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Notes on Improper Integrals MT105 Prof Hiedberg Fall 2005 In class I mentioned that by de nition 00 B at dac lim at die 00 A ioo A Bgtltxgt where A and B independently approach foo and 00 respectively provided the limit exists and is finite If this limit does not exist or does exist but is not finite then we say that the improper integral does not exist or does not converge Here are some examples Ezrample 1 fjooox dx We have 00 B an die lim x dx foo A ioo A Bgtltxgt B 2 at lim 7 A ioo 2 Bgtltxgt A lim B27A2 Agt7ltxgt Bgtltxgt This last limit does not exist for example if we take B 2A then the quantity goes to 00 while if we take A 2B then it goes to foo So this improper integral does not exist Note The symmetric limit 0 lim an die Cgtltxgt C C 0 This symmetric limit is called the Cauchy prin C does exist since the integral is L cipal value In this example the Cauchy principal value exists even though the improper integral do es not Ezrample 2 ff 6quot dx 0 B 6quot die lim 6quot die 00 A ioo A Bgtltxgt Taking A negative and B positive we have B 0 B 6quot die 6 die 67 die A A 0 1 remember that inc when x S 0 Evaluating these two integrals we see that 6quot die Alim 17 6 7673 1 7 3110 2 Here the limit exists as AB independently approach their limiting values and the im proper integral does exist ExampleRemark 5 There are many improper integrals that can be evaluated using the theory of functions of a complex variable MT460 That is one looks at fz where z is allowed to be complex It turns out that integration theory is very rich in this setting and it allows one to evaluate improper integrals Here are a few examples dac 727139 7006l 37 0 2 77139 W day 7 OO 1 ac 22 50 These two are very surprisingiafter all what do the functions being integrated have to do with the number 7r But if you have a calculator that does numerical integration then you can check that these formulas are believable by computing the definite integrals which arise in the limit for specific A B such as A 71000 B 3000 Using complex variables one can also show that gt0 00 Sin d 0 x 2 Note the contrast though we showed in class that flee dmi did not exist the extra factor of sinac oscillates and that damps down growth of the integral At least in this case the appearance of 7r in the evaluation seems reasonable First Applications of Integration by Parts Lecture in MT105 92105 Prof Friedberg We will give two applications of Integration by Parts Remember that we have the version for de nite integrals Application One Let 1 Jn e mdx n012 0 Since ic e c 2 0 on the interval 01 we see that J7 2 0 Since e c S 1 on the interval we have 1 Oangxndx 0 n1 We conclude that hm J7 0 But what is J First 1 1 J0 e mdic7e c 17e 1 0 0 Second we can get a recursion relation using Integration by Parts Set it ix i e m so 010 ie m to see that for all n at least 1 1 J7 ic e c die 0 1 1 iic e c 7 nx 717e m die 0 0 7e 1 an1 well actually use this below with n replaced by n 1 which is OK since it is true for all n at least 1 J7 751 n 1J Using this relation we can easily determine J7 for any particular 11 For example J17e 11J0172e 1 J27e 12J1275e 1 J37e 13J26716e 1 1 Using the factorial function we can write the general formula Recall that for n 2 1 an integer 11 means the product of the first 11 positive integers For example 4 4 32 l 24 and 6 6 5 4 3 2 l 720 We also declare that 0 l for reasons we shall see later Here is the general formula for J 1 gtkn J7 n 7 nleil M k0 Here written out 171 1 1 1 kzoauia m How does one prove this Well we know that the actual J7 satisfies J0 l 7 6 1 and J7 76 1 n 1J so if we verify that the function given by the right hand side of satisfies the same initial condition and same recursion relation then it must be correct for all n 0 l 2 More formally this is checked by induction see page 88 of the text This means that we check for n 0 easy and we also show that if we assume and the recursion relation then we may deduce n1 by combining them Here7s how J7 7671 n 1J by 7 1 75 1nl nl7nle 1 by k0 7671 n l 7 n lle 1 Zn 1 since 11 l n 1 71 n 11 4 n 1151 lt2 3 where the last line comes since 7n ll67111 76 1 But this last line shows us that n1 is valid So we conclude by induction that holds for all n 0 12 This is just a very quick introduction to proof by induction It is taught in much more detail in MT216 l7m not expecting you to master it in our class But it should seem reasonable that since the right hand side of satisfies the same initial condition and recurrence that J7 does it must actually equal Jn Here is a nice application of our work Let us recall now that J 7 0 as n 7 oo 1 7 71 Jninl 16 E k0 and since 11 gets large rapidly as 11 gets large the term in brackets above must go to 0 1 39 7 71 i Whine l e E kl 0 2 Since In fact we7ve proved more not only must this limit be 0 but it must approach it so rapidly that even multiplying it by n we still get 0 Rearranging this we discover that 77 1 gtkgtk enlgr o k0 This formula allows one to calculate an approximation to e very rapidly We will find a generalization of formula gtk gtk later in the course Application Two Next well talk about another function the Gamma function This function is important in mathematics and some parts of physics It is defined for x gt 0 any real number by PW e t t 1 dt 0 Notice that as it varies it is the function that you are integrating that changes In a similar way to Jm we can check that Pl l done in class as first example of an improper integral and PW l x To see this recursion we must first write B PW 1 lim e t t c dt Bgtltxgt 0 and then apply integration by parts Let ut t c and it e t Then lBP gives B PW l Blim iBme B ictmile t dt H00 0 Since lim Bme B 0 Bgtltxgt for any at the formula PW l P follows Just as before we can use this recursion to evaluation PW for all positive integers in We see that r21r11 r32r221 r43r3321 r54r44321 Its not dif cult to give a proof by induction of the formula 4 Pnn1h which makes sense provided 11 is an integer at least 2 But if we apply this for n 1 it tells us that 0 Pl and since Pl l we see that 0 l is a natural definition as it allows us to extend formula 4 to n l You can also play With formula 4 for fractional exponents For example Pl2 e tt 12 dt 2 6 2 du 0 0 Where the second equality comes from making the substitution u 1512 Thus 44 77 2 r32 512 a du 0 This last integral is the area under half the bell curve In fact 12 isn7t something people usually write since we think of factorial as a product of falling terms Instead they discuss values of the Gamma function From the Wallis formula to the Gaussian distribution Mark Reeder Department of Mathematics Boston College Chestnut Hill MA 02467 September 5 2005 Contents t t 1 The Wallis formula for W Wallis and coin tossing The Gaussian integral and factorial 0f Wallis and the binomial distribution The function ml The Wallis formula for W 21 24 And he made a molten sea ten cubits from the one brim to the other it was round all about and his height was ve cubits and a line of thirty cubits did compass it round about 1 Kings 7 v23 W 31413 m o wusz zsnlwm This ancient text says that W 30 Actually much better approximations were known prior to this for example the Egyptians used the approximation 316 Now of course billions of digits of W are known The next 57 of them are ayouzzszsnll9716939937510582097 These digits do not repeat themselves and have no recognized pattern However John Wallis 16161703 Savilian professor of Geometry at Oxford discovered that there is a pattern if we write W as a product of fractions instead of a sum of powers of 10 Here is Wallis formula for W W i 2 2 4 4 6 6 2 1 3 3 7 This is an in nite product which is de ned as a limit of nite products Let E 224466 2n 2n WRZ21335539 2W12W1 SO 22 8 2244 128 r272 27777 2 Ill 13 3 666666 U 1335 8 Wallis formula says lim W2 W Ram Let s see what happens as n gets bigger starting with n 3 using a computer Obviously Wallis found his formula with no more than a quill pen some expen sive parchment and his wits The computation of the W n s continues as follows 512 7 32768 7 131072 m2 m m3 W W i 784637716923335095479473677900958302012794430558004314112 48 251048655304760134193016160220151596401393008816948260625 1 Not too good We should not expect that one of mankind s rst formulas for W should also converge rapidly that must wait until the 20th century Here is the biggest one I could do on my computer 117250 3138458897671616 It s getting there but in its own sweet time When we say limit goes to in nity we mean it Nevertheless we can prove that Wallis formula is true 2 Start with the formulas W 2462n It six1mlxdx 2 f0 3574271 quot2 1352n 1W It sinmx dx 7 2quot f0 2462n 2 If you didn t understand the derivations of these formulas take them on faith for now and we ll go through them in detail later To prove Wallis formula we start by showing that the I get smaller as the subscripts increase The function sin 3 takes values between 0 and 1 when x is in our interval of integration 0 That is 2 0 g sinx g 1 forx 6 0 This implies that for any positive integer k we have sinklr1 x g sink 3 3 There is therefore less area under the graph of sin r1 3 than under sink 3 So if we integrate both sides of 3 from 0 to WB we get 7r2 7r2 sink r1 x dx 3 sink x dx 0 0 Recall that these integrals are 1H1 11 respectively So we have Ik1 S 119 In other words if we increase the subscripts the 1 s get smaller So 12n2 S 12n1 S 12w Using the formulas 2 for 127 12 12 respectively this chain of inequalities says 1 32n 12n 1 2 4 271M271 2 If lt 24 2n lt 13 2n 1 2 352n1 242n 2 Multiply each term in the chain of inequalities by Almost everything cancels 1n the outer two terms and W2 appears on the 1ns1 e and we get 7 ltJlt1 4 271 W Taking the limit we have limH00 1 and is trapped inside so W2 W gt 1 as well Thus lim Wyn W 7100 as we wished to show How did Wallis think of this The above proof cleaned up by hindsight and added knowledge came much after Wallis who arrived at his formula for W by a wild and creative path guided by guessing intuition and lots of computations Let us see how he thought about it First of all in Wallis time they did not use the letter W to denote the ratio of the circumference over the diameter Here Wallis studies the number which he calls 13 In our terminology 1 0 1 x212dxg 5 The idea is to compute the integral by some other means to get a formula for Wzl Wallis contemplates this integral and notices that if he switched the 2 and the 1 2 he would get a much easier integral 1 1392 2 1 1392 2 1 1 1 x dx1 2x xdx1 2i i 0 0 3 2 6 Then Wallis sees that for any integers p 1 he can similarly compute the integral 1 f 1 4610qu 6 0 by expanding the integrand He does this for various 9 q and arrives at the rst table below The integral 6 is always 1 divided by an integer and he just writes the integer so the entry in row 19 and column 1 in the table below is 71 10 46Wquot 146 7 Actually Wallis computed many more entries than this because by examining this table he hoped to discover a general formula in terms of p and q for the integral 6 Then if this formula still held true even if p and 1 were not integers he could substitute 9 q 12 since those are the numbers in his original integral 5 and he would have a formula for 13 hence a formula for W However things did not go so smoothly Now you may have already noticed the pattern in Wallis table do you see Pascal s triangle The entry in row 19 column 1 is 71 P 13 f1 1 v y 1 W 1 ix 8 am p 0 x x For example the numbers in the second row are 1 lrzyq 5m 1W1 392 9 and in the third row we have 1 aw glq1q2q3 10 and so on But remember that Wallis wants 9 and q to be halfintegers And he does not know how to take the factorial of a fraction where would he stop so it looks as though his idea is doomed It is at this point that Wallis shows his courage as a mathematician He does not give up and begins work on a new table by optimistically adding new columns and rows for p etc and similarly for 1 If only Wallis can numerically compute 15 which is 13 in the next table he ll be done But he can t compute 1 because formula 8 says that l l 2gt2 11 1 55 32 which doesn t seem to make any sense Still he tries to ll in his new table as much as he can He can ll in the entries as long as p is an integer since formulas like 9 and 10 make sense for any 1 For example 1 1 1 15 3923 l 2 g 1 5 5 5 231 5 7 7 1 7 7 7 r13 65 lt2 53 16 Wallis is guessing here he does not know for sure that formulas 9 and 10 actually give the value of the integral when p or q are not integers Luckily they do Then he notices the formula am 19 awn 11 which plays the same role as our reduction formula did in the more modern proof Again he only knows I l for integers you can check it yourself using factorials and Wallis just assumes l l is true for nonintegers Formula 1 l lets him move two steps to right in any row remember that the steps are now by halves and this allows him to ll in the blank spaces in table two using 13 In row 19 g for example he gets lPqllfl1lil23 lilll lfil lf l1 lf z yiw Just as we did Wallis remembers these entries are reciprocals of integrals and the integrals get smaller as 1 increases so the entries get larger as 1 increases So for example we have Elt246 KI lt 154 3 Which after remembering that 13 i can be written W lt7lt 3963 7 Thus we see Wallis formula emerging The factor says this approximation of g is accurate to within a factor of If you go farther out in the row the error becomes smaller than any given number Thus Wallis did not actually prove his formula because he made some unjustifed assumptions along the way However the assumptions turn out to be true once you know how to do the factorial of g which we shall learn and Wallis intuition was correct The derivation of his formula that we gave at the beginning is a real proof The reduction formulas Let s make sure we know how to calculate the integrals 1 used in our proof of Wallis formula At the beginning it doesn t matter if k is even or odd Recall that V 7r2 1k sinkx dx 0 Use integration by parts with u sinkquot1 42 do sin xdx k 2 12 du 1 sm xcosx dx v cosx You should be able to see that avg2 cowquotsinkquotl 331 1 So my 7r2 1k sinkx dx 0 1 sink zxcos x cos x dx 0 0 A72 1 sinkquot2 xcos2x dx 0 A72 1 f0 sinkquot2 x1 sin2 3 dx 13 A72 1 sink zx sink 3 dx 0 7r2 7r2 1 sink zx dx 1 sinkxdx 0 0 1k2 1k So 1k 1k2 1k Add 1Ik to both sides and get ka 1k2 This gives us k 1 k 1 To get it started we have my 7 7r2 Il dx 7 11 sinxdx 1 0 3 0 Then the reduction formula takes over 1k 2 reduction formula 2 1 1 7 3 1 2 I I 77 I I 7 2 2 0 2 2 3 3 1 3 4 1 3 1 7 5 1 4 2 1 1777 17 1 77 4 42422 0 and so on It should be clear now what the pattern is but if you want to be fussy let Jl 3 J 1 1 and for larger subscripts I i ez 135727 L1 14 1352n 1W JZn 2462n 5 note that Jk J 2 in both cases Thus the J s begin the same way as the 1 s and have the same recursion so they stay the same as the 1 s and In J for all n 2 0 Exercise 11 Use u WB x and the identity cosx 8111W2 to show that V KZ xz cask x dx sink x dx 0 0 Exercise 12 Use the identity coszx 1 sin2 3 to calculate sin6 42 Lbs x dx 0 answer L 218 110 The same method works when the powers on sinx and cos x are both even If say cos 3 appears with odd power you can split off a cos 3 write the rest of the cos x s in terms of sin2 3 and use the substitution u sin 3 instead of 1 Exercise 13 Recall our homework problem about fol x 1 xb fol x 1 aquot If a or I is an integer you can expand 1 x in one of these integrals and integrate termby term for example 2 3 392 l 3 392 2 l 392 r 392 2 2 16 2 faquot 1 46 dx 4 1 42 dx at 2x 42 dx flt o o a 9 But how to do it if neither 1 nor 3 are integers Let x sin2 9 and show that l 7r2 xaU 46 cosh E sin2M1 E if 0 0 Now you can use the method of exercise 2 Use this to calculate l x521 M3 dx 0 Exercise 14 Let p gt 1 be a constant Show that 00 1 A72 f 72 dx cosh 2 E if 0 1 33 p 0 Exercise 15 Calculate f0 1 x2o2 Exercise 16 Prove the formula used to make Wallis rst table v v 1 xlpq dx 19 q 0 where p and q are positive integers Hint Think of p as xed and let Aq and Bq be the left and right sides respectively First show that Al B 1 Then show that I 1 14 714 and 7B q 91 ql q 91 ql Exercise 17 Use 4 and the fact that W lt 4 to show that 2 ltW Unltn1 Check this with 117250 as computed above Exercise 18 Here is a similar way to approximate ca with fractions Let 5 Ln Omarquot dx 1 Use integration by parts to show that L ca nan Then show that L gt 0 by looking at the graph of lnx and conclude that nLWl gt ca Calculate a few Ln s and approximate ca Exercise 19 What number is this 2 2 W 24 4 466818 1 Perhaps you would like to use a machine and guess The correct guess can be proved using Stirling s formula 2 Wallis and cointossing The number W is used for more than measuring circles because it appears in many dilTerent areas of mathematics Likewise the Wallis formula for W has many ap plications In this section we show how Wallis is related to sequences of random 01 events like cointossing This includes a probabalistic interpretation of the integrals 1 used to prove Wallis formula First we need a bit of background on binomial coe icients Recall that the factorial of a positive integer n is de ned as n Mn 1n 221 We also de ne 0 1 Why de ne 0 this way For now just accept that we de ne 0 1 to make the formulas come out right We will give a better reason in the next section There are many interpretations of nl It is the number of ways to 10 u put n letters in n mailboxes o arrange n people in a row 0 paint n houses with n colors 0 marry n men to n women 0 permute n distinct objects Now if n and k are integers with 0 g k g n we de ne the binomial coef cient by n n klm k and call it n choose k because is the number of ways to choose k objects from n objects From a group of n people you can form possible teams of k members For example from a class of 30 people you can make l l a i 142506 5 5 possible basketball teams Proof make the class line up in all 30 possible ways and each time take the rst ve for your team You will get all possible teams this way but you will get the same team several times so we have to divide 30 by the number of times each team occurs We get the same team from di erent lines by either permuting the rst 5 members of the line or permuting the 25 members in the rest of the line Thus atotal of 5325 lines give the same team There are many other interpretations of binomial coe icients For example in algebra we have the binomial theorem R 1 x n n xquot lt k because the coe icient of xk in 1 x is the number of ways to choose k x s from the n factors 1 The Wallis formula has to do with the following interpretation of If we label our n objects as 1 2 n then a choice of k objects can be expressed as a sequence of k 1 s and n k 0 s where the 1 s correspond to the chosen objects For example the 6 teams of two from a group of 4 are the sequences 1100 1010 1001 0110 0101 0011 ll Such seqences are also the outcomes of an experiment of n coin tosses where 1 means heads and 0 means tails For example if we toss the coin 4 times and get heads tails tails heads this outcome is 1001 Thus when we toss the coin n tkir esllthe number of possible khead outcomes is The probability of getting ea s 1s number ofpossible k head outcomes 7 1 n k In particular the probability of getting k heads from 2k tosses is the number we will call Pk de ned by number of all possible outcomes 2 22k 242k2 V 1 2k 112342k 12k 1352k 1 Pkg k T T We have seen this number before The formula for 12k in 2 may be written as 2 W 7 singquot x dx Pk W 0 This means that Pk is also the average of the function sinzk x on the interval 0 Recall that in the proof of Wallis formula we used the fact that sinzkx gt 0 as k gt Hence lim Pk 0 kao This means that the probability of getting half heads in a large even number of tosses is essentially nil This may seem strange since halfheads is the most likely outcome But there are more and more outcomes that take their share of the probability We will examine this more closely in later sections Here we want to show how the Wallis formula amounts to knowing Pk for large k Recall that Wallis says that lim W kgtco where W233 3 quot 3 quot F 133557 2k 12k1 We have changed n to k to conform to the present notation Note that 2P m 2k1 12 So Wallis formula for W can be written as 2 lim I m2k 15 kgtco 7 This means that for large k we have the approximation P c k 02k 1 where c x BW is a constant Thus Wallis tells us how fast the odds of getting half heads goes to zero Exercise 21 Explain why it is a good idea to de ne 0 1 by giving a coin tossing interpretation of and Exercise 22 Use the formula for to prove that garter Exercise 23 Explain in two ways why n n n 2quot 3 o 1 n rst using the formula for 1 x then using the coin tossing interpretation Exercise 24 The triangle erroneously attributed to Pascal it was known to the Chinese over 1000 years earlier has in the k entry of row n from the top Starting at the top of Pascal s triangle and moving downward at each step what is the number of ways to get to the entry Hint interpret a choice of path as a seqence of cointosses 3 The Gaussian integral and factorial of We have been talking about factorials of integers which are the building blocks of binomial coe icients But we also saw that Wallis pursuit of his formula for W led him to repeated confrontations with the strange number It seemed that Wallis eventually managed to avoid having to know but the fact that it did not go away 13 easily hints that the value of g really does lie at the heart of Wallis formula In this section we will show that this is the case by going in the other direction we will use Wallis formula for W to compute What follows uses only math that we already know but it may seem tricky That s because it took 200 years to discover It is also not very well known I found it by accident in a paper of Thomas Stieltjes 1856 1894 who was a Dutch mathematician remembered today mainly for the Stieltjes integral which you would encounter in graduate school Before going into the tricky part let s examine the dif culty of computing Recall that for a positive integer n the factorial n is de ned as the product 7 Llnn 121 16 of all positive integers g n Formula 16 only makes sense if n is a positive integer The rst step due to Euler is to nd a dilTerent formula for n that makes sense even if n is not a positive integer Then we can just plug n into this formula to compute Right The rst observation is that n can be de ned recursively by the two rules 1 1 717171 1 17 Next Euler observed that 00 1 dx 1 0 and that integration by parts with u x and do Vida shows that 00 00 We dx n xn lca dx 0 0 So the integral xnca Hix satis es the same recursion formulas 17 as n hence must equal ml 00 713 We ldx 18 0 This is a formula for n that does not depend on n being a positive integer We take 18 as a new de nition of nl It agrees with the old de nition 17 when n is a positive integer but it can also be used for other H We just have to make sure the integral converges There is no problem with the limit oc since Fl crushes any power of x as x 1 But at the limit 0 there could be a problem if n lt 0 since 0 1 we have the approximation xnca w x for 3 near 0 which shows 14 that flee xnca x dx converges only when n gt 1 So we can use formula 18 to compute n for any real number n gt 1 For example formula 18 says that 00 00 0 f 4600 dx f 1 dx 1 0 0 so we get 0 1 from the formula instead of by at as before Most interesting to us now however is that formula 18 says 00 V 00 461267 dx and ail267 dx 0 0 All very nice but what are the values of these integrals They are actually famous integrals in disguise used in many areas of mathematics Let s work on the second one starting with the substitution u 3312 We get dx Bu du so 00 V 00 2 00 2 3 ail267 dx 2 u lca u du of du 0 0 700 so whatever it is the number 3 is the area under the whole graph of 2 which is the famous Bell Curve used in probability And 00 2 g 2 f w dx 19 0 is half of this area We keep talking about 3 without actually computing it That s because the integral 19 cannot be computed by nding an antiderivative of 7 12 go on try it If only there were an extra 3 we could do it Namely if instead of the integral in 19 we had 00 2 f 4667 dx 0 then taking u 332 would turn it into 00 2 00 x 7 1 r 7 l I C dx 2 c du 2 0 0 If we had an extra 332 and did u 332 again we d get 00 2 00 Jrl ii 12ru 711 fro dx Z u c alu 2 2 0 0 15 which is back to the hard integral we started with A more clever idea is to use integration by parts with 7 2 u 3 do are 1 since we just integrated dv However this will lead to the same hard integral again But at least it we get back to the same hard integral and not some new one To analyze why some of these integrals are hard and some are easy let us de ne 00 2 GR xnca dx 0 The integral Go ISO 2013 is the famous Gaussian integral So far we know that l 11 1053 011 02551 Even though we know it won t work completely lets try integration by parts on G for n 2 2 and see what happens With u at and do xca dx we get 00 2 GR We dx 0 1 7 7200 71 1 00 7 7xquot 16 1 l xquot 26 0 2 0 392 771 1 2 Gniz 7 7 2 s1nce n 2 2 and hian00 x 16 1 0 So we have the recurs1on formula 771 1 2 0 GR 20 with the initial values Go hard GI 1 easy This means the even G s are hard and the the odd G s are easy 12k2k2 1sz72 21 2 1 2 1 k2 W2 klPkGo where we recall from the previous section that 1 2k 132k 1 Pk 7 4quot k 2 4 2k is the probability of getting k heads in 12k cointosses So all the even 0 boil down to the single hard integral Go On the other hand for the odd G s we get an actual answer 2k sz1 39sz71 722i 2k 2 2 2 39 sza 22 k What have we achieved so far Almost nothing apparently We have just been analyzing our dif culties We want to compute Go and we have just seen that all the integrals 12k boil down to G0 whereas the integrals G2 can be computed exactly But now comes the tricky part that eluded Wallis and Euler and was nally discovered by Stieltj es The idea is to express the hard integral 12k in terms of the easy integrals 1 and 121 This cannot be done by means of an equality but instead using an inequality We will show that lt 07407 for all n 2 1 23 The idea of 23 seems to me a spark of genius I cannot explain how Stieltjes thought of it but once thought of it is not hard to prove Think where have we seen something like B2 The quadratic formula of course The quadratic polynomial that gives rise to the terms in 23 is pm Ham 2th GM Remember that the G s are just numbers which we happen not to know com pletely and they are the coef cients in the polynomial pt Of course the G s are integrals 00 2 00 2 00 2 174 xn lca dx GR xnca dx G7 xn ca dx 0 0 0 so pm is a sum of integrals Let s combine these into one integral note that t is a constant with respect to dx so can be moved inside the integrals 00 00 00 pm t2 xn lca dx 1 xnca dx xn ca dx 0 0 0 00 Haquotquot71 2M xn ya dx 0 00 4 f at t xzca 2rix 0 This integral is the worst one yet but we are not going to compute it Just note that the integrand as a function of x is Z 0 and is equal to zero at no more than two points So there is positive area under the integrand and the integral is positive Hence pm gt 0 for all t This means pm has no real roots Now if a quadratic polynomial Atz 28f C has no real roots then from the quadratic formula we must have B2 lt 0 So GuilGnl lt 0 18 which is the inequality 23 that we wanted to prove Where does this brilliant step take us We now have one equality easy n 1 GR 2 67372 24 and one inequality brilliant lt GuilGnh and believe it or not the hard work is over We are just going use 24 and 25 and then Wallis to get Go Applying 25 to n 12k 1 and then H 2k we get 12k 1 2 12k 1 a g 2 02k7102kb 26 2 121 f5 02k02k2 Now plug our computations 21 and 22 6 kiPkGO sz1 into 26 to get v 2 v v lt 2k1ltk3PkGOV lt 2Ai1 A i 1 392 2 Dividing everything by kl2V we get 2k1 1 lt 22k1P fG lt 2k 27 We have seen that Wallis formula can be written 1 2 z V 2 7 7 13310109 1Pk W and clearly I I 12k 7 2k So taking k gt oc in 27 we get meaning that V GO 7 and we are done Done with what Let s review Recall that Go is the Gaussian integral and is also 1 2 00 2 Go 1 dx 3 0 By computing that Go 2 we have shown that 00 2 r1 7 l 7 f c dx 2 0 In other words the area under the whole graph of the Bell Curve 7 12 is exactly W This computation combined the work of Wallis Euler and Stieltjes from the 17 18 and 19 centuries respectively The limits 0 and ac were essential We never computed the antiderivative f 7 12 dx Since W is involved you might guess that G is somehow related to the area of a circle This is true and leads to another way to compute G0 which is easier than what we just did and was known long before Stieltjes but requires double integrals and is beyond our course In the other direction knowing Go allows one to compute the volume of a sphere in any dimension This requires even more multiple integrals Since all the even G s boiled down to G0 we have actually computed many other integrals We leave this to the exercises as Exercise 31 Use the recursion formula n n n 1 to compute 12 1312 and in terms of Exercise 32 Give a formula for for any integer k 2 0 Exercise 33 We have seen that Go What is the analogous formula for 12k Hint use equation 24 In the remaining exercises make a substitution to turn the integral into the factorial inte gral then compute it Leave all answers as fractions Exercise 34 Calculate 3360 dx 20 Exercise 35 Calculate 413667412 dx Exercise 36 Calculate xEca xg dx Exercise 37 Calculate 3 2 dx 00 71112 0 6 dx 1 1s a pos1t1ve constant 0017 7 1 Exercise 310 Show that f0 6 1 dx pl Exercise 38 Calculate Exercise 311 Calculate fol 71 4 Wallis and the binomial distribution If we graph the outcomes of n coin tosses we get pictures like for n 4 and 21 for n 6 where the number of dots in the k column is the number of out comes with kheads These look like discrete versions of bell curves In the previ ous section we saw that the middle column which is the most likely outcome and which grows as n increases nevertheless becomes a negligible proportion of the total number of dots Moreover Wallis told us precisely how fast this proportion goes to zero In this section we will see that Wallis also tells us the parameters of the bell curve which approximates the binomial distributions above As we have mentioned the basic bell curve is the graph of 7 12 This has a maximum of at x 0 and in ection points at l1 The latter are the points where the graph of 7 12 begins to are outwards The point 0 is the mean and 1 is the standard deviation of this bell curve However we need bell curves with an arbitrary mean it and standard deviation 7 Such a curve is the graph of 1 2 exp j 28 2 a 22 This is basically the just the graph of 7 12 but shifted to have its maximum at p and stretched to have its in ection points at p l 7 Exercise 41 Show that the function in 28 has its maximum at p and in ection points at p l 7 Exercise 42 Use our formula for the Gaussian integral ff 7 12 dx W to show that 2 co 1 exp 5 dx m 2W 00 2 a 1 1 1t 2 7mm 7 Ix12W i 392 0 is called the Gaussian or norma distribution It approximates in an easily understood visual way many different occurences of discrete random behavior that may be very dif cult to compute one at a time You just have to adjust it and a to the case at hand For example it is very dif cult to compute the binomial coef cients necessary to determine the exact proportion of outcomes with k heads from n coin tosses if n is large Instead we can use the approximation 1 n 1 gt 1 k lt 2 H w map 2 l a j gt IL large 29 We just have to determine 1 and a The maximum of the right side of 29 which is u should be the most likely outcome of the left side which is 712 so The function 1t T That was easy What about the standard deviation 7 By now it is clear that the Wallis formula knows everything about large binomial coef cients so it is no surprise that Wallis will tell us 7 Taking k p 712 in we get the approximation 1 IL 1 2 M Ix12W 23 for large n On the other hand by Wallis see equation 15 we have 1 n 1 2 2quot M W xTLll 7m for large n So we should have 1 7 2 0 2w V W meaning that a In summary for large n we have the approximation as functions of k 1 n 1 1 k p 2 n wmexp 2 a J gt where 5 07 30 This is one of many instances of a discrete function being approximated by a continuous function It may seem paradoxical but the latter is easier to work with as we will see in the next chapter 5 The function erfx If you make a large number n of coin tosses we have seen that the probability of getting any particular outcome is almost nil One is more interested in the probability that a certain range of outcomes will occur For example if we toss a coin 100 times the probability of getting exactly 50 heads is almost zero but what is the probability of getting between 48 and 52 heads To answer this we could compute 5 very large binomial coef cients add them up and diVide by 2 00 It is much easier to use our formula 30 in the preVious section The probability of getting between a and 3 heads in n coin tosses is exactly 1 b 2 3 exp dx 31 l a To handle this integral we de ne the error function 1 x 4 v mllx tigrit l Here x can be any real number So ml is the part of the area under a certain bell curve between 0 and x Exercise 51 Show that ml has the following properties 1 ml0 0 2 limxg00 mlx 3 mlx is an odd function That is ml x mlx 4 ml is always increasing is concave up for x lt 0 and concave down for x gt 0 Exercise 52 Show that 1 b 1 k pz b a p 7 7 j 7 f 7 f 7 txp 21 J x m m Thus if you make a large number n of tosses the probability of getting be tween a and 3 heads is approximately ml M ml U MZQ 0 a a 392 392 You can look up values of ml in tables or on your calculator just as you would with trig or log functions Example If we toss 100 times what is the approximate probability of getting between 48 and 52 heads We have n 100 it 50 7 5 so the approximate probability is 2 F0 48 quot0 ml ml 2ml4 2 3108 25 With these 100 tosses what is the approximate probability of getting at least 40 heads Here n Ma are unchanged but a 40 I 30 so the approximate probability is 39 390 till F0 ml erf mlbo erK B w 4773 Exercise 53 For n 100 tosses nd the approximate probabilities of getting a between 45 and 55 heads answer 6826 b between 50 and 60 heads 0 at least 45 heads Exercise 54 For n 10 000 tosses nd the approximate probabilities of getting a between 4950 and 5050 heads answer 6826 again b between 4900 and 5100 heads 0 no more than 4500 heads Exercise 55 In this problem you are not given n so you won t know it and I either Nevertheless please nd the approximate probabilities of getting a between it 7 and it 7 heads b between it 20 and it 20 heads Besides coin tossing the same method can compute approximate probabilities in any situation where there are two possible results equally likely and the trial is repeated a large number of times 26 Exercise 56 A story problem The town of Bumpkin has the shape of a triangle with BankBumpkin at the northern peak in Upper Bumpkin where all the money resides The streets of Bumpkin were made long ago by wellorganized cows and the map of Bumpkin looks like this BB Below Upper Bumpkin lies Middle Bumpkin and even further south is Lower Bumpkin not shown where the map is very complicated indeed Jack and Jill grew up in Lower Bumpkin After Jack fell down and broke his crown he was never the same and after a series of Failures in Life poor Jack had turned to a life of crime And so one day Jack went north to rob BankBumpkin He tied up everyone in the bank took all the cash he could nd and dashed out the door making a run for it down the streets toward the labyrinth of Lower Bumpkin Jack knew that if he got far enough south the cops would never nd him Unfortunately one of the tellers had recognized Jack and had gotten loose and called the police As for Jill who came tumbling down after Jack in that famous accident she had made a full recovery and being very clever went on to become a Pure Math ematician But since Jill was not interested either in teaching or in practical applications of math she was forced to support herself as a police dispatcher which in the usually quiet town of Bumpkin allowed plenty of time for research and many of her mathematical discoveries were made while contemplating the map of Bumpkin on the station wall And of course it was Jill who answered the phone after the bank robbery Jill knew Jack pretty well from the old days She knew he could run fast and that he was thinking only of getting to Lower Bumpkin as quickly as possible Also he was surely panicked and making a random choice at each intersection though always heading in a southerly direction So Jill temporarily shelving her disdain for what everyone else called the real world as though they knew what that meant she would snort to herself performed some brief calculations which were amusing in and of themselves She gured that by the time the police got rolling old Jack would be nearing his 100th intersection But which one There 27 were 101 possibilities and not nearly that many cops on the Bumpkin beat so Jill suggested a deployment of just enough police to have a 95 percent chance of catching Jack With one uniform at each intersection how many police did she deploy 28
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