mm-285 ==== Subject: Y proportional to X^n - find n (again)This is a repost of a question that was asked a while ago...=I have some experimental data that follows Y X60.07 6001957.019 4735453.13 3747249.37 2792543.77 1994761.36 57048I expect to find that Y is proportional to X^nI known that n 'should' lie at a value of about 0.4 and by trial and errorwith a calculator i found that when n = 0.37, I get: Y X X^n60.07 60019 58.6157.019 47354 53.6853.13 37472 49.2349.37 27925 44.1643.77 19947 38.9961.36 57048 57.52 So for experimental data thats not a bad correlation considering that Y vsXis not quite linear.So the question is how do I solve Y = X^n mathematically to find n in eachcase??------- -I was advised that.... -------If Y=c*X^n, then log(Y) = n*log(X)+log(c), so in practisewhat you would do is plot on a graph log(X) against log(Y)-- ------ I did as suggested and established that the slope of the log,log graphcomes out at about 0.29 and the intersect is at about 0.9 so if Y = antilog C x X^slope of the trend. This gives a correlation ofY = 8.14 X^0.29 but when this is plotted it in no way resembles my obtained data. I expected to find that n = about 0.37 because by trial and error Y =X^0.37 gives a good resemblence to the experimental data. can anyone see where I'm going wrong. SS.(I used Matlab to do my calculations....heres the code if it helps.) ³ ³ ==== %thermolabclear all, close all%these are the results obtained in the labRe = [60018.87 47353.94 37471.92 27925.42 19946.73]Nu = [60.07 57.74 53.13 49.37 43.77]subplot(2,1,1),plot(Re,Nu),hold on,title('experimentalvalues'),xlabel('Re'),,hold off, ylabel('Nu')%Linearise the trend - least squares fit.s=polyfit(Re,Nu,1);f=polyval(s,Re);subplot(2,1,2),plot(Re, f,'ro-'),hold off,title('linearised experimentalvalues'),,xlabel('Re'),,hold off, ylabel('Nu')pauseclose%now plot log re vs log nuR = log(Re)N = log(Nu)plot(R,N),hold on%linearise the trend again.p=polyfit(R,N,1);hhh=polyval(p,R);plot(R,hhh,'ro-'), legend('as recieved','linearised'),title('Graph showingLog(Re) vs Log Nu for experimentally obtained values')xlabel('Log(Re)'),,hold off, ylabel('Log(Nu)')%find slope and intercept of the linearised trenddisp('the gradient M and the Intersect C is ')%p= [M,C] %(slope,intercept)M = p(1) %comes out at about 0.29C = p(2) %comes out at about 0.9%put text on graph.gtext([ 'The slope of the linearised trend is ' num2str(M)])gtext([ 'The intersect of the linearised trend is ' num2str(C)])pause%my values using my calculated exponent.If my understanding is correct then%Nu = antilog C x Re^slope of the trend.plot(Re,8.14.*Re.^0.29,'r'),hold on%plot expermental valuesplot(Re,Nu,'m'),hold onlegend('my calculated values','experimental values')plot(Re,Re.^0.37,'b') ³ ³ ==== ³ ³ ==== Subject: Re: Y proportional to X^n - find n (again)> This is a repost of a question that was asked a while ago...> I have some experimental data that follows> Y X> 60.07 60019> 57.019 47354> 53.13 37472> 49.37 27925> 43.77 19947> 61.36 57048> I expect to find that Y is proportional to X^n> I known that n 'should' lie at a value of about 0.4 and by trial > and error with a calculator i found that when n = 0.37, I get:I see a possible source of confusion here. The curve you seem tobe fitting by trial and error is Y = X^n. The curve you are fitting using Matlab (below) is Y = exp(c)*X^n. (When you did your Matlab fit, you left the last data point out.Maybe that was intentional, but I just thought I'd mention it.)Something no one else in this thread has mentioned yet isweighting your data. I'm assuming that the data for your X andY values have similar precisions. Even so, when you take theirlogarithms, the transformed data will have different precisions,and the most correct answer will take this into account.If your reading error for the Y values is delY, then thereading error for V =df ln(Y) is delV = del(ln(Y)) = delY/Y.This may not make much difference to you, since your valuesare all pretty close in size. On the other hand, it's not much work to include the weights in your calculation.If Y = X^n is the curve you want to fit to, there are least-square-fit formulas for that, too.Define U = ln(X) and V = ln(Y). then V - n*U = 0We want to minimize sum{ (V - n*U)^2 } with respect to n, or with weighting sum{ (V - n*U)^2/(delY/Y)^2 }Differentiating with respect to n, setting to zero, and solvingfor n: n = sum{ Y^2*(V*U) }/sum{ Y^2*(U^2) }Jim Burns> Y X X^n> 60.07 60019 58.61> 57.019 47354 53.68> 53.13 37472 49.23> 49.37 27925 44.16> 43.77 19947 38.99> 61.36 57048 57.52> So for experimental data thats not a bad correlation considering that Y vs> X> is not quite linear.> So the question is how do I solve Y = X^n mathematically to find n in each> case??> I was advised that....> ---> If Y=c*X^n, then log(Y) = n*log(X)+log(c), so in practise> what you would do is plot on a graph log(X) against log(Y)> I did as suggested and established that the slope of the log,log graph> comes out at about 0.29 and the intersect is at about 0.9> so if Y = antilog C x X^slope of the trend. This gives a correlation of> Y = 8.14 X^0.29> but when this is plotted it in no way resembles my obtained data.> I expected to find that n = about 0.37 because by trial and error Y => X^0.37 gives a good resemblence to the experimental data.> can anyone see where I'm going wrong.> SS.> (I used Matlab to do my calculations....heres the code if it helps.)> ³ ³ ==== > %thermolab> clear all, close all> %these are the results obtained in the lab> Re = [60018.87 47353.94 37471.92 27925.42 19946.73]> Nu = [60.07 57.74 53.13 49.37 43.77]> subplot(2,1,1),plot(Re,Nu),hold on,title('experimental> values'),xlabel('Re'),,hold off, ylabel('Nu')> %Linearise the trend - least squares fit.> s=polyfit(Re,Nu,1);> f=polyval(s,Re);> subplot(2,1,2),plot(Re,f,'ro-'),hold off,title('linearised experimental> values'),,xlabel('Re'),,hold off, ylabel('Nu')> pause> close> %now plot log re vs log nu> R = log(Re)> N = log(Nu)> plot(R,N),hold on> %linearise the trend again.> p=polyfit(R,N,1);> hhh=polyval(p,R);> plot(R,hhh,'ro-'),legend('as recieved','linearised'),title('Graph showing> Log(Re) vs Log Nu for experimentally obtained values')> xlabel('Log(Re)'),,hold off, ylabel('Log(Nu)')> %find slope and intercept of the linearised trend> disp('the gradient M and the Intersect C is ')> %p= [M,C] %(slope,intercept)> M = p(1) %comes out at about 0.29> C = p(2) %comes out at about 0.9> %put text on graph.> gtext([ 'The slope of the linearised trend is ' num2str(M)])> gtext([ 'The intersect of the linearised trend is ' num2str(C)])> pause> %my values using my calculated exponent.If my understanding is correct then> %Nu = antilog C x Re^slope of the trend.> plot(Re,8.14.*Re.^0.29,'r'),hold on> %plot expermental values> plot(Re,Nu,'m'),hold on> legend('my calculated values','experimental values')> plot(Re,Re.^0.37,'b')> ³ ³ ==== Subject: Re: Y proportional to X^n - find n (again)>This is a repost of a question that was asked a while ago...> ³ ³ ==== have some experimental data that follows> Y X>60.07 60019>57.019 47354>53.13 37472>49.37 27925>43.77 19947>61.36 57048>I expect to find that Y is proportional to X^n>I known that n 'should' lie at a value of about 0.4 and by trial and error>with a calculator i found that when n = 0.37, I get:> Y X X^n>60.07 60019 58.61>57.019 47354 53.68>53.13 37472 49.23>49.37 27925 44.16>43.77 19947 38.99>61.36 57048 57.52> So for experimental data thats not a bad correlation considering that Y vs>X>is not quite linear.>So the question is how do I solve Y = X^n mathematically to find n in each>case??>----- >I was advised that....> --->If Y=c*X^n, then log(Y) = n*log(X)+log(c), so in practise>what you would do is plot on a graph log(X) against log(Y)>- ---> -- --> I did as suggested and established that the slope of the log,log graph>comes out at about 0.29 and the intersect is at about 0.9> so if Y = antilog C x X^slope of the trend. This gives a correlation of>Y = 8.14 X^0.29> but when this is plotted it in no way resembles my obtained data.> I expected to find that n = about 0.37 because by trial and error Y =>X^0.37 gives a good resemblence to the experimental data.> can anyone see where I'm going wrong.Doing a least squares regression on the log-log data (with a handcalculator), I get a slope of .296912623309 and an intercept of847618738581; close to what you got. However, the logs are base-e,not base-10, so the .847618738581 turns to a factor of 2.33408216925.Trying 2.334 X^.2969, I get Y X 2.334 X^.2969 60.07 60019 61.20 57.019 47354 57.04 53.13 37472 53.22 49.37 27925 48.77 43.77 19947 44.13 61.36 57048 60.29Looks like a decent fit to me.Rob Johnson take out the trash before replying ³ ³ ==== ³ ³ ==== Subject: Re: Y proportional to X^n - find n (again)>Doing a least squares regression on the log-log data (with a hand>calculator), I get a slope of .296912623309 and an intercept of>847618738581; close to what you got. However, the logs are base-e,>not base-10, so the .847618738581 turns to a factor of 2.33408216925.>Trying 2.334 X^.2969, I get> Y X 2.334 X^.2969> 60.07 60019 61.20> 57.019 47354 57.04> 53.13 37472 53.22> 49.37 27925 48.77> 43.77 19947 44.13> 61.36 57048 60.29>Looks like a decent fit to me.The intercept should be .847618738581; the decimal point got missed.Rob Johnson take out the trash before replying ==== Subject: Re: Y proportional to X^n - find n (again)> This is a repost of a question that was asked a while ago...> ³ ³ ==== I have some experimental data that follows> Y X> 60.07 60019> 57.019 47354> 53.13 37472> 49.37 27925> 43.77 19947> 61.36 57048Y = X^n => log Y = n log X => n = log Y / log X> I expect to find that Y is proportional to X^n> I known that n 'should' lie at a value of about 0.4 and by trial and error> with a calculator i found that when n = 0.37, I get:> Y X X^n> 60.07 60019 58.61> 57.019 47354 53.68> 53.13 37472 49.23> 49.37 27925 44.16> 43.77 19947 38.99> 61.36 57048 57.52> So for experimental data thats not a bad correlation considering that Y vs> X> is not quite linear.> So the question is how do I solve Y = X^n mathematically to find n in each> case??> I was advised that....> ---> If Y=c*X^n, then log(Y) = n*log(X)+log(c), so in practise> what you would do is plot on a graph log(X) against log(Y)It's more usual to plot log(Y) against log(X).> I did as suggested and established that the slope of the log,log graph> comes out at about 0.29 and the intersect is at about 0.9> so if Y = antilog C x X^slope of the trend. This gives a correlation of> Y = 8.14 X^0.29> but when this is plotted it in no way resembles my obtained data.> I expected to find that n = about 0.37 because by trial and error Y => X^0.37 gives a good resemblence to the experimental data.I get the same; plotting the data confirm that they're nowhere near thebest fit line.Time for the ad hoc solution: for each data point, compute n = logY/logXand take the average of these; then plot Y = (Y[1]/X[1]^n)*X^n and seewhat you get.> (I used Matlab to do my calculations....heres the code if it helps.)Really? I used awk and gnuplot.-- P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. ³ ³ ==== I have some experimental data that follows> Y X> 60.07 60019> 57.019 47354> 53.13 37472> 49.37 27925> 43.77 19947> 61.36 57048> I expect to find that Y is proportional to X^n> I known that n 'should' lie at a value of about 0.4 and by trial and error> with a calculator i found that when n = 0.37, I get:> Y X X^n> 60.07 60019 58.61> 57.019 47354 53.68> 53.13 37472 49.23> 49.37 27925 44.16> 43.77 19947 38.99> 61.36 57048 57.52> So for experimental data thats not a bad correlation considering that Y vs> X> is not quite linear.> So the question is how do I solve Y = X^n mathematically to find n in each> case??> I was advised that....> ---> If Y=c*X^n, then log(Y) = n*log(X)+log(c), so in practise> what you would do is plot on a graph log(X) against log(Y)> I did as suggested and established that the slope of the log,log graph> comes out at about 0.29 and the intersect is at about 0.9> so if Y = antilog C x X^slope of the trend. This gives a correlation of> Y = 8.14 X^0.29> but when this is plotted it in no way resembles my obtained data.> I expected to find that n = about 0.37 because by trial and error Y => X^0.37 gives a good resemblence to the experimental data.> can anyone see where I'm going wrong.> SS.> (I used Matlab to do my calculations....heres the code if it helps.)> ³ ³ ==== ³ ³ ==== ³ ³ ==== > %thermolab> clear all, close all> %these are the results obtained in the lab> Re = [60018.87 47353.94 37471.92 27925.42 19946.73]> Nu = [60.07 57.74 53.13 49.37 43.77]> subplot(2,1,1),plot(Re,Nu),hold on,title('experimental> values'),xlabel('Re'),,hold off, ylabel('Nu')> %Linearise the trend - least squares fit.> s=polyfit(Re,Nu,1);> f=polyval(s,Re);> subplot(2,1,2),plot(Re,f,'ro-'),hold off,title('linearised experimental> values'),,xlabel('Re'),,hold off, ylabel('Nu')> pause> close> %now plot log re vs log nu> R = log(Re)> N = log(Nu)> plot(R,N),hold on> %linearise the trend again.> p=polyfit(R,N,1);> hhh=polyval(p,R);> plot(R,hhh,'ro-'),legend('as recieved','linearised'),title('Graph showing> Log(Re) vs Log Nu for experimentally obtained values')> xlabel('Log(Re)'),,hold off, ylabel('Log(Nu)')> %find slope and intercept of the linearised trend> disp('the gradient M and the Intersect C is ')> %p= [M,C] %(slope,intercept)> M = p(1) %comes out at about 0.29> C = p(2) %comes out at about 0.9> %put text on graph.> gtext([ 'The slope of the linearised trend is ' num2str(M)])> gtext([ 'The intersect of the linearised trend is ' num2str(C)])> pause> %my values using my calculated exponent.If my understanding is correct then> %Nu = antilog C x Re^slope of the trend.> plot(Re,8.14.*Re.^0.29,'r'),hold on> %plot expermental values> plot(Re,Nu,'m'),hold on> legend('my calculated values','experimental values')> plot(Re,Re.^0.37,'b')> ³ ³ ==== Did you look herehttp://mathworld.wolfram.com/ LeastSquaresFittingExponential.htmlor maybehttp://www.wam.umd.edu/~toh/spectrum/ CurveFitting.htmlBill ==== Subject: very basic question plz helpHi I am currently an A-level maths student in Englnd I am 17 years oldand just need a little basic help with my understanding ofdifferentiating an equation wih a square using the doubledifferentiating from first principles formula thanx.Also can someone explain to me the basic conditional probabilityformula for a intersection b and also a union b. ==== Subject: Re: very basic question plz helpZarar Mohammed> Hi I am currently an A-level maths student in Englnd I am 17 years old> and just need a little basic help with my understanding of> differentiating an equation wih a square using the double> differentiating from first principles formula thanx.If f(x) = x^2 then f(x+h) - f(x) = 2xh - h^2. Divide by h and take thelimit at h=0, showing f ' (x) = 2x.Double differentiating? You can show that the partial derivatives ofxy are y and x, and get the derivative of x^2 that way, but I'm not sureif that was the question.LH ==== Subject: Re: very basic question plz helpLarry Hammick> If f(x) = x^2 then f(x+h) - f(x) = 2xh - h^2. Divide by h and take the> limit at h=0, showing f ' (x) = 2x.Should say f(x+h) - f(x) = 2xh + h^2, of course.LH ³ ³ ==== ³ ³ ==== Subject: basic queryHi I am currently an A-level maths student in Englnd I am 17 years oldand just need a little basic help with my understanding ofdifferentiating an equation wih a square using the doubledifferentiating from first principles formula thanx.Also can someone explain to me the basic conditional probabilityformula for a intersection b and also a union b. ³ ³ ==== ³ ³ ==== Subject: Re: expectation of the product of two dependent random variables> exp(p times X times Y)e^(pXY) ==== Subject: Re: Candidate looking for EmploymentHi Uncle Al, You said, Uncle Al's computer's clock is set 58 seconds fast You must mean 58 minutes slow.Your clock says 1 pm when other's say 2 pm. ==== Subject: Re: Candidate looking for EmploymentIn sci.math, Jeff Relf<1em5n85mxv2nz.dlg@x.Jeff.Relf>:> Hi Uncle Al, You said, Uncle Al's computer's clock is set 58 seconds fast > You must mean 58 minutes slow.> Your clock says 1 pm when other's say 2 pm.Daylight Savings time is either 2 or 3 weeks from now(depending on locale). :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. ==== Subject: Antidiagonal, Infinity by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2DDCCt25554;The antidiagonal argument's result in binary is necessary to accept its con-clusion. That is because each of those infinite strings has a representation in binary. Even though you can contrive an antidiagonal in for example deci-mal, the listed element is counterexample in binary.It's easy to construct functions that _don't_ map a set to another. The point of using the binary case of the antidiagonal is that the elements, being as-sumed in the presentation of the argument to be a complete list, may be re-ordered thus that an antidiagonal is forthwith constructed and that element has an alternate representation that is equal in its value via dual represen-tation of some rationals that satisfies being different at each place in its expansion yet still is the same. The sequences are different, the number is the same.Any possible mapping between the two sets of the antidiagonal argument is to be allowed. To be quite blunt, professors of the antidiagonal argument in terms of the reals and the naturals who prop its staven sides with more than two elements have lied to themselves, and perhaps others, or been incompetent.That's not to say that they're mistaken in general about sets infinite and nonstandard measure theory, it's just that the conclusion that the antidiagonal argument directly displays the impossibility of any bijective mapping between the reals and naturals without resort to displaying its effect with regards to the set and its powerset and compositionally to the set and some other set that is not its powerset and has dual representation of its elements as binary or as well decimal or other real base sequences has simply been mistaken. They were probably informed of the result and accepted it, that is not necessarily a fault except in their logical reasoning ability and comprehension of this very small subset of mathematical enquiry.The argument about the set and exactly its own powerset is quite the different argument. With f(x)=x, S(0)=N, with f(x)=x+1, S(N)=0. Zero is the empty set.The empty set is bewilderingly overloaded in its meaning, and lack of it. In-vert the matrix.What I like about one criticism of the nested intervals argument is that it is specifically monotonically mapping the integers to each element of an interval of the reals. The Equivalency Function is one such example of a function, I've been getting to thinking about a wide variety of variations upon that theme.The evaluation of any integral of any function symmetric about the origin is zero. The extension of the nested intervals argument may be that the mappings of the naturals to the reals demand to be of that form.The theory progresses.Warm regards, Ross F. ==== Subject: Re: Antidiagonal, Infinity> The antidiagonal argument's result in binary is necessary to accept its con-> clusion. That is because each of those infinite strings has a representation > in binary. Even though you can contrive an antidiagonal in for example deci-> mal, the listed element is counterexample in binary.This appears to be arguinig that if there is a flawed proof of a theroem, then no valid proof can establish that theorem. Cantor's diagonal proof of the uncountability of the reals is valid without modification for any base greater than three, and can easily be modified in bases two and three to be valid.That Ross doesn't want it to be valid is irrelevant.> It's easy to construct functions that _don't_ map a set to another. The > point > of using the binary case of the antidiagonal is that the elements, being as-> sumed in the presentation of the argument to be a complete list, may be re-> ordered thus that an antidiagonal is forthwith constructed and that element > has an alternate representation that is equal in its value via dual represen-> tation of some rationals that satisfies being different at each place in its > expansion yet still is the same. The sequences are different, the number is > the same.The dual representation problem is easily avoided by choosing the diagonal to differ at a pair of digits from the listed number using only pairs not subject to dual representation ambiguity.This has several times been pointed out to Ross, but he continues to opt for ignorance rather that enlightenment. [a great deal of garbage snippped]³ ³ ==== Subject: Re: Antidiagonal, InfinityIn sci.math, Ross A. Finlayson> The antidiagonal argument's result in binary is necessary to accept its> conclusion. That is because each of those infinite strings has> a representation in binary. Even though you can contrive an> antidiagonal in for example decimal, the listed element is> counterexample in binary.One does not have to use the strict diagonal. If one is forcedto use binary for some whimsical reason in Cantor's proof,one can do the following.Assume as per usual the ordering of the real number line [0,1),and the bijection:0 <-> 0.000000000000.....1 <-> 0.011010101101.....2 <-> 0.100101001011.....3 <-> 0.001100110101.....4 <-> 0.011011010111.....5 <-> 0.110110110111.....etc. The numbers on the right are whimsical.Simply take two's:0 <-> 0.[00] 00 00 00 00 00.....1 <-> 0. 01 [10] 10 10 11 01.....2 <-> 0. 10 01 [01] 00 10 11.....3 <-> 0. 00 11 00 [11] 01 01.....4 <-> 0. 01 10 11 01 [01] 11.....5 <-> 0. 11 01 10 11 01 [11]....Now construct the number x in the usual way, exceptthat one appends pairs of digits to x:00 -> 0101 -> 1010 -> 0011 -> 01The exact mapping doesn't matter as long as one doesn't have 11,to avoid the usual 0.1111... = 1.0000... problem.And now one has a constructed Cantorian diagonal number:x ??? 0.010010011001...The astute will note that all I've done really is convert theentire problem into base 4 -- but so what? It's still valid. :-)[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless.³ ³ ==== Subject: Re: Antidiagonal, Infinity> In sci.math, Ross A. Finlayson> > The antidiagonal argument's result in binary is necessary to accept its> conclusion. That is because each of those infinite strings has> a representation in binary. Even though you can contrive an> antidiagonal in for example decimal, the listed element is> counterexample in binary.> One does not have to use the strict diagonal. If one is forced> to use binary for some whimsical reason in Cantor's proof,> one can do the following.> Assume as per usual the ordering of the real number line [0,1),> and the bijection:> 0 <-> 0.000000000000.....> 1 <-> 0.011010101101.....> 2 <-> 0.100101001011.....> 3 <-> 0.001100110101.....> 4 <-> 0.011011010111.....> 5 <-> 0.110110110111.....> etc. The numbers on the right are whimsical.> Simply take two's:> 0 <-> 0.[00] 00 00 00 00 00.....> 1 <-> 0. 01 [10] 10 10 11 01.....> 2 <-> 0. 10 01 [01] 00 10 11.....> 3 <-> 0. 00 11 00 [11] 01 01.....> 4 <-> 0. 01 10 11 01 [01] 11.....> 5 <-> 0. 11 01 10 11 01 [11]....> Now construct the number x in the usual way, except> that one appends pairs of digits to x:> 00 -> 01> 01 -> 10> 10 -> 00> 11 -> 01> The exact mapping doesn't matter as long as one doesn't have 11,> to avoid the usual 0.1111... = 1.0000... problem.Actually, one must avoid both 11 AND 00, as both can lead to numbers with dual binary representations, whereas 01 and 10 cannot.> And now one has a constructed Cantorian diagonal number:> x ??? 0.010010011001...> The astute will note that all I've done really is convert the> entire problem into base 4 -- but so what? It's still valid. :-)> [rest snipped]³ ³ ==== Subject: Re: Antidiagonal, Infinity > One does not have to use the strict diagonal. If one is forced to> use binary for some whimsical reason in Cantor's proof, one can do> the following.Alternatively, whenever you come to an entry in the list that has tworepresentations in binary (or whatever base you're using), justdiagonalize it twice; i.e., insert the alternative representation intothe list.³ ³ ==== Subject: Re: Antidiagonal, Infinity> The antidiagonal argument's result in binary is necessary to accept its con-> clusion. That is because each of those infinite strings has a representation> in binary. Even though you can contrive an antidiagonal in for example deci-> mal, the listed element is counterexample in binary.According to you the sets {1,2,3,4,...} and {1,10,11,100,...}do not have the same cardinality.But clearly, each of the elements of the first set has a representationin binary that happens to match the corresponding element of thesecond set.So you are still just as stupid as you were 3 years ago.Dirk Vdm³ ³ ==== Subject: Re: Conencted Small GraphsX-ID: Ekl2X8ZHrejAOXQoRpEFymP1FbMYs7ZlWSkDkwE2zArFXogOZ0zecFFabio Rojas schrieb:> Not sure if this was posted the first time...> Is there a reference that tells me the % of connected graphs with 30 nodes or> less? How about equivalence classes of graphs? Formula would be nice, if> it exists.> The table would look like:look athttp://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA.cgi?Anum=A001349andhttp://www.research.att.com/ cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum= A000088hthKlaus> Nodes Percent Connected> 3 50%> .....> 30 .000003%> Feel free to email me or post a citation.> Fabio³ ³ ==== Subject: Re: When Cayley transform is orthogonal ?>Let A be a n x n real matrix such that det(I-A)=/=0 .>The Cayley transformation of A is C(A)=(I-A)^{-1}(I+A).>How we can describe all matrices A for which C(A) is orthogonal ?All real antisymmetric matrices. Note that I-C(A) = -2(I-A)^{-1}A,and I+C(A) = 2(I-A)^{-1}, so that A = -(I+C(A))^{-1}(I-C(A)), providedC(A) does not have -1 as an eignevalue. It now follows from the orthogonality of C(A) that A is antisymmetric. More generally, ifC(A) has -1 as an eigenvalue, then C(A) is of the form B diag(-I,D) B^{-1}, for an orthogonal matrix B, and orthogonal matrices I and D whose dimensions sum to n, and where -1 is not an eigenvalue of D. LetB^{-1} A B = (A_1 A_2) (A_3 A_4),so that (-I O) = (I-A_1 -A_2 )^{-1} (I+A_1 A_2 ) ( O D) (-A_3 1-A_4) ( A_3 I+A_4),so that (-I+A_1 -A_2 D ) = (I+A_1 A_2 ) ( A_3 D-A_4 D) ( A_3 I+A_4).This can't hold unless C(A) does not have -1 as an eigenvalue. In otherwords, C(A) is orthogonal iff A is antisymmetric, and in that case, C(A) does not have -1 as an eigenvalue.David McAnally But I'm always true to you, darlin', in my fashion, Yes, I'm always true to you, darlin', in my way. -- Lois Lane----³ ³ ==== Subject: inverse trigonometric functions of complex numbersAnyone know how to take inverse trigonometric functions of complexnumbers/variables using the Google calculator?Thx.Andy Eppink³ ³ ==== Subject: fibonacci equationCan any one resolve this equationfib(n) + 5 = m^2fib(n) is fibonacci number³ ³ ==== Subject: Re: fibonacci equation>Can any one resolve this equation>fib(n) + 5 = m^2>fib(n) is fibonacci numberI assume m is an integer.In order for fib(n)+5 to be a square mod 8, we need for n to becongruent to 4 or 10 mod 12. If n is divisible by 4, however,fib(n) is divisible by 3, and fib(n)+5 is not a square mod 3,which rules out n being congruent to 4 mod 12.If n=-2, the natural extension of fib(n) makes fib(-2)=-1, and-1+5 is a square, but I doubt you want to consider -1 to bea Fibonacci number.For n>0, I've tried n=10, 22, 34,..., 502 using Pari-GP andnot found any solutions, so I rather suspect there aren'tany. I can't see offhand how to prove it, though.Keith Ramsay³ ³ ==== Subject: Re: fibonacci equation>Can any one resolve this equation>fib(n) + 5 = m^2>fib(n) is fibonacci number> I assume m is an integer.> In order for fib(n)+5 to be a square mod 8, we need for n to beWhere did mod 8 come from?> congruent to 4 or 10 mod 12. If n is divisible by 4, however,> fib(n) is divisible by 3, and fib(n)+5 is not a square mod 3,> which rules out n being congruent to 4 mod 12.> If n=-2, the natural extension of fib(n) makes fib(-2)=-1, and> -1+5 is a square, but I doubt you want to consider -1 to be> a Fibonacci number.> For n>0, I've tried n=10, 22, 34,..., 502 using Pari-GP and> not found any solutions, so I rather suspect there aren't> any. I can't see offhand how to prove it, though.³ ³ ==== Subject: Re: Design Question> Of course,> the best way to separate signal from noise,> is to cross-correlate the effect signal> with the cause signal.> A Google search on MTI RADAR might> show you how auto-correlation is done with analog circuits.> Correlation involves summing the product of two strings of data.> If you multiply two random strings of data, and sum the products,> you end up with zero in the case of pure noise.> The larger the sum of the products,> the greater the correlation between the two strings.> Note that the time delay (Phase in the case of periodic signals)> has to be adjusted to make the data strings overlap,> as there is a time interval or phase difference between a cause and an> effect.Tom,He said to extract and amplify and not detect. And he didn't say thathe had a reference. So, how does correlation help?Fred³ ³ ==== Subject: Re: Design Question> Of course,> the best way to separate signal from noise,> is to cross-correlate the effect signal> with the cause signal.> A Google search on MTI RADAR might> show you how auto-correlation is done with analog circuits.> Correlation involves summing the product of two strings of data.> If you multiply two random strings of data, and sum the products,> you end up with zero in the case of pure noise.> The larger the sum of the products,> the greater the correlation between the two strings.> Note that the time delay (Phase in the case of periodic signals)> has to be adjusted to make the data strings overlap,> as there is a time interval or phase difference between a cause and an> effect.> Tom,> He said to extract and amplify and not detect. And he didn't say that> he had a reference. So, how does correlation help?The poster stated:I need to extract and amplify only the components of two signals that areexactly in phase. Out of phase signals of any amplitude are noise. I amthinking this is sort of the opposite of common mode rejection. Should I bepursuing an electronic (hardware) solution or a math (DSP) solution? Isthere some way to utilize the phase part of an FFT?I suggest that if he has two signals,and wants to amplify the in phase component,that one of the signals qualifies as a reference.For example, the old MTI RADAR systems,used the preceding signal as a reference,to subtract out the clutter in the succeeding signal.Correlation would minimize the noise,and artifact unrelated to the two signals,and provide a cleaner signal to use foranalysis, driving output devices, etc.Even if he needs power amplification,he would be able to use a more efficientdevice (Speaker, ear phone, actuator, servo, etc.)if he minimizes the apparent power drains,cause by noise, spikes, and out of phase conditions.--Tom Potter http://tompotter.us³ ³ ==== Subject: Re: Design Question>>Of course,>>the best way to separate signal from noise,>>is to cross-correlate the effect signal>>with the cause signal.>>A Google search on MTI RADAR might>>show you how auto-correlation is done with analog circuits.>>Correlation involves summing the product of two strings of data.>>If you multiply two random strings of data, and sum the products,>>you end up with zero in the case of pure noise.>>The larger the sum of the products,>>the greater the correlation between the two strings.>>Note that the time delay (Phase in the case of periodic signals)>>has to be adjusted to make the data strings overlap,>>as there is a time interval or phase difference between a cause and an>>effect.> Tom,> He said to extract and amplify and not detect. And he didn't say that> he had a reference. So, how does correlation help?> FredFred,Without a reference, how does he determine phase? On reading theoriginal post I was tempted to ask, Exactly in phase with what?. Notwanting to play the curmudgeon, I forbore. I ask it now.Jerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash]³ ³ ==== Subject: Re: Design Question> Tom,> He said to extract and amplify and not detect. And he didn't saythat> he had a reference. So, how does correlation help?> Fred> Fred,> Without a reference, how does he determine phase? On reading the> original post I was tempted to ask, Exactly in phase with what?. Not> wanting to play the curmudgeon, I forbore. I ask it now.Jerry,Indeed! See my other post in response to the OP.My purpose here was to question correlation as being pertinent.But, maybe nothing is pertinent, eh?Fred³ ³ ==== Subject: OT: Design Question, namely exact ...>>Fred,>>Without a reference, how does he determine phase? On reading the>>original post I was tempted to ask, Exactly in phase with what?. Not>>wanting to play the curmudgeon, I forbore. I ask it now.> Jerry,> Indeed! See my other post in response to the OP.> My purpose here was to question correlation as being pertinent.> But, maybe nothing is pertinent, eh?> FredI was approached at a neighborhood party by someone who wanted a holedrilled dead center in an aluminum cup. (3 diameter cylinder, 3deep with flat bottom; .25 wall thickness all around.) He had had itmade in a university shop a few weeks before and needed to add the holein the middle of the flat bottom. (Or does one subtract a hole? Aninteresting question!) He asked me because I'm known to have a small (6swing over the ways) lathe and to be accommodating. I adapted a reticulemicroscope to ride on the carriage for centering critical work.Me: How accurate does it have to be?He: Absolutely perfect. There's no room for error.Me: There's no way to do that.He: Why not?Me: First of all, if it was turned from solid stock a few weeks ago, it doesn't have a center any more. Aluminum, especially 4041, warps after being machined. Second, my equipment isn't perfect. I can tell when a round is a tenth off center, but I can't correct that.He: A tenth of what?Me: A mil.He: What's a mil?Me: A thousandth of an inch.He: That's pretty small.Me: That depends on what you work with. A piece of newsprint is about 3 mils thick. A telescope mirror is finished to about 1/200th.He: Oh hell, I can't see that small!In the end, I slapped it on the three-jaw chuck. He was happy.Jerry-- Engineering is the art of making what you want from things you can get.[OSlash][OSlash][OSlash][OSlash]³ ³ ==== Subject: Re: Design Question> Hi Ron : I can't help thinking you are going to have problems determining> what exactly in phase means. Is there any way you can expand your problem> description with some sort of mathematical definition of how similar these> components need to be before you stop ignoring them?I think most respondents have assumed that the 2 components are of the samefrequency or at least very close to the same frequency. I don't think most of thesuggestions made so far will work if components means something else.-jim-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----http://www.newsfeeds.com - The #1 Newsgroup Service in the World!-----== Over 100,000 Newsgroups - 19 Different Servers! =-----³ ³ ==== Subject: Re: skewed dualities: functions and sets> the adjoint functor theorem of category theory explains this, in a> way. it says that left adjoint functors preserve colimits (such as> unions) and, dually, right adjoint functors preserve limits (such as> intersections). a functor is a left adjoint functor if it has a right> adjoint functor, and vice versa. inverse image is both a left> adjoint and a right adjoint functor. the left adjoint that it has (by> virtue of being a right adjoint) is called image, while the right> adjoint that it has (by virtue of being a left adjoint) is probably> not well enough known even to have a familiar name; i'll call itanti-image for now. the anti-image of a subset s of the domain of a> function is the subset of the co-domain containing just those points> whose inverse images are subsets of s. anti-image preserves> intersections but not unions.zzzzzzzzzzzzz ...³ ³ ==== Subject: Re: skewed dualities: functions and sets>> the adjoint functor theorem of category theory explains this, in a>> way. it says that left adjoint functors preserve colimits (such as>> unions) and, dually, right adjoint functors preserve limits (such as>> intersections). a functor is a left adjoint functor if it has a right>> adjoint functor, and vice versa. inverse image is both a left>> adjoint and a right adjoint functor. the left adjoint that it has (by>> virtue of being a right adjoint) is called image, while the right>> adjoint that it has (by virtue of being a left adjoint) is probably>> not well enough known even to have a familiar name; i'll call it>anti-image for now. the anti-image of a subset s of the domain of a>> function is the subset of the co-domain containing just those points>> whose inverse images are subsets of s. anti-image preserves>> intersections but not unions.>zzzzzzzzzzzzz ...Categories for the Sleeping Mathematician?Lee Rudolph³ ³ ==== Subject: Re: skewed dualities: functions and sets|| the adjoint functor theorem of category theory explains this, in a| way. it says that left adjoint functors preserve colimits (such as| unions) and, dually, right adjoint functors preserve limits (such as| intersections). a functor is a left adjoint functor if it has a right| adjoint functor, and vice versa. inverse image is both a left| adjoint and a right adjoint functor. the left adjoint that it has (by| virtue of being a right adjoint) is called image, while the right| adjoint that it has (by virtue of being a left adjoint) is probably| not well enough known even to have a familiar name; i'll call it| anti-image for now. the anti-image of a subset s of the domain of a| function is the subset of the co-domain containing just those points| whose inverse images are subsets of s. anti-image preserves| intersections but not unions.||zzzzzzzzzzzzz ...as i said, you probably have to learn a fair amount of category theoryto appreciate this, so you're only demonstrating that you're an idiot.-- [e-mail address jdolan@math.ucr.edu]³ ³ ==== Subject: monoids and groupLet M be a monoid. We can find a group G and a morphisme i : M -> G that solve the universal problem : for all morphism f : M -> H where H is a group, there exist a unique morphism f' : G -> H with f' o i = f.If M is regular and commutatif, i is injective. Because it is the classical construction of fraction group.If M is free, i is injective. Because G is the free group with the same generators.If M is not regular, i is not injective.Find a non commutative and non free but regular monoid with i not injective.³ ³ ==== Subject: Re: monoids and group> Let M be a monoid. We can find a group G and a morphisme i : M -> G that > solve the universal problem : for all morphism f : M -> H where H is a > group, there exist a unique morphism f' : G -> H with f' o i = f.> If M is regular and commutatif, i is injective. Because it is the > classical construction of fraction group.> If M is free, i is injective. Because G is the free group with the same > generators.> If M is not regular, i is not injective.> Find a non commutative and non free but regular monoid with i not > injective.There's an example due to Malcev: the monoid generated by eight elementsa_1, a_2, a_3, a_4, b_1, b_2, b_3, and b_4 and by the relations a_1a_2 == a_3a_4, a_1b_2 = a_3b_4, and b_1a_2 = b_3a_4.Don't ask me way. I took this example from Jacobson's Basic Algebra I.It's exercise 7 from section 2.9 (at least, from the second edition).Best regards,Jose Carlos Santos³ ³ ==== Subject: Re: monoids and groupJos.8e Carlos Santos a .8ecrit dans le message de> Let M be a monoid. We can find a group G and a morphisme i : M -> G that> solve the universal problem : for all morphism f : M -> H where H is a> group, there exist a unique morphism f' : G -> H with f' o i = f.> If M is regular and commutatif, i is injective. Because it is the> classical construction of fraction group.> If M is free, i is injective. Because G is the free group with the same> generators.> If M is not regular, i is not injective.> Find a non commutative and non free but regular monoid with i not> injective.> There's an example due to Malcev: the monoid generated by eight elements> a_1, a_2, a_3, a_4, b_1, b_2, b_3, and b_4 and by the relations a_1a_2 => = a_3a_4, a_1b_2 = a_3b_4, and b_1a_2 = b_3a_4.> Don't ask me way. IIt will be a good exercice> took this example from Jacobson's Basic Algebra I.> It's exercise 7 from section 2.9 (at least, from the second edition).Seems a good book ...> Best regards,> Jose Carlos Santos³ ³ ==== Subject: Re: Shafer-Dempster vs. Bayes>Dear Friends,>I am doing some research about treating uncertainties in engineering>calculations. As one method I found the Shafer-Dempster Theory of evidence.>I read that resulting from that theory, the Bayesian approach of treating>uncertain data is a special case of the Shafer-Dempster Theory and that is>the point where I got confused. Can anybody explain me that? I also would>like to know what the advantages and disadvantages of both approaches are.As an engineer who found it useful, I recommendMyron Tribus' book: Rational Descriptions Decisions and Designshttp://www.bayesian.org/books/tribus.htmlRemarkably, its over 20 years old and still in reprint.In that book, evidence is mentioned as an alternative formulation ortransformation of probability or odds, expressing degree of certainty.As Tribus presents it, evidence and Baysian probabilities should leadto equivalent results.Apparently Tribus pre-dates Shafer-Dempster Theory which upon a quickcasual look, provides a basis for interpretation of the math whichviews probability as a subset of belief. The Bayesian approach can besaid to view degree of subjective certainty as a subset ofprobability. As an engineer, I could care less, provided the resultsare rational.If Tribus' book is not easily found, perhaps http://xyz.lanl.gov/abs/hep-ph/9512295Probability and Measurement Uncertainty in Physics - a Bayesian Primerby G. D'Agostini will help.(quoting the abstract:)Bayesian statistics is based on the subjective definition ofprobability as {it ``degree of belief''} and on Bayes' theorem, thebasic tool for assigning probabilities to hypotheses combining {it apriori} judgements and experimental information. John Baileyhttp://home.rochester.rr.com/jbxroads/mailto.htmlSubject : Re: Cantor's Diagonal ArgumentIn sci.logic, |-|erc<40525513$0$31905$ afc38c87@news.optusnet.com.au>:> oo> ____|mn> / /_/ / _> / K-9/ /_/ - www.YeOldeCoffeeShoppe.com -> /____/_____ Prob(exists j, for i = 1..n, b_i = c(j)_i)=1 -> Prob(exists j,>> for i = 1..n+1, b_i = c(j)_i)=1>> By induction on n :>> for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n>> This suggests NotEqual(b, c(j)) will not halt.> It won't halt anyway; that's part of the problem.> Assume one is testing all b's. A UTM that enumerates>> all b's can be created (the problem then becomes>> slightly recursive!) and one can then run NotEqual(b, c(j))>> on all j, while running simultaneously the b-generator UTM>> and doing the comparisons with oodles of threads.> A pretty and complex problem -- but this problem will>> not halt regardless, as there are an infinite number>> of b's to test.> That's not a concern. The initial formulation of Cantor's proof is that it> finds ONE b! We take that number and however many digits we compare> its computable. Its effective to demonstrate a proof that shows that> the 1 formulated number is on the list.>> b is only computable if the list of numbers is computable, which>> one has to assume if only because the mapping c(j) is specified.> Ouch, I thought I defined the mapping given the usage of a UTM. See> It was a long procedure I outlined, basically to overcome non halting> programs in the list of all functions' outputs. Say all programs halt,> then UTM(1) = some number, UTM(2) = some number, UTM(3) = some number.> The set of numbers output from UTM applied to N in turn, contains all> computable numbers. All I did was list the ones that halt 1st,> testing digit by digit, to guarantee that a digit is present on the> diagonal.That's fine for computable numbers, and I agree; the number of suchnumbers is countable. (It's an easy proof, similar to the oneyou outlined; a von Neumann/TM/engine can only have a finite numberof states, and therefore the number of such is countable.)The problem is that not all real numbers are computable via thesemethods...in fact, from the looks of it they're not computableat all.pi is a computable number; many series are available forit and one could in theory generate a Turing machine with ascratch tape to spit out digits.One of them is based on John Machin's formula (I knew it hadto have a name!):pi/4 = 4 arctan(1/5) - arctan(1/239)and arctan expressed as the infinite seriesarctan(y) = y - y^3/3 + y^5/5 - y^7/7 ...>> Or as you originally put it :>>Your problem is slightly different; you need to show that for *all* b>> which is actually where my proof begins, covering the extended problem> of multiple b's.>> For a decimal matrix there are 8 to 9 possibilites for each digit of b,> the number of b's is just 9^digits to test. Parallel testing is no problem,> but not necessary, if the Cantonian formulation is invalid for one of the> contenders the others will likely fail too.>> Go for it.> Ouch!! Doesn't this mean b = c(j)? and is it correctly derived?> for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n> This is equivalent to : for all n, b_n = c(j)_nYou have to prove that there is such a j.>> It won't halt anyway; that's part of the problem.>> This is only a problem for multiple c(j), we can consider b a constant.> //Globals> b = Cantor(C, random)> cj = FindMatch(C, b, 1)>> DiagNotEqual () {> DigitsEqual (b, cj, 1)> Return TRUE> }>> DigitsEqual (DE1, DE2, digit) {> if (DE1_digit = DE2_digit) {> DigitsEqual (DE1, FindMatch(C, b, digit+1), digit+1)> }> }>> Here FindMatch will tally through computable numbers until it finds> a number that matches the initial digits of b.>> j is allowed change to a different number during testing. This algorithm> will still determine whether a list of numbers does not contain a new> test number. The algorithm will not halt on the list of computables> against a new number however it is constructed.>> Unfortunately the algorithm is too giving, it would never determine> that pi was off the list of rationals, it would keep constructing> larger fractions to continue the comparison.>> To me its enough that any number off the list is nonsensical,> it only works when that number is infinitely long, which voids> it from any mathematical construction.>> It doesn't work anyway as an algorithm. It only works as a>semialgorithm; the routine halts and announces failure, but>> goes into an infinite loop upon success.> sprung, I was lazy with the construction. FindMatch goes to the> next number if there is no match, fails the next digit comparison,> falls out of the loop as not equal. I should write a general> function that works with any list and a number.>> But that's not a problem, mathematically. All you have to do>> is prove that DiagNotEqual(b, c) never halts for a>> Cantorian-constructed b.> I think I came close, but if it proves diag is computable it also> proves pi is rational.pi is neither rational nor algebraic. Its irrationality was provedby Johann Heinrich Lambert in 1761. Its transcendentality was proved byCarl Louis Ferdinand von Lindemann in 1882, based largely on results byCharles Hermite.> I'd had to overhaul my assumptions at> this point, work 'can be calculated' into my defn of computable.> Herc-- #191, ewill3@earthlink.netIt's still legal to go .sigless. ³ ³ ==== Subject: Re: Cantor's Diagonal Argument> Ouch!! Doesn't this mean b = c(j)? and is it correctly derived?> for all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n> This is equivalent to : for all n, b_n = c(j)_n> You have to prove that there is such a j.note the exists jfor all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_nadmitedly this is different to :exists j, for all n, b_n = c(j)_nI showed that no matter how many digits you examine, the diagup to that many digits is covered on the list.Herc³ ³ ==== Subject: Re: finding a Maclaurin seriesIn sci.math, Lucas:> Could someone show me how to find the Maclaurin series for some> function, either sin x, cos x, or ln (1+x)? thankshttp://en.wikipedia.org/wiki/Maclaurinor simply Google for Colin Maclaurin.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.³ ³ ==== Subject: Re: SCHOENFELDS RANDOM THEOREMIn sci.math, Servo:> [...]>>Sorry about horning in on your thread, but the poster seems to be>>avoiding my replies. I wonder why. :-)>> I wonder too:-) Still, I've seen many cases in which the order in>> which posts appear on one server differ (quite significantly, at>> times) from they time ordering on another. So, give it some more>> time.>>I killfiled Servo if that is what you are wondering.> There is no accounting for taste.> True. GR_Learner is the one who responded to Evil Al's 100 linesROTFLMAO!> Imagine the fun he'll have when /Mein Kampf/ is published in comic> book form.They did for the Bible. Somebody out there no doubt will try.[rest snipped]-- #191, ewill3@earthlink.netIt's still legal to go .sigless .³ ³ ==== Subject: Re: SCHOENFELDS RANDOM THEOREM[...]> True. GR_Learner is the one who responded to Evil Al's 100 linesROTFLMAO!> Imagine the fun he'll have when /Mein Kampf/ is published in comic> book form.> They did for the Bible. Somebody out there no doubt will try.> [rest snipped]Don't tell Evil Al. Killing off the firstborn of a whole nationmay not have occurred to him yet.Servo ³ ³ ==== Subject: Re: SCHOENFELDS RANDOM THEOREM> SCHOENFELDS RANDOM THEOREM:> With probability 1, all finite sequences of integers contained by> [n,m] occur infinitely in an infinite sequence of random integers> bound by [n,m].Could you please define contained by [n,m] and bound by [n,m]?³ ³ ==== Subject: Re: sine seriesIn sci.math, Taranjeet Singh> The series> sin x = x - (x^3)/3! + (x^5)/5! - ...> doesnot works when x is large, eg. x = 80> Is there any series which is generalized for computing sine for any x?The standard representation (IEEE-754, IINM) for a floating-pointnumber can be expressed as2^(f) * (1 + m/2^53)/2where f is from -0x400 to +0x3ff and 0 <= m < 2^53, if I'vedone this right.Since m is an integer the error is at most 2^(-54+f), but iff > about 50 you might as well give up as the precisionwill be lost. :-) However, x=80 is far below that point(f is 6).If you have x represented as a large integer you might havebetter luck, and you'll then get successive rationalapproximations, depending on your math system (I use GP/Pari).It will just take a fair number of terms, as the k'th termisx^(2k+1)/(2k+1)!and therefore the terms only start getting smaller if k > (x-1)/2.Of course the simplest method is to find y = floor(x / (2*pi)) andthen compute z = x - 2*pi*y, basically normalizing the problem.-- #191, ewill3@earthlink.netIt's still legal to go .sigless.³ ³ ==== Subject: Anyone seen thisAnyone seen this . How do you do it?http://digicc.com/fido/³ ³ ==== Subject: Re: Anyone seen this> Anyone seen this . How do you do it?> http://digicc.com/fido/The difference of the original number and its jumbled version will always bea multiple of 9.³ ³ ==== Subject: Partitions into h powers of a fixed numberGiven a rational number r>1, and a natural number h, does there existan upper bound to the number of ways of representing a natural numbern as the sum of exactly h powers of r?What if we loosen the restrictions on r and n and allow them to be anypositive real numbers?My interest in this was sparked after seeing a paper due to Mahlerproving that the number of ways of writing a natural number as a sumof powers of a fixed natural number b C_n has ln C_n asymptotic to (lnn)^2/2lnb.So it seems natural to ask for a estimates of the number of ways ofrepresenting a number as the sum of exactly some fixed number ofpowers.If we assume that the number r is rational, say r=p/q, it isstraightforwardto prove that to get more than one representation of some naturalnumber n, we must have both p and q leq h (which in fact implies q isleq h-1, since we assume r>1). In particular, if h=2, we must have r=2, and we quickly see that wecannot have more than one representation.For higher numbers, things are not quite so simple, and we can seestraight away that for h=3, we can get two different reps, forexample, 2^2+1+1=2+2+2.But we still have that the number of reps is less than 2 by quicklyconsidering the possible values of r, that is 2, 3, 3/2.I haven't looked at numerical bounds for higher numbers.Does anyone have any ideas for how to tackle it, or has anyone seenanything related to this anywhere? Any help would be appreciated.David³ ³ ==== Subject: Re: Partitions into h powers of a fixed number> Given a rational number r>1, and a natural number h, does there exist> an upper bound to the number of ways of representing a natural number> n as the sum of exactly h powers of r?> What if we loosen the restrictions on r and n and allow them to be any> positive real numbers?> My interest in this was sparked after seeing a paper due to Mahler> proving that the number of ways of writing a natural number as a sum> of powers of a fixed natural number b C_n has ln C_n asymptotic to (ln> n)^2/2lnb.> So it seems natural to ask for a estimates of the number of ways of> representing a number as the sum of exactly some fixed number of> powers.> If we assume that the number r is rational, say r=p/q, it is> straightforward> to prove that to get more than one representation of some natural> number n, we must have both p and q leq h (which in fact implies q is> leq h-1, since we assume r>1).> In particular, if h=2, we must have r=2, and we quickly see that we> cannot have more than one representation.> For higher numbers, things are not quite so simple, and we can see> straight away that for h=3, we can get two different reps, for> example, 2^2+1+1=2+2+2.> But we still have that the number of reps is less than 2 by quickly> considering the possible values of r, that is 2, 3, 3/2.> I haven't looked at numerical bounds for higher numbers.> Does anyone have any ideas for how to tackle it, or has anyone seen> anything related to this anywhere? Any help would be appreciated.> DavidActually, the proof I gave was bunkum. But a very similar argument works for ran integer.³ ³ ==== Subject: Re: Partitions into h powers of a fixed number> Given a rational number r>1, and a natural number h, does there exist> an upper bound to the number of ways of representing a natural number> n as the sum of exactly h powers of r?> What if we loosen the restrictions on r and n and allow them to be any> positive real numbers?> My interest in this was sparked after seeing a paper due to Mahler> proving that the number of ways of writing a natural number as a sum> of powers of a fixed natural number b C_n has ln C_n asymptotic to (ln> n)^2/2lnb.> So it seems natural to ask for a estimates of the number of ways of> representing a number as the sum of exactly some fixed number of> powers.> If we assume that the number r is rational, say r=p/q, it is> straightforward> to prove that to get more than one representation of some natural> number n, we must have both p and q leq h (which in fact implies q is> leq h-1, since we assume r>1).> In particular, if h=2, we must have r=2, and we quickly see that we> cannot have more than one representation.> For higher numbers, things are not quite so simple, and we can see> straight away that for h=3, we can get two different reps, for> example, 2^2+1+1=2+2+2.> But we still have that the number of reps is less than 2 by quickly> considering the possible values of r, that is 2, 3, 3/2.> I haven't looked at numerical bounds for higher numbers.> Does anyone have any ideas for how to tackle it, or has anyone seen> anything related to this anywhere? Any help would be appreciated.> DavidSo, I've found a solution to this for the rational case since postingit. Here's a sketch:Let r=p/q.Proceed by induction. Assume that for k = 1, ...., h-1, the statementis true, that is there exists an upper bound to the number of ways ofrepresenting any natural number as the sum of k powers of r.Suppose r satisfiesr^n_1 + ... r^n_h = r^m_1 + ... + r^m_hThen after perhaps some cancellation, we can convert this to q^a_1 p^b_1 + ... + q^a_g-1 p^b_g-1 + p^l = q^c_1 p^d_1 + ... q^c_gp^d_g,where g leq h, and a_i + b_i = c_i + d_i =l , for all i between 1 andg.Also for arguments sake assume that a_1 geq a_2 etc. and the same forthe c_i.If g f a.e.The function:C: E -> lim int(f_n, E), where C is defined over the measurable sets E (of ameasure space (X,A,u) over whom the f_n and f are defined), is completelyadditive.The proof uses the following trisection:Considering (E_i) a countable family of disjoint measurable sets of X,abs(C(E) - sum(C(E_i),i=1..m)) <=abs(C(E) - int(f_n, E)) +abs(int(f_n, E) - sum(int(f_n, E_i), i=1..m)) +abs(int(f_n, U (E_i, i =1..m)) - C(U (E_i, i=1..m)).The proof says:For e>0, the first and the third terms on the right are less than e/2 forall n>=n_0, where n_0 is sufficiently large, depending on e, but not on m.Then we have the result when m->+oo.My problem is that I dont see why n_0 doesn't depend on m (and consequentlywhy we can take m->+oo).Thks for any helpJ.S³ ³ ==== Subject: Re: A question in measure theory>I'm going through Avner Friedman's book Foundations of modern analysis and>i'm stuck on a detail.>Lemma 2.6.3 asserts:>Let (f_n) be a sequence of integrable simple functions that is Cauchy in>the mean and such that there exists a measurable function f, f_n -> f a.e.>The function:>C: E -> lim int(f_n, E), where C is defined over the measurable sets E (of a>measure space (X,A,u) over whom the f_n and f are defined), is completely>additive.>The proof uses the following trisection:>Considering (E_i) a countable family of disjoint measurable sets of X,>abs(C(E) - sum(C(E_i),i=1..m)) <=>abs(C(E) - int(f_n, E)) +>abs(int(f_n, E) - sum(int(f_n, E_i), i=1..m)) +>abs(int(f_n, U (E_i, i =1..m)) - C(U (E_i, i=1..m)).>The proof says:>For e>0, the first and the third terms on the right are less than e/2 for>all n>=n_0, where n_0 is sufficiently large, depending on e, but not on m.>Then we have the result when m->+oo.>My problem is that I dont see why n_0 doesn't depend on m (and consequently>why we can take m->+oo).Took me a minute - then I realized we hadn't used the fact that (f_n)is Cauchy in mean (which presumably means that int|f_n1 - f_n2| -> 0.)I imagine you see why the first term is < e/2 for all large enough n,independent of m. There exists N, depending only on e, suchthat int |f_n1 - f_n2| < e/2 for all n1, n2 > N. It follows from thisthat if A is any set then |int_A f_n - C(A)| <= e/2 for all n > N.>Thks for any help>J.S************************³ ³ ==== Subject: Re: A question in measure theory> independent of m. There exists N, depending only on e, such> that int |f_n1 - f_n2| < e/2 for all n1, n2 > N. It follows from this> that if A is any set then |int_A f_n - C(A)| <= e/2 for all n > N.Dang, I could have searched forever: I had forgotten that Cauchy in themean hypothesis too ...³ ³ ==== Subject: Re: what are holomorphic/mero sections|I wanted to know how would one define meromorphic and holomorphic sectionsBy the definition of vector bundle the base space can be covered by open setsU on which the bundle becomes trivial, in the sense that there is a coordinatechart on U with coordinates (x1,...,xn) and the elements of the bundle over Ucanbe put into one-to-one correspondence with points (x1,...,xn; y1,...,ym) where(x1,...,xn) are the coordinates of a point of U, and where the addition andscalarmultiplication of elements of the bundle at U correspond to addition and scalarmultiplication of elements (y1,...,ym) of R^m. I'm being a little sketchy andmynotation is lame but I hope you're familiar with this part.A section is called holomorphic if given any such trivialization, the imageof the section under the trivialization is a section of R^m over the image of Ugiven by m holomorphic functions y1,...,ym of x1,...,xn, and likewise formeromorphic sections.This only makes sense if the coordinate charts on the base space are assumedto be consistent up to biholomorphism, of course, because otherwise a sectionholomorphic in one trivialization will generally not be holomorphic in anotherone.In other words, the base space should be complex analytic.Keith Ramsay³ ³ ==== Subject: Re: what are holomorphic/mero sections> Dear NG,> I wanted to know how would one define meromorphic and holomorphic sections (of > Sincerely,> Jose CapcoIf M is your comlex varietie and V is your n-vector bundle p:V ---> M a section is a function s:M --- V such that s.p = identiti in M, so...S is holomorphic if in the chrts s is holomorphic. That is, given amap of varieties f:X --- Y, f is holomorphic (meromorphic)if for all xin X, there exist a chart in X; (U,g) and x in U, and a a chart (V,h)of f(x)in V such tahat the mapg^(-1)f h from C^(dim X) to C^(dim Y) is holomorphic (meromorphic);Like a V is a complex varietie with te structur induced by M, and asection s is a function from M to V... s is holomorhic (meromorphic)as in the last deffinition; Rogelio.³ ³ ==== Subject: Re: a digital game|Oops! Forgot about Mr. Nobel's grudge against mathematicians.It should perhaps be pointed out that Nobel's reason for notgiving a prize in mathematics wasn't, contrary to a popularurban legend, due to a mathematician having an affair withhis wife or girlfriend. This has been debunked various times.As far as I know there's no evidence of his having a grudgeagainst mathematicians.Nobel claimed he was choosing less theoretical areas for hisprizes. He seems to have been interested in areas that couldhelp offset the hazards of dynamite to society.Keith Ramsay³ ³ ==== Subject: Re: 5 circlesto construct the pentagramma mirificum.if the i'th great circle interscts the (i-1) and (i+1) circlesat 90 degrees, what is the angle of intersectionof those two? the imaginary circles of the @=120 case sounds interesting.> I formulated it this way because the usual determinant> approach gives the unique value @=120 deg...but you> can't draw it, some circles will always have radii> being NOT positive real.> I see! - So you were using distance geometric methods, too. > Are you interested in the Maple worksheet with the calculations?³ ³ ==== Subject: Re: Strange Complex Variables ProblemErr, I should read my postings before I send them. Sorry about theinaccuracy again. In your counterexample:If D is the unit disk minus the origin, then F(z) = z, F'/F = 1/z,which I believe _is_ square integrable over D. It has a finite L2 norm= integral(over D) { 1/|z|^2 dxdy } . To see this, we just transformthis to polar coordinates and since integral(over (0, 1)) { dx/x^2 }is finite and we can integrate this integral around the whole annulusas theta goes from 0 to 2pi, which is also finite.So maybe this problem is solvable the way it's stated.³ ³ ==== Subject: Re: Strange Complex Variables Problem> If D is the unit disk minus the origin, then F(z) = z, F'/F = 1/z,> which I believe _is_ square integrable over D. It has a finite L2 norm> = integral(over D) { 1/|z|^2 dxdy } . To see this, we just transform> this to polar coordinates and since integral(over (0, 1)) { dx/x^2 }> is finite and we can integrate this integral around the whole annulus> as theta goes from 0 to 2pi, which is also finite.integral(over (0, 1)) { dx/x^2 } = oo. But that's not the integral you want anyway. What happened to the r*dr from polar coordinates? integral(over D) { 1/|z|^2 dxdy } actually works out to 2Pi*integral(over (0, 1)) { dx/x }. But this = oo also.³ ³ ==== Subject: Re: Strange Complex Variables Problem> Answer me a question. Is this a problem like in a book> somewhere? I know that you know the answer to _that_> question... if it is a problem in the book could you tell> us _exactly_ how the problem reads?Sorry, David, I wish I could, but I am in the same boat that you are.This problem was given to me (in fact the whole class) by my complexvariables professor (Paul Garabedian), and it's infuriatingly unclearwhat the problem even asks. I asked him to clarify some points whichmattered (like, is the center of the target annulus the origin), andeven that he was reluctant to do.I'm trying to figure out a version of this problem that I can actuallysolve.>F'/F is certainly square-integrable. > No, it's not true that F'/F is certainly square-integrable!> I explained this a few days ago. Suppose that D is the unit disk> minus the origin. Then D is certainly doubly-connected, and> your annulus is D itself; F(z) = z, and then F'/F is> _not_ square-integrable over D.This hit me like a nail on the nose. You're absolutely right! Thenwhat the heck is he asking?? What problem could he possibly be asking?Regarding your corrections, you're right, I have a habit of sayingapproximately the right thing and assuming you know what I mean. WhenI said uniformly convergent I meant uniformly convergent on compactaof D.Is this statement true: analytic functions which are defined on thewhole bounded region D, with a finite L2 norm, form a hilbert spaceunder the L2 norm? If the functions are defined everywhere on D then aCauchy sequence of these wrt the L2 norm should converge uniformly onevery compact set of D so it will converge to a function defined onall of D, hence they form a hilbert space -- is this argument valid?I'm doing my first year of Ph. D math, and the way this professorteaches complex variables has always been a bit weird. Last semesterhe went off on a bunch of tangents including this, sort of of amateurfunctional analysis thing. He told us to prove that |f(z)| <= L2(f,D) / (sqrt(pi)*r) for all z in D, where r is the smallest distance tothe boundary of D. That I could do from what I knew of complexvariables. This time he assigned a problem where I am not even surebasically,1. Analytic functions form a Hilbert Space H on a domain D in somesense.2. We want to consider the subset J of H with functions whose integralaround the hole is i*2PI and the map F mapping D onto an annulus(which I assume is centered around the origin).3. And we want to show that F'/F has an L2 norm (over D) that is <=the L2 norms of all functions in J.And it looks like he somehow wants to tie this in with the RiemannMapping Theorem which we will get to later in class. He keeps sayingthat the riemann mapping minimizes the L2 norm of something, but againhe's very tough to understand because he's not being precise at all!What you said though is a great counterexample. If D is the disk (0,r) minus the origin, then F'/F = 1/z doesn't even have a finite L2norm!Do you have any idea what possible minimize L2 norm statement mightbe provable here? I'd appreciate any ideas.Sincerely,Greg Magarshak³ ³ ==== Subject: Re: Strange Complex Variables Problem>> Answer me a question. Is this a problem like in a book>> somewhere? I know that you know the answer to _that_>> question... if it is a problem in the book could you tell>> us _exactly_ how the problem reads?>Sorry, David, I wish I could, but I am in the same boat that you are.>This problem was given to me (in fact the whole class) by my complex>variables professor (Paul Garabedian), and it's infuriatingly unclear>what the problem even asks. I asked him to clarify some points which>mattered (like, is the center of the target annulus the origin), and>even that he was reluctant to do.If this is that Paul Garabedian he knows a lot more aboutthis than I do...>I'm trying to figure out a version of this problem that I can actually>solve.>F'/F is certainly square-integrable. > No, it's not true that F'/F is certainly square-integrable!> I explained this a few days ago. Suppose that D is the unit disk>> minus the origin. Then D is certainly doubly-connected, and>> your annulus is D itself; F(z) = z, and then F'/F is>> _not_ square-integrable over D.>This hit me like a nail on the nose. You're absolutely right! Then>what the heck is he asking?? What problem could he possibly be asking?>Regarding your corrections, you're right, I have a habit of saying>approximately the right thing and assuming you know what I mean. When>I said uniformly convergent I meant uniformly convergent on compacta>of D.>Is this statement true: analytic functions which are defined on the>whole bounded region D, with a finite L2 norm, form a hilbert space>under the L2 norm? If the functions are defined everywhere on D then a>Cauchy sequence of these wrt the L2 norm should converge uniformly on>every compact set of D so it will converge to a function defined on>all of D, hence they form a hilbert space -- is this argument valid?It's true - I don't know whether I'd call it an argument, there'sa few things that need to be proved there.(And the converse is still false: uniform convergence on compactsets does not imply convergence in L2.)>I'm doing my first year of Ph. D math, and the way this professor>teaches complex variables has always been a bit weird. Last semester>he went off on a bunch of tangents including this, sort of of amateur>functional analysis thing. He told us to prove that |f(z)| <= L2(f,>D) / (sqrt(pi)*r) for all z in D, where r is the smallest distance to>the boundary of D. Right - that's why convergence in L2 implies uniform convergenceon compact sets (given a compact set K in D there exists r > 0such that every point of K is at distance at least r from theboundary.)>That I could do from what I knew of complex>variables. This time he assigned a problem where I am not even sure>basically,>1. Analytic functions form a Hilbert Space H on a domain D in some>sense.>2. We want to consider the subset J of H with functions whose integral>around the hole is i*2PI and the map F mapping D onto an annulus>(which I assume is centered around the origin).>3. And we want to show that F'/F has an L2 norm (over D) that is <=>the L2 norms of all functions in J.On general hilbert-space grounds this is the same as sayingthat F'/F is orthogonal to all the functions which have integral0 around the hole. Oh - I just realized that those are the sameas the functions which are derivatives, and it seems possiblethat I see how to do the problem based on that:Without giving everything away: Let's use z* to mean the complexconjugate of z, and say is the integral of f(g*) over D.Let O be the set of all f in H such that the integral of f aroundthe hole is 0.You need to show that = 0 for all f in O. If f is inO then there exists g analytic in D with f = g', and at leastif D is a nice enough domain g will also lie in yourHilbert space (possibly g always lies in H, I'm not sure).So you need to show that = 0. Multiply the integrandby F'/F', to get |F'|^2 to appear. Now make a change ofvariables, using F to rewrite the integral as an integralover the annulus. Look carefully at what you're integratingover that annulus. (Note that in the annulus a functionis a derivative if and only if there is no 1/z in its Laurentexpansion...) Looks like it comes out to 0.>And it looks like he somehow wants to tie this in with the Riemann>Mapping Theorem which we will get to later in class. He keeps saying>that the riemann mapping minimizes the L2 norm of something, but again>he's very tough to understand because he's not being precise at all!Hmm, you said something about PhD? Maybe he's trying toget you warmed up for life in the real world, where you have tofirst figure out the question and then the answer.Mentioning the RMT rings a bell, although very faintly; I'dalready tried to recall what L2 thing the Riemann mapminimizes without success.Say D is the unit disk. Let J be the set of all functions f withf'(0) = 1. Then f(z) = z has the minimal L2 norm in J. I couldmake a wild guess on that basis what the Riemann mapwas supposed to minimize, but it would just be a wildguess.>What you said though is a great counterexample. If D is the disk (0,>r) minus the origin, then F'/F = 1/z doesn't even have a finite L2>norm!>Do you have any idea what possible minimize L2 norm statement might>be provable here? I'd appreciate any ideas.If the annulus has positive inner radius and finite outer radius thenF'/F _is_ in L2.>Sincerely,>Greg Magarshak************************³ ³ ==== Subject: Stats QuestionI have two groups (control and experiment). I want to give them a pre-test,perform the experiment, then give them a post-test. What stats test would Iuse to compare the increase of scores? At first i thought it would be atwo-tailed t-test, but I'm not 100% sure.³ ³ ==== Subject: Re: Question on modular arithmeticHmm, thanks for the response, I'll have to look at it again when I'mmore awake.The algorithm I'm working on is for base conversion (I'm currentlywriting a java program for it). I defined a Modular class (it takes anumber, x, and subtracts the mod, y, until x-y<0), and the actualalgorithm effectively takes (for 135 in base 10 to base 4, forexample)x_a=x_(a-1)-(x_(a-1) mod y)sum(10^k(x_k mod y)), where the x_0 is 135, any y is 4Thus, you get x_0=135, x_1=33, x_2=8, x_3=2, x_4=0 (end algorithm);and (x_0 mod 4)=3, (x_1 mod 4)=1, (x_2 mod 4)=0, (x_3 mod 4)=2The series ends up being the following:10^0(3)+10^1(1)+10^2(0)+10^3(2), or '2013'.I'm trying to figure out how to express this series in a singlestatement, if possible.(Note, this is for positive bases only at the moment, I'm working onmy negabase converter, for positive and negative representations,utilizing a NegaModular class)³ ³ ==== Subject: Re: Question on modular arithmetic ETAuAhUAmQWXOBv638kbq9DRVAuIqFIh2kUCFQCk3aoJ99g4gYM+ KNuMbjJH2JXCAg== I hope you ARE awake enough to figure out that for my construction youneed an odd modulus for this thing to work. It looks like you're usingeven moduli, and then all bets are off.--OL³ ³ ==== Subject: Re: Closure property question> Consider a collection of sets F. Define a set C_n of the form> oo oo> C_n = U / (B_njk), where B_njk belongs to F for all j and k.> j=1 k=1> {C_n} is the collection of all such sets. Question: Is {C_n} closed> under countable unions? I believe it is since the resulting iterated> unions can be expressed as one union, and the resulting set is some> C_k. Is this correct, or is there some other reason {C_n} is closed> under unions?> {C_n} would not necessarily be closed under countable intersections? > Is this correct?As Tobias has pointed out, the meaning of n here is mysterious. IfAgapito means to let B be an arbitrary function from N^3 to F, then the collection is not necessarily closed under unions: Let B_njk= {1,2,...,n, k} for a counterexample.Based on Agapito's earlier version of this post, I am guessing what hemeans to ask is the following:Let F be a field (of sets, not necessarily a sigma-field). For eachcountable collection b = {B_j: j in N} of countable collections ofsets in F, let c(b) = union(j=1..infty, intersection(B_j)). Let C be the image of c. Is C necessarily closed under countable unionsand/or intersections?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu³ ³ ==== Subject: Re: Closure property question> Consider a collection of sets F. Define a set C_n of the form> oo oo> C_n = U / (B_njk), where B_njk belongs to F for all j and k.> j=1 k=1> {C_n} is the collection of all such sets. Question: Is {C_n} closed> under countable unions? I believe it is since the resulting iterated> unions can be expressed as one union, and the resulting set is some> C_k. Is this correct, or is there some other reason {C_n} is closed> under unions?> {C_n} would not necessarily be closed under countable intersections?> Is this correct? All help appreciated.> Probably I do not understand your question correctly, because this does not> make too much sense to me:> E_njk is indexed by the three parameters n, j and k. But what does a> specific value of n have to do with all the other values of n?For each n, a mapping from (j,k) into F is generated. The collection{C_n} contains one set of the indicated form (union of intersections),for each of the mappings. Therefore, any union of C_n's represents aunion of unions of intersections.> Why should the resulting set be some C_k?> For example, set E_njk={1,...,n}. Then C_n={1,...,n}. But the union of all> C_n is N. Good example. In this case set F would have to contain all positiveintegers. One of our mappings would result in B_njk=N and C_n=N, forall n,j,and k. So the union of all C_n's would clearly belong to{C_n}. Does this make sense?> The same thing applies for intersections: set> E_njk=(N{1,...,n}). Then the intersection of all C_n is empty, though none> of the C_n is.Here, if F is a field, then it contains the empty set. Again if weset B_njk=0 for all n,j,and k, then C_n =0 for all n, and {C_n} isclosed under intersections. Can a counterexample be produced to showthat this is not the case in general?Many thanks for taking time to respond.³ ³ ==== Subject: Re: Closure property question> Then C_n = Q for all n and also their union /{ C_n | n in N } = Q> which is neither closed nor open in the reals.> He didn't say closed but closed under (ie: stable under) and from what> I remember all his assertions seemed true to me.Requires aleph_0 perfect.³ ³ ==== Subject: Re: ZZCrap> In sci.math, ZZBunker> > :>> In sci.math, ZZBunker>> :>> In sci.math, ZZBunker>> :>>to isn't a verb.>> The last time I heard that to wasn't an intransitive>> verb was the last time that math wasn't spelled>> in the moronic fashion maths.> What's math?>> Nobody knows, Since like we've been telling morons>> for several thousand years, now: >> Maths is philosophy not science.>> See Plato for philosophy and festive pillar work.> Math is an abstract model, nothing more. Occasionally it's>> useful; we've even harnessed the Queen of Uselessness,>> number theory, into modern cryptography. :-) Who says>> primes aren't useful? :-)> However, '1' cannot be captured in a butterfly net. 'pi',>> 'e', and i=sqrt(-1) are even more elusive. Good luck>> finding a Taylor or MacLaurin series running around in the wild.>> Nobody ever said that you can find *anything* related> to moronic *Calculus* running around in the wild.> We have merely stated that you can find *real* > *Turing machines* running around in the wild.> Models only, and imperfect ones at that. The ideal Turing>> machine has an infinite length tape (most of which is blank).>> All modern machines have finite tapes -- if they have tapes>> at all.> The ideal Turing machine never needed an infinite> tape to begin with. Since neither Turing nor any> other mathematician before him proved anything> conclusive about the word finite, other than> their token prayers to Euclid.> True; running a Turing machine with an infinite empty tape> would be slightly pointless, even were the machine the> simple one:> S=0: S <= 0, '0', Right> S=0: S <= 1, '0', Right> S=1: S <= 0, '1', Right> However, from a theoretical standpoint one has to specify> something along the lines of has a tape longer than needed.> As for finite: everything's finite. Even the symbol '+oo'> is a token of the concept we think of as infinity; it is> not, however, infinite in itself.> (The only infinite thing is God, if such exists at all, and> even that's highly debatable -- but probably not here. :-) )>> There are examples of Nature taking advantage of Fibonacchi>> numbers (e.g., sunflower seed spirals, plant stalk branchings) but >> that doesn't mean Fibonacchi numbers are out there.> The main requirement for math is that it be self-consistent.>> The *only* requirement for math is that it> exist in a non-existent vaccuum.> Other than that it's the most trivial> thing in the universe.> Math does not exist except as a series of thought patterns>> and as ink on paper and bits on storage devices and traveling>> through cables and air -- and those are also imperfect models.> Since models of math are also math,> we have never claimed that cables> are even approximately anything > other than exactly what they are, > which is cables. But when you > get somebody other than a > retarded mathematican using cables, you also> get a free subscription to > GPS TV, with the cable. So > again whatever math dorks > consider perfect is up to> math dorks. The rest of us are> still going to Mars. No > matter what Feynmann & heads Inc. say.> Cables can be (at least) the following:> [1] Information conduits. No they can't. Since they are only called cables by *philosophers*. Unfortunately for philosophers, real Engineers call them cable-grams.> [2] Power conduits. No they can't be again. Since power conduits are only called cables by philosophers. And again, the Engineers who are living call them high-tension lines.> [3] Garrotes. But since the only the Mafia and philosophers call Garrotes cables, and Engineers call then shoe-strings, that's obviously the reason that the Mafia can still only find gainful employment working for Saddam Hussein, the Mayor of Chicago, The New York Yankees, or the French. Whereas most Engineers still work for the NASA & The Air Force.> [4] Structural supports (e.g., suspension bridges) But they're still only called Structural supports by San Franciso Engineers. But, East Coast engineers with real jobs still call them 65 MPH Highways.> [5] Directional indicators (i.e., paths) But they're still only called Directional indicators by Mathematicians and Casino owners.³ ³ ==== Subject: Re: Matrix norms.>>Is it possible to define an unbounded norm on matrices?> You must not be asking exactly the question you meant> to ask: _every_ matrix norm is unbounded!You are correct. my question was incorrectly stated. What I wanted to ask is is it possible to define a normed space on matrices such that the matrices will no longer be bounded operators.³ ³ ==== Subject: Re: Matrix norms.> You are correct. my question was incorrectly stated. What I wanted to> ask is is it possible to define a normed space on matrices such that> the matrices will no longer be bounded operators.No, matrices form a finite-dimensional vector space, so all norms areequivalent. See e.g.http://planetmath.org/encyclopedia/EquivalentNorms.html-- I'm on warm milk and laxativesCherry-flavored antacidsreverse my forename for mail! - saibot³ ³ ==== Subject: Re: Matrix norms.By the way, is this correct: a norm F is a function from a vectorspace X to [0, +infinity), not [0, +infinity], which satisfies:1. F(cx) = |c|F(x) for all x in X2. F(x+y) <= F(x) + F(y) for all x in X, y in Y3. F(x) = 0 => x = 0i.e. a norm is never infinite! If it's infinite then it's undefined.-Greg³ ³ ==== Subject: Re: Matrix norms.>Is it possible to define an unbounded norm on matrices?> Perhaps that I am missing something here, but the only vector>> that I know in which you can define a bounded norm is {0}.>>Hm, in the (german) terminology I know, a linear map f:V->U, where V and U>are normed spaces, is said to be bounded iff there is a c>0 such that>||f(x)|| <= c ||x||Of course. But the question was not about bounded linear maps,the question was about bounded _norms_.>The smallest such c then is the norm of f. It is bounded iff it is>continuous.No, the norm of f is _finite_ iff f is continuous.>BTW, what is the english noun for continuous?Continuity ==== Subject: Re: Matrix norms.>Is it possible to define an unbounded norm on matrices?> Perhaps that I am missing something here, but the only vector>> that I know in which you can define a bounded norm is {0}.>> Hm, in the (german) terminology I know, a linear map f:V->U, where V and U> are normed spaces, is said to be bounded iff there is a c>0 such that> ||f(x)|| <= c ||x||> The smallest such c then is the norm of f. It is bounded iff it is> continuous.> BTW, what is the english noun for continuous?It is also bounded, in the case of linear operators. However, noticewhen we are dealing with an unbounded linear operator we just say thatit is unbounded, not that its norm is unbounded.Best regards,Jose Carlos Santos³ ³ ==== Subject: simplify a fractionI'm stuck!!I've got a fraction...(S^2 + S +1) / (S^2 +S)I want to simplify this to make the denominator = to 1.I know that if I divide top and bottom by (S^2 +S) this will make thedenominator = 1.what will the numerator be??SS³ ³ ==== Subject: Re: simplify a fraction> I'm stuck!!> I've got a fraction...> (S^2 + S +1) / (S^2 +S)> I want to simplify this to make the denominator = to 1.> I know that if I divide top and bottom by (S^2 +S) this will make the> denominator = 1.> what will the numerator be??> SSThat is like asking how to reduce the improper fraction 11/2 to a whole number. It is not possible.It is, however possible to reduce the improper fraction 11/2 to a mixed number:11/2 = 10/2 + 1/2 = 5 + 1/2In a similar way, the improper rational expression reduces to a polynomial expression plus a proper rational expression: (S^2 + S +1) / (S^2 +S) = [(S^2 + S ) / (S^2 +S)] + [1 / (S^2 +S)] = 1 + [1 / (S^2 +S)]³ ³ ==== ==== Subject: Re: Sum where some terms are added, others subtracted> Here is one of those results which seem like they should be> common-knowledge, yet I have never seen them before.> Let a(k,m) = 1 if k <= sqrt(m);> Let a(k,m) = -1 if k > sqrt(m).> If m = any positive integer, then:> sum{k=1 to m} floor(m/k) a(k,m)> always equals floor(sqrt(m))^2,> which is simply m if m is a square, of course.> (unless I am wrong.)>...So,...If {x} = fractional part of x,then:sum{k=1 to m} {m/k} a(k,m) =m *(2*H(floor(sqrt(m))) -H(m)) - floor(sqrt(m))^2,of course.So, therefore,limit{m-> oo}(1/m) sum{k=1 tom} {m/k} a(k,m) = c - 1,where c = Euler's constant.Leroy Quet³ ³ ==== Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairsath: meganewsservers.com>) Spin a pencil, etc does NOT fail due to the quantum nature of>) reality. Space itself is not quantized.>You're going to have to measure it. There's your quantum nature.>SaSW, WillemIf you're actually measuring a physical quantityin the real world, you're likely to find that the measurementhas only a finite precision and an uncertainty in any case,unless the quantity you're measuing comes in discrete chunks.George³ ³ ==== Subject: Sum Of Powers Of Divisor-FunctionFirst, some specific identities, then the general result, then some limits.Each of these is probably very well known, or at least easily found.But they seem very deserving of being even more well known, so I will post these.Let d(n) be the number of positive divisors of n.Then, for m = any positive integer:sum{k|m^2} d(k) = d(m) d(m^2).sum{k|m^6} d(k)^2 = d(m^3) d(m^4) d(m^6).sum{k|m^2} d(k)^3 = d(m)^2 d(m^2)^2.-Generally,sum{k|m} d(k)^r =product{p=primes,p|m} (sum{k=1 to a(p,k)+1} k^r),where p^a(p,k) = highest power of prime p which divides k.-2 limits:For r > 1,limit{n-> oo} sum{k|m^n} 1/d(k)^r =zeta(r)^b(m),where b(m) is the number of distinct primes dividing m.For r >= 0,limit{n-> oo}(sum{k|m^n} d(k)^r )/(n^{(r+1)b(m)}) = (product{p|m} a(p,m)^(r+1))/(r+1)^b(m).I may have made an error, but the above results are easily found (if right).Leroy Quet³ ³ ==== Subject: Ellipse ProblemGiven an ellipse with major axis of length M, minor axis of length m,rotated around the angle theta and centered at the origin, I want to findthe coordinates (x, y) of the rightmost point on its perimeter. (And perhapsthe topmost point, but it should be a similar approach.) After tinkeringaround with this for some time, I'm not sure the plan of attack. I'm after aformula I can convert into a graphics algorithm, but I'm still curious tosee how the solution is derived. Can someone help?-jk³ ³ ==== Subject: Re: Ellipse Problemcontrapositive> Given an ellipse with major axis of length M, minor axis of length m,> rotated around the angle theta and centered at the origin, I want to find> the coordinates (x, y) of the rightmost point on its perimeter. (Andperhaps> the topmost point, but it should be a similar approach.) After tinkering> around with this for some time, I'm not sure the plan of attack. I'm aftera> formula I can convert into a graphics algorithm, but I'm still curious to> see how the solution is derived. Can someone help?I would leave the ellipse aligned with the axes (equation xx + ayy = c) andfindpoints at which the slope has the desired value cot(theta).LH³ ³ ==== Subject: Calculus of Variations [Q]X-No-Confirm: yesI'm reading Structure and Interpretation of Classical Mechanics bySussman and Wisdom. On p. 93, during their discussion of how toderive Lagrange's equations for the case of constrained motions,the author's make a statement that doesn't make sense to me. Theystart out by discussing the case in which the permissible configurationpaths q must satisfy a constraint of the form phi(t, q(t), (D q)(t)) = 0,where t is time and D denotes derivative. They claim that theproof in this constrained case is very similar to the proof (sketch)they gave for the unconstrained case. In this earlier proof, akey step was arguing that since the integral of ({ (@_1 L o Gamma[q]) - D (@_2 L o Gamma[q]) }*eta)(t)from t_1 to t_2 is 0 for all permissible variations eta, theintegrand above must be identically 0.[NOTATIONAL ASIDE: I use @ for the partial derivative symbol andthe letter o to indicate functional composition; L is a Lagrangianfunction for the physical system under consideration; Gamma is apath dependent map that takes a path q: R -> R^n and returns themap t |-> (t, q(t), (D q)(t)); eta is a perturbating map with thesame range and domain as q, and with the additional property thateta(t_1) = eta(t_2) = 0; the * before the eta denotes the productof a row/covariant vector and a column/contravariant vector.]For the constrained motion case, the difference is that eta, inaddition to satisfying eta(t_1) = eta(t_2) = 0, must also be suchthat the perturbed path q + eta satisfies the constraint on themotion. The authors claim that the expression above must beidentically zero in this case as well, because, as they write inthe footnote on p. 93, [g]iven any acceptable variation [eta], we may make another acceptable variation by multiplying the given one by a bump function that emphasizes any particular time interval.I've seen this kind of argument before; it is essentially thereasoning the authors used when they proved the unconstrained motioncase. But in this case I just don't see what justifies the assertionthat by multiplying an acceptable variation by a bump function oneends up with another acceptable variation. After all, at thispoint in the argument, the constraint is pretty abstract (nothingmore than the (phi o Gamma[q])(t) = 0 expression above). Withso little information about the constraint, what justifies theassumption that a suitable bump function exists, that when multipliedby the origianl variation will yield a path that satisfies theconstraint?Any clues would be much appreciated!jill-- To s&e^n]d me m~a}i]l r%e*m?ov[e bit from my a|d)d:r{e:s]s.³ ³ ==== Subject: Re: Sum, Over Some Prime Multiples, Is Always m> Let P be any subset of the primes.> (example: P contains all primes of even-index,> or those congruent to 1 (mod 6).)> Let A be the set of positive integers: including 1 and every positive> integer which is a multiple of only the primes in P, and A contains> no member which is divisible by a prime not in P.> Let B be the set of positive integers: including 1 and every positive> integer which is a multiple of only the primes *not* in P, and B> contains no member which is divisible by a prime *in* P.> (Yes, there are positive integers in neither set, as long as P does> not contain every prime.)> Let g(x) be the number of distinct elements of A which are <= x,> for x = a positive real.> So then, for m = any positive integer:> ---> / g(m/k)> ---> k=elements of B, k <= m> always = m.> In linear-mode:> sum{k=elements of B, k<=m} g(m/k)> always equals m.> (Right?)Here is a more general theorem:For any sequence {a(j,k)}, defined for each positive integer k and j,and for any positive integer m:--- --- / / a(j,k)--- ---1<=k<=m 1<=j<=m/kk=element of B j=element of A=---/ a(j(n),n/j(n)), ---1<=n<=mwhere j(n) is the highest divisor of n which is an element of A.And the integer sequences A and B are as defined in my original post.Linear-mode:sum{1<=k<=m,k=element of B} sum{1<=j<=m/k,j=element of A} a(j,k)=sum{1<=n<=m} a(j(n),n/j(n))(Easy, but interesting, since the double sum reduces to a single sum.)-So, a few things:Sequence A can be defined as:sum{j=elements of A} 1/j^x= product{p=primes in P} 1/(1 -1/p^x),and B can be defined as:sum{k=elements of B} 1/k^x= product{q=primes not in P} 1/(1 -1/q^x),for all x's where the sums and products converge.-As you may have figured out, the result in the original post is whena(j,k) = 1 for all j and k.Another example:(a(j,k) = ln(jk). Raise e to both sides.)product{1<=k<=m,k=element of B} (product{1<=j<=m/k,j=element of A} j ) *k^g(m/k)= m!,no matter which primes are in P.And, as in original post,g(x) is the number of elements of A which are <= x.Leroy Quet³ ³ ==== Subject: Holomorphic functions PC^1 PC^2I am looking for all the holomorhic functions from the comlexprojectives spacesPC^2 to PC^1.I think that we have only 2 (essentialy different) because If f isholomorphic in the point p, then it is holomorhic localy in p, thatmeans that if I take an open chart in p (C^2 for example) and otherin f(p) we have holomorphic functon from C^2 C and this give mea C-morphism from the fields C((z)) to C((w,s)), so only twopossibilities.... am I right? or do you have a hint?Sorry for my english!³ ³ ==== Subject: Re: Holomorphic functions PC^1 PC^2>I am looking for all the holomorhic functions from the comlex>projectives spaces>PC^2 to PC^1.>I think that we have only 2 (essentialy different) because If f is>holomorphic in the point p, then it is holomorhic localy in p, that>means that if I take an open chart in p (C^2 for example) and other>in f(p) we have holomorphic functon from C^2 C and this give me>a C-morphism from the fields C((z)) to C((w,s)), so only two>possibilities.... am I right? or do you have a hint?>Sorry for my english!I'm not sure what you are looking for (but the problem is not withyour English, it is with my understanding). I assume you are notlooking for actual *functions* (in the set-theoretic sense) fromPC^2 to PC^1 which are holomorphic (with respect to the given complex structures on PC^2 and PC^1), since it is easy to see that any such function f is constant (if f takes on the value zin PC^1 then either f^{-1}(z) is all of PC^2, and we're done,or f^{-1}(z) is an algebraic curve in PC^2 [it's certainly ananalytic curve, and it's compact, so a theorem of Chow says it'salgebraic]; but in the latter case, if f also takes on the valuez' not equal to z, then the algebraic curves f^{-1}(z) andf^{-1}(z') must have a non-empty intersection, which is absurdsince f is a function in the set-theoretical sense). On the other hand, if you are looking for a meromorphic function on PC^2(so that there can be points of indeterminacy; for example,f(z_1:z_2:z_3)=z_1/z_2 is a meromorphic function which isn'tdefined at (0:0:1)), that is (after some more appeals to what a more sophisticated person might call GAGA, I wouldn't know) a rational function on PC^2l (which is what you're mention ofthe fields C((z)) and C((w,s)) suggests), then I don't see whatequivalence relation (of any interest) gives you only twopossibilities. I mean, for instance, any polynomial P(w,s)defines (if I understand the terms correctly) a C-morphism fromC((z)) to C((w,s)), by declaring z = P(w,s), doesn't it?And the classification of polynomial maps from C^2 to C,even up to topological equivalence (conjugation by homeomorphismsof domain and range), much less algebraic equivalence (conjugationby polynomial isomorphisms of domain and range), is *vastly*complicated. Ask Pepe Seade for some references (and tell himhi from me), I'm sure he knows. Or look at Arnaud Bodin's papershttp://arxiv.org/abs/math.AG/0009002 and http://arxiv.org/abs/math.AG/0210321 , and their references.Finally, if your question is really the one in the subject headerand not the one in the body of the text, then I can tell you lotsmore (but not everything, which I think is not known by anyoneyet) if you ask again.Lee Rudolph³ ³ ==== Subject: Re: Die Cantor Die> When trying to distinguish between productive lines of inquiry,> and mental masturbation, it's a good idea to consider the possible> applications of your inquiry.> What applications have you generated so far?The intended application was to come up with a foundationfor mathematics which could also serve as a foundation forartificial intelligence. This requires making a connectionbetween mathematical ideas and computation.However, I encountered so much ignorant resistance to myideas, such as is clearly seen in this discussion, that Igave up. Now I just talk about it, for reasons that arenot entirely clear to me.³ ³ ==== Subject: Re: Die Cantor Die> When trying to distinguish between productive lines of inquiry,> and mental masturbation, it's a good idea to consider the possible> applications of your inquiry.> What applications have you generated so far?> The intended application was to come up with a foundation> for mathematics which could also serve as a foundation for> artificial intelligence. This requires making a connection> between mathematical ideas and computation.It also presumes a connection, not presently in evidence, between Petry and intelligence.> However, I encountered so much ignorant resistance to my> ideas, such as is clearly seen in this discussion, that I> gave up. Now I just talk about it, for reasons that are> not entirely clear to me.Nor to us.³ ³ ==== Subject: Re: Die Cantor Die>When trying to distinguish between productive lines of inquiry,>and mental masturbation, it's a good idea to consider the possible>applications of your inquiry.>>What applications have you generated so far?> The intended application was to come up with a foundation> for mathematics which could also serve as a foundation for> artificial intelligence. This requires making a connection> between mathematical ideas and computation.> However, I encountered so much ignorant resistance to my> ideas, such as is clearly seen in this discussion, that I> gave up. Now I just talk about it, for reasons that are> not entirely clear to me.I take it from the above that you haven't actually produced anything besides a great deal of discussion as of yet. For myself, I have less resistence to your ideas than a desire to see you start doing something with them. I don't know what the results will be, but it seems clear to me that if there will be anything useful, you will have an easier time generating results either by studying the work of others before you or by producing some results now.As for AI, neural nets, fuzzy logic/set theory, and many other ideas seem to be producing interesting results. These have involved some new math, and some old. They have not overthrown the usual foundations of arithmatic to accomplish these results, however.-- Will Twentymanemail: wtwentyman at copper dot net³ ³ ==== Subject: A brief note on pens and coathangers I was recently compelled by the university to turn my thoughtsfor a moment away from the pure realm of mathematics, and to suffermyself for a while to consider the crude and clumsy applied sciences. Recognizing me as a genious of rare calibre, when the school wasasked on short notice to provide an original exhibition for a sciencefair, naturally the physics department turned in their desperation tome for help. It has been observed many times, and used to illustrate manysubtle and unintuitive propositions of quantum mechanics, that (leftto themselves and in the absense of an observer) coathangers willmultiply, while pens will decay at a rate which closely resemblesradioactive decay. And so I devised the following experiment: into a lead box,airtight, I inserted a large sample of ordinary pens, a sample ofordinary coathangers of equal size, and an equally large sample oflong, thin pens, carefully twisted so as to also serve as coathangers. These items were carefully made up of disparate chemicalsubstances, so that they might be distinguished from one another nomatter what permutations or transformations they would endure. At the end of a month's time, the contents of the box wereanalyzed: whence, by carefully examining the chemical compounds, itwas discovered that many of the ordinary pens had decayed: but, beingconfounded by the confines of the box, these suffered themselves tojoin the pen-coathangers, twisting themselves appropriately so as tofit in; simultaneously, the ordinary coathangers, seeking to multiplybut being also confounded by the same box, balanced this effect bydrawing upon the pen-coathangers. Using a mathematical model, it was demonstrated that if thisexperiment were repeated with arbitrarily large samples, and allowedto sit unobserved for arbitrarily large periods of time, the limitingprocess would be an unchanged number of pen-coathangers, a completeabsense of pens, and an exact doubling of coathangers. This limitwould never actually be obtained by any finite amount of time,however, lest the theorem of conservation of rate of change of pensand coathangers be violated. These principles were put to practical use for the sake oftransporting matter from one location to another: the items at theirsource were enclosed within a large container made entirely ofinterwoven pens, next to a giant pile of coathangers; while at thedestination, an enormous pile of pen-coathangers was gathered. As youcan guess, the the container made of pens was eventually teleportedinto the remote pile of pen-coathangers, an event which was balancedout, of course, by an equal number of the pen-coathangers untwistingthemselves and teleporting to the pile of ordinary coathangers. Inthis way, the university was able to transport a small-sizedautomobile from New York, NY to Beijing, China, with almost no energyexerted. My good colleagues from the physics department are workingclosely with the department of defense to determine whether this newtechnology might prove useful for national defense; I assure you thatI, however, am glad to be rid of the project, that I might return toconsiderations of mathematics.Your friend and correspondent,Nathan the GreatAge 11 :-)³ ³ ==== Subject: Filling In Rectangles: GameHere is a dumb game (which might help kids learn multiplications anddivisions and about primes, etc.):2 players.2 decks of cards, n cards each, the cards in both decks labeled withthe integers from 1 to n (one integer per card).You also have an m-by-m grid, where m, I suggest, is in the vicinityofjust over n^(3/2)/2, but m is up to the players.Players take turns drawing 2 cards, one card from each deck.Players, after drawing the cards, take the product, k, of the 2 cards'values.Players fill in any previously unfilled grid-squares so as to form arectangle of area k.(And the width/height of the rectangle need not be the same as thecards' numbers, unless 1 and a prime {or another 1} are drawn.)Players do not have to fill any rectangles if they volutarily skiptheir turn or if no rectangle of area k can be filled.Play continues until either the grid is filled completely or until thecards have all been drawn, whichever comes first.3 variations on how to determine winner: 1) Winner has most number of grid-squares filled with their color atgame's end.2) Winner has most number of rectangles at game's end filled withtheir color.3) Winner is last player to fill in a rectangle.And, if you want to, maybe you can simply use a standard card-deckinstead.And you draw 2 cards at a time, assigning appropriate numerical valuesto Jack, Queen, King, and Ace, of course.But how would each winner-determination variation affect strategy?And what would you suggest for an m based on n?Leroy Quet³ ³ ==== Subject: Re: Why are there 63360 inches in a mile?Originator: baez@math-cl-n03.math.ucr.edu (John Baez)>If I remember correctly, in one of Feynman's stories he describes>being challenged by a fellow with an abacus. The last calculation>suggested by the guy with the abacus is extracting the cube root>of 1729. According to his story, Feyman recognizes this right away>as one more than the number of cubic inches in a cubic foot,>getting the integer part right away, and then he gets the next digit>with an approximation. I think in the movie Infinity (where he's>played by Matthew Broderick) they have him getting more digits.>That particular choice of number to take the cube root of seems>almost too convenient, if you ask me.Feynman *and* Ramanujan got to act like geniuses by rememberingthat 1729 was one more than 12^3? This is getting ridiculous!Maybe the abacus guy had this number chosen ahead of timeand the trick backfired.