mm-23218 === Subject: Re: Uniform Convergence > I'm afraid that there are several (probably typing) mistakes in your working, > and I cannot see how the result helps us at all. Let me rephrase your method in terms of some simpler functions. Let g_c(x) = c^x, and let f_c(x) = (c^x - 1)/x. Let h > 0. There exists x in (0,h) with g_c'(x) = (g_c(h) - g_c(0))/(h-0) = f_c(h) There exists y in (0,x) with g_c''(y) = (g_c'(x) - g_c'(0))/(x-0) so that g_c'(x) - g_c'(0) = x * g_c''(y) Let M be the maximum of g_c'' on (0,x). Now x * g_c''(y) <= M * x < M * h. And: |f_c(h) - g_c'(0)| > = |g_c'(x) - g_c'(0)| > = |x * g_c''(y)| > < |Mh| So given any e > 0, we can set d = e / |M|, and then |h - 0| < d => |f_c(h) - g_c'(0)| < e The problem is that, for uniformity, we require our d to not be dependent on > h. However d depends on M, which depends on x, which depends on h (since the > choice of x using MVT depended on our choice of h). But how big do you think M can get if c is assumed to lie in a compact subinterval of (0,oo)? Answer: not so big. In fact there is uniform bound on M for any compact interval of c's in (0,oo). That proves uniform convergence on each such interval (which is the best you can hope for) and will give you what you want. I can give you details on the above if you need them, but I thought of an easier way to see that g_c'(0) is a continuous function of c. Fix a > 0. Suppose a <= b < c. Then for 0 < h < 1, 0 < (c^h - 1)/h - (b^h - 1)/h = (c^h - b^h)/h = h*d^(h-1)*(c-b)/h = d^(h-1)*(c-b) < (1/a)*(c-b). We have used the MVT in getting the second equality; note b < d < c. Taking limits as h -> 0+, we have 0 <= g_c'(0) - g_b'(0) <= (1/a)*(c-b). This shows g_c'(0) is a continuous (in fact uniformly continuous) function of c on [a, oo). Hence g_c'(0) is continuous on (0,oo). === Subject: Re: Trigonometry problem with two unknowns... <3i5ju7Fjil7sU1@individual.net > You don't have enough information to determine e and f. To see this, > call the leftmost point in your picture A and the topmost point B so > that a is the length of line AB. Put a point of a compass on A and the > pencil on B and swing a clockwise arc. Imagine moving the point B down > this arc and moving the attached lines with it. This only changes the > lengths of e and f leaving all your other measurements the same. So > they clearly don't determine unique values for e and f. I finally did it (using the compass way :-). Here is how: The lowest point (between b and c) is A: > xa=0 > ya=0 > The leftmost point is B: > xb=cos(alpha+delta)*b > yb=sin(alpha+delta)*b > The center point (between c, d and e) is D: > xd=cos(delta)*c > yd=sin(delta)*c > And the point of our compass needle is C: > xc=cos(phi+beta)*e+xd > yc=sin(phi+beta)*e+yd I have changed the drawing from my original posting a bit: phi is the angle > between d and e. And I'm calculating e and phi now instead of e and f.. but > that doesn't matter, because f can be calculated if you know e, d and phi. > And there is angle delta between 0Á and c. I can say that a^2 = (xc-xb)^2 + (yc-yb)^2 a^2 = (cos(phi+beta)*e + cos(delta)*c - cos(alpha+delta)*b)^2 + > (sin(phi+beta)*e + sin(delta)*c - sin(alpha+delta)*b)^2 after squaring, e can be factored out: 0 = e^2 + e*((2*c*sin(delta)-2*b*sin(alpha+delta))*sin(phi+beta)+ ...) - a^2 This is a quadratic equation which can be solved for e. Now all that has to be done is to solve e(alpha1, beta1) - e(alpha2, beta2) = 0 for phi. Where alpha1 and beta1 are the angles from the first measurement and alpha2 > and beta2 are the angles from the second measurement. I don't think solving that equation analytically is possible, but it works > fine iteratively with newtons method. > I've no doubt that you can find A solution from two measurements (where > one exists), but are you completely sure that the solution you've found > is the ONLY solution? I'm not necessarily saying you're wrong, it's > just when I looked at it I had some doubts about this. Just to clarify my comments some more. There are various related questions here: 1. Given a, b, c, d, gamma, arbitrarily-chosen alpha1, alpha2, and corresponding beta1, beta2, is it always possible to determine a unique e, f? Answer: I think not. 2. Given a, b, c, d, gamma, arbitrarily-chosen alpha1, alpha2, and corresponding beta1, beta2, is it always possible to determine a unique PLAUSIBLE e, f (i.e. that does not change the configuration of your diagram by having lines cross over)? Answer: possibly not. If I get the energy I'll see if I can find an example where you can't. 3. Given a, b, c, d and gamma, is it always possible to CHOOSE an alpha1, alpha2 so that a unique plausible e, f can be found? Answer: don't know; this question is much more difficult. Initially I thought yes, but my logic was faulty. === Subject: Re: Trigonometry problem with two unknowns... <3i5ju7Fjil7sU1@individual.net 1. Given a, b, c, d, gamma, arbitrarily-chosen alpha1, alpha2, and > corresponding beta1, beta2, is it always possible to determine a unique > e, f? Answer: I think not. 2. Given a, b, c, d, gamma, arbitrarily-chosen alpha1, alpha2, and > corresponding beta1, beta2, is it always possible to determine a unique > PLAUSIBLE e, f (i.e. that does not change the configuration of your > diagram by having lines cross over)? Answer: possibly not. If I get the > energy I'll see if I can find an example where you can't. Well, I had a look at this. This is what I found (or didn't find). It's easy to find cases when, given a, b, c, d, gamma, alpha1, alpha2, beta1, beta2, there is more than one solution for e, f. The most solutions I've found is four (but I have not shown that four is the maximum). However, I have not been able to find a case where there are two solutions (or more) both adhering to the configuration of the original diagram. Typically, although all the line lengths and angles are of course correct, the lines cross over each other, or the triangle bounded by d, e, f is on the wrong side of line d, or whatever. Incidentally, I don't think that your method (Markus), neat though it is, will find all these misconfigured solutions. This is because you assume that e and f being constant implies phi is constant, which is not true if the lines are allowed to go anywhere (we could have phi1 = 360 deg. - phi2). I mention this as a curiosity: I'm sure your solution is fine for your purposes. The nearest I've come to a valid multiple solution are cases when the configuration is OK except that the quadrilateral bounded by a, e, c, b can become concave at the point between a, b. I discarded these because they didn't seem to be in the spirit of the original problem. I've been searching randomly for these multiple solutions; I left the computer churning for several hours, but obviously the fact that I haven't found any doesn't mean there aren't any! It does mean that they're at least rare though. (Assuming my program is working OK!!!!) Interesting problem! === Subject: Re: Trigonometry problem with two unknowns... <3i5ju7Fjil7sU1@individual.net > You don't have enough information to determine e and f. To see this, > call the leftmost point in your picture A and the topmost point B so > that a is the length of line AB. Put a point of a compass on A and the > pencil on B and swing a clockwise arc. Imagine moving the point B down > this arc and moving the attached lines with it. This only changes the > lengths of e and f leaving all your other measurements the same. So > they clearly don't determine unique values for e and f. I finally did it (using the compass way :-). Here is how: The lowest point (between b and c) is A: > xa=0 > ya=0 > The leftmost point is B: > xb=cos(alpha+delta)*b > yb=sin(alpha+delta)*b > The center point (between c, d and e) is D: > xd=cos(delta)*c > yd=sin(delta)*c > And the point of our compass needle is C: > xc=cos(phi+beta)*e+xd > yc=sin(phi+beta)*e+yd I have changed the drawing from my original posting a bit: phi is the angle > between d and e. And I'm calculating e and phi now instead of e and f.. but > that doesn't matter, because f can be calculated if you know e, d and phi. > And there is angle delta between 0Á and c. I can say that a^2 = (xc-xb)^2 + (yc-yb)^2 a^2 = (cos(phi+beta)*e + cos(delta)*c - cos(alpha+delta)*b)^2 + > (sin(phi+beta)*e + sin(delta)*c - sin(alpha+delta)*b)^2 after squaring, e can be factored out: 0 = e^2 + e*((2*c*sin(delta)-2*b*sin(alpha+delta))*sin(phi+beta)+ ...) - a^2 This is a quadratic equation which can be solved for e. Now all that has to be done is to solve e(alpha1, beta1) - e(alpha2, beta2) = 0 for phi. Where alpha1 and beta1 are the angles from the first measurement and alpha2 > and beta2 are the angles from the second measurement. I don't think solving that equation analytically is possible, but it works > fine iteratively with newtons method. > I've no doubt that you can find A solution from two measurements (where > one exists), but are you completely sure that the solution you've found > is the ONLY solution? I'm not necessarily saying you're wrong, it's > just when I looked at it I had some doubts about this. Just to clarify my comments some more. There are various related > questions here: 1. Given a, b, c, d, gamma, arbitrarily-chosen alpha1, alpha2, and > corresponding beta1, beta2, is it always possible to determine a unique > e, f? Answer: I think not. > That was badly worded. Is it always possible to determine a unique e, f sounds like I might be debating whether there is ANY solution for e, f. It seems fairly clear that there are some values of the variables for which NO solution exists. What I actually meant was Given that there exists at least one solution (e.g. a solution you find with your numerical method), is that solution unique? And similarly for the other questions. Maybe I should read this stuff before I hit send. Groan. === Subject: MONOTONICITY vs. CONTINUITY Let f be monotonic on [a,b], and g be continuous on [a,b]. Moreover, g(a) Let f be monotonic on [a,b], and g be continuous on > [a,b]. Moreover, g(a) there exists some c in (a,b) such that f(c)=g(c). You can't. It's not true. Counterexample: let a= 0, b= 1, g(x)= 1- x, f(x)= 1/4 if 0<= x<= 1/2, 3/4 if 1/2<= x<= 1. === Subject: Re: MONOTONICITY vs. CONTINUITY > Let f be monotonic on [a,b], and g be continuous on > [a,b]. Moreover, g(a) there exists some c in (a,b) such that f(c)=g(c). You can't. It's not true. Counterexample: let a= 0, b= 1, g(x)= 1- x, f(x)= 1/4 if 0<= x<= 1/2, 3/4 if 1/2<= x<= 1. That doesn't satisfy g(a) < g(b). -- Stefan Holm Don't taunt the fear demon. === Subject: Re: MONOTONICITY vs. CONTINUITY > Let f be monotonic on [a,b], and g be continuous on >> [a,b]. Moreover, g(a)> there exists some c in (a,b) such that f(c)=g(c). >> You can't. It's not true. Counterexample: let a= 0, b= 1, g(x)= 1- >> x, f(x)= 1/4 if 0<= x<= 1/2, 3/4 if 1/2<= x<= 1. That doesn't satisfy g(a) < g(b). -- > Stefan Holm > Don't taunt the fear demon. The proof is almost identical to a simple completeness of the intermediate value. If of course, as you point out, f is increasing. Take greatest let k = sup { x in [a,b] : st g(x) <= f(x) } Clearly non empty g(a) < f(a) Real numbers are complete so k is in [a,b]. Now assume g( k) < f(k). 0 < e = f(k) - g(k). so exist d > 0 st. | g( k + d) - g( k) | < e = f(k) - g(k) <= f(k+d) - g(k) so g(k+d) - g(k) +g(k) <= f(k+d) hence contradiction k isn't an upper bound, g(k+d) <= f(k+d) or g(k) > f(k) 0 < E = g(k) - f(k) | g(k) - g(k - d)| < E = g(k) - f(k) <= g(k) - f(k - d) g(k-d) >= f(k-d) so k isn't the least upper bound. === Subject: Re: MONOTONICITY vs. CONTINUITY > Let f be monotonic on [a,b], and g be continuous on [a,b]. Moreover, > g(a) f(c)=g(c). > === Subject: Re: looking for help with a proof for number theory class show that if n and m are odd and not divisible by 3, then 24 | n^2 - m^2 > The numbers relatively prime to 24 meet the conditions for m and n (1,5,7,11,13,17,19,23). The square of each is = 1 mod 24. Therefore the difference of the squares is divisible by 24. === Subject: Re: looking for help with a proof for number theory class in alt.math.undergrad: > show that if n and m are odd and not divisible by 3, then 24 | n^2 - m^2 (1) Show that if n is not divisible by 3, then n^2 is of the form 3k + 1. Conclude that 3 | n^2 - m^2. (2) If n = 2k + 1 and m = 2j + 1, then n^2 - m^2 = 4k^2 + 4k - 4j^2 - 4j = 4(k^2 - j^2 + k - j), so 4 | n^2 - m^2. Factor the expression in parentheses to find another factor of 2. Brian