mm-2689 === Subject: Re: example of series of positive terms that converges Find an example of a series of positive terms that > converges despite the fact that > lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. Start with a positive series that converges nicely, e.g., a_n = 1/2^n. Pick a subsequence a_n_k in such a way that the gaps n_(k+1) - n_k increase very rapidly. For each k, replace a_n_(k+1) by k*a_n_k. Show that the resulting series still converges. [...] Brian But if (a_n)= 1/2^n, then |a_(n+1)/a_n|= |(1/(2^(n+1)))/(1/2^n)|= |(2^n)/(2^(n+1))|= |2^-1|= 1/2. So how do I take the lim sup of this? and how does it equal oo??? I think I'm confused... === Subject: Re: example of series of positive terms that converges in alt.math.undergrad: > Find an example of a series of positive terms that > converges despite the fact that > lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. >> Start with a positive series that converges nicely, e.g., >> a_n = 1/2^n. Pick a subsequence a_n_k in such a way that >> the gaps n_(k+1) - n_k increase very rapidly. For each k, >> replace a_n_(k+1) by k*a_n_k. Show that the resulting >> series still converges. > But if (a_n)= 1/2^n, then > |a_(n+1)/a_n|= |(1/(2^(n+1)))/(1/2^n)|= |(2^n)/(2^(n+1))|= > |2^-1|= 1/2. (Note that since you're dealing with positive series, you don't really need the absolute values.) > So how do I take the lim sup of this? You did it fine here, but that's the lim sup of the original sequences whose terms are a_n. You missed the fact that we're no longer looking at the original sequence, but rather at one in which an increasingly sparse subsequence has been modified. Here it is again, with a little more detail. Let n_k (k > 0) be an increasing sequence of positive integers such that the gaps n_(k+1) - n_k increase very rapidly. (For an extreme example, imagine taking n_k = k!.) Pick a subsequence a_n_k as explained above, and define a new sequence b_n by b_n_(k+1) = k*a_n_k for each k, and b_m = a_m if m is not in the subsequence of n_(k+1)'s. Now for each k you have b_n_(k+1)/b_n_k = k*a_n_k/a_n_k = k, while if m is not one of the subscripts n_k, then b_(m+1)/b_m = a_(m+1)/a_m = 1/2. Clearly the lim sup of the b_n's is infinity. But if the gaps n_(k+1) - n_k get big quickly enough, the sum of the b_n's will still be finite. (There's still some work needed in that last step, of course.) [...] Brian === Subject: Re: example of series of positive terms that converges Find an example of a series of positive terms that > converges despite the fact that > lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. >> Start with a positive series that converges nicely, e.g., >> a_n = 1/2^n. Pick a subsequence a_n_k in such a way that >> the gaps n_(k+1) - n_k increase very rapidly. For each k, >> replace a_n_(k+1) by k*a_n_k. Show that the resulting >> series still converges. > But if (a_n)= 1/2^n, then > |a_(n+1)/a_n|= |(1/(2^(n+1)))/(1/2^n)|= |(2^n)/(2^(n+1))|= > |2^-1|= 1/2. (Note that since you're dealing with positive series, you don't really need the absolute values.) > So how do I take the lim sup of this? You did it fine here, but that's the lim sup of the original sequences whose terms are a_n. You missed the fact that we're no longer looking at the original sequence, but rather at one in which an increasingly sparse subsequence has been modified. Here it is again, with a little more detail. Let n_k (k > 0) be an increasing sequence of positive integers such that the gaps n_(k+1) - n_k increase very rapidly. (For an extreme example, imagine taking n_k = k!.) Pick a subsequence a_n_k as explained above, and define a new sequence b_n by b_n_(k+1) = k*a_n_k for each k, and b_m = a_m if m is not in the subsequence of n_(k+1)'s. Now for each k you have b_n_(k+1)/b_n_k = k*a_n_k/a_n_k = k, while if m is not one of the subscripts n_k, then b_(m+1)/b_m = a_(m+1)/a_m = 1/2. Clearly the lim sup of the b_n's is infinity. But if the gaps n_(k+1) - n_k get big quickly enough, the sum of the b_n's will still be finite. (There's still some work needed in that last step, of course.) I understand this, but I need help finding a concrete example of a concrete series... not a proof... === Subject: Re: example of series of positive terms that converges in alt.math.undergrad: >> Find an example of a series of positive terms that >> converges despite the fact that >> lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. > Start with a positive series that converges nicely, e.g., > a_n = 1/2^n. Pick a subsequence a_n_k in such a way that > the gaps n_(k+1) - n_k increase very rapidly. For each k, > replace a_n_(k+1) by k*a_n_k. Show that the resulting > series still converges. > But if (a_n)= 1/2^n, then > |a_(n+1)/a_n|= |(1/(2^(n+1)))/(1/2^n)|= |(2^n)/(2^(n+1))|= > |2^-1|= 1/2. >> (Note that since you're dealing with positive series, you >> don't really need the absolute values.) > So how do I take the lim sup of this? >> You did it fine here, but that's the lim sup of the original >> sequences whose terms are a_n. You missed the fact that >> we're no longer looking at the original sequence, but rather >> at one in which an increasingly sparse subsequence has been >> modified. >> Here it is again, with a little more detail. Let n_k >> (k > 0) be an increasing sequence of positive integers such >> that the gaps n_(k+1) - n_k increase very rapidly. (For an >> extreme example, imagine taking n_k = k!.) >> Pick a subsequence a_n_k as explained above, and define a >> new sequence b_n by >> b_n_(k+1) = k*a_n_k for each k, and >> b_m = a_m if m is not in the subsequence of n_(k+1)'s. >> Now for each k you have >> b_n_(k+1)/b_n_k = k*a_n_k/a_n_k = k, >> while if m is not one of the subscripts n_k, then >> b_(m+1)/b_m = a_(m+1)/a_m = 1/2. >> Clearly the lim sup of the b_n's is infinity. But if the >> gaps n_(k+1) - n_k get big quickly enough, the sum of the >> b_n's will still be finite. (There's still some work needed >> in that last step, of course.) > I understand this, but I need help finding a concrete > example of a concrete series... not a proof... I'm afraid that you *don't* understand it: what I've given you is a *very* detailed hint for finding a concrete example of a concrete series with the desired properties, and the only proof involved here is proving that the series that you find actually has the desired properties. I don't at the moment see how I can say much more without simply doing the problem for you. I'll think about it, but for now I can only suggest that you take another look at my last post, bearing in mind that it *is* (almost) a recipe for finding a concrete example. Brian === Subject: Re: example of series of positive terms that converges Find an example of a series of positive terms that >> converges despite the fact that >> lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. How about (a_n)= (3+ (-1)^n)^(-n)? === Subject: Re: example of series of positive terms that converges in alt.math.undergrad: > Find an example of a series of positive terms that > converges despite the fact that > lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. > How about (a_n)= (3+ (-1)^n)^(-n)? It appears to work; all you have to do now is prove (a) that the series SUM[n=0 to inf; a_n] is finite, and (b) that the lim sup of a_{n+1}/a_n is infinite. (In fact for (a) you can actually evaluate the sum exactly.) Brian === Subject: Re: example of series of positive terms that converges Find an example of a series of positive terms that > converges despite the fact that > lim sup (n->oo) |(a_(n+1))/(a_n)|=oo. > How about (a_n)= (3+ (-1)^n)^(-n)? It appears to work; all you have to do now is prove (a) that the series SUM[n=0 to inf; a_n] is finite, and (b) that the lim sup of a_{n+1}/a_n is infinite. (In fact for (a) you can actually evaluate the sum exactly.) Brian The series is convergent because 0 < a_n <= 2^(-n). If n is even, a_(n+1)/a_n = 2^(-n-1)/4^(-n) = 2^(n-1) --> oo. === Subject: dyadic series question Prove that if the terms of a sequence decrease monotonically and converge to 0 then the series [sum](a_k) converges if and only if the associated series (a_1)+ 2(a_2)+ 4(a_4)+ 8(a_8)+ ... = [sum](2^k)*(a_(2^k)). I know that since the series decreases monotonically, then (a_1)>= (a_2)>= .... And I know the associated series is called the dyadic series. === Subject: Newton's Method Using MATLAB/MATHEMATICA Hi guys my Optimization class prof. wants us to use Newton's Method to solve some nonlinear optimization problems, and he wants us to program it using Matlab or mathematica. problem is no one knows how to use either and he wont teach us how to program!!! he says we should already know, even though the only programming class req. for math majors is an easy C++ intro class. anyways, the problem is this: Apply Newton's method to find all three solutions of: f(x) = x^3 - 5x^2 - 12x + 19 = 0 if anyone can help with some matlab code as to how to do this using newton's method i would be very very grateful!!! === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA > my Optimization class prof. wants us to use Newton's Method to solve some > nonlinear optimization problems, and he wants us to program it using > Matlab or mathematica. Problem is no one knows how to use either Hi. Here's the solution using Mathematica. Ouch. The professor wants you to learn Mathematica in a few hours?? That seems rather hard. Anyway... f[x_] := x^3 - 5*x^2 - 12*x + 19 The Newton equation: Simplify[x - f[x]/ Derivative[1][f][x]] (19 + 5*x^2 - 2*x^3)/ (12 + 10*x - 3*x^2) Make it a function. Note the . in the first number 19. We want to use machine precision. ff[x_] := (19. + (5 - 2*x)* x^2)/(12 + (10 - 3*x)*x) Lets pick a guess for starting value, say 5. FixedPointList[ff, 5] {5, 8.153846153846155, 6.918560378169316, 6.472953958451311, 6.410871476670647, 6.409698867110431, 6.409698452136048, 6.409698452135993, 6.409698452135993} FixedPointList[ff, 3] {3, 0.6666666666666666, 1.1901709401709402, 1.1556353312572267, 1.1555460134394535, 1.1555460128138022, 1.1555460128138022} FixedPointList[ff, -3] {-3, -2.6222222222222222, -2.566437743644001, -2.5652450059063234, -2.5652444649499073, -2.5652444649497963, -2.565244464949796} We can solve these equations. Looks like our Newton method worked well. NSolve[x^3 - 5*x^2 - 12*x + 19 == 0] {{x -> 6.409698452135987}, {x -> -2.565244464949797}, {x -> 1.1555460128138026}} Good luck! HTH. :>) Note: Your Newton equation will look like x - f[x] / f'[x] when switched to Standard Form in Mathematica. -- Dana DeLouis > Hi guys > my Optimization class prof. wants us to use Newton's Method to solve some > nonlinear optimization problems, and he wants us to program it using > Matlab or mathematica. problem is no one knows how to use either and he > wont teach us how to program!!! he says we should already know, even > though the only programming class req. for math majors is an easy C++ > intro class. anyways, the problem is this: > Apply Newton's method to find all three solutions of: > f(x) = x^3 - 5x^2 - 12x + 19 = 0 > if anyone can help with some matlab code as to how to do this using > newton's method i would be very very grateful!!! === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA f(x1, x2) = .... = 0 f(x1, x2) = ...... = 0 Now it wants me to solve those two equations (whatever they may be) now i can define the two equations, but how to i define the gradient? it would be a 2x2 matrix, but i cant find out to do that in mathematica's help === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA for reference, the prob is this: Solve: f(x1, x2) = x1^2 + x2^2 - 1 =0 f(x1, x2) = 5*x1^2 - x2 - 2 = 0 === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA > Apply Newton's method to find all three solutions of: > f(x) = x^3 - 5x^2 - 12x + 19 = 0 > if anyone can help with some matlab code as to how to do this using > newton's method i would be very very grateful!!! As f(1) = 1, one root may be close to x = 1. === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA >> Apply Newton's method to find all three solutions of: >> f(x) = x^3 - 5x^2 - 12x + 19 = 0 >> if anyone can help with some matlab code as to how to do this using >> newton's method i would be very very grateful!!! > As f(1) = 1, one root may be close to x = 1. f(1)=3 === Subject: (p-1)! question This is the actual question (its for homework but as of yet I've not really made any headway): Use the fact that p - j = -j(mod p) to show that if p is an odd prime, p = 2k + 1, then (p-1)! = (-1)^k [ ( (p-1) / 2)! ]^2 (mod p) or 2 k [[ ( p-1 ) ]] ( p - 1)! = (-1) [ ( --- ) ! ] (mod p) [[ ( 2 ) ]] The proof of Wilson's theorem seems similar but that doozy on the right - where did they get that? -e === Subject: Re: (p-1)! question > This is the actual question (its for homework but as of yet I've > not really made any headway): > Use the fact that > p - j = -j(mod p) > to show that if p is an odd prime, p = 2k + 1, then > (p-1)! = (-1)^k [ ( (p-1) / 2)! ]^2 (mod p) (p-1)! = [1*2*...*k] * [(k+1)*(k+2)*...*(2k)] = [1*2*...*k] * [(p-k)*(p-(k-1))*...*(p-1)]; Now use the fact at the top on the factors in the second square bracket. [...] Brian === Subject: Re: (p-1)! question This is the actual question (its for homework but as of yet I've >> not really made any headway): >> Use the fact that >> p - j = -j(mod p) >> to show that if p is an odd prime, p = 2k + 1, then >> (p-1)! = (-1)^k [ ( (p-1) / 2)! ]^2 (mod p) > (p-1)! = [1*2*...*k] * [(k+1)*(k+2)*...*(2k)] > = [1*2*...*k] * [(p-k)*(p-(k-1))*...*(p-1)]; > Now use the fact at the top on the factors in the second > square bracket. > [...] thanx! -e === Subject: Re: 2nd order DE w/initial conditions problem help <17296995.1131899951547.JavaMail.jakarta@nitrogen.mathforum.org I apologize for not using spaces. > I did get the answer that you provided. I'm confused at how to > integrate sqr(2y^3 + c) to find y in order to solve for the initial > conditions. As the equation is gone, I'm excused from making correction and/or answer. > Also, Not sure if your response is rude or sarcastic. Are you directing this to me? With removal of context, following a thread gets difficult. Even using Google, it can get bothersome. Lots of use don't use Google, so you may want to take that into accout, so please don't use the quick reply feature. Find and use the quote facitlity. Here's my standard reply. I've not the time nor the inclination for the clerical chore of back tracking the thread to reconstruct the line of thought. If you want intelligent and well thought answers, instead of sloppy inaccurate answers from what I remember, then include the context pertinent to your reply and to whom your are talking. Please learn and use better math group manners as demonstrated by other participants of this newsgroup and as described at http://oakroadsystems.com/genl/unice.htm#quote If you're posting from Mathforum or Google, it is requested you use the quote feature. Many of us use different news browsers than you. === Subject: Diophantine equation I`m trying to find all the integer solutions to: x^2+4=y^3 using as an aid, the field Z[i] (gaussian integers). My thoughts. I write the equation as (x+2i)(x-2i)=y^3. I would like to say that the ideals (x+2i) and (x-2i) are coprime, but I`m not sure of that. In any way I consider the ideal (x+2i,x-2i)=(d), with d a generator. If d divides x+2i and x-2i it also divides 4i, which factors into i(1+i)^4. Here I`m a bit stuck. Where I want to go to is to show that x+2i must be equal to a third power in Z[i], but I`m not even sure that's right. I'd appreciate any help. I want to try to show that Y^3 is some third power in Z[i]. === Subject: One-to-One I need to prove that if g composite f is one-to-one, then f is one-to-one. I have found the theorem stating the opposite, if two one-to-one functions can be composed then their composition is one-to-one. === Subject: Re: One-to-One days. My association with the Department is that of an alumnus. >I need to prove that if g composite f is one-to-one, then f is one-to-one. >I have found the theorem stating the opposite, if two one-to-one >functions can be composed then their composition is one-to-one. Well, the theorem you found is true, but it is not ->quite<- the opposite of what you want to prove. What you want to prove says nothing about g; but that's because there is nothing that ->can<- be said about g. Consider the following: f:X->Y, g:Y->Z, where X = {x}, Y = {y,y'}, and Z={z}. Define f(x)=y, and g(y)=g(y')=z. Then g composite x is the map from X to Z that maps x to z, which is one-to-one; but g is not one to one. Now, back to your original problem: Let f:A->B, g:B->C. And we assume that g composite f is one-to-one. What does that mean? It means that: (*) if g(f(a)) = g(f(a')) for a and a' in A, then a=a'; or equivalently, that (**) If a and a' are in A, and a is different from a', then g(f(a)) is different from g(f(a'')). You want to prove that f is one-to-one. that means that you want to prove one of the following statements (which are logically equivalent): (i) If a and a' are in A, and f(a)=f(a'), then a=a'; or (ii) If a and a' are in A, and a is different from a', then f(a) is different from f(a'). The easiest (in my humble opinion), is the first: suppose f(a)=f(a'). Can you think of some way of applying (*) or (**) to conclude that a=a'? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: One-to-One >I need to prove that if g composite f is one-to-one, >then f is one-to-one. >I have found the theorem stating the opposite, if two >one-to-one functions can be composed then their >composition is one-to-one. Assume f is not one-to-one. Then there exist x_1, x_2 with x_1 <> x_2 and f(x_1) = f(x_2). But if f(x_1) = f(x_2), then also g(f(x_1)) = g(f(x_2)), thus, g o f is not one-to-one. Contradiction. Best wishes Torsten. === Subject: Re: One-to-One days. My association with the Department is that of an alumnus. >>I need to prove that if g composite f is one-to-one, >then f is one-to-one. >>I have found the theorem stating the opposite, if two >one-to-one functions can be composed then their >composition is one-to-one. >Assume f is not one-to-one. >Then there exist x_1, x_2 with x_1 <> x_2 >and f(x_1) = f(x_2). >But if f(x_1) = f(x_2), then also >g(f(x_1)) = g(f(x_2)), >thus, g o f is not one-to-one. >Contradiction. Why contradiction? Why not just the contrapositive? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Sequence and Series hi guys, 1) suppose I have this sequence( NOT series) (n©Ö-1)n! / n(n+1)! , n=1,2..... can I use ratio test to solve it? 2) for this series, determine if it converges or diverges. how do you solve it? infinity Sum of (2 + sin n)/ Á.94(n-1) n=2 === Subject: Re: Sequence and Series >hi guys, >suppose I have this sequence( NOT series) >(n©Ö-1)n! / n(n+1)! , n=1,2..... >can I use ratio test to solve it? I don't know what you mean by 'solve'. If you want to calculate lim (n->oo) (n©Ö-1)n! / n(n+1)!, you should write n^2 - 1 as (n+1)*(n-1) in the numerator. >for this series, determine if it converges or diverges. >how do you solve it? >infinity >Sum of (2 + sin n)/ ?(n-1) >n=2 You know the comparison test for series ? Since |sin n| <= 1, (2 + sin n)/(n-1) >= 1/(n-1) for all n >= 2. Best wishes Torsten. === Subject: Symbolic Logic Are the following logical axioms: (1)Az[Ay(Py --> Py) --> (Pc --> Pc)] (2) AxEyPxy --> EyPyy (3) [AxPx --> AzPz) --> Py] --> [AxPx --> (AzPz --> Py)] Where A is the universal quantifier for all and E is the existential quantifier there exists. === Subject: Re: Stupid question days. My association with the Department is that of an alumnus. >> In lieu of bounded, compact may used, or rather subcompact, >> ie contained in a compact set. Subcompact and bounded >> are equivalent only for complete metric spaces. >> That would make any Banach space finite dimensional. >Are all infinite dimensional Banach spaces compact? As it appeared in this month's Monthly (vol. 112, no. 9, November 2005, pp. 816), proof by Sharham Biglari of the University of Lepzig: THEOREM. Let X be a real normed vector space. If the closed unit disk D in X is compact, then X is finite dimensionl. Proof. Let S be the sphere that bounds D. This is a closed subset of D, so it is compact; by Hahn-Banach, there exists for each x in S a continuous linear transformation pi_x:X-->R such that pi_x(x) is not zero. Then S subset Union pi_x^{-1}(R0) where the union is taken over all x in S. By compactness of S, we can pick a finite subcover, so there exist x1,..., xn in S such that S is contained in the union of pi_{xj}^{-1}(R0). The continuous linear map f:X-->R^n defined by x|-> (pi_(x1)(x),...,pi_(xn)(x)) is an R-monomorphism, so the space X is finite dimensional. QED If subcompact and bounded are equivalent in any complete metric space, then by this theorem, since the unit disc is trivially bounded in any Banach space, it would be subcompact, so the closed unit disk would be compact, so the Banach space would be finite dimensional. Since, of course, there are infinite-dimensional Banach spaces, that means that complete metric space is not sufficient for 'subcompact' and 'bounded' to be equivalent. Your statement could be read as saying it is, though I guess that formally it just says that complete metric space is ->necessary<- for the equivalence. Is that what you meant? -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: expectations of discrete r.v. Here's a simple question, but not for me! The distribution function of a random variable X is given by: F(x) = {0 if x < -3 {3/8 if -3 <= x < 0 {1/2 if 0 <= x < 3 {3/4 if 3 <= x < 4 {1 if x => 4 Calculate E(X) and E(X^2 - 2|X|). Here's what I did. E(X) = x * p(x) So, 0 + (-3)(3/8) + (-2)(3/8) + (-1)(3/8) + (1)(1/2) + (2)(1/2) + (3)(3/4) + (4)(1) = 4 But the answer in the back of the book is 5/8. If someone could help out it would be great. I'm sure if I find out how to get E(X), then calculating E(X^2 - 2|X|) should not be that difficult. === Subject: Re: expectations of discrete r.v. in alt.math.undergrad: > Here's a simple question, but not for me! > The distribution function of a random variable X is given by: > F(x) = {0 if x < -3 > {3/8 if -3 <= x < 0 > {1/2 if 0 <= x < 3 > {3/4 if 3 <= x < 4 > {1 if x => 4 > Calculate E(X) and E(X^2 - 2|X|). > Here's what I did. > E(X) = x * p(x) No, E(X) is the _sum_ of all terms of the form x * p(x) (as you realized when you set it up below). > So, 0 + (-3)(3/8) + (-2)(3/8) + (-1)(3/8) + (1)(1/2) + (2)(1/2) + (3)(3/4) + (4)(1) = 4 But F(x) is the distribution function, not the probability density function, so F(x) isn't the probability that X = x. F(x) jumps from 0 up to 3/8 at x = -3, so p(-3) = 3/8. Then F(x) is constant from -3 up to but not including 0, so p(x) = 0 for -3 < x < 0. At x = 0, however, F(x) takes another jump, from 3/8 up to 1/2; this means that p(0) must be 1/2 - 3/8 = 1/8. Similarly, p(3) = 3/4 - 1/2 = 1/4, and p(4) = 1 - 3/4 = 1/4. (Check: 3/8 + 1/8 + 1/4 + 1/4 = 1, so we've accounted for the total probability.) Hence E(X) = (-3)(3/8) + (0)(1/8) + (3)(1/4) + (4)(1/4) = -9/8 + 3/4 + 1 = 5/8. > But the answer in the back of the book is 5/8. > If someone could help out it would be great. I'm sure if > I find out how to get E(X), then calculating E(X^2 - 2|X|) > should not be that difficult. Right: just remember to use the probability density function, not the distribution. Brian === Subject: distribution function - need help OK. This question is giving me some problems. Let p(x) = 3/4(1/4)^x, x = 0, 1, 2, 3... be the probability mass function of a random variable X. Find the distribution function of X. Any help would be appreciated. === Subject: Re: distribution function - need help >OK. This question is giving me some problems. >Let p(x) = 3/4(1/4)^x, x = 0, 1, 2, 3... be the probability mass function of a random variable X. Find the distribution function of X. >Any help would be appreciated. Hint: Look up geometric progression in your algebra book. --Lynn === Subject: another discrete r.v. problem In a small town there are 40 taxis, numbered 1 to 40. Three taxis arrive at random at a station to pick up passengers. What is the probability that the number of at least one of the taxis is less than 5? Answer is 0.277 but I don't know how to get it. === Subject: Re: another discrete r.v. problem in alt.math.undergrad: > In a small town there are 40 taxis, numbered 1 to 40. > Three taxis arrive at random at a station to pick up > passengers. What is the probability that the number of > at least one of the taxis is less than 5? > Answer is 0.277 but I don't know how to get it. When you're asked for the probability of at least one of something, it's usually a good idea to look at the complementary problem: what's the probability that *none* of the taxis has a number less than 5, i.e., that *all* of them have numbers 5 or more? Look at the first arrival: the probability that its number is 5 or more is 36/40 = 0.9. Suppose that its number is in fact 5 or more. Now there are 39 taxis out there, 35 of which still have numbers of 5 or more, so the probability that the second arrival has a number of 5 or more is 35/39. Similarly, if both of the first two arrivals have numbers of 5 or more, the probability that the third arrival has such a number is 34/38. These probabilities are assumed to be independent, so the probability that all three taxis have numbers of 5 or more is (36/40)(35/39)(34/38); rounded to three decimal places this is 0.723. The probability that at least one of the taxis has a number less than 5 is the probability that they *don't* all have numbers of 5 or more, which is 1 - 0.723 = 0.277 (to three decimal places). Brian