mm-584 === Subject: Re: Consecutive Terms in math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i89LFHv24876; Can't you somehow find a way to change the method into an equation? There just has to be an equation for regardless of complexity... === Subject: Re: Consecutive Terms in math days. My association with the Department is that of an alumnus. >Can't you somehow find a way to change the method into an equation? Not off the top of my head. >There just has to be an equation for regardless of complexity... Not true. There doesn't have to be an equation to solve the problem. There are problems that cannot be solved through equations of the kind you are envisioning, and there is no reason to simply assume this isn't one of them. -- 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: subbasis for the topology of a set alt.math.undergrad: You snipped a bit too much here, so I'm restoring some context. You'd said that you were trying to prove this: > If X is a set and G an arbitrary set of parts of X, there > is exactly one topology T(G) on X such that G is a > subbasis for T(G) (the topology``generated'' by G). I responded: > Let's take a closer look at what you're trying to prove. > The claim is that there is one and only one topology > having G as a subbase. This means that you have two > things to prove: > (1) There is some topology on X that has G as a subbase. > (2) If T_1 and T_2 are topologies on X that both have > G as a subbase, then T_1 = T_2. > Here (1) says that G is a subbase for at least one > topology on X, and (2) says that G is a subbase for at > most one topology on X; putting the two together gives > the desired result. I then continued as you've quoted: > Now in fact (1) isn't quite true as you've stated the > theorem: in order to prove (1), you have to know that G > covers X, i.e., that every point of X is in at least one > member of G. (Another way to say this: U{A : A in G} = X.) > I don't know whether you've misstated the problem, or > whether it was poorly stated in your source. However, (2) > is still true, and I suspect that it's the main point of the > exercise. > perhaps i left something out. that was a passage (not a theorem) > of a book called ``topology'' by Klaus Janich which I might have > omitted something important. [...] > and here's the passage: > ``Definition (subbasis): Let X be a topological space. A > set S of open sets is called a subbasis for the topology > if every open set is a union of finite intersections of > sets in S. This is one of several equivalent definitions; I prefer a slightly different version, but this is correct. (Without the requirement that S consist of open sets it would not be correct.) > Of course the word ``finite'' here does not mean that the > intersection should be a finite set, but that is the > intersection of finitely many sets. I know: I'm a topologist. > This includes the intersection of zero sets (that is, an > empty family of sets), which by a meaningful convention > is defined to be equal to the whole space Ah, okay; that convention is not universally used, which is why some authors explicitly require that the union of S be the set X. If you're using it, then (1) is indeed true as stated. > With these conventions we then have that if X is a set and > S and arbitrary set of parts of X, there is exactly one > topology T(S) on X such that S is a subbasis for T(S) > (the topology generated by S). It consists exactly of > the unions of finite intersections of sets in S.'' > i am trying to show that the last paragraph is true. before, let > me give you in my words, what I understand as a subbasis. > let X be a topological space. let S be a collection of open > sets. S is a subbasis for the topology of X if and only if > n > B = { / s_i such that s_i, ..., s_n is in S} > i=1 > is a basis for the topology of X. is this correct? Yes. > and what I need to show to prove that passage from klaus > janich is (as you said) that there is some topology on X > that has S as a subbasis and if T_1 and T_2 are > topologies on X such that both have S as subbasis, then > T_1 = T_2. Yes. > proof: > we must show that there is T_1(S) on X such that S is a > subbasis. this is first thing. we also must show that if > T_2(S) is a topology on X such that S is a subbase for > open sets of (X,T_2), then T_2 = T_1. > i'm trying to use ``S as a subbase'' instead of show that > T_1 is a subset of T_2 and T_2 is a subset of T_1 as you > advised. i think it's easier this way. > define T_1(S) = {/_{i in I} b_i | for every i in I, b_i > is in B}. claim T_1(S) is a topology. let A_j be in > T_1(S) for every j in J. now, I must show that /_{j in > J} A_j is in T_1(S). so, for every j in J, there's an > I_j such that for every i in I_j, there is B_i ``with > respect to j'' in B such that A_j = /_{i in I_j} B_i > ``with respect to j''. > now, if I show that /_{j in J} A_j is in T_1(S), i > believe that means that T_1(S) = T_2(S). Not at all: it wouldn't even be a complete proof that T_1(S) is a topology on X. You'd still have to show that and X belong to T_1(S) and that if V and W are in T_1(S), then V / W is also in T_1(S). And then you'd still have to show that if T_2 is another topology on X having S as a subbase, then T_2 = T_1(S). > i tried my best with the notation. sorry if it's hard to > read. What you did is understandable, though you got a bit tangled up at the end where you really need double indices -- something like B_j_i, or the like. I find it easier without the indices. Let B = {/F : F is a finite subset of S}, which is your B defined with slightly different notation, and let T = {/A : A is a subset of B}, which is your T_1(S) defined with slightly different notation. You need to show first that T is a topology on X. The details will depend on just which of the various equivalent definitions you're using, but basically you have to show that X and belong to T and that T is closed under pairwise intersections and arbitrary unions. Your conventions on empty unions and intersections take care of the first two conditions: is a finite subset of S, so X = /257 is in B and hence in T, and is a subset of B, so = / is also in T. Suppose that V and W are in T. Then there are subsets B(V) and B(W) of B such that V = /B(V) and W = /B(W). Let p be any point in V / W. Then there are sets b(V) in B(V) and b(W) in B(W) such that p is in b(V) / b(W); let b(p) = b(V) / b(W), which is clearly a subset of V / W. Since b(V) is in B, there is a finite subset F(V) of S such that b(V) = /F(V); similarly, there is a finite subset F(W) of S such that b(W) = /F(W). But then b(p) = /(F(V) / F(W)), where F(V) / F(W) is a finite subset of S, so b(p) is in B. We've now shown that for each p in V / W there is a b(p) in B such that p is in b(p), and b(p) is a subset of V / W. It follows that V / W = /{b(p) : p is in V / W}, which by definition is in T. The proof that T is closed under arbitrary unions is trivial. Let U be any subset of T. For each u in U there is a subset B(u) of B such that u = /B(u). Let B(U) = /{B(u) : u is in U}; this is of course a subset of B, and /U = /{u : u is in U} = /{/B(u) : u is in U} = /B(U), which by definition is in T. This shows that T, the family of unions of finite intersections of sets in S, is a topology on X. To show that it's unique, suppose that T' is another topology having S as subbase. By definition this means (1) that S is a subset of T' (because the members of a subbase are required to be open in the space), and (2) that every member of T' is a union of finite intersections of sets in S. Clearly (2) implies that T' is a subset of T. Moreover, since T' is a topology on X, (1) implies first that B is a subset of T' (since T' is closed under finite intersections) and then that T is a subset of T' (since T' is closed under arbitrary unions). This shows that T' = T and establishes the desired result. (This was done a little hurriedly, so there might be a typo or two.) Brian === Subject: What are the best math symbol and geometry drawing programs? What are the best math symbol and geometry drawing programs? Please? -- Casey === Subject: Re: 25 of March and 28 of October >Hello people.. I recently found out that what happens is that 25 of >March and 28 of October have the same day, every year. e.g if 25 of >March 1980 is Monday , then 28 of October is Monday also.. >I' m trying for the proof but I have difficulties. >Can you please give me some help? >Have you ever seen that with some other dates maybe? Our calendar may be weird, but it is consistent. The interval between _any_ two dates is the same each year -- unless the interval includes leap year day. As a result, there are only 14 possible calendar years, seven for non leap years and seven for leap years. The seven in each case are for years starting on Sunday, Monday, etc. The set of universal calendars used to be printed in our local phone book, with an index telling which years used each calendar. If the interval you give is a multiple of seven, so they come out on same day, then it would also be true for 24 march and 27 October, 23 march & 26 October, etc, as well as your 25 march & 21 Oct, etc. etc. bob === Subject: Re: Precalculus Help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8A14Sc12914; The question asks Determine the quadrant(s) in which (x,y) is located so that the condition(s) is (are) satisfied 15. x>0 and y<0 21. (x,-y) is in the second quadrant could u please answer these problems and show me how to do them. I also need the solution to this as soon as possible please. === Subject: Re: Precalculus Help! alt.math.undergrad: >The question asks Determine the quadrant(s) in which (x,y) is located >so that the condition(s) is (are) satisfied >15. x>0 and y<0 >21. (x,-y) is in the second quadrant >could u please answer these problems and show me how to do them. I >also need the solution to this as soon as possible please. The best way to solve these is to read the section of your textbook that defines quadrants more carefully, and work through the examples. Your 15. is a direct reference to the definition of quadrants. For your 21., ask yourself where (x,y) must be if (x,-y) is in the second quadrant. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8A14Sp12922; you are faced with opening a combination safe numbered 0-59...you don't if it is a 3 or 4 number combination or if you dial left or right, if you take 15 seconds to try each combination how many days would it take you to try all of them???? === Subject: Re: help! X-NFilter: 1.2.0 >you are faced with opening a combination safe numbered 0-59...you >don't if it is a 3 or 4 number combination or if you dial left or >right, if you take 15 seconds to try each combination how many days >would it take you to try all of them???? How many three digit combinations to the left are there? How many four digit ones? How many total (think symmetry)? 4* total = minutes minutes/60 = hours hours/24 = days < === Subject: Re: ma 214 >f(t) = e^((-t^2)/2); f'(t) = -t.f(t); f(t) = -f(t) + t^2 f(t) >a(t) = f(t) integeral(0,t) f(t) dt > Not correct. > a(t) = f(t).integeral(0,t) dt/f(t) What's not correct about it? You've misquoted me and as the context is missing, you're claim lacks validation. Clarify dt/f(t). d(t/f(t)) ? >a'(t) = -t.a(t) + f(t)(f(t) - 1) >a(t) = -a(t) - t.a'(t) - 2t.f(t)^2 + t.f(t) > a'(t) = -t.a(t) + f(t)/f(t) = -t.a(t) + 1 > a(t) = -a(t) - t.a'(t) > and also > f(t) = -f(t) - t.f'(t) Doesn't mean anything out of context. === Subject: 2 questions _Algebra. I have following two difficult (to me) problems. Would you please give me some hint or sketch of proof? [1] If a group G has at least one element x s.t. Z(G)={g in G : x*g=g*x} then, Prove G is Abelian group. [2] For a group G, If m is a divisor of |G| and H_1,...,H_n is all subgroups of G with order m, Then, Prove H_1/H_2 ... /H_n is normal subgroup of G. === Subject: Re: 2 questions _Algebra. alt.math.undergrad: > I have following two difficult (to me) problems. > Would you please give me some hint or sketch of proof? > [1] > If a group G has at least one element x s.t. > Z(G)={g in G : x*g=g*x} then, > Prove G is Abelian group. Clearly x is in Z(G). But Z(G) is the centre of G, so x commutes with ... ? > [2] > For a group G, > If m is a divisor of |G| and > H_1,...,H_n is all subgroups of G with order m, > Then, Prove H_1/H_2 ... /H_n is normal subgroup of G. For any x in G and any subgroup H of G order m, x^(-1) * H_k * x (i.e., {x^(-1)*h*x : h in H}) is a subgroup of G of order m. Suppose that h is in H = H_1 / H_2 / ... / H_n, but that for some x in G x^(-1)*h*x is not in H. Then there is at least one k such that x^(-1)*h*x is not in H_k, and hence h is not in x * H_k * x^(-1). But this is a contradiction, because ... . Brian === Subject: sequences ahhhhhh!!!!! help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8AEhNN19469; 4 8 16 32 - + - + -- + -- 1 5 25 125 find ths sum to infinite I have the formula to work out the sum to infinite but i cant work out what the common difference of the terms are. Sn to infinite = a1/(1-r) (for infinite sequences when -14 8 16 32 >- + - + -- + -- >1 5 25 125 >find ths sum to infinite >I have the formula to work out the sum to infinite but i cant work out >what the common difference of the terms are. >Sn to infinite = a1/(1-r) >(for infinite sequences when -1sn= sum of the first n numbers >a1 is the first number in the sequence >r= the common ratio of the sequence >i am sure this is the correct formulae? but what is the common >difference if you can help me i would be very grateful as i am >struggling although it may be a very simple question?????? >sam This is, as you suspect, a geometric series but the way it is written may be misleading you! Notice that it starts at with 4 rather than 1: 4+ 8/5+ 16/25+ ... If you take a 4 out of each term: 4(1)+ 4(2/5)+ 4(4/25)+ ... See what happens? This is simply 4(2/5)^0+ 4(2/5)+ 4(2/5)^2+ ..., a geometric series with common ration (NOT common difference) 2/5. In any geometric series you can find the common ratio by dividing a term by the preceding term: 8/5 divided by 4 = 2/5 16/25 divided by 8/5= 2/5, 34/125 divided by 16/25= 2/5. So: a1= 4 and r= 2/5 (which IS between -1 and 1). Can you find the sum now? === Subject: Re: sequences ahhhhhh!!!!! help > 4 8 16 32 > - + - + -- + -- > 1 5 25 125 It's better to write this on one line, since some of us normally use proportional fonts: 4/1 + 8/5 + 16/25 + 32/125 + ... > find ths sum to infinite > I have the formula to work out the sum to infinite but i cant work out > what the common difference of the terms are. You don't want the common *difference*, since this isn't an arithmetic series; it's a geometric series, and you want the common *ratio*. If you have a sequence a_1, a_2, a_3, ..., the first ratio is a_2/a_1, the second ratio is a_3/a_2, the third ratio is a_4/a_3, and so on. If these are all the same, that number is the common ratio: it's common to every pair of consecutive terms. In your case the first ratio is (8/5)/(4/1) = (8/5)*(1/4) = 2/5. The second ratio is (16/25)/(8/5) = (16/25)*(5/8) = 2/5, and you can easily check that the third is also 2/5. Assuming that the terms continue to grow in the same fashion, the common ratio is therefore 2/5, and the sum is (4/1)/(1 - 2/5) = 4/(3/5) = 4*(5/3) = 20/3. [...] Brian === Subject: Re: Domain of inequality quadratic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8AGWoI29278; Ya but what's the domain then? {x|x...} === Subject: Re: Domain of inequality quadratic > Ya but what's the domain then? {x|x...} ÔDomain' is usually used in the context of functions. Given just the inequality you were given (or rather relayed to us) I would rather ask for the Ôsolution,' ie the set of all x values that make the inequality true. As suggested, when you move everything to one side and factor you get: (n + 5)(n - 4) < 0 So, answer the question already asked of you and you should have a clear idea how to proceed. What can you say about *that* inequality? Is not the product of two factors negative only when they differ in sign? Temporarily solve the related Ôequation' (n+5)(n-4)=0. Use those two solutions to divide the number line into three distinct intervals. Pick a number in each interval and determine the sign of each factor when evaluated at that particular x-value. The solution set to the original inequality will therefore be the interval(s) where the two factors differ in sign. -- Darrell === Subject: Re: Domain of inequality quadratic <...> > Temporarily solve the related Ôequation' (n+5)(n-4)=0. Use those two > solutions to divide the number line into three distinct intervals. Pick a > number in each interval and determine the sign of each factor when evaluated > at that particular x-value. n-value, that is, for your particular inequality. -- Darrell === Subject: Re: Domain of inequality quadratic > Ya but what's the domain then? {x|x...} Your equation is in n so why {x|x..}? it should be {n|n...} Where does the curve n^2 + n - 20 cross the x-axis? While it is below the x-axis n is in the domain. While it is above or on the x-axis n is not in the domain. Solving the quadratic tells you where the critical points are. === Subject: Problem I cannot solve by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8AHesN02702; A biased four-sided die is rolled once. Random variable N is defined to be the down-face value and is described by the PMF: [Eth] P(N) = N/10 for N = 1,2,3,4 [Eth] P(N) = 0 for all other values of N Based on the outcome of this roll, a coin is supplied for which, on any ßip, the probability of an outcome of heads is (N+1)/2N. The coin is ßipped once and the outcome of this ßip completes the experiment. Determine: [Eth] The Expected Value and Variance of the random variable N. [Eth] The conditional PMF, conditional expected value and conditional variance for random variable N given that the coin came up heads. [Eth] If we define the events: Event A: Value of down face on die roll is either 1 or 4. Event H: Outcome of coin ßip is heads Are the events A and H independent? === Subject: integral of y dx by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8B19YC07434; How do you solve this dy = integral of y dx. === Subject: Re: integral of y dx >How do you solve this dy = integral of y dx. Generally useful thing to try get everything to do with y on one side and everything to do with x on the other side of the equal sign (algebra is your friend) Then, what does it look like you might do to find what y really is in terms of x? That's only a hint, it seems like students are more impressed when they can figure out a problem on their own, with a suitable hint. === Subject: Re: integral of y dx > How do you solve this dy = integral of y dx. If this means what I think it means, you have y'' = y or y = a*exp(x) + b*exp(-x) -- Paul Sperry Columbia, SC (USA) === Subject: Re: integral of y dx > How do you solve this dy = integral of y dx. You can't, it's not a meaningful equation. For example how do you solve dy = x^2 ? === Subject: Outcome of trial by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8BCSFv27427; Friends... I am going to civil trial (honest; this is not a hypothetical). My attorney is arguing a legal point with 5 variables (A B C D E). To win we need prove only one point. For opposing counsel to win, they must prove all 5 points (already you can see that the odds will be wildly in my favor). The probability that opposing counsel will win each point is... A = 95% B = 95% C = 75% D = 50% E = 25% I have been extremely generous to opposing counsel. Question: What is the probability that opposing counsel will win all five points? Question: What is the probability that my attorney will win just one point? Paul P. Los Angeles === Subject: Re: Outcome of trial >I am going to civil trial (honest; this is not a hypothetical). My >attorney is arguing a legal point with 5 variables (A B C D E). To >win we need prove only one point. For opposing counsel to win, they >must prove all 5 points (already you can see that the odds will be >wildly in my favor). That does not follow AT ALL. It depends on what the five points are. >The probability that opposing counsel will win each point is... >A = 95% >B = 95% >C = 75% >D = 50% >E = 25% Let me see: you have numerical probabilities of winning each of five arguments, and this is not a hypothetical but a real-life situation. Uh-huh. Fer sure. >Question: What is the probability that opposing counsel will win all >five points? It is impossible to say, because it depends on a key issue: are the points independent? Or does winning one point make another point easier (or harder) to win? If the points are independent, you multiply the five probabilities. But in a real-life situation (such as you claim), five points in a legal argument are not usually independent in a mathematical sense. >Question: What is the probability that my attorney will win just one >point? That's the same as the probability that the other side will win four and lose one. Again, you can't answer this with the given information. It depends on whether the five points are independent. If the five points are independent (again, not terribly likely), you look at A and say, They have a .95 probability of winning and therefore we have a .05 probability of winning. The probability that we win A and they win BCDE is .05 * .95 * .75 * .50 * .25. Then you make a similar computation for each of the other four points, and add up the five numbers you obtained by those multiplications. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/