mm-3599 === Subject: Re: Induction > Take well-ordered set T which has smallest element 0. Let T have the > additional property that all elements in it, expect 0, have immediate > predecessor. If Q is a property that > i) 0 has the property Q, Q(0). > ii) if x in T has the property Q then the successor of x, S(x), has the > property Q, Q(S(x)). > Then property Q holds for every element in T. > Yes, and here's why. Let B = {t in T : ~Q(t)}, the set of Actually I made this proof before the original post but somehow felt unsure about it. Don't it have an hidden assumption that induction principle indeed works if T is infinite? Or am I missing something very elementary here? > elements of T that don't have property Q. If B is > non-empty, then it has a smallest element, b, in the > well-ordering of T. Since 0 has the property Q, clearly b > is not 0, and therefore b = S(t) for some t in T. But t < b Here for some t in T suggests that we are working with finite sets...? > (where I use '<' for the well-ordering on T), and b was the > <-minimal element of B, so t is not in B, and therefore > Q(t). But then (ii) implies Q(b), a contradiction. It > follows that B must be empty, i.e., that Q(t) for each t in > T. > Now, if T is infinite, does this mean that all elements in > the infinite set T has the property Q [...] > Yes: see above. This gets pretty abstract and difficult... === Subject: Re: Induction On 22 May 2006 02:50:37 -0700, un student On 21 May 2006 01:13:53 -0700, un student >> Take well-ordered set T which has smallest element 0. Let T have the >> additional property that all elements in it, expect 0, have immediate >> predecessor. If Q is a property that >> i) 0 has the property Q, Q(0). >> ii) if x in T has the property Q then the successor of x, S(x), has the >> property Q, Q(S(x)). >> Then property Q holds for every element in T. >> Yes, and here's why. Let B = {t in T : ~Q(t)}, the set of >Actually I made this proof before the original post but somehow felt >unsure about it. Don't it have an hidden assumption that induction >principle indeed works if T is infinite? >Or am I missing something very elementary here? >> elements of T that don't have property Q. If B is >> non-empty, then it has a smallest element, b, in the >> well-ordering of T. Since 0 has the property Q, clearly b >> is not 0, and therefore b = S(t) for some t in T. But t < b >Here for some t in T suggests that we are working with finite >sets...? ??? Why do you think that? >> (where I use '<' for the well-ordering on T), and b was the >> <-minimal element of B, so t is not in B, and therefore >> Q(t). But then (ii) implies Q(b), a contradiction. It >> follows that B must be empty, i.e., that Q(t) for each t in >> T. >> Now, if T is infinite, does this mean that all elements in >> the infinite set T has the property Q [...] >> Yes: see above. >This gets pretty abstract and difficult... Not difficult at all - there's just something basic you're seriously misunderstanding. Hence the question above. ************************ === Subject: Re: Induction On 22-May-2006, Jim Heckman additional property that all elements in it, expect 0, have immediate > predecessor. If Q is a property that > i) 0 has the property Q, Q(0). > ii) if x in T has the property Q then the successor of x, S(x), has the > property Q, Q(S(x)). > Then property Q holds for every element in T. > Now, if T is infinite, does this mean that all elements in the infinite > set T has the property Q or that for every finite subset T_m = {x | x < > m, x in T}, m in N+, all elements x in T_m has the property Q? > There is no such T. Any well-ordered set all of whose elements > except 0 have an immediate predecessor is finite. D'oh. Except for sets order-isomorphic to omega, of course. But then it's just a question of ordinary mathematical induction. -- Jim Heckman === Subject: Re: Induction Take well-ordered set T which has smallest element 0. Let T have the > additional property that all elements in it, expect 0, have immediate > predecessor. If Q is a property that <.. > Now, if T is infinite, does this mean that all elements in the infinite > set T has the property Q or that for every finite subset T_m = {x | x < > m, x in T}, m in N+, all elements x in T_m has the property Q? > There is no such T. Any well-ordered set all of whose elements > except 0 have an immediate predecessor is finite. I see, never thought about that. What if we drop immediate predecessor and state that every element has exactly one successor? Would it change the situation? Does the induction principle need the immediate predecessor in order to work? Induction principle seems like an easy way to proove things for finite sets, but I don't understand how to extend it to infinite sets. (or can it be extended?) > D'oh. Except for sets order-isomorphic to omega, of course. But > then it's just a question of ordinary mathematical induction. Emh, ok, I take your word for it :) (don't have a clue what order-isomorphic to omega means...) === Subject: Re: Induction Take well-ordered set T which has smallest element 0. Let T have the >> additional property that all elements in it, expect 0, have immediate >> predecessor. If Q is a property that ><..> Now, if T is infinite, does this mean that all elements in the infinite >> set T has the property Q or that for every finite subset T_m = {x | x < >> m, x in T}, m in N+, all elements x in T_m has the property Q? >> There is no such T. Any well-ordered set all of whose elements >> except 0 have an immediate predecessor is finite. >I see, never thought about that. It's not true. Also not relevant. >What if we drop immediate >predecessor and state that every element has exactly one successor? >Would it change the situation? >Does the induction principle need the immediate predecessor in order >to work? For induction _in_ the form that it's being stated and used here yes, we need that every element have an immediate predecessor. There are other ways of formulating induction that do not require this, but we should start by getting this version straight. > Induction principle seems like an easy way to proove things >for finite sets, but I don't understand how to extend it to infinite >sets. (or can it be extended?) Nobody has any idea where you get the idea that something here only works for finite sets. Look. We're given that T is well-ordered, and that every element of T other than the smallest element 0 has an immediate predecessor. That says that for every x in T except for x = 0 there exists y in T such that x = S(y), where S(y) denotes the successor of y. We're also given that 0 has property P, and that P(y) implies P(S(y)) (that is, if y has property P then S(y) has property P.) It follows that every element of T has property P. Proof: Let A be the set of elements of T which do not have property P. We need to show that A is empty. So we assume that A is not empty, and derive a contradiction. Now since A is nonempty and T is well-ordered A has a smallest element. Say x is the smallest element of A. There are two possibilities, x = 0 or x > 0. But x = 0 is impossible: x is in A, so x does not have property P, but we're given that 0 _does_ have property P. So x is not equal to 0. So x > 0. Then by assumption there exists y with x = S(y). This implies that y < x. Since x is the smallest element of A, y is not an element of A. Since y is not an element of A, y has property P. But now this implies that x also has property P, since x = S(y). Which contradicts the fact that x is an element of A. QED. There's the proof. Nothing there has anything at all to do with any set being finite. >> D'oh. Except for sets order-isomorphic to omega, of course. But >> then it's just a question of ordinary mathematical induction. >Emh, ok, I take your word for it :) (don't have a clue what >order-isomorphic to omega means...) ************************ === Subject: Re: Induction Nobody has any idea where you get the idea that something > here only works for finite sets. Gosh, it was a stupid stupid misthought (and no, I'm not telling where idea came from :)... I just completely mixed two separate things in one proof I saw). I guess its time for vacation... === Subject: Re: Induction Except for sets order-isomorphic to omega, of course. But > then it's just a question of ordinary mathematical induction. > Emh, ok, I take your word for it :) (don't have a clue what > order-isomorphic to omega means...) f:X -> Y is an order isomorphism when f surjection, X,Y (partially) ordered sets and for all x,y in X, (x <= y iff f(x) <= f(y)). Exercise: order ismorphisms are bijections (even when between partially ordered sets). === Subject: Re: Geometry Problem > Hi All: > Solving a geometry problem I run into some > difficulties. > There is a triangle ABC with its incircle, O as a > center. The points of > tangency with AC and BC are M and N respectively. BO > intersects MN at P. > My text book starts solving the problem which is > larger, saying that the > trick to solve it is realizing that angle APB is 90 > degrees. > I had to run over complex trigonometric substitutions > to show that this > angle is indeed 90 degrees. Is there any other > simpler way to demonstrate > it? (angle MNO) = (angle ACB)/2 (angle NPB) = Pi/2 - (angle ACB)/2 - (angle CBA)/2 = = (angle BAC)/2 = (angle MAO) (angle MAO) + (angle MPO) = Pi Therefore the points A, M, P and O are cyclic (?) and === Subject: Re: Geometry Problem > Hi All: > Solving a geometry problem I run into some > difficulties. > There is a triangle ABC with its incircle, O as a > center. The points of > tangency with AC and BC are M and N respectively. BO > intersects MN at P. > My text book starts solving the problem which is > larger, saying that the > trick to solve it is realizing that angle APB is 90 > degrees. > I had to run over complex trigonometric substitutions > to show that this > angle is indeed 90 degrees. Is there any other > simpler way to demonstrate > it? > (angle MNO) = (angle ACB)/2 > (angle NPB) = Pi/2 - (angle ACB)/2 - (angle CBA)/2 = > = (angle BAC)/2 = (angle MAO) > (angle MAO) + (angle MPO) = Pi > Therefore the points A, M, P and O are cyclic (?) and === Subject: Re: riemann integrable >I have this problem, >Let A be a bounded subset of R, and suppose f: A -> R is a bounded Riemann integrable function. > If f(x) > or = 0 for all x in A and the (intergral A) f = 0, prove the set A of 0 = { x in A : f(x) DNE 0 } has measure zero. >I know to start it by Defining A of n = {x in A : f(x) > 1/n} Let's call that set A_n. >however I am lost after that.. Show that A_n has measure zero, show that the set where f > 0 is the union of the A_n, for n = 1, 2, ..., and prove or use the previously-proved fact that a countable union of sets of measure 0 has measure 0. ************************ === Subject: Re: riemann integrable a .8ecrit : >I have this problem, >Let A be a bounded subset of R, and suppose f: A -> R is a bounded Riemann integrable function. > If f(x) > or = 0 for all x in A and the (intergral A) f = 0, prove the set A of 0 = { x in A : f(x) DNE 0 } has measure zero. >I know to start it by Defining A of n = {x in A : f(x) > 1/n} > Let's call that set A n. >however I am lost after that.. > Show that A n has measure zero, show that the set where f > 0 is > the union of the A n, for n = 1, 2, ..., and prove or use the > previously-proved fact that a countable union of sets of measure > 0 has measure 0. > ************************ > BTW the exact same demonstration works for positive Lebesgue-integrable functions, a mutch more general case... === Subject: Re: need help checking a basic integration formula via differentiation Ken === Subject: Re: need help checking a basic integration formula via differentiation On Mon, 22 May 2006 03:36:23 EDT, lingyai My text (Forgotten Calculus, Bleau) states the following integration formula: >integral [dx / (x ln x)] = ln |ln x| + C. >In the course of a problem where I've applied the formula, I realised I can't prove it. >Here's my attempt, for what it's worth. >(Re notation: >exponents are denoted with the ^ sign, e.g x squared = x^2; >first derivatives are denoted with the ' symbol, e.g. the 1st derivative of f(x) = f'(x)) >f(x) = ln |ln x| + C >I want to differentiate this via the product rule, so I rewrite the function as >f(x) = (ln x^0) * (ln x) + C Others have already pointed out that this 'rewrite' is not correct. >Using the product rule, >f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ] >= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)] >= [ ln / x ] + [0] >= ln / x. >I don't see how ln / x = ln |ln x|. >Any advice? Others have also pointed out that all you need to do is apply the chain rule. As a learning tool, you might want to check out the following site: For the question you posted, click on 'derivatives' and then enter your function, written as ln[ ln[x] ] The step-by-step result should make clear to you how the chain rule is used to find the derivative. === Subject: Re: need help checking a basic integration formula via differentiation On Mon, 22 May 2006 03:36:23 EDT, lingyai My text (Forgotten Calculus, Bleau) states the following integration formula: >integral [dx / (x ln x)] = ln |ln x| + C. >In the course of a problem where I've applied the formula, I realised I can't prove it. >Here's my attempt, for what it's worth. >(Re notation: >exponents are denoted with the ^ sign, e.g x squared = x^2; >first derivatives are denoted with the ' symbol, e.g. the 1st derivative of f(x) = f'(x)) >f(x) = ln |ln x| + C >I want to differentiate this via the product rule, so I rewrite the function as >f(x) = (ln x^0) * (ln x) + C That rewriting is wrong. f(x) is not what the rewrite says, f(x) is ln(|ln(x)|). For started restrict to the case x > a, so f(x) = ln(ln(x)). >Using the product rule, >f'(x) = [ (ln x^0) * (ln x)' ] + [(ln x^0)' * (ln x) ] >= [ (ln 1) * (1/x)] + [ (0/1) * (ln x)] >= [ ln / x ] + [0] >= ln / x. >I don't see how ln / x = ln |ln x|. >Any advice? ************************ === Subject: Re: need help checking a basic integration formula via differentiation > I'm stuck on something simple, I'd be grateful for any help. > My text (Forgotten Calculus, Bleau) states the following integration formula: > integral [dx / (x ln x)] = ln |ln x| + C. > In the course of a problem where I've applied the formula, I realised I can't prove it. > Here's my attempt, for what it's worth. > (Re notation: > exponents are denoted with the ^ sign, e.g x squared = x^2; > first derivatives are denoted with the ' symbol, e.g. the 1st derivative of f(x) = f'(x)) > f(x) = ln |ln x| + C > I want to differentiate this via the product rule, so I rewrite the function as > f(x) = (ln x^0) * (ln x) + C Why are you using the product rule here? The chain rule seems definitely more appropriate: f[g[x]]' = f'[g[x]] * g'[x]. Moreover, since the function contains an absolute value and to be rigorous, you should considered two cases: x > 1 and 0 < x < 1. That is, if x > 1 we have Log[x} > 0, therefore f[x] = Log[Log[x]]; and if 0 < x < 1, we have Log[x] < 0, therefore f[x] = Log[-Log[x]]. Then, applying the chain to both cases should be straightforward: in the first case, g[x] = Log[x]; and g[x] = -Log[x] in the second case. HTH, Jean-Marc === Subject: Re: diff eq. >Can someone please take the time to check my work: >Find a general solution to the DE using the method of variation of >parameters: >y + 2y' + y = e^-t >homogeneous: y + 2y' +y = 0 >auxiliary equation: r^2 + 2r + 1 = 0 > ( r + 1)^2 = 0 > r = -1 (repeated root) >y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions. >homogeneous solution: y(t) = c1 y1(t) + c2 y2(t) > = c1 e^-t + c2 te^-t >particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t) > = v1(t) e^-t + v2(t) te^-t Just a suggestion. Instead of using variation of parameters, why don't you use the method of undetermined coefficients, or better yet, the method of annihilators if you have studied that? See if you can see why you would expect a particular solution of the form C*t^2*e^(-t). --Lynn === Subject: Re: diff eq. >> Can someone please take the time to check my work: >> Find a general solution to the DE using the method of variation of >> parameters: >> y + 2y' + y = e^-t >> homogeneous: y + 2y' +y = 0 >> auxiliary equation: r^2 + 2r + 1 = 0 >> ( r + 1)^2 = 0 >> r = -1 (repeated root) >> y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions. >> homogeneous solution: y(t) = c1 y1(t) + c2 y2(t) >> = c1 e^-t + c2 te^-t >> particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t) >> = v1(t) e^-t + v2(t) te^-t >> v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt >> W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t) >> y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t >> W = - e^(-2t)( 2t - 1) > I get a different, simpler value for the Wronskian. is this a correct Wronskian calculation then: W = e^-t (-te^-t + e^-t) - te^-t * -e^-t = -te^(-2t) + e^(-2t) - te^(-2t) = -2te^(-2t) + e^(-2t) = - e^(-2t) (2t -1) I am having a problem with another similar exercise as well, and if I am doing these wrong that would explain a lot. >> v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|) >> v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2 >> ln >> |2t -1| >> y(t) = particular solution + homogeneous solution: >> y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1| >> + >> c1 e^-t + c2 te^-t > My particular solution is (x^2/2)*e^(-x), which works in the original > equation, but yours does not. >> help >> means a lot. === Subject: Re: diff eq. >> Can someone please take the time to check my work: >> Find a general solution to the DE using the method of variation of >> parameters: >> y + 2y' + y = e^-t >> homogeneous: y + 2y' +y = 0 >> auxiliary equation: r^2 + 2r + 1 = 0 >> ( r + 1)^2 = 0 >> r = -1 (repeated root) >> y1(t) = e^-t and y2(t) = te^-t are two linearly independant solutions. >> homogeneous solution: y(t) = c1 y1(t) + c2 y2(t) >> = c1 e^-t + c2 te^-t >> particular solution: y(t) = v1(t) y1(t) + v2(t) y2(t) >> = v1(t) e^-t + v2(t) te^-t >> v1(t) = int[ (-te^-2t) / W(y1(t), y2(t))] dt >> W(y1(t), y2(t)) = y1(t) y2'(t) - y2(t)y1'(t) >> y1'(t) = -e^-t and y2'(t) = -t e^-t + e^-t >> W = - e^(-2t)( 2t - 1) > I get a different, simpler value for the Wronskian. > is this a correct Wronskian calculation then: > W = e^-t (-te^-t + e^-t) - te^-t * -e^-t > = -te^(-2t) + e^(-2t) - te^(-2t) = -2te^(-2t) + e^(-2t) ^ Sign error---------------------- ^, neg times neg is pos. > = - e^(-2t) (2t -1) > I am having a problem with another similar exercise as well, and if I am > doing these wrong that would explain a lot. >> v1(t) = int[ t / (2t -1) ] dt = 1/4( 2t -1 + ln |2t-1|) >> v2(t) = int[ e^(-2t) / (-e^(-2t)(2t-1)] dt = - int[ dt / (2t-1) ] = -1/2 >> ln >> |2t -1| >> y(t) = particular solution + homogeneous solution: y(t) = (e^-t / 4)( 2t -1 + ln |2t-1|) + (te^-t / 2) ln |2t - 1| + c1 e^-t + c2 te^-t > My particular solution is (x^2/2)*e^(-x), which works in the original > equation, but yours does not. >> help means a lot.