du t cos(t) d, v e /. F (t) uv u dv sin(t) e / sin(t) e / v du t tf (t). cos(t)e / d As, e / blows up while sin(t) stys bounded, so sin(t)/e / goes to. Therefore F (t) tf (t). We know the solutions to this differentil eqution: constnt multiples of e t /. So cos(t)e / d Ce t / for some constnt C. To find C, set t. The left side is e / d, which is π/ by (4.). The right side is C. Thus C π/, so we re done: for ll rel t, π cos(t)e / d / e t. emrk 6.1. If we wnt to compute G(t) sin(t)e / d, with sin(t) in plce of cos(t), then in plce of F (t) tf (t) we hve G (t) 1 tg(t), nd G(). From the differentil eqution, (e t / G(t)) e t /, so G(t) e t / t e / d. So while cos(t)e / d π e t /, the integrl sin(t)e / d is impossible to epress in terms of elementry functions. 7. Logs in the denomintor, prt I Consider the following integrl over [, 1], where t > : 1 t 1 log d. Since 1/ log s +, the integrnd vnishes t. As 1, ( t 1)/ log t. Therefore when t is fied the integrnd is continuous function of on [, 1], so the integrl is not n improper integrl. The tderivtive of this integrl is 1 t log log d 1 t d 1 t + 1,
8 8 KEITH CONAD which we recognize s the tderivtive of log(t + 1). Therefore 1 t 1 log d log(t + 1) + C for some C. To find C, let t +. On the right side, log(1 + t) tends to. On the left side, the integrnd tends to : ( t 1)/ log (e t log 1)/ log t becuse e 1 when. Therefore the integrl on the left tends to s t +. So C, which implies (7.1) 1 t 1 log d log(t + 1) for ll t >, nd it s obviously lso true for t. Another wy to compute this integrl is to write t e t log s power series nd integrte term by term, which is vlid for 1 < t < 1. Under the chnge of vribles e y, (7.1) becomes ( (7.) e y e (t+1)y) dy log(t + 1). y 8. Logs in the denomintor, prt II We now consider the integrl d F (t) t log for t > 1. The integrl converges by comprison with d/ t. We know tht t t 1 the integrl diverges to : d b log lim d b log b lim log log b lim b log log b log log. So we epect tht s t 1 +, F (t) should blow up. But how does it blow up? By nlyzing F (t) nd then integrting bck, we re going to show F (t) behves essentilly like log(t 1) s t 1 +. Using differentition under the integrl sign, for t > 1 ( ) F 1 (t) t t d log t ( log ) d log d t t+1 t t 1 t. We wnt to bound this derivtive from bove nd below when t > 1. Then we will integrte to get bounds on the size of F (t).
9 DIFFEENTIATING UNDE THE INTEGAL SIGN 9 For t > 1, the difference 1 t is negtive, so 1 t < 1. Dividing both sides of this by 1 t, which is negtive, reverses the sense of the inequlity nd gives 1 t 1 t > 1 1 t. This is lower bound on F (t). To get n upper bound on F (t), we wnt to use lower bound on 1 t. Since e + 1 for ll (the grph of y e lies on or bove its tngent line t, which is y + 1), for ll. Tking 1 t, e log (log ) + 1 (8.1) 1 t (log )(1 t) + 1. When t > 1, 1 t is negtive, so dividing (8.1) by 1 t reverses the sense of the inequlity: 1 t t 1 log t. This is n upper bound on F (t). Putting the upper nd lower bounds on F (t) together, (8.) 1 1 t < F (t) log t for ll t > 1. We re concerned with the behvior of F (t) s t 1 +. Let s integrte (8.) from to, where 1 < < : dt 1 t < F (t) dt Using the Fundmentl Theorem of Clculus, log(t 1) < F (t) so ( log + 1 ) dt. 1 t ((log )t log(t 1)) log( 1) < F () F () (log )( ) + log( 1). Mnipulting to get inequlities on F (), we hve (log )( ) log( 1) + F () F () < log( 1) + F () Since > 1 for 1 < <, (log )( ) is greter thn log. This gives the bounds Writing s t, we get log( 1) + F () log F () < log( 1) + F () log(t 1) + F () log F (t) < log(t 1) + F (), so F (t) is bounded distnce from log(t 1) when 1 < t <. In prticulr, F (t) s t 1 +.,
10 1 KEITH CONAD 9. Smoothly dividing by t Let h(t) be n infinitely differentible function for ll rel t such tht h(). The rtio h(t)/t mkes sense for t, nd it lso cn be given resonble mening t t : from the very definition of the derivtive, when t we hve Therefore the function h(t) t r(t) h(t) h() h (). t { h(t)/t, if t, h (), if t is continuous for ll t. We cn see immeditely from the definition of r(t) tht it is better thn continuous when t : it is infinitely differentible when t. The question we wnt to ddress is this: is r(t) infinitely differentible t t too? If h(t) hs power series representtion round t, then it is esy to show tht r(t) is infinitely differentible t t by working with the series for h(t). Indeed, write h(t) c 1 t + c t + c 3 t 3 + for ll smll t. Here c 1 h (), c h ()/! nd so on. For smll t, we divide by t nd get (9.1) r(t) c 1 + c t + c 3 t 3 +, which is power series representtion for r(t) for ll smll t. The vlue of the right side of (9.1) t t is c 1 h (), which is lso the defined vlue of r(), so (9.1) is vlid for ll smll (including t ). Therefore r(t) hs power series representtion round (it s just the power series for h(t) t divided by t). Since functions with power series representtions round point re infinitely differentible t the point, r(t) is infinitely differentible t t. However, this is n incomplete nswer to our question bout the infinite differentibility of r(t) t t becuse we know by the key emple of e 1/t (t t ) tht function cn be infinitely differentible t point without hving power series representtion t the point. How re we going to show r(t) h(t)/t is infinitely differentible t t if we don t hve power series to help us out? Might there ctully be counteremple? The solution is to write h(t) in very clever wy using differentition under the integrl sign. Strt with h(t) t h (u) du. (This is correct since h().) For t, introduce the chnge of vribles u t, so du t d. At the boundry, if u then. If u t then 1 (we cn divide the eqution t t by t becuse t ). Therefore Dividing by t when t, we get h(t) 1 h (t)t d t 1 h (t) d. r(t) h(t) h (t) d. t The left nd right sides don t hve ny t in the denomintor. Are they equl t t too? The left side t t is r() h (). The right side is 1 h () d h () too, so (9.) r(t) 1 1 h (t) d
11 DIFFEENTIATING UNDE THE INTEGAL SIGN 11 for ll t, including t. This is formul for h(t)/t where there is no longer t being divided! Now we re set to use differentition under the integrl sign. The wy we hve set things up here, we wnt to differentite with respect to t; the integrtion vrible on the right is. We cn use differentition under the integrl sign on (9.) when the integrnd is differentible. Since the integrnd is infinitely differentible, r(t) is infinitely differentible! Eplicitly, nd nd more generlly r (t) r (t) r (k) (t) In prticulr, r (k) () 1 k h (k+1) () d h(k+1) () k+1. vh (t) d vh (t) d k h (k+1) (t) d. 1. Counteremples We hve seen mny emples where differentition under the integrl sign cn be crried out with interesting results, but we hve not ctully stted conditions under which (1.) is vlid. Something does need to be checked. In [8], n incorrect use of differentition under the integrl sign due to Cuchy is discussed, where divergent integrl is evluted s finite epression. Here re two other emples where differentition under the integrl sign does not work. Emple 1.1. It is pointed out in [3, Emple 6] tht the formul sin d π, which we discussed t the end of Section 3, leds to n erroneous instnce of differentition under the integrl sign. ewrite the formul s (1.1) sin(ty) y dy π for ny t >, by the chnge of vribles ty. Then differentition under the integrl sign implies which doesn t mke sense. cos(ty) dy, The net emple shows tht even if both sides of (1.) mke sense, they need not be equl. Emple 1.. For ny rel numbers nd t, let t 3 f(, t) ( + t, if or t, ), if nd t. Let F (t) 1 f(, t) d.
12 1 KEITH CONAD For instnce, F () 1 f(, ) d 1 d. When t, 1 t 3 F (t) ( + t ) d 1+t t 3 t t3 u u du (where u + t ) u1+t ut t 3 (1 + t ) + t3 t t (1 + t ). This formul lso works t t, so F (t) t/((1 + t )) for ll t. Therefore F (t) is differentible nd F (t) 1 t (1 + t ) f(, t) d. Since f(, t) for ll t, f(, t) is differen f(, t). For, f(, t) is differentible in t nd for ll t. In prticulr, F () 1. Now we compute t f(, t) nd then 1 tible in t nd t t Combining both cses ( nd ), (1.) t f(, t) ( + t ) (3t ) t 3 ( + t )t ( + t ) 4 t ( + t )(3( + t ) 4t ) ( + t ) 4 t (3 t ) ( + t ) 3. f(, t) t { t (3 t ), ( +t ) 3 if,, if. In prticulr t t f(, t). Therefore t t the left side of the formul d dt 1 is F () 1/ nd the right side is 1 f(, t) d t 1 f(, t) d. t t f(, t) d. The two sides re unequl! The problem in this emple is tht tf(, t) is not continuous function of (, t). Indeed, the denomintor in the formul in (1.) is ( + t ) 3, which hs problem ner (, ). Specificlly, while this derivtive vnishes t (, ), it we let (, t) (, ) long the line t, then on this line tf(, t) hs the vlue 1/(4), which does not tend to s (, t) (, ). Theorem 1.3. The eqution d dt b f(, t) d b f(, t) d t
13 DIFFEENTIATING UNDE THE INTEGAL SIGN 13 is vlid t t t, in the sense tht both sides eist nd re equl, provided the following two conditions hold: f(, t) nd tf(, t) re continuous functions of two vribles when is in the rnge of integrtion nd t is in some intervl round t, there re upper bounds f(, t) A() nd tf(, t) B(), both being independent of t, such tht b A() d nd b B() d eist. Proof. See [4, pp ]. If the intervl of integrtion is infinite, b A() d nd b B() d re improper. In Tble 1 we include choices for A() nd B() for ech of the functions we hve treted. Since the clcultion of derivtive t point only depends on n intervl round the point, we hve replced trnge such s t > with t c > in some cses to obtin choices for A() nd B(). Section f(, t) rnge t rnge t we wnt A() B() n e t [, ) t c > 1 n e c n+1 e c 3 e t sin (, ) t c > e c e c e 4 t (1+ ) 1 [, 1] t c t c 5 n e t t c > 1 n e c n+ e c 6 cos(t)e / [, ) ll t e / e / 7 t 1 log (, 1] < t < c 1 1 c log t log [, ) t c > 1 t > 1 log 9 k h (k+1) (t) [, 1] t < c m Tble 1. Summry y c h(k+1) (y) 1 c m y c h(k+) (y) Corollry 1.4. If (t) nd b(t) re both differentible, then d dt b(t) (t) f(, t) d b(t) (t) if the following conditions re stisfied: t f(, t) d + f(b(t), t)b (t) f((t), t) (t) there re α < β nd c 1 < c such tht f(, t) nd t f(, t) re continuous on [α, β] (c 1, c ), we hve (t) [α, β] nd b(t) [α, β] for ll t (c 1, c ), there re upper bounds f(, t) A() nd t f(, t) B() for (, t) [α, β] (c 1, c ) such tht β α A() d nd β α B() d eist. Proof. This is consequence of Theorem 1.3 nd the chin rule for multivrible functions. Set function of three vribles I(t,, b) b f(, t) d for (t,, b) (c 1, c ) [α, β] [α, β]. (Here nd b re not functions of t.) Then (1.3) I (t,, b) t b I f(, t) d, t I (t,, b) f(, t), (t,, b) f(b, t), b
14 14 KEITH CONAD where the first formul follows from Theorem 1.3 (its hypotheses re stisfied for ech nd b in [α, β]) nd the second nd third formuls re the Fundmentl Theorem of Clculus. For differentible functions (t) nd b(t) with vlues in [α, β] for c 1 < t < c, by the chin rule d dt b(t) (t) f(, t) d d I(t, (t), b(t)) dt I dt (t, (t), b(t)) t dt + I d (t, (t), b(t)) dt + I db (t, (t), b(t)) b dt b(t) f t (, t) d f((t), t) (t) + f(b(t), t)b (t) by (1.3). (t) A version of differentition under the integrl sign for t comple vrible is in [5, pp ]. Emple 1.5. For prmetric integrl t f(, t) d, where is fied, Corollry 1.4 tells us tht (1.4) d dt t f(, t) d t f(, t) d + f(t, t) t provided the hypotheses of the corollry re stisfied: (i) there re α < β nd c 1 < c such tht f nd f/t re continuous for (, t) [α, β] (c 1, c ), (ii) α β nd (c 1, c ) [α, β], nd (iii) there re bounds f(, t) A() nd t f(, t) B() for (, t) [α, β] (c 1, c ) such tht the integrls β α A() d nd β α B() d both eist. Let s pply this to the integrl F (t) t log(1 + t) 1 + d where t >. Here f(, t) log(1 + t)/(1 + ) nd t f(, t) 1 (1+t)(1+ ). Fi c > nd use α c 1 nd β c c. Then we cn justify (1.4) for (, t) [, c] (, c) using A() log(1 + c )/(1 + ) nd B() c/(1 + ), so F (t) t t (1 + t)(1 + ) d + log(1 + t ) 1 + t ( 1 t 1 + t 1 + t + t + ) 1 + d + log(1 + t ) 1 + t. Splitting up the integrl into three terms, ( 1 F t (t) log(1 + t) t 1 + t rctn() + log(1 + ) ) t (1 + t ) log(1 + t ) 1 + t + t rctn(t) 1 + t + log(1 + t ) (1 + t ) + log(1 + t ) 1 + t t rctn(t) 1 + t + log(1 + t ) (1 + t ). Clerly F (), so by the Fundmentl Theorem of Clculus t t ( y rctn(y) F (t) F (y) dy 1 + y + log(1 + ) y ) (1 + y dy. ) + log(1 + t ) 1 + t
15 DIFFEENTIATING UNDE THE INTEGAL SIGN 15 Using integrtion by prts on the first integrnd with u rctn(y) nd dv t t t F (t) uv log(1 + y ) v du + (1 + y ) dy rctn(y) log(1 + y t ) t log(1 + y ) t (1 + y ) dy + 1 rctn(t) log(1 + t ). y 1+y dy, log(1 + y ) (1 + y ) dy Thus for < t < c we hve t log(1 + t) 1 + d 1 rctn(t) log(1 + t ). Since c ws rbitrry, this eqution holds for ll t > (nd trivilly t t too). Setting t 1, 1 log(1 + ) 1 + d 1 π log rctn(1) log The Fundmentl Theorem of Algebr By differentiting under the integrl sign we will deduce the fundmentl theorem of lgebr: nonconstnt polynomil p(z) with coefficients in C hs root in C. The proof is due to Schep [7]. Arguing by contrdiction, ssume p(z) for ll z C. For r, consider the following integrl round circle of rdius r centered t the origin: I(r) π dθ p(re iθ ). This integrl mkes sense since the denomintor is never, so 1/p(z) is continuous on C. Let f(θ, r) 1/p(re iθ ), so I(r) π f(θ, r) dθ. We will prove three properties of I(r): (1) Theorem 1.3 cn be pplied to I(r) for r >, () I(r) s r, (3) I(r) I() s r + (continuity t r ). Tking these for grnted, let s see how contrdiction occurs. For r >, Since for r > we hve I (r) π I (r) f(θ, r) dθ r π f(θ, r) dθ r π p (re iθ )e iθ p(re iθ ) θ f(θ, r) p (re iθ ) p(re iθ ) ireiθ ir f(θ, r), r π 1 ir θ f(θ, r) dθ 1 f(θ, r) ir θπ θ dθ. 1 ir ( 1 p(r) 1 ). p(r) Thus I(r) is constnt for r >. Since I(r) s r, the constnt is zero: I(r) for r >. Since I(r) I() s r + we get I(), which is flse since I() π/p(). It remins to prove the three properties of I(r). (1) Theorem 1.3 cn be pplied to I(r) for r > :
16 16 KEITH CONAD Since p(z) nd p (z) re both continuous on C, the functions f(θ, r) nd (/r)f(θ, r) re continuous for θ [, π] nd ll r. This confirms the first condition in Theorem 1.3. For ech r > the set {(θ, r) : θ [, π], r [, r ] is closed nd bounded, so the functions f(θ, r) nd (/r)f(θ, r) re both bounded bove by constnt (independent of r nd θ) on this set. The rnge of integrtion [, π] is finite, so the second condition in Theorem 1.3 is stisfied using constnts for A(θ) nd B(θ). () I(r) s r : Let p(z) hve leding term cz d, with d deg p(z) 1. As r, p(re iθ ) / re iθ d c >, so for ll lrge r we hve p(re iθ ) c r d /. For such lrge r, I(r) π dθ p(re iθ ) π dθ c r d / 4π c r d, nd the upper bound tends to s r since d >, so I(r) s r. (3) I(r) I() s r + : For r >, π ( 1 (11.1) I(r) I() p(re iθ ) 1 ) π dθ I(r) I() 1 p() p(re iθ ) 1 p() dθ. Since 1/p(z) is continuous t, for ny ε > there is δ > such tht z < δ 1/p(z) 1/p() < ε. Therefore if < r < δ, (11.1) implies I(r) I() π ε dθ πε. 1. An emple needing chnge of vribles Our net emple is tken from [1, pp. 78,84]. For ll t, we will show by differentition under the integrl sign tht cos(t) (1.1) 1 + d πe t. Here f(, t) cos(t)/(1 + ). Since f(, t) is continuous nd f(, t) 1/(1 + ), the integrl eists for ll t. The function πe t is not differentible t t, so we shouldn t epect to be ble to prove (1.1) t t using differentition under the integrl sign; this specil cse cn be treted with elementry clculus: d 1 + rctn π. The integrl in (1.1) is n even function of t, so to compute it for t it suffices to tret the cse t >. 1 Let cos(t) F (t) 1 + d. If we try to compute F (t) for t > using differentition under the integrl sign, we get ( ) (1.) F (t)? cos(t) sin(t) t 1 + d 1 + d. Unfortuntely, there is no upper bound tf(, t) B() tht justifies differentiting F (t) under the integrl sign (or even justifies tht F (t) is differentible). Indeed, when is ner lrge odd multiple of (π/)/t, the integrnd in (1.) hs vlues tht re pproimtely /(1 + ) 1/, which is not integrble for lrge. Tht does not men (1.) is ctully flse, lthough if we weren t lredy told the nswer on the right side of (1.1) then we might be suspicious bout 1 A reder who knows comple nlysis cn derive (1.1) for t > by the residue theorem, viewing cos(t) s the rel prt of e it.
17 DIFFEENTIATING UNDE THE INTEGAL SIGN 17 whether the integrl is differentible for ll t > ; fter ll, you cn t esily tell from the integrl tht it is not differentible t t. Hving lredy rised suspicions bout (1.), we cn get something relly crzy if we differentite under the integrl sign second time: F (t)? cos(t) 1 + d. If this mde sense then (1.3) F ( + 1) cos(t) (t) F (t) 1 + d cos(t) d???. All is not lost! Let s mke chnge of vribles. Fiing t >, set y t, so dy t d nd F (t) cos y dy 1 + y /t t t cos y t + y dy. This new integrl will be ccessible to differentition under the integrl sign. (Although the new integrl is n odd function of t while F (t) is n even function of t, there is no contrdiction becuse this new integrl ws derived only for t >.) Fi c > c >. For t (c, c ), the integrnd in t cos y t + y dy is bounded bove in bsolute vlue by t/(t + y ) c /(c + y ), which is independent of t nd integrble over. The tprtil derivtive of the integrnd is (y t )(cos y)/(t + y ), which is bounded bove in bsolute vlue by (y + t )/(t + y ) 1/(t + y ) 1/(c + y ), which is independent of t nd integrble over. This justifies the use differentition under the integrl sign ccording to Theorem 1.3: for c < t < c, nd hence for ll t > since we never specified c or c, ( ) F t cos y y t (t) t t + y dy (t + y cos y dy. ) We wnt to compute F (t) using differentition under the integrl sign. For < c < t < c, the tprtil derivtive of the integrnd for F (t) is bounded bove in bsolute vlue by function of y tht is independent of t nd integrble over (eercise), so for ll t > we hve F (t) t ( t cos y t + y ) dy t ( t t + y It turns out tht ( /t )(t/(t + y )) ( /y )(t/(t + y )), so F ( ) t (t) y t + y cos y dy. ) cos y dy. Using integrtion by prts on this formul for F (t) twice (strting with u cos y nd dv ( /y )(t/(t + y )), we obtin ( ) ( ) F t t (t) y t + y sin y dy t + y cos y dy F (t). The eqution F (t) F (t) is second order liner ODE whose generl solution is e t + be t, so cos(t) (1.4) 1 + d et + be t
18 18 KEITH CONAD for ll t > nd some rel constnts nd b. To determine nd b we look t the behvior of the integrl in (1.4) s t + nd s t. As t +, the integrnd in (1.4) tends pointwise to 1/(1 + ), so we epect the integrl tends to d/(1 + ) π s t +. To justify this, we will bound the bsolute vlue of the difference cos(t) 1 + d d 1 + N cos(t) d by n epression tht is rbitrrily smll s t +. For ny N >, brek up the integrl over into the regions N nd N. We hve cos(t) 1 cos(t) d N 1 + d + N 1 + d t 1 + d d t N N ( π ) 1 + d + 4 rctn N. Tking N sufficiently lrge, we cn mke π/ rctn N s smll s we wish, nd fter doing tht we cn mke the first term s smll s we wish by tking t sufficiently smll. eturning to (1.4), letting t + we obtin π + b, so (1.5) cos(t) 1 + d et + (π )e t for ll t >. Now let t in (1.5). The integrl tends to by the iemnn Lebesgue lemm from Fourier nlysis, lthough we cn eplin this concretely in our specil cse: using integrtion by prts with u 1/(1 + ) nd dv cos(t) d, we get cos(t) 1 + d 1 t sin(t) (1 + ) d. The bsolute vlue of the term on the right is bounded bove by constnt divided by t, which tends to s t. Therefore e t + (π )e t s t. This forces, which completes the proof tht F (t) πe t for t >. 13. Eercises t sin 1. From the formul e d π rctn t for t >, in Section 3, use chnge sin(b) of vribles to obtin formul for e d when nd b re positive. Then use differentition under the integrl sign with respect to b to find formul for e cos(b) d when nd b re positive. (Differentition under the integrl sign with respect to will produce formul for e sin(b) d, but tht would be circulr in our pproch since we used tht integrl in our derivtion of the formul for t sin e d in Section 3.)
19 . From the formul with > implies DIFFEENTIATING UNDE THE INTEGAL SIGN 19 t sin e d π rctn t for t >, the chnge of vribles y ty sin(y) e dy π rctn t, y so the integrl on the left is independent of nd thus hs derivtive. Differentition under the integrl sign, with respect to, implies e ty (cos(y) t sin(y)) dy. Verify tht this ppliction of differentition under the integrl sign is vlid when > nd t >. Wht hppens if t? sin(t) 3. Show ( + 1) d π (1 e t ) for t > by justifying differentition under the integrl sign nd using (1.1). ( ) t cos 1 t 4. Prove tht e d log for t >. Wht hppens to the integrl 1 + t s t +? 5. Prove tht for > nd b >. 6. Prove tht log(1 + t ) 1 + d π log(1 + t) for t > (it is obvious for t ). Then deduce (e e t ) d log(1 + ) b + sign. This is (7.) for t > 1. Deduce tht d π log(1 + b) b log t for t > by justifying differentition under the integrl (e e b ) d 7. In clculus tetbooks, formuls for the indefinite integrls n sin d nd n cos d log(b/) for positive nd b. re derived recursively using integrtion by prts. Find formuls for these integrls when n 1,, 3, 4 using differentition under the integrl sign strting with the formuls cos(t) d sin(t), sin(t) d cos(t) t t for t >. 8. If you re fmilir with integrtion of complevlued functions, show for y tht e (+iy) d π. In other words, show the integrl on the left side is independent of y. (Hint: Use differentition under the integrl sign to compute the yderivtive of the left side.)
20 KEITH CONAD Appendi A. Justifying pssge to the limit in sine integrl In Section 3 we derived the eqution (A.1) t sin e d π rctn t for t >, which by nive pssge to the limit s t + suggests tht (A.) To prove (A.) is correct, we will show (A.3) sin d sin d π. sin d eists nd then show the difference t sin e d (1 e t ) sin tends to s t +. The key in both cses is lternting series. On the intervl [kπ, (k + 1)π], where k is n integer, we cn write sin ( 1) k sin, so convergence of sin d lim b sin b d is equivlent to convergence of the series k (k+1)π kπ sin d d (k+1)π ( 1) k k kπ This is n lternting series in which the terms k (k+1)π kπ k+1 (k+)π (k+1)π sin d (k+1)π kπ sin( + π) + π sin sin (k+1)π d d. d re monotoniclly decresing: kπ sin + π d < k. By simple estimte k 1 kπ for k 1, so k. Thus sin d k ( 1)k k converges. To show the right side of (A.3) tends to s t +, we write it s n lternting series. Breking up the intervl of integrtion [, ) into union of intervls [kπ, (k + 1)π] for k, (A.4) (1 e t ) sin d ( 1) k I k (t), where I k (t) k (k+1)π kπ (1 e t sin ) d. Since 1 e t > for t > nd >, the series k ( 1)k I k (t) is lternting. The upper bound 1 e t < 1 tells us I k (t) 1 kπ for k 1, so I k(t) s k. To show the terms I k (t) re monotoniclly decresing with k, set this up s the inequlity (A.5) I k (t) I k+1 (t) > for t >. Ech I k (t) is function of t for ll t, not just t > (note I k (t) only involves integrtion on bounded intervl). The difference I k (t) I k+1 (t) vnishes when t (in fct both terms re then ), nd I k (t) (k+1)π kπ e t sin d for ll t by differentition under the integrl sign, so (A.5) would follow from the derivtive inequlity I k (t) I k+1 (t) > for t >. By chnge of vribles y π in the integrl for I k+1 (t), (k+1)π (k+1)π I k+1 (t) e t(y+π) sin(y + π) dy e tπ e ty sin y dy < I k (t). kπ This completes the proof tht the series in (A.4) for t > stisfies the lternting series test. kπ
21 DIFFEENTIATING UNDE THE INTEGAL SIGN 1 If we truncte the series k ( 1)k I k (t) fter the Nth term, the mgnitude of the error is no greter thn the bsolute vlue of the net term: N ( 1) k I k (t) ( 1) k 1 I k (t) + r N, r N I N+1 (t) (N + 1)π. k Since 1 e t t, N ( 1) k I k (t) k k (N+1)π (1 e t sin (N+1)π ) d t d t(n + 1)π. Thus (1 e t ) sin d t(n + 1)π + 1 (N + 1)π. For ny ε > we cn mke the second term t most ε/ by suitble choice of N. Then the first term is t most ε/ for ll smll enough t (depending on N), nd tht shows (A.3) tends to s t +. eferences [1] W. Appel, Mthemtics for Physics nd Physicists, Princeton Univ. Press, Princeton, 7. []. P. Feynmn, Surely You re Joking, Mr. Feynmn!, Bntm, New York, [3] S. K. Goel nd A. J. Zjt, Prmetric Integrtion Techniques, Mth. Mg. 6 (1989), [4] S. Lng, Undergrdute Anlysis, nd ed., SpringerVerlg, New York, [5] S. Lng, Comple Anlysis, 3rd ed., SpringerVerlg, New York, [6] M. ozmn, Evlute Gussin integrl using differentition under the integrl sign, Course notes for Physics 4 (UConn), Spring 16. [7] A.. Schep, A Simple Comple Anlysis nd n Advnced Clculus Proof of the Fundmentl Theorem of Algebr, Amer. Mth. Monthly 116 (9), [8] E. Tlvil, Some Divergent Trigonometric Integrls, Amer. Mth. Monthly 18 (1), [9]
200506 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 256 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE  Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationMODULE 3. 0, y = 0 for all y
Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More information6 Energy Methods And The Energy of Waves MATH 22C
6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine soclled volumes of
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationModule Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials
MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More informationReview Problems for the Final of Math 121, Fall 2014
Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationFUNCTIONS AND EQUATIONS. xεs. The simplest way to represent a set is by listing its members. We use the notation
FUNCTIONS AND EQUATIONS. SETS AND SUBSETS.. Definition of set. A set is ny collection of objects which re clled its elements. If x is n element of the set S, we sy tht x belongs to S nd write If y does
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationAll pay auctions with certain and uncertain prizes a comment
CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 12015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More informationCHAPTER 11 Numerical Differentiation and Integration
CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationTreatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.
The nlysis of vrince (ANOVA) Although the ttest is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the ttest cn be used to compre the mens of only
More informationBabylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity
Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse rel number to its binry representtion,. convert binry number to n equivlent bse number. In everydy
More informationThe Fundamental Theorem of Calculus
Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk Te Funmentl Teorem of Clculus EXAMPLE: If f is function wose grp is sown below n g() = f(t)t, fin te vlues of g(), g(), g(), g(3), g(4), n g(5).
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls : The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N):  counting numers. {,,,,, } Whole Numers (W):  counting numers with 0. {0,,,,,, } Integers (I): 
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationDistributions. (corresponding to the cumulative distribution function for the discrete case).
Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationExponential and Logarithmic Functions
Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationThinking out of the Box... Problem It s a richer problem than we ever imagined
From the Mthemtics Techer, Vol. 95, No. 8, pges 568574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is soclled becuse when the sclr product of two vectors
More informationHarvard College. Math 21a: Multivariable Calculus Formula and Theorem Review
Hrvrd College Mth 21: Multivrible Clculus Formul nd Theorem Review Tommy McWillim, 13 tmcwillim@college.hrvrd.edu December 15, 2009 1 Contents Tble of Contents 4 9 Vectors nd the Geometry of Spce 5 9.1
More informationIntroduction to Integration Part 2: The Definite Integral
Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the
More informationPROBLEMS 13  APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS  APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More information19. The FermatEuler Prime Number Theorem
19. The FermtEuler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout
More informationThe Riemann Integral. Chapter 1
Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd
More informationWarmup for Differential Calculus
Summer Assignment Wrmup for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationEuler Euler Everywhere Using the EulerLagrange Equation to Solve Calculus of Variation Problems
Euler Euler Everywhere Using the EulerLgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch
More information15.6. The mean value and the rootmeansquare value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style
The men vlue nd the rootmensqure vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time
More informationg(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany
Lecture Notes to Accompny Scientific Computing An Introductory Survey Second Edition by Michel T Heth Boundry Vlue Problems Side conditions prescribing solution or derivtive vlues t specified points required
More informationBasically, logarithmic transformations ask, a number, to what power equals another number?
Wht i logrithm? To nwer thi, firt try to nwer the following: wht i x in thi eqution? 9 = 3 x wht i x in thi eqution? 8 = 2 x Biclly, logrithmic trnformtion k, number, to wht power equl nother number? In
More informationRIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS
RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is
More informationThe Definite Integral
Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know
More informationCURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.
CURVES ANDRÉ NEVES 1. Problems (1) (Ex 1 of 1.3 of Do Crmo) Show tht the tngent line to the curve α(t) (3t, 3t 2, 2t 3 ) mkes constnt ngle with the line z x, y. (2) (Ex 6 of 1.3 of Do Crmo) Let α(t) (e
More informationSolving BAMO Problems
Solving BAMO Problems Tom Dvis tomrdvis@erthlink.net http://www.geometer.org/mthcircles Februry 20, 2000 Abstrct Strtegies for solving problems in the BAMO contest (the By Are Mthemticl Olympid). Only
More informationReal Analysis and Multivariable Calculus: Graduate Level Problems and Solutions. Igor Yanovsky
Rel Anlysis nd Multivrible Clculus: Grdute Level Problems nd Solutions Igor Ynovsky 1 Rel Anlysis nd Multivrible Clculus Igor Ynovsky, 2005 2 Disclimer: This hndbook is intended to ssist grdute students
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationObject Semantics. 6.170 Lecture 2
Object Semntics 6.170 Lecture 2 The objectives of this lecture re to: to help you become fmilir with the bsic runtime mechnism common to ll objectoriented lnguges (but with prticulr focus on Jv): vribles,
More information4 Approximations. 4.1 Background. D. Levy
D. Levy 4 Approximtions 4.1 Bckground In this chpter we re interested in pproximtion problems. Generlly speking, strting from function f(x) we would like to find different function g(x) tht belongs to
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationThe Fundamental Theorem of Calculus for Lebesgue Integral
Divulgciones Mtemátics Vol. 8 No. 1 (2000), pp. 75 85 The Fundmentl Theorem of Clculus for Lebesgue Integrl El Teorem Fundmentl del Cálculo pr l Integrl de Lebesgue Diómedes Bárcens (brcens@ciens.ul.ve)
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationRotating DC Motors Part II
Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors
More information1. In the Bohr model, compare the magnitudes of the electron s kinetic and potential energies in orbit. What does this imply?
Assignment 3: Bohr s model nd lser fundmentls 1. In the Bohr model, compre the mgnitudes of the electron s kinetic nd potentil energies in orit. Wht does this imply? When n electron moves in n orit, the
More informationLECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.
LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 6483.
More informationHow To Understand The Theory Of Inequlities
Ostrowski Type Inequlities nd Applictions in Numericl Integrtion Edited By: Sever S Drgomir nd Themistocles M Rssis SS Drgomir) School nd Communictions nd Informtics, Victori University of Technology,
More informationExam 1 Study Guide. Differentiation and Antidifferentiation Rules from Calculus I
Exm Stuy Guie Mth 2020  Clculus II, Winter 204 The following is list of importnt concepts from ech section tht will be teste on exm. This is not complete list of the mteril tht you shoul know for the
More informationRadius of the Earth  Radii Used in Geodesy James R. Clynch February 2006
dius of the Erth  dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.
More informationPhysics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2.
Physics 6010, Fll 2010 Symmetries nd Conservtion Lws: Energy, Momentum nd Angulr Momentum Relevnt Sections in Text: 2.6, 2.7 Symmetries nd Conservtion Lws By conservtion lw we men quntity constructed from
More informationAA1H Calculus Notes Math1115, Honours 1 1998. John Hutchinson
AA1H Clculus Notes Mth1115, Honours 1 1998 John Hutchinson Author ddress: Deprtment of Mthemtics, School of Mthemticl Sciences, Austrlin Ntionl University Emil ddress: John.Hutchinson@nu.edu.u Contents
More informationNumerical Methods of Approximating Definite Integrals
6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils,
More information3 The Utility Maximization Problem
3 The Utility Mxiiztion Proble We hve now discussed how to describe preferences in ters of utility functions nd how to forulte siple budget sets. The rtionl choice ssuption, tht consuers pick the best
More informationINTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović
THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp. 15 29 INTERCHANGING TWO LIMITS Zorn Kdelburg nd Milosv M. Mrjnović This pper is dedicted to the memory of our illustrious professor of nlysis Slobodn
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives
More informationPentominoes. Pentominoes. Bruce Baguley Cascade Math Systems, LLC. The pentominoes are a simplelooking set of objects through which some powerful
Pentominoes Bruce Bguley Cscde Mth Systems, LLC Astrct. Pentominoes nd their reltives the polyominoes, polycues, nd polyhypercues will e used to explore nd pply vrious importnt mthemticl concepts. In this
More information0.1 Basic Set Theory and Interval Notation
0.1 Bsic Set Theory nd Intervl Nottion 3 0.1 Bsic Set Theory nd Intervl Nottion 0.1.1 Some Bsic Set Theory Notions Like ll good Mth ooks, we egin with definition. Definition 0.1. A set is welldefined
More informationImproper Integrals. Dr. Philippe B. laval Kennesaw State University. September 19, 2005. f (x) dx over a finite interval [a, b].
Improper Inegrls Dr. Philippe B. lvl Kennesw Se Universiy Sepember 9, 25 Absrc Noes on improper inegrls. Improper Inegrls. Inroducion In Clculus II, sudens defined he inegrl f (x) over finie inervl [,
More informationRoots of Polynomials. Ch. 7. Roots of Polynomials. Roots of Polynomials. dy dt. a dt. y = General form:
Roots o Polynomils C. 7 Generl orm: Roots o Polynomils ( ) n n order o te polynomil i constnt coeicients n Roots Rel or Comple. For n n t order polynomil n rel or comple roots. I n is odd At lest rel root
More information10.6 Applications of Quadratic Equations
10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More information