mm-551 === Subject: Re: PROBLEM (a^3 + b^3 = c^3) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LHXt610784; > Does anyone know what three numbers you would plug in to have that > formula come through. You cant use 0 and it has to be a whole number >There aren't any. This is a well studied problem. >-- >-- Geo. Michael Henry >No! Bad dog! I said sit! anonymous LOL! A great big elephant sits in this room. Who's going to say it first? === Subject: Re: PROBLEM (a^3 + b^3 = c^3) > Does anyone know what three numbers you would plug in to have that > formula come through. You cant use 0 and it has to be a whole > number > >There aren't any. This is a well studied problem. > LOL! A great big elephant sits in this room. Who's going to > say it first? Are you perhaps referring to Fermat's Last Theorem? But the OP's question concerned just the case n=3. The proof that there are no solutions in that particular case was known centuries before the proof of the general theorem. David === Subject: Re: PROBLEM (a^3 + b^3 = c^3) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LJxIK23944; > Does anyone know what three numbers you would plug in to have that > formula come through. You cant use 0 and it has to be a whole > number > > >There aren't any. This is a well studied problem. > LOL! A great big elephant sits in this room. Who's going to > say it first? >Are you perhaps referring to Fermat's Last Theorem? But the OP's question >concerned just the case n=3. The proof that there are no solutions in that >particular case was known centuries before the proof of the general >theorem. >David Um, yes, I know that. (IIRC, it was Euler who gave the first proof for this case, and Fermat made reference to this specific case in the famous quote in Latin where he claims to have a proof for the general case.) This is a well-studied problem sounded so tongue-in-cheek that I had to laugh. === Subject: Conics-Ellipses by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LISIp15715; Please help!! Halley's comit has an elliptical orbit with the sun at one focus. The eccentricity of the orbit is approximately 0.97. The length of the major axis of the orbit is about 36.23 astronomical units. (astronimical unit is about 93 million miles). Find an equation for the orbit. Place the center of the orbit at the orgin and place the major axis on the x axis. Respond ASAP!! PLEASE! === Subject: Re: Conics-Ellipses >Please help!! >Halley's comit has an elliptical orbit with the sun at one focus. The >eccentricity of the orbit is approximately 0.97. The length of the >major axis of the orbit is about 36.23 astronomical units. >(astronimical unit is about 93 million miles). Find an equation for >the orbit. Place the center of the orbit at the orgin and place the >major axis on the x axis. >Respond ASAP!! PLEASE! Hej, hej Kristy You know that the long axis 2a = 36.23 u, so a = 18.115 If the ellipse has focuses F and G, we have: FG = 2c Now 1. e = c/a is the excentricity, you have e and a so you can calculate c 2. b^2+c^2 = a^2 having a and c, you can calculate b 3. the equation of the ellipse is: x^2/a^2 + y^2/b^2 = 1 helps? Hans === Subject: Re: The true #0 !? > It would now seem to me that: > > R#=Real Number > oo=infinite > > 0*oo= R# >_Which_ real number? > Any and all (sorta like an answer set or domain) Tim Twentyman's objection is understandable. However, there is nothing wrong whatsoever with dealing with an arithmetic in which the objects are sets of numbers, rather than individual numbers themselves. Interval arithmetic is an important example of such an arithmetic. If we were to be technically correct (or pedantic, depending on one's point of view), eschewing any abuse of notation, then presumably you should have written {0}*{oo} = R since the objects we're dealing with are sets. However, in interval arithmetic, degenerate intervals [x, x] are commonly written as merely x, rather than {x}, and this does not lead to confusion normally. I think the same slight abuse of notation could be permitted here as well, allowing us to write simply x, instead of {x}. I shall do that henceforth. But even then there's something slightly wrong with 0*oo = R, it seems to me. Suppose the sets with which we're working have as elements members of the one-point extension of the reals, and let's denote that extension as R*. Then I suggest that what you should actually want is 0*oo = R*. This is virtually the same thing as is done in interval arithmetic sometimes, except that there the two-point extension is normally used. available on the web. > R#/0= oo (I think I may need absolute value or a (+-)infinite) Maybe you really meant x/0 = oo for all x in R*. [Note that no absolute value or +- is needed if we're dealing with the one-point extension.] However, this does not work well for x = 0, it seems to me. I would suggest x/0 = oo for all nonzero x in R*, 0/0 = R*. [This, as well as what I suggest below, is essentially what Walster does, except that he's working with the two-point extension instead.] > R#/oo= 0 I suggest x/oo = 0 for all finite x in R*, oo/oo = R*. > It sorta makes sense... Yes, it does indeed make sense if, say, you're familiar with so-called indeterminate forms of limits. BTW, when you're making up examples concerning division, I'd recommend staying away from doughnuts. I think it would be easier for most people to think about dividing up, say, a line segment of a certain length. David W. Cantrell === Subject: Re: Integral of (1/x-1)^.5 dx > I meant.... > sqrt[(1/x) - 1] dx > No one in my calculus class, including my professor, could do this > problem. > The answer in the back of the book is... > x sqrt[(1/x)-1]-arctan[sqrt[(1/x-1]]+C > Also you have to integration by parts to do the integral. Or sqrt(x-x^2) + arcsin(2*x-1)/2 + C === Subject: Re: Integral of (1/x-1)^.5 dx by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LHXtB10761; >I meant.... >sqrt[(1/x) - 1] dx >No one in my calculus class, including my professor, could do this >problem. >The answer in the back of the book is... >x sqrt[(1/x)-1]-arctan[sqrt[(1/x-1]]+C >Also you have to integration by parts to do the integral. Something is wrong here: the derivative of the answer in the back of the book isn't sqrt[(1/x) - 1], unless I've fouled up big time. If the book is in error, the publisher should be contacted. Here's what I did: sqrt[(1/x) - 1] = (sqrt[x - x^2])/x. Complete the square to get: sqrt[1/4 - (x - (1/2))^2]/x. Set x - 1/2 = (1/2)sin t to get (sqrt[1/4 - (x - (1/2))^2])/x dx = (1/2) cos^2(t)/(1 + sin t) dt Multiply numerator and denominator of the last integrand by 1 - sin t to get (1/2)(1 - sin t) dt. Integrate and substitute back in to get (1/2)( arcsin(2x - 1) ) - sqrt[x - x^2] + C as final answer. The derivative checks out (after a little algebra). Your professor couldn't do this? Or didn't get the answer in the back? Todd Trimble === Subject: Re: Integral of (1/x-1)^.5 dx >I meant.... >sqrt[(1/x) - 1] dx >No one in my calculus class, including my professor, could do this >problem. >The answer in the back of the book is... >x sqrt[(1/x)-1]-arctan[sqrt[(1/x-1]]+C >Also you have to integration by parts to do the integral. > Something is wrong here: the derivative of the answer in the back > of the book isn't sqrt[(1/x) - 1], unless I've fouled up big time. FWIW, according to the almighty integrals.wolfram.com :-), the integral of sqrt[(1/x) - 1]dx is ln[sqrt(x - 1) + sqrt(x)] sqrt(x) sqrt[(1/x) - 1] { sqrt(x) - ----------------------- } sqrt(x - 1) I leave it to someone with more time on their hands than I to see if that simplifies to the book's answer. -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: Integral of (1/x-1)^.5 dx by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LIsGB17847; >I meant.... >sqrt[(1/x) - 1] dx >No one in my calculus class, including my professor, could do this >problem. >The answer in the back of the book is... >x sqrt[(1/x)-1]-arctan[sqrt[(1/x-1]]+C >Also you have to integration by parts to do the integral. >Something is wrong here: the derivative of the answer in the back >of the book isn't sqrt[(1/x) - 1], unless I've fouled up big time. >If the book is in error, the publisher should be contacted. >Here's what I did: sqrt[(1/x) - 1] = (sqrt[x - x^2])/x. Complete >the square to get: > sqrt[1/4 - (x - (1/2))^2]/x. >Set x - 1/2 = (1/2)sin t to get > (sqrt[1/4 - (x - (1/2))^2])/x dx = (1/2) cos^2(t)/(1 + sin t) dt >Multiply numerator and denominator of the last integrand by >1 - sin t to get (1/2)(1 - sin t) dt. Integrate and substitute >back in to get > (1/2)( arcsin(2x - 1) ) - sqrt[x - x^2] + C >as final answer. The derivative checks out (after a little >algebra). >Your professor couldn't do this? Or didn't get the answer in >the back? >Todd Trimble Whoops -- book's answer is ok. And I had a sign mistake in my answer: should be ... + sqrt[x - x^2]. Otherwise the two answers, while different in appearance, are reconcilable. === Subject: Re: chainrule > Hello. > I've heard that you should always judge the quality of a calculus book by > looking at the proof of the chainrule (meaning that many authors have > incorrect proof). > So, what is a correct proof, and have anyone seen any incorrect proofs (that > looks good, but has some ßaws)? > Here is one proof that I've found. What can you say about it? > y=f(g(x)), y=f(u), u=g(x) > dy/dx = dy/dx = dy/du * du/dx > Let dx->0 and we get > lim dy/dx = lim dy/du * lim du/dx > x->0 x->0 x->0 > y'(x)=f'(u)*u'(x) > This assumes that du != 0 for all small dx Since that need not be the case, and du = g(x + dx) - g(x) can be zero for arbirarily small dx, the proof is ßawed. And this is the ßaw that occurs in most texts that have a ßaw. IIRC, the Apostol Calculus series avoids that error. And Courant almost certainly avoids it. === Subject: China girl in Paris charset=Windows-1252
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=== Subject: Hard probability problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8LLi8i01218; Does anyone have any ideas on how to solve this: If we have an empty room, and every week there is a .6 probability for a person to join, .2 probabillity for a person to leave, and .2 probability for no change (only one action can be performed per week), what is the probability that there are at least 40 people in the room after 104 weeks? Mike === Subject: Re: Hard probability problem X-RFC2646: Original > Does anyone have any ideas on how to solve this: > If we have an empty room, and every week there is a .6 probability for > a person to join, .2 probabillity for a person to leave, and .2 > probability for no change (only one action can be performed per week), > what is the probability that there are at least 40 people in the room > after 104 weeks? > Mike It's not quite true in the first week, Since no one is there, there can not be a .2 probability of someone leaving. Also I assume you mean that .2 probability for a person to leave means a .2 probability that anyone in the room leaves - i.e. only one person can leave per week. Have you studied Markov Chains and Random walks? This is the kind of area this falls into. Bill === Subject: Re: Hard probability problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8M1Fdd19703; Hi Bill, I have not studied randoma walks and markov chain. I have looked them up on mathworld.wolfram, but that didn't give me much info. Regarding the first week, you can assume that there are 2 people in the room. Could you maybe give me some clues on how I should start thinking about this problem? I appreciate your help. Mike > Does anyone have any ideas on how to solve this: > If we have an empty room, and every week there is a .6 probability for > a person to join, .2 probabillity for a person to leave, and .2 > probability for no change (only one action can be performed per week), > what is the probability that there are at least 40 people in the room > after 104 weeks? > Mike >It's not quite true in the first week, Since no one is there, there can not be >a .2 probability of someone leaving. Also I assume you mean that .2 >probability for a person to leave means a .2 probability that anyone in the >room leaves - i.e. only one person can leave per week. Have you studied Markov >Chains and Random walks? This is the kind of area this falls into. >Bill === Subject: Re: Hard probability problem X-RFC2646: Original > Hi Bill, > I have not studied randoma walks and markov chain. I have looked them > up on mathworld.wolfram, but that didn't give me much info. > Regarding the first week, you can assume that there are 2 people in > the room. > Could you maybe give me some clues on how I should start thinking > about this problem? > I appreciate your help. You changed the initial problem. Is this a homework problem or something you are doing on your own? In any case, I really can't give you much additional help because that is what this is - random walk/Markov chain. If this is something you are studying on your own, you might research discrete Markov chains and transition matrices. If this is homework, you might give me some suggestions as to what areas you are studying now and that might trigger something. But I would not count on it. :) Or perhaps someone else has a different approach. Bill > Mike > Does anyone have any ideas on how to solve this: > If we have an empty room, and every week there is a .6 probability > for > a person to join, .2 probabillity for a person to leave, and .2 > probability for no change (only one action can be performed per > week), > what is the probability that there are at least 40 people in the > room > after 104 weeks? > Mike >It's not quite true in the first week, Since no one is there, there > can not be >a .2 probability of someone leaving. Also I assume you mean that .2 >probability for a person to leave means a .2 probability that anyone > in the >room leaves - i.e. only one person can leave per week. Have you > studied Markov >Chains and Random walks? This is the kind of area this falls into. >Bill === Subject: Re: Hard probability problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8M1Fdq19713; Hi Bill, I have not studied random walks and markov chain. I looked them up online at mathworld and did not find much there. Could you maybe give me some clues on how i should start thinking this problem? Mike > Does anyone have any ideas on how to solve this: > If we have an empty room, and every week there is a .6 probability for > a person to join, .2 probabillity for a person to leave, and .2 > probability for no change (only one action can be performed per week), > what is the probability that there are at least 40 people in the room > after 104 weeks? > Mike >It's not quite true in the first week, Since no one is there, there can not be >a .2 probability of someone leaving. Also I assume you mean that .2 >probability for a person to leave means a .2 probability that anyone in the >room leaves - i.e. only one person can leave per week. Have you studied Markov >Chains and Random walks? This is the kind of area this falls into. >Bill === Subject: Re: Solving for the n root. >a^n = 2a+n+1 >n^a = 2n+a+1 >n>a > >How do you solve and what is the solution? > n = 3, a = 2. This was solved by inspection, really, not by > any systematic analysis. > Another integer solution is n = 1, a = -2. I suspect that it may be the > only other real solution. Perhaps it would interesting to note that if the inequality is weakened to n >= a, we get two more real solutions: n = a = 2.40199262... (obtained by solving x^x = 3x + 1 numerically) and n = a = 0 (assuming that 0^0 is taken to be 1, as is often done). David Cantrell > One question is: what are a, n? Integers, real numbers, or what? > I expect that the solution above is the unique real solution in > any case, but it's probably not hard in the integer case to put > small bounds on the size of a and n (e.g. |a|, |n| < 4) and then > since this leaves only finitely many possibilities, one can argue > by process of elimination if nothing else. In the real case, > something more sophisticated may be required to demonstrate > uniqueness. It looks messy. === Subject: problem about differentiation of matrix Given a 1xn matrix A and nx1 matrix B, what is d(AB)^2/dA ? Can anyone give me some cue? === Subject: Re: problem about differentiation of matrix by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8MFw4q00967; >Given a 1xn matrix A and nx1 matrix B, what is >d(AB)^2/dA ? >Can anyone give me some cue? Ignore the fact that A and B are matrices and just do the derivative as if they were numbers. What is the derivative of (xy)^2 with respect to x? In any algebra in which the derivative can be defined, the product rule and chain rule are true. === Subject: Re: For whose want to know there level of knowledges in analisis === Subject: Re: For whose want to know there level of knowledges in analisis > For continuous(only) function in(a,b) every point > from (a,b) is a local extremum.Proof that f=const. >If f isn't constant on (a, b), then there are c and d in >(a, b) such that c < d, and either f(c) < f(d), or f(c) > >f(d). Assume that f(c) < f(d); the other case is similar >(or if f(c) > f(d), you can look at the function -f instead). >Define g(x) = max{f(y) : y in [c, d]}. Note that g is a >non-decreasing function on [c, d], f(x) <= g(x) for all x in >[c, d], g(c) = f(c), g(d) = f(d), and g is continuous on [c, d]. g is a constant function with, I presume, domain (a,b). >If f has a local maximum at some u in (c, d), then there is a v > u >such that g is constant on [u, v]. If f has a local minimum at >some u in (c, d), then there is a v < u such that g is constant on >[v, u]. g is constant on all of (a,b). >Let U be the union of all open intervals in [c, d] on which >g is constant; the last two sentences of the previous >paragraph show that U is dense in [c, d] (i.e., every point >of [c, d] is a limit point of U). Define an equivalence >relation, ~, on U: for x, y in U with x < y, x ~ y if and >only if g is constant on [x, y]. The equivalence classes of >this relation are pairwise disjoint open intervals, and of >course g is constant on each of them; let E be the set of >these equivalence classes. >Each of these equivalence classes must contain a rational >number, so there can be only countably many equivalence >classes. But g is continuous, so the range of g contains >the interval [g(c), g(d)], which is uncountable. Codomain g, I presume, is R. Range of g is a single point. >Pick a y in [g(c), g(d)] that is not the value of g on any >member of E; there must be some x in [c, d] such that g(x) = y. >Clearly x is not in any member of E; indeed, if V is in E, >then x cannot even be an endpoint of V. But this >contradicts the fact, previously established, that g is >constant on a closed interval having x as one endpoint. Would you clarify how is g defined? ---- === Subject: Re: For whose want to know there level of knowledges in analisis === > Subject: Re: For whose want to know there level of knowledges in analisis > For continuous(only) function in(a,b) every point > from (a,b) is a local extremum.Proof that f=const. >If f isn't constant on (a, b), then there are c and d in >(a, b) such that c < d, and either f(c) < f(d), or f(c) > >f(d). Assume that f(c) < f(d); the other case is similar >(or if f(c) > f(d), you can look at the function -f instead). >Define g(x) = max{f(y) : y in [c, d]}. Note that g is a >non-decreasing function on [c, d], f(x) <= g(x) for all x in >[c, d], g(c) = f(c), g(d) = f(d), and g is continuous on [c, d]. > g is a constant function with, I presume, domain (a,b). Obviously that should have been g(x) = max{f(y) : y in [c, x]}. [...] Brian === Subject: Re: i need a riddle solved!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8MBiv906363; Riddle: Presto...Abracadabra...First one then the other. I'm _ _ _ _ _ _ _ _ _ _ _ _ _ _ === Subject: Re: i need a riddle solved!!! > Riddle: > Presto...Abracadabra...First one then the other. > I'm _ _ _ _ _ _ _ _ _ _ _ _ _ _ o f f t o p i c p o s t e r -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Discrete Mathematical structures by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8ME4cO21612; CAN ANYONE HELP TO THESE HOMEWORK QUESTIONS... 1) A = B = R A relation R: A .b3_ B defined by R(a) = ea Determine whether that relation as defined is also a function; and if so, what is the range of the function? 2) A = R B = Z a function f is defined f: A .b3_ B so that f(a) = .b3.b2 a .b3 called the ßoor function. Find the images produced by the following inputs: a) f( 3.67) = ____________ b) f( -3.67) = ____________ c) f( - .b3 21) = __________ d) f( .b3 17 ) = ____________ 3) A = { 1,2,3,4} B = { a,b,c,d} f : A .b3_ B f = { (1,a), (2,a), (3,c), (4,d) } Is this function; one-to-one, onto, both; or neither? Give supporting evidence for your decisions. 4) Using the same function and sets from part three now determine the inverse of f. Is f V1 also a function? Explain your choice of answer. 5) Let set A be a finite set of four elements ; A = { a,b,c,d} define A* as the set of all finite sequences of elements taken from A. Let l : A* .b3_ Z be a function that matches each string w with the integer number of elements in that string. This function l(w) has outputs of the length of those strings. a) Verify the l(w) is everywhere defined on A*. b) Determine whether l(w) is onto or not and give supporting evidence. c) Prove that l(w) is not one-to-one. 6) Show that f( n ) = n100 , is O(g) for the function g(n) = 2n ; but that f(n) is not g. 7) Order the following functions by -classes from lowest or slowest growing to fastest growing. F1 (n) = 5 n lg (n) F2 (n) = 6 n2 + 4n V6 F3 (n) = 2.5n F4 (n) = lg(n4) F5 (n) = 12,271 F6 (n) = -31n F7 (n) = n ! F8 (n) = .b3 n { 5n 8) Which of these following functions are permutations of R ? f : R .b3_ R a) f(a) = a3 b) f(a) = e a c) f(a) = 2a + 5 For problems 9 - 12 we will use the set A = { 1,2,3,4,5} and the four permutations of A P1 = .b3 1 2 3 4 5 .b3.83 P2 = .b3 1 2 3 4 5 .b3.83 P3 = .b3 1 2 3 4 5 .b3.83 P4 = .b3 1 2 3 4 5 .b3.83 .b3 4 5 2 1 3 .b3 .b3 3 1 2 4 5 .b3 .b3 5 4 3 2 1 .b3 .b3 4 3 1 2 5 .b3 9) Compute: a) P1-1 = _______________ b) P3 o P2 = _______________ c) P2-1 o P2 = _________________ 10) How many permutations of set A are possible? How many even? How many odd? 11) a) Write P3 as a product of disjoint cycles: b) Write P2 as a product of transpositions: c) Determine which of our four given permutations are even and which are odd: 12) Compute these products : a) ( 1 , 5, 3) o ( 2, 4) = b) ( 2, 5, 4 ) o ( 1 , 3) = c) (1, 4) o ( 3, 5) = 13) If we assume that 10,370 account records need to be stored using a hashing function h ; which takes the first three digits of an account number as one value and the last four as a second value adds these two and applies the Mod- 93 function to find a location list. a) How many linked lists will we need ? b) If an approximately even distribution is achieved, roughly how many records will be stored in each lsit? c) Compute the location number for the following accounts; 4738124 1253135 3086291 14) Prove rule #3 (na) is lower than (nb) if and only if 0The weather forecastor reports that the porbability of rain tomorrow >is 10%. >How do you interpret the probability? > What that really means is Looking over our data bank of past > weather, we find that, of all days when our various indications > (temperature, air pressure, wind speed, weather upwind, etc) we the > same as now, it rained on 10% of those days. Another variable is the size of the area covered by the forecast: do they count it as an occurrence of rain if there was rain anywhere in that area? What if it rained for only 3 seconds? It doesn't seem like a well-defined probability. -- john