A122 === Subject: Zoomable ListPlot I made a small function to produce a zoomable listplot, hold alt down and draw a rect to zoom, shift click to return. I'm not very familiar with the eventhandlers, maybe someone else can do this much better, without redrawing the plot as many times? I would also like to be able to turn data series on and off by clicking the legend, but I have no idea on how to do that... PListPlot[fun_, opts : OptionsPattern[{Options[ListPlot], Legend -> {}, Frame -> True, ColorScheme -> 1}]] := DynamicModule[{p1, p2, pr = Automatic, pt = {{0, 0}, {0, 0}}, pl = {}, mylegend, is, style}, mylegend[data_] := Framed[Grid[ MapIndexed[{Item[ Graphics[{ColorData[OptionValue[ColorScheme], First@#2], Thick, Line[{{0, 0}, {1, 0}}]}, AspectRatio -> .1, ImageSize -> 30], Alignment -> {Center, Center}], Item[Style[#1, FontSize -> 16, FontFamily -> Helvetica], Alignment -> {Left, Bottom}]} &, data]], Background -> RGBColor[1, 1, 0.99999]]; is[x_] := {x, x/GoldenRatio}; EventHandler[ Dynamic[ ListPlot[fun, PlotRange -> pr, Epilog -> {pl, Inset[mylegend[OptionValue[Legend]], ImageScaled[{1 - 0.02, 1 - 0.02}], {Right, Top}]}, Evaluate@ FilterRules[{opts}, Options[ListPlot]], MaxPlotPoints -> ControlActive[1000, Infinity]], SynchronousUpdating -> Automatic], MouseDown :> (If[CurrentValue[AltKey], p1 = MousePosition[Graphics]]; If[CurrentValue[ShiftKey], pr = Automatic]; ), MouseUp :> (If[CurrentValue[AltKey], p2 = MousePosition[Graphics]; pt[[1, 1]] = p1[[1]]; pt[[1, 2]] = p2[[1]]; pt[[2, 1]] = p1[[2]]; pt[[2, 2]] = p2[[2]]; pr = pt; pl = {}; ] ), MouseDragged :> (If[CurrentValue[AltKey], p2 = MousePosition[Graphics]; pt[[1, 1]] = p1[[1]]; pt[[1, 2]] = p2[[1]]; pt[[2, 1]] = p1[[2]]; pt[[2, 2]] = p2[[2]]; pl = {Line[{p1, {p2[[1]], p1[[2]]}, p2, {p1[[1]], p2[[2]]}, p1}]}; ]; ) , PassEventsDown -> True ] ] === Subject: Text-based interface: Editing line input I have been enjoying mathematica, for a bit. While some might consider me a minimalist, I enjoy the text-based interface over the notebook style usage. However I severely miss the ability to move the cursor left and right for editing a single line. The documentation itself suggests pasting lines, but most of my pasting requires a little editing too. So instead of pasting whole lines of input, I resort to partial line paste, adding different text, then pasting the remaining relevant portion of a line. Is there any way to be able to freely move the cursor on a line? I have obviously tried the arrow keys, and also have tried emacs style movements as well with no luck (ctrl-b ctrl-f for backwards and forewards). Any suggestions? === Subject: ListContourPlot3D, no output I am trying to use ListContourPlot3D on a table of the form {{x,y,z,f},...}. Execution does not give any error number and I get only a system of axis, with all the ranges {0,1}. I made sure that the table is properly formatted. I am really puzzled. Adi === Subject: Alternating sums of large numbers I'm using a Mathematica 6, and I've faced with the following problem. Here is a copy of my Mathematica notebook: -------------------------------------------------------- NN = 69.5; n = 7; coeff = (Gamma[2*NN - n + 1]*(2*NN - 2*n))/n!; f[z_] := Sum[((-1)^([Mu] + [Nu])*Binomial[n, [Mu]]*Binomial[n, [Nu]]*Gamma[2*NN - 2*n + [Mu] + [Nu], z])/(Gamma[2*NN - 2*n + [Mu] + 1]*Gamma[2*NN - 2*n + [Nu] + 1]), {[Mu],0, n}, {[Nu], 0, n}]; Plot[coeff*f[z], {z, 0, 100}] -------------------------------------------------------- As you can see, I want to calculate a double alternating sum, consisting of large terms (products of Gamma-functions and binomial coefficients). Then I want to plot the result, in dependence on parameter z, which takes part in the summation as an argument of the incomplete Gamma-function, Gamma[2*NN - 2*n + [Mu] + [Nu], z]. Apart from this, I have another parameter, n, which is an upper limit for both of sums, and also takes part in Gamma functions. When this parameter grows, the expression next to summation also increases. At some point, Mathematica begins to show very strange results - and my question is actually about this. For instance, if the parameter n=5, everything is O.K., the plot shows a smooth curve. When we set n=6, there appears a little noise at 60 Hello everybody, > I have a list of {x,y} data and from a theoretical calculation I kno= w > that the data should be fitted by an equation, for example: > y=ax+bx+c; where a, b, c are constant parameters. > The question is how i can find the best fit for my data finding the > value of this parameters. > Do mathematica has a specific function for that? Try searching the documentation for fitting ... the first result, Fit, > does exactly what you describe here. when I try to do the FindFit command, the parameters have to be positive, so i was searching for the optimal values of the parameter that fit the data. How I can modify my code in order to find the best parameters fitting the data?I tried to do the Norm between the value of the data and the value of the equation but i cant do more. Remove[Global`*] data = {{1, 1}, {28, 0.719188377}, {54, 0.35746493}, {81, 0.182114228}, {117, 0.166082164}, {260, 0.132765531}}; express = (1 - k*x)*(1 - k*x/q)*(1 - p*k*x/q) this is the equation with which i want to fit the data (1 - k x) (1 - (k x)/q) (1 - (k p x)/q) {xx, yy} = data[Transpose] {{1, 28, 54, 81, 117, 260}, {1, 0.719188, 0.357465, 0.182114, 0.166082, 0.132766}} Try1 f1 = FindFit[ data, (1 - k*x)*(1 - k*x/q)*(1 - p*k*x/q), {{k, 0.01}, {p, 1.5}, {q, 1}}, x, MaxIterations -> 200] {k -> 0.00586032, p -> 2.86841, q -> 2.86841} here we have the parameters fitting the curve expr1 = express /. f1 (1 - 0.00586032 x) (1 - 0.00586032 x) (1 - 0.00204306 x) yyfit = expr1 /. x -> xx {0.986295, 0.658775, 0.415684, 0.230288, 0.0751917, 0.128567} yy - yyfit {0.0137055, 0.0604133, -0.0582186, -0.0481737, 0.0908904, 0.00419858} Norm[%] 0.133516 Norm[yy - ((express /. f1) /. x -> xx)] 0.133516 f11 = {k, p, q} /. Table[FindFit[ data, (1 - k*x)*(1 - k*x/q)*(1 - p*k*x/q), {{k, k0}, {p, 1.5}, {q, 1}}, x, MaxIterations -> 200], {k0, 0.01, 1, 0.1}] // TableForm f11 = Table[ FindFit[data, (1 - k*x)*(1 - k*x/q)*(1 - p*k*x/q), {{k, k0}, {p, 1.5}, {q, here i am trying with k from 0.01 to 1 1}}, x, MaxIterations -> 200], {k0, 0.01, 1, 0.1}] Norm[yy - ((express /. #) /. x -> xx)] & /@ f11 {0.133516, 0.133516, 0.133516, 0.133516, 0.162362, 0.133516, 0.162362, 0.133516, 0.133516, 0.133516} % // Min 0.133516 Here i find the minimum for Norm and find the position at which i have the best value of the fitting Position[%%, %] {{4}, {6}, {8}} % // Union {0.133516, 0.133516, 0.133516, 0.133516, 0.162362} === Subject: Transformations in expressions Hi all, How can I to transform a specific expression to other by substitution of a part of it. For example, I want to transform the expression Log[x^2+y^3]/ Sin[x^2+y^3] in Log[z]/Sin[z] by substitution x^2+y^3 --> z. And other question: Is the new expression dependent of z. I see it is, but how can I to programar it. === Subject: Re: Debugger > Welcome to the club! > I gave up on debugger a long time ago. > WRI has developed a (supposedly) good functionality, but they did not > bother to document it well. > Why should we, the users, waste time trying to figure out how it > works? > You will find some explanations (by WRI people) on this group about > Debug, but not much. > I suggest you use ON[] and Off[] and sprinkle the code you are testing > with Print commands to see the value of variables. > This way you clearly see the behind-the-scenes actions of your code. > Trace gives you the same information, but the output is not so well > formatted and is difficult to read. > Workbench is supposed to be the Integrated development environment > (IDE) for Mathematica, but well-known mathgroupers have told me that - at the > moment - it is more an headhache, rather than an Aspirin. > I too gave up on the debugger a long time ago for all the reasons debugger based on TraceScan, and described it at a developer conference. Because it had to execute many Mathematica steps for every step of the program, it was unfortunately rather slow, but it had more features than the built-in offering, and was used by quite a few people. Among other extra features: 1) It recognised that certain 'steps' should be skipped. For example, rather than stopping on a complicated CompoundExpression construct, it skipped that step and went to the first part of the CE. This made it much more practical to use. 2) It kept a step count, which it would print out in various circumstances. The idea was, that if you performed one debug session and got too far, you could note the step count, and debug the program again up to, say count-100. Again, this made it feasible to debug far more complicated examples. 3) There were various tools to let the user hide certain parts of the code from the debugger. For example, you could specify that the debugger should not step into code that was not in Global` context. I hope WRI make the notebook debugger into a real feature for the next release. I don't really want to use the Workbench just to get a better version of the debugger that is meant to be present in the notebook interface! David Bailey http://www.dbaileyconsultancy.co.uk === Subject: Problems with Frame in ListPlots with Legends I have a problem with using Frame in a ListPlot if it has a legend. Needs[PlotLegends] a = ListPlot[lowData, Joined -> True, PlotRange -> {{1, 3}, {.5, 1}}, PlotStyle -> {Hue[0], Hue[1/3], Hue[2/3]} ] b = ListPlot[lowData, Joined -> True, Frame -> {{True, False}, {True, False}}, FrameLabel -> {X, Y}, PlotRange -> {{1, 3}, {.5, 1}}, PlotStyle -> {Hue[0], Hue[1/3], Hue[2/3]} ] Plot a and Plot b both work. Plot b has a frame and frame labels. I can add a legend to Plot a. Adding a legend to Plot b works, but gives errors. l = {{{Hue[0], 1}, {Hue[1/3], 2}, {Hue[2/3], 3}}, LegendPosition -> {1, 0}, LegendSize -> {0.5, 0.5}}; ShowLegend[a, l] (* This works *) ShowLegend[a, l] (* Makes a plot with a legend, but gives errors *) The errors are a set of Axes::axes: {{True,False},{True,False}} is not a valid axis specification. I can't work out what is going on. It works with Frame->True, but I only want axes on the left and at the bottom, and I need frame labels. It is a shame that the PlotLegend option does not seem to work with ListPlot, although the documentation says it should. Any help much appreciated. Neil. === Subject: Simulate a finite-state markov process I've seen a message in comp.soft-sys.math.mathematica : I found it very interesting and tried to develop a similar program using the same idea. The problem is that I'm not that good with programming... Suppose I have the following Emission matrix: {{0.470949,0.260239,0.268812},{0.529412,0.470588,0},{0.625,0,0.375}, {0.722222,0,0.277778},{0.446667,0.24,0.313333},{0.5,0.5,0}, {0.481301,0.252846,0.265854},{0.471374,0.257634,0.270992}, {0.47293,0.269108,0.257962},{0.52027,0.239865,0.239865}, {0.595238,0.214286,0.190476},{0.444444,0.296296,0.259259}, {0.478528,0.263804,0.257669},{0.507874,0.233071,0.259055}, {0.434251,0.308869,0.256881},{0.504274,0.262108,0.233618}, {0.419753,0.320988,0.259259},{0.473684,0.315789,0.210526}, {0.512012,0.231231,0.256757},{0.5,0.19469,0.30531}, {0.390244,0.195122,0.414634},{0.510949,0.19708,0.291971}, {0.464286,0.25,0.285714},{0.285714,0,0.714286}} The alphabet is {A,B,C} And the following Transition Matrix: {{0.470949,0,0,0,0,0,0.260239,0,0,0,0,0,0,0.268812,0,0,0,0,0,0,0,0,0,0}, {0.529412,0,0,0,0,0,0,0.470588,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0.625,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.375,0,0,0,0,0,0,0,0}, {0.722222,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.277778,0,0,0,0,0,0}, {0.446667,0,0,0.24,0,0,0,0,0,0,0,0,0,0,0,0.313333,0,0,0,0,0,0,0,0}, {0.5,0,0,0,0.5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0.481301,0,0,0,0,0,0,0,0.252846,0,0,0,0,0,0.265854,0,0,0,0,0,0,0,0,0}, {0.471374,0,0,0,0,0,0,0,0,0.257634,0,0,0,0,0,0.270992,0,0,0,0,0,0,0,0}, {0.47293,0,0,0,0,0,0,0,0,0,0,0,0.269108,0,0,0,0.257962,0,0,0,0,0,0,0}, {0.52027,0,0,0,0,0,0,0,0,0,0,0,0.239865,0,0,0,0,0.239865,0,0,0,0,0,0}, {0.595238,0,0,0,0.214286,0.190476,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0.444444,0.259259,0,0,0,0,0,0,0.296296,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0.478528,0,0,0,0,0,0,0,0,0,0.257669,0,0.263804,0,0,0,0,0,0,0,0,0,0,0}, {0.507874,0,0,0,0,0,0,0.233071,0,0,0,0,0,0,0,0,0,0,0.259055,0,0,0,0,0}, {0.434251,0,0,0,0,0,0,0.308869,0,0,0,0,0,0,0,0,0,0,0,0.256881,0,0,0,0}, {0.504274,0,0,0,0,0,0,0.262108,0,0,0,0,0,0,0,0,0,0,0,0,0.233618,0,0,0}, {0.419753,0,0,0,0.320988,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.259259,0,0,0,0}, {0.473684,0,0,0,0.315789,0,0,0,0,0,0,0,0,0,0,0,0,0,0.210526,0,0,0,0,0}, {0.512012,0,0,0,0,0,0,0.231231,0,0,0,0,0,0,0,0,0,0,0,0,0,0.256757,0,0}, {0.5,0,0,0,0.19469,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.30531,0,0}, {0.390244,0,0.195122,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.414634,0,0}, {0.510949,0,0,0,0,0,0,0,0,0,0,0.19708,0,0,0,0,0,0,0,0,0,0,0.291971,0}, {0.464286,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0.285714,0.25}, {0.285714,0.714286,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}} How can I develop a program to generate x sequences of a specified length? By the way, does anyone know how to calculate the probability of a specific sequence to occur within n generated sequences?? Edson Ferreira === Subject: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]? >> I don't think there is a way to simplify one of these expressions >> into the >> other, but one can use Mathematica as an aid in proving that they >> are equal >> (for real x and y). One way to do this is: If this is case, I therefore might have to rely on hand calculation. > But I would think Mathematica should be improved so that mathematical > expressions can be manipulated in more versatile ways. Peng I always agree that Mathematica should be improved (although perhaps I put the limit the point where it would start putting human mathematicians out of their jobs). However, that is easy to say, the hard thing is to find the right algorithms and to implement them. Algorithmic simplification of expressions is a very complex problem and only for certain types of expressions (like polynomials, rational functions, exponentials, logs and a few others) sufficiently general methods exist (and this ignores the not exactly trivial problem of defining when one expression is to be considered simpler than another). But the situation is much worse when you wish to convert from one form to another an expression, when the conversion is only valid under certain assumptions. If you can suggest a concrete general algorithm that you think should be implemented in Mathematica and is not implemented now I am sure WRI will be happy to oblige but just saying it should be improved isn't very helpful. You can be sure that there are people who are trying to do that all the time. Andrzej Kozlowski === Subject: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]? I don't think there is a way to simplify one of these expressions into the other, but one can use Mathematica as an aid in proving that they are equal (for real x and y). One way to do this is: expres = {ArcCos[x/Sqrt[x^2 + y^2]], Pi/2 - ArcTan[x/Abs[y]]}; Subtract @@ Assuming[Element[x | y, Reals], Simplify[Sin /@ expres]] 0 Subtract @@ Assuming[Element[x | y, Reals], Simplify[Cos /@ expres]] 0 So for real x and y, the two expressions have equal sines and cosines. That means that they must differ by an integer multiple of 2Pi. However, since the difference is a continuous function of x and y, it has to be constant. Now, putting (for example) x=0, y=1, we see that the constant must be zero, hence they are equal. Andrzej Kozlowski ArcCos[x/Sqrt[x^2+y^2]] and Pi/2-ArcTan[x/Abs[y]] are the same. But I can not get the first expression be simplified to the second > one. And the following command can not be simplified to zero in > mathematica as well. FullSimplify[ArcCos[x/Sqrt[x^2 + y^2]] - ( Pi/2 - ArcTan[x/Abs[y]]), > Element[x, Reals] && Element[y, Reals]] I'm wondering if there is any walkaround to do this simplification. Peng > === Subject: Re: Problem with replacement rules > I use Solve[expression(x,y),x]. Then, x is obtained as a one-to-one function of y. I want to define the function x(y) using the output of Solve. How should I do? In[1]:= (* Solve the expression for x and save the result in sol *) sol = Solve[a * x + y == 7, x] (* Define x as a function of y *) x[y_] = x /. sol[[1]] (* Use the function *) x[10] Out[1]= {{x -> (7 - y)/a}} Out[2]= (7 - y)/a Out[3]= -(3/a) -- Jean-Marc === Subject: Re: Problem with replacement rules makeFunction[expr_, x_] := Module[{var}, var = Complement[Variables[expr], {x}]; Function @@ {var, x /. Solve[expr == 0, x][[1]]} ] and f = makeFunction[2 + x - y^2, x]; Plot[f[z], {z, -2, 2}] ?? > I use Solve[expression(x,y),x]. Then, x is obtained as a one-to-one function of y. I want to define the function x(y) using the output of Solve. How should I do? > === Subject: Re: How can I get list form like this expr = Part[f, #] & /@ Range[5] // Quiet % /. f -> {a, b, c, d, e} {f[[1]], f[[2]], f[[3]], f[[4]], f[[5]]} {a, b, c, d, e} -- David Park djmpark@comcast.net http://home.comcast.net/~djmpark/ As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? === Subject: Re: How can I get list form like this > As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? Map a pure function such as f[[#]]&. For instance, depending on whether the expression holds some values before the mapping or not, In[1]:= f = {1.1, 1.2, 1.3}; (f[[#1]] & ) /@ Range[3] Out[2]= {1.1, 1.2, 1.3} In[3]:= (HoldForm[g[[#1]]] & ) /@ Range[3] g = {1, 2, 3}; ReleaseHold[%%] Out[3]= {HoldForm[g[[1]]], HoldForm[g[[2]]], HoldForm[g[[3]]]} Out[5]= {1, 2, 3} -- Jean-Marc === Subject: Re: How can I get list form like this Hold[Part[##]] & @@@ Table[{f, i}, {i, 1, 5}] As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? === Subject: Re: How can I get list form like this Hi. The answer is Quiet[f[[#]] & /@ Range[10]] but will it really works as you want? === Subject: Re: How can I get list form like this Part[f, #] & /@ Range[85] returns a list in the form you requested, but produces warnings because f is not defined. If you have f defined, you will get the parts. As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's= too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? === Subject: Re: How can I get list form like this As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? > You _could_ write f[[#]]& /@ Range[n], but Take[f,n] is shorter and most propably faster. Peter === Subject: Re: How can I get list form like this As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? > If you need to slice lists, use one of the following: f[[ Range[n] ]] f[[ 1;;n ]] f[[ ;;n ]] (The last two will only work in Mathematica 6) === Subject: Re: How can I get list form like this Use a simple pure function: f[[##]]&/@Range[n] Look at the command Function in the help file for more detail about this construction. Januk As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; > Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too > tedious to type one by one*) > It means the 1st,2nd,=85=85part of list f. > Looks similar to each other, however I can't figure it out. > How to do this ? === Subject: Re: Comparison of coefficients Coefficients[]. === Subject: Re: Problem with replacement rules Clear[x]; x[y_] = x /. Solve[y == x/a + b, x][[1]] -a (b-y) x[2] -a (b-2) I use Solve[expression(x,y),x]. Then, x is obtained as a one-to-one function of y. I want to define the function x(y) using the output of Solve. How should I do? -- === Subject: Re: How can I get list form like this f = a + b + c + d + e; f[[#]] & /@ Range[Length[f]] {a,b,c,d,e} or more simply List @@ f {a,b,c,d,e} Or, if as you state, f is a List, then a list of its parts is just f f = {a, b, c, d, e} {a,b,c,d,e} (f[[#]] & /@ Range[Length[f]]) == List @@ f == f True As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too tedious to type one by one*) It means the 1st,2nd,=85=85part of list f. Looks similar to each other, however I can't figure it out. How to do this ? === Subject: Re: Trying to speed up a function Hi Abhishek, no mistery here. By definition Table[expr,{n}] completely evaluates expr n times. Therefore, it is a good idea to take out redundant calculations. Note also that Table has the attribute HoldAll and does therefore not evaluate its arguments. Daniel > Say I have a previously defined matrix 'vecs' and I want to calculate > certain sums of columns. I find it easier to write in terms of the > Transpose but it seems that when I put Transpose[vecs] in the > definition of a table, mathematica evaluates the transpose over and > over to get at every element of the matrix. Which takes much longer > than doing it in 2 steps (2nd method). I only found this out by trial > and error so I am curious about the explanation for this behaviour and > how to avoid it in general. BTW I am aware that the expression can be written in 1 line without > using Transpose also, but I'd still like to know how to tell that if a > sub-expression in a function is going to be evaluated repeatedly. > Abhishek In[15]:= Dimensions[vecs] Out[15]= {1620, 1620} In[23]:= Table[Total[Transpose[vecs][[n]][[811 ;; 1620]]], {n, 1, > 1620}]; // Timing Out[23]= {27.844, Null} In[24]:= vecst = Transpose[vecs]; > Table[Total[vecst[[n]][[811 ;; 1620]]], {n, 1, 1620}]; // Timing Out[25]= {0.016, Null} Internet: === Subject: Re: Error in file / load entire notebook into memory mathematica files are XML I think; i.e. they're text with extra kind-of flags in it. So the file will open using a good text editor. It should be possible to identify what is input and what is output. So what I'd do is eliminate the output then try to fix the input so that it looks right compared with another valid mathematica file. Then save and run the file. I've done this before with a student's exam file that got corrupted - it works. You're still left with the versions 5/6 issue however. good luck Peter 2008/9/9 Adam Weyhaupt : > I am teaching Calc I this semester and am having my students do > Mathematica labs. The labs are written in Mathematica 6, but we have both > Mathematica 6 and Mathematica 5.2 on campus, so I can't guarantee that the labs were never opened in 5.2. On a few (about 3 files out of 60), I am > getting the following error. This occurs in a pop up dialog box > after scrolling down through the file. There appears to be an error in the file named below. You should > close it without saving, then check the Load entire notebook into > memory checkbox in the Notebook Options dialog. File name: /Users/ > adam/Desktop/downloads/lab1_exercises_1_-2.nb. On two of these labs, the file seemed to be badly damaged --- I > couldn't read most of it. On a third, I can see no visual or > computational problem with the file. Do you have any thoughts about what might be causing these errors? > Have you seen this before? I can't find the load entire notebook > into memory option; do you have any suggestions about where to find it? I'm using Mathematica 6.0.2, Mac OS X 10.4.11, 2.2 GHz Intel Core 2 > Duo. (The files were also opened and worked on on Win XP, > Mathematica 6.0.0, I think. There could have been other machines > involved.) > http://www.siue.edu/~aweyhau/ > Science Building 1316 (618) 650-2220 === Subject: Re: About NDSolve k = .3; Clear[g]; > g[b_?NumberQ] := > g[b] = {x[t], v[t], b} /. First[NDSolve[{x'[t] == v[t] , > v'[t] == -x[t]^3 - k*v[t] + b*Cos[t], x[0] == 1, v[0] == 0}, {x, > v}, {t, 0, 2000}, MaxSteps -> Infinity]] > pts = Table[ > Evaluate[Table[g[b], {b, 0.2, 0.6, .01}]], {t, 1950, 2000}]; If I use h[s_]:= s^3 and replace x[t]^3 by h[x[t]], NDSolve in pts complains. Why can't I do this? > Piecewise or Which (e.g., Which[x[t] < -1, -1, x[t] > 1, 1, True, x[t]]) on the place of x[t]^3 also gives problems. > Any suggestions? > P_ter > I tried this, but didn't get any errors. === Subject: Re: How I can fit data with a parametric equation? > Hello everybody, > I have a list of {x,y} data and from a theoretical > calculation I know > that the data should be fitted by an equation, for > example: y=ax+bx+c; where a, b, c are constant parameters. The question is how i can find the best fit for my > data finding the > value of this parameters. Do mathematica has a specific function for that? > Dino > Given A your collection of {x,y} data, just : FindFit[A, a x + b, {a, b}, x] (careful, y = ax + bx + c should just be y = (a + b)x + c = dx + c !) Luca P. === Subject: Re: How can I do a grep -v equivalent in Import[]? > Valid point: > This will likely line wrap: --snip-- > CPU NFS CIFS HTTP Total Net kB/s Disk kB/s Tape kB/s Cache > Cache CP CP Disk FCP iSCSI FCP kB/s > in out read write read write age > hit time ty util in out > 24% 0 741 0 767 9381 6144 88815 0 0 81011 9 > 98% 0% - 31% 0 26 0 0 > --snip-- I'm trying to avoid importing through a pipe and I'm trying to avoid > preprocessing the data (for no real reason other than trying to get a handle > on data processing in Mathematica). I've done the following: importTemp = Import[ToFileName[Directory[], input.txt], {Lines}]; > regexMatch = ^d{2}.*; > sysstat = StringCases[importTemp, RegularExpression[regexMatch]]; which gets me ALMOST what I'm looking for.... I end up with the the > non-matched lines being empty data sets which Part:partw doesn't like. e.g. output is like this: {{}, {}, {24}, {29}, ... -jbl > I am still not 100% sure I know what you want to do, but assuming you want just the first two digits of lines that start with 2 digits, I am still not 100% sure I know what you want to do, but assuming you want just the first two digits of lines that start with 2 digits, tmp2 = Select[importTemp, StringMatchQ[#, DigitCharacter ~~ DigitCharacter ~~ __] &]; Map[StringTake[#, 2] &, tmp2] Alternatively, you could clean up the output that you have obtained already: {{}, {}, {24}, {29}} /. {} :> Sequence @@ {} With all the new string matching functionality, it is rarely necessary to resort to C preprocessing, unless speed is critical (i.e. your input file is very big!). David Bailey http://www.dbaileyconsultancy.co.uk === Subject: Problem with replacement rules I use Solve[expression(x,y),x]. Then, x is obtained as a one-to-one function of y. I want to define the function x(y) using the output of Solve. How should I do? === Subject: Rules from TreeForm Hi all, I am wanting a simple way to get the ordered list of rules that are internally generated using TreeForm, given the user has input an expression to that function. TreeForm usually generates a GraphPlot (or LayeredGraphPlot?) image. I am after the rules underpinning that graph visualisation. (BTW I searched this group but could not find a matching solution to this problem. Apologies if I have missed something obvious.) === Subject: Re: How I can fit data with a parametric equation? >Hello everybody, I have a list of {x,y} data and from a theoretical >calculation I know that the data should be fitted by an equation, >for example: >y=ax+bx+c; where a, b, c are constant parameters. >The question is how i can find the best fit for my data finding the >value of this parameters. >Do mathematica has a specific function for that? Yes, FindFit. Assume data is a list of {x,y} pairs. Then FindFit[data, a x + b, {a,b}, x] will find values for a and b that represent the best fit (least squares) line. FindFit supports non-linear models as well === Subject: Re: How can I do a grep -v equivalent in Import[]? >I've done the following: >importTemp = Import[ToFileName[Directory[], input.txt], >{Lines}]; regexMatch = ^d{2}.*; sysstat = >StringCases[importTemp, RegularExpression[regexMatch]]; >which gets me ALMOST what I'm looking for.... I end up with the the >non-matched lines being empty data sets which Part:partw doesn't >like. >e.g. output is like this: >{{}, {}, {24}, {29}, ... This last is easily dealt with using DeleteCases That is if StringCases[importTemp, RegularExpression[regexMatch]] produces the output you showed then DeleteCases[StringCases[importTemp, RegularExpression[regexMatch]],{}] will produce {{24}, {29}, ... === Subject: Re: How to simplify ArcCos[x/Sqrt[x^2+y^2]] to Pi/2-ArcTan[x/Abs[y]]? >> ArcCos[x/Sqrt[x^2+y^2]] >> and >> Pi/2-ArcTan[x/Abs[y]] >> are the same. >> But I can not get the first expression be simplified to the second >> one. And the following command can not be simplified to zero in >> mathematica as well. >> FullSimplify[ArcCos[x/Sqrt[x^2 + y^2]] - ( Pi/2 - ArcTan[x/Abs[y]]), >> Element[x, Reals] && Element[y, Reals]] >> I'm wondering if there is any walkaround to do this simplification. Try Assuming[x > 0 && y > 0, > FullSimplify[ > TrigToExp[ArcCos[x/Sqrt[x^2 + y^2]] - (Pi/2 - ArcTan[x/Abs[y]])]]] (1/2)*I*(Log[1/(I*x + y)] + Log[I*x + y]) > Obviously Mathematica isn't making a guess to which leaf of the complex > Log your algebraic complex travel history brought even if you arrived on > the positive real line. What I want is to simplify ArcCos[x/Sqrt[x^2 + y^2]] as Pi/2 - ArcTan[x/Abs[y]]. The two expressions are the same for any real x and y (maybe also true for complex x and y?). The use of such manipulation is useful when I want to convert the expression to CForm and eventually integrated in a C++ program, as the second expression is less likely to under or overflow than the first one. But I just don't find a way to do such manipulation in Mathematica. The more general question is how to manipulate an arithmetic expression to make it less likely to over or underflow. Another example is that the abs of a complex number is never computed as sqrt(x*x+y*y), it is always computed as |x| *sqrt(1+std::pow(x/y,2)) if |x| >= |y|. or |y| *sqrt(1+std::pow(y/x,2)) if |x| < |y|. I doubt Mathematica could solve this problem in general. But can I make it solve at least those particular cases? === Subject: How can I get list form like this As we know f/@Range[n] gives a list {f[1],f[2],f[3],=85=85}; Here what I want is a list form like this{f[[1]],f[[2]],=85=85}(*It's too tedious to type one by one*) It means the 1st,2nd,=85=85part of list f. Looks similar to each other, however I can't figure it out. How to do this ? === Subject: Re: Real and Complex Roots presented in a single plot > z[x_] = 1.3 Sin[1.7* x] + 0.6 Sin[4* x] ; > Plot[z[x], {x, 0, 18}] In the above plot we can see all the real roots of z very > approximately at {2.1,3.4, 5.5, 7.6, 9, 11, 13.1, 14.5, 16.5}, as the > curve crosses x-axis. We can also recognize and see the real parts of all the > complex roots of z where the curve is nearest to x -axis at {1.4, > 4.2, 6.6, 9.6, 12.3, 15.3, 17.7}. They are near to x-values where the > local maxima/minima occur.But we cannot 'see' their complex parts, as > they need to be computed. After computation of all real and complex roots I would like to > represent all roots, real and complex roots alike, on an ( x - y ) > Argand diagram (in a 2D plot) with the real part on x - axis and > imaginary part on y - axis. How to do this ? Narasimham This computes all roots with nonnegative imaginary part. You can flip across the real axis to get the conjugates. ListPlot[ Map[{Re[#], Im[#]} &, x /. N[{ToRules[ Reduce[{z[x] == 0, -100 <= Re[x] <= 100, 0 <= Im[x] <= 1/2}, x]]}]]] I capped the imaginary part at 1/2 because FindInstance tells me I can do so. In[27]:= FindInstance[{z[x] == 0, Im[x] > 1/2}, x] Out[27]= {} You can also get a hint of this from plotting the absolute value, letting the imaginary part of x go from 0 to 1. Plot3D[Abs[z[x + I*y]], {x, 0, 18}, {y, 0, 1}] Daniel Lichtblau Wolfram Research === Subject: Re: Real and Complex Roots presented in a single plot > z[x_] = 1.3 Sin[1.7* x] + 0.6 Sin[4* x] ; > Plot[z[x], {x, 0, 18}] In the above plot we can see all the real roots of z very > approximately at {2.1,3.4, 5.5, 7.6, 9, 11, 13.1, 14.5, 16.5}, as the > curve crosses x-axis. We can also recognize and see the real parts of all the > complex roots of z where the curve is nearest to x -axis at {1.4, > 4.2, 6.6, 9.6, 12.3, 15.3, 17.7}. They are near to x-values where the > local maxima/minima occur.But we cannot 'see' their complex parts, as > they need to be computed. After computation of all real and complex roots I would like to > represent all roots, real and complex roots alike, on an ( x - y ) > Argand diagram (in a 2D plot) with the real part on x - axis and > imaginary part on y - axis. How to do this ? Assuming I have correctly understood what you are looking for, visualizing real as well as complex roots can be achieve by formatting the list of solutions as a list of pairs {Re, Im} that can be displayed with ListPlot. For instance, z[x_] = 1.3 Sin[1.7*x] + 0.6 Sin[4*x] g = Plot[z[x], {x, 0, 18}] sols = Reduce[Rationalize[z[x], 0] == 0, x] // N Select[List[sols /. C[1] -> 0 // ToRules], (0 <= Re[x /. #] <= 18 &)] pts = Through[{Re, Im}[x /. %]] // Transpose p = ListPlot[pts, PlotStyle -> {Orange, PointSize[Large]}] Show[g, p] HTH, -- Jean-Marc === Subject: Re: Real and Complex Roots presented in a single plot Here's some suggestions. First, since the roots of the function include complex numbers, it's better to call the input variable z, and then use a different name for the function. Second, you might have better luck if you rationalize the coefficients, solve exactly, then take the numerical values of the exact roots found. That way you won't have to do lots of FindRoot instances. Third, it seems to me that plotting the real and imaginary parts of each root separately, on the x- and y-axes, respectively, will not be very enlightening unless all you're interested in is those real and imaginary parts: you won't be able to tell which real part goes with which imaginary part. So I also suggest that you plot the complex points themselves. If you want, in addition, to plot the real and imaginary parts of each on an axis, that should now be simple to add to the plot. Here's one way I would do it; undoubtedly there's a simpler way than I show to obtain the {x,y} pairs from the real and non-real roots. f[z_] := (13/10) Sin[(17/10) z] + (6/10) Sin[4 z] (* find the roots *) roots = z /. Solve[f[z] == 0, z] nRoots = N /@ roots (* massage roots to get {x,y} coordinates for each *) coords = nRoots /. Complex[a_, b_] -> {a, b} xyPairs = Replace[coords, x_?(Length[#] == 0 &) -> {x, 0}, 1] (* plot the roots *) ListPlot[xyPairs, PlotStyle -> PointSize[Medium]] With David Park's Presentation package, this is MUCH easier, as no massaging of the roots is needed: << Presentations`Master` Draw2D[{PointSize[Medium], ComplexPoint /@ roots}, Axes -> True, AspectRatio -> 0.5] > z[x_] = 1.3 Sin[1.7* x] + 0.6 Sin[4* x] ; > Plot[z[x], {x, 0, 18}] In the above plot we can see all the real roots of z very > approximately at {2.1,3.4, 5.5, 7.6, 9, 11, 13.1, 14.5, 16.5}, as the > curve crosses x-axis. We can also recognize and see the real parts of all the > complex roots of z where the curve is nearest to x -axis at {1.4, > 4.2, 6.6, 9.6, 12.3, 15.3, 17.7}. They are near to x-values where the > local maxima/minima occur.But we cannot 'see' their complex parts, as > they need to be computed. After computation of all real and complex roots I would like to > represent all roots, real and complex roots alike, on an ( x - y ) > Argand diagram (in a 2D plot) with the real part on x - axis and > imaginary part on y - axis. How to do this ? Narasimham === Subject: Re: Real and Complex Roots presented in a single plot Narasimham, I'll do your problem using the Presentations package. Needs[Presentations`Master`] First I find your notation just a little jarring. z is usually used for the complex variable, x is usually the real part of a complex number and I'll use f for the function. f[z_] := 1.3 Sin[1.7 z] + 0.6 Sin[4 z] The easiest way to see the zeros is to make a contour plot of the modulus of the function. All the zeros occur in a narrow strip about the x-axis. With[ {zmin = -I, zmax = 18 + I, contourlist = {0, 0.1, 0.2, .5, 1, 2}}, Draw2D[ {ComplexCartesianContour[f[z], {z, zmin, zmax}, Abs, Contours -> contourlist, ColorFunctionScaling -> False, ColorFunction -> (ContourColors[contourlist, ColorData[SolarColors]][#] &), PlotPoints -> {50, 15}, MaxRecursion -> 3, PlotRange -> {0, 4}]}, AspectRatio -> Automatic, Frame -> True, FrameLabel -> {Re, Im}, RotateLabel -> False, PlotLabel -> Row[{Modulus of , f[z]}], ImageSize -> 900] ] The following gives a closeup of the region that contains the first seven zeros. With[ {zmin = -.5 - I, zmax = 5 + I, contourlist = {0, 0.1, 0.2, .5, 1, 2, 10}}, Draw2D[ {ComplexCartesianContour[f[z], {z, zmin, zmax}, Abs, Contours -> contourlist, ColorFunctionScaling -> False, ColorFunction -> (ContourColors[contourlist, ColorData[SolarColors]][#] &), PlotPoints -> {30, 15}, MaxRecursion -> 3, PlotRange -> {0, 11}]}, AspectRatio -> Automatic, PlotRange -> {{-.5, 5}, {-1, 1}}, Frame -> True, FrameLabel -> {Re, Im}, RotateLabel -> False, PlotLabel -> Row[{Modulus of , f[z]}], ImageSize -> 900] ] I might be happy with those diagrams, but if we want numerical root values we can click the root locations off the diagram, using DrawingTools, and then refine the values with FindRoot. roots = z /. FindRoot[ f[z], {z, ToComplex@#}] & /@ {{-0.005546, -0.02102}, {1.285, 0.3771}, {1.279, -0.3678}, {2.081, 0.01109}, {3.462, -0.008173}, {4.194, -0.3485}, {4.213, 0.345}} // Chop {0, 1.27946+ 0.374308 I, 1.27946- 0.374308 I, 2.09155, 3.43663, 4.19786- 0.355809 I, 4.19786+ 0.355809 I} The following then plots just the roots in the Argand plane. Draw2D[ {ComplexCirclePoint[#, 3, Black, Red] & /@ roots}, AspectRatio -> Automatic, PlotRange -> {{-.5, 5}, {-1, 1}}, Frame -> True, FrameLabel -> {Re, Im}, RotateLabel -> False, PlotLabel -> Row[{Some Roots of , f[z]}], BaseStyle -> {FontSize -> 12}, ImageSize -> 900] -- David Park djmpark@comcast.net http://home.comcast.net/~djmpark/ > z[x_] = 1.3 Sin[1.7* x] + 0.6 Sin[4* x] ; > Plot[z[x], {x, 0, 18}] In the above plot we can see all the real roots of z very > approximately at {2.1,3.4, 5.5, 7.6, 9, 11, 13.1, 14.5, 16.5}, as the > curve crosses x-axis. We can also recognize and see the real parts of all the > complex roots of z where the curve is nearest to x -axis at {1.4, > 4.2, 6.6, 9.6, 12.3, 15.3, 17.7}. They are near to x-values where the > local maxima/minima occur.But we cannot 'see' their complex parts, as > they need to be computed. After computation of all real and complex roots I would like to > represent all roots, real and complex roots alike, on an ( x - y ) > Argand diagram (in a 2D plot) with the real part on x - axis and > imaginary part on y - axis. How to do this ? Narasimham > === Subject: Re: How I can fit data with a parametric equation? > Hello everybody, > I have a list of {x,y} data and from a theoretical calculation I know > that the data should be fitted by an equation, for example: y=ax+bx+c; where a, b, c are constant parameters. The question is how i can find the best fit for my data finding the > value of this parameters. Do mathematica has a specific function for that? Try searching the documentation for fitting ... the first result, Fit, does exactly what you describe here. === Subject: Re: How I can fit data with a parametric equation? > I have a list of {x,y} data and from a theoretical calculation I know > that the data should be fitted by an equation, for example: y=ax+bx+c; where a, b, c are constant parameters. The question is how i can find the best fit for my data finding the > value of this parameters. Do mathematica has a specific function for that? http://reference.wolfram.com/mathematica/ref/FindFit.html http://reference.wolfram.com/mathematica/guide/CurveFittingAndApproximateFun ctions.html -- Jean-Marc === Subject: Re: How I can fit data with a parametric equation? Fit fits well! !(Fit[yourlist, {1, x, x^2}, x]) === Subject: Re: How I can fit data with a parametric equation? Look up Fit and FindFit. Hello everybody, I have a list of {x,y} data and from a theoretical calculation I know that the data should be fitted by an equation, for example: y=ax+bx+c; where a, b, c are constant parameters. The question is how i can find the best fit for my data finding the value of this parameters. Do mathematica has a specific function for that? === Subject: Re: How to change the FrontEnd Encoding? > I'm using Mathematica 6 on Linux and I would want to know if there is > the net and only found a way to change the Kernel encoding. Most modern linux distributions use UTF-8 as its base encoding but the > FrontEnd is fixed in ISO8859-1. Therefore I'm having some problems > when I open files in directories that have characters with tildes or > something, like /home/f=F3=F3. In the open dialog the FrontEnd does recognize these directories. But > the window title is not displayed correctly and if I try to save the > file with Save as, the open dialog throws me a lot of errors saying > Could not read directory /home/f~~ I am not hundred percent sure that this is what you are looking for and if the option is available on Linux (I use Mac OS X), but open the option inspector (menu Format -> Option Inspector) and under Global Preferences, look for Formatting Options -> Font Options -> CharacterEncoding. In the drop down menu select the encoding scheme you wish. === Subject: Re: Computing Euler Rotation angles(?) If you have Mathematica it has a command, RotationTransform, that will provide the transformation function to rotate one vector into the direction of a second vector, which is what I think you are looking for. If you are looking for Euler angles the Mathematica Presentations package from my web site has an EulerAngles routine that will calculate both sets of Euler angles for all possible xyz rotation sequences. There is also a routine that will extract the rotation direction and rotation angle from any 3D rotation matrix. I'm sure there are many other pieces of software that will do the same things. http://home.comcast.net/~djmpark/ > Hello - I would like to compute the Euler rotation angles (rotX, rotY, rotZ) > required to rotate one vector A to another B. Actually what I'm trying > to do is to aim a newly created directional light -- that points in > the direction A = (0,0,-1) by default -- in the same direction as the > camera view direction B. === Subject: Re: HELP: Compounding equation that sum all the FUTURE VALUES. > Can you convert this to an excel formula where NP is dynamic? :) I'm sure that Excel has a built-in function to do what you want... It's the accumulated value of an annuity, isn't it? Check out FV or FVAL.. === Subject: Spiral problem The following appeared on one of my relative's math tests which was left on the kitchen table. There are 25 multiple choice (A,B,C,D or E) problems. The problems are supposed to be quickly solvable by a 17 year old and I can do most of them. To show the level, one of the questions says the reciprocal of 3/10 is 1+1/x what is x? I can do that in a few seconds. But the following problem is giving me brain ache. The question shows a grid of numbers arranged in a spiral with 1 in the centre square. 2 is to the right of 1. 3 is above 2. 4 is above 1. 5 is above 6. 6 is to the left of 1. 7 is below 6. 8 is below 1. 9 is below 2. 10 is to the right of 9. 11 is above 10. Numbers up to 17 are shown with a down arrow showing how to continue. And here is the question. What is the sum of the number that appears directly above 2007 and the number that appears directly below 2007. A) 4014 B) 4016 C) 4018 D) 4020 E) 4022 Maybe I'm missing the obvious but I can't see how to be sure of the answer quickly. If 2007 is in a vertical column with n+1 and n-1 above and below then the answer is A but is that a correct assumption? If I'd been doing the test I'd have answered A and moved on. === Subject: Re: Spiral problem >The following appeared on one of my relative's math tests which was left on >the kitchen table. There are 25 multiple choice (A,B,C,D or E) problems. >The problems are supposed to be quickly solvable by a 17 year old and I can >do most of them. To show the level, one of the questions says the reciprocal >of 3/10 is 1+1/x what is x? I can do that in a few seconds. >But the following problem is giving me brain ache. >The question shows a grid of numbers arranged in a spiral with 1 in the >centre square. 2 is to the right of 1. 3 is above 2. 4 is above 1. 5 is >above 6. 6 is to the left of 1. 7 is below 6. 8 is below 1. 9 is below 2. 10 >is to the right of 9. 11 is above 10. Numbers up to 17 are shown with a down >arrow showing how to continue. >And here is the question. What is the sum of the number that appears >directly above 2007 and the number that appears directly below 2007. >A) 4014 ÊB) 4016 ÊC) 4018 ÊD) 4020 ÊE) 4022 >Maybe I'm missing the obvious but I can't see how to be sure of the answer >quickly. >If 2007 is in a vertical column with n+1 and n-1 above and below then the >answer is A but is that a correct assumption? If I'd been doing the test I'd >have answered A and moved on. I get E, arguing that 2007 = 45^2 - 18, therefore the numbers > directly above and below it are 43^2 - 17 and 47^2 - 19, with > sum: (45 - 2)^2 + (45 + 2)^2 - 36 Why didn't you just add 43^2 - 17 and 47^2 - 19? > = 2*(45^2) + 2*(2^2) - 36 = > 4050 - 28 = 4022. To the OP: in such a spiral, the even squares form the upper left corners of the spiral rings and the odd squares form the lower right corners of the spiral ring. 2007 falls between 44^2 (1936) and 45^2 (2025), 1936 | | | | |---------------2025 and 2007 is more than halfway between 1936 and 2025, so it's past the lower left corner, so it's on a horizontal part of the spiral ring. 1936 | | | | |----2007----------2025 which means the numbers above and below it are on different rings, eliminating (n+1 plus n-1) as a solution. Where Angus got his numbers is from realizing that the numbers kitty-corner to 45^2 are 43^2 and 47^2 1936 | | | | 1849 |----2007----------2025 2209 with offsets 1 less and 1 greater, respectively. 1936 | | | | 1832------1849 |----2007----------2025 2190--------------2209 1832 + 2190 = 4022 -- > Angus Rodgers > Contains mild peril === Subject: Re: More with Quadratic Diophantine Theorem posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) > i think you would be surprised at gauss' elementariness Yeah, hero worship is fun an easy, especially with a dead guy. He's > safe, right? I think you would be surprised if he were alive to consider what you > and your society do with current mathematical results of an elementary > nature, like in my original two posts in this thread. Do you really think Gauss would accept you as part of math society? For what? Because you'll say nice things about him? Do you think Gauss would like you or something? gauss was a paranoid asshole who screwed his kids over and spent so much effort inflating himself crying me me me that many of his contemporaries disliked him that said he was a damn fine mathematician no hero though i figured that if you spent some time reading him you too could be that paranoid asshole inflating yourself with me me me and finally be a fair mathematician to boot -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- === Subject: Re: More with Quadratic Diophantine Theorem > i think you would be surprised at gauss' elementariness > Yeah, hero worship is fun an easy, especially with a dead guy. ÊHe's > safe, right? > I think you would be surprised if he were alive to consider what you > and your society do with current mathematical results of an elementary > nature, like in my original two posts in this thread. > Do you really think Gauss would accept you as part of math society? > For what? ÊBecause you'll say nice things about him? > Do you think Gauss would like you or something? gauss was a paranoid asshole > Ê who screwed his kids over > Ê and spent so much effort inflating himself > Ê Ê crying me me me > that many of his contemporaries disliked him that said > Ê he was a damn fine mathematician no hero though i figured that if you spent some time reading him > Ê you too could be that paranoid asshole > Ê inflating yourself with me me me > and finally be a fair mathematician to boot I'm contemplating this awesome result linking all quadratic Diophantine equations in 2 variables and wondering how far it goes. You're mumbling. History will remember me. But what about you? Oh, you don't care, right? Hundreds of years from now, some people may be wondering about who I am and what kind of person I was, where one may hero worship, and another may mouth off about my many failings. But who will even know about you? James Harris === Subject: Re: More with Quadratic Diophantine Theorem [...] > > I'm contemplating this awesome result linking all quadratic > Diophantine equations in 2 variables and wondering how far it goes. There's the Perfect cuboid problem which no one has solved. Any Perfect cuboid (if there is one) is an Euler brick. http://mathworld.wolfram.com/PerfectCuboid.html I wonder if a perfect cuboid exists... What are the chances? David Bernier > You're mumbling. > > History will remember me. But what about you? > > Oh, you don't care, right? > > Hundreds of years from now, some people may be wondering about who I > am and what kind of person I was, where one may hero worship, and > another may mouth off about my many failings. > > But who will even know about you? > > > James Harris === Subject: Re: More with Quadratic Diophantine Theorem posting-account=Yn5cwwoAAADntcMuRwk-EwLg-DMZ_hXN Gecko/20070509 Camino/1.5,gzip(gfe),gzip(gfe) >(for an example of what i came to dislike >http://r6.ca/blog/20051210T202900Z.html >) How Dedekind Screwed Up a Hundred Years of Mathematics [...] My friend suggested that for most of the 3600 year history > of mathematics, mathematics was done constructively. I've been suspecting something of the kind, and am hoping to sneak > up on the philosophical problem of constructivism via the history > of mathematics (as my mind remains blank on the philosophy). [...] But Dedekind wasnÕt happy with the equivalence relations > in the theory and he created the set-theoretic formulation that > we suffer with today. I hear the sound of an axe being ground, and ... [...] Once again, Dedekind wasnÕt happy with this, and in 1872, he > proposed his own notion of real numbers based on set-theory by > Dedekind cuts of the rational numbers. A bit of sneaky rhetoric there. I doubt that the objective issues > are reducible to Dedekind's subjective happiness or unhappiness, > although undoubtedly his subjectivity would have given him (and, > via his writings, others) a handle on the objective issues. But > I must reserve judgement until I understand those issues better. Also, there /was/ no set theory on which to base the theory of cuts > (or, for that matter, ideals); rather, surely it was these convincing > applications of a still non-existent theory which helped to bring the > theory into being? Whether Dedekind himself would recognise ZFC, for > example, as his child is another question. of course you are right about there being no set theory and the overreaching imagination of his desires and the axe-to-grind style and i don't agree with the idea that it was somehow dedekind's fault mathematics was screwed up for a hundred years (though i think it was screwed up) it's a mentality that gets annoyed with questions of fundaments and moves to belittlement or a mentality that seeks certainty but doesn't want to obvious mythology so substitutes other unobservables like bivalence or ... there's a hundred and one reasons why it was screwed up and dedekind was just one of the authorities that made authority appeal easier i linked to the text mostly to point to the issues that dedekind created and didn't intend all of the rhetoric (just some of it :) ) [as an interesting aside dedekind's construction of the reals is not constructively equivalent to cauchy's i actually suspect the initial acceptance of equivalence may have helped confuse constructive foundations enough to delay it beyond the classical formalisation] -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- === Subject: Re: More with Quadratic Diophantine Theorem just because the ideals were not made to solve Fermat's Last Theorem, they don't work?... there's a good case-book that uses the golden section as the exemplar. I agree, though, it is best to avoid them til you might need them; I certainly don't need'em. > [as an interesting aside > Ê Êdedekind's construction of the reals > Ê Êis not constructively equivalent to cauchy's Êi actually suspect the initial acceptance of equivalence > Êmay have helped confuse constructive foundations > Ê Êenough to delay it beyond the classical formalisation] thus: an aether simply provides a prealistical model of the phenomenon of waves; one need not deal with photons at all, since they are just formally dual to waves, not at all pardoxical, as shown by Dirac. far too much of theory rests upon Einstein, saying that's impossible, which is the reality behind Michelson-Morley and those who carried the examination further -- but, herr doktor-professor Albert said, Nein!, when he was once in his dysused office at Caltech. let us remember him for those things which he was really good at, not the Big Bang Constant and its infernal klingons (fridge magnets) ?! > I don't need any aether to stick a magnet to a fridge, light up an thus: so, to ask an ignorant question of personal interest, how much of this could be used toward a new or old proof of quadratic reciprocity? > That Ac and -Dc must be quadratic residues, mod D and A, respectively, > is a trivial result: > Let Ax^2 - Dy^2 = c. [1] > Take this mod A: > -Dy^2 = c mod A > (Dy)^2 = -Dc mod A > Therefore, -Dc is a quadratic residue mod A. > Similarly, taking [1] mod D: > Ax^2 = c mod D > (Ax)^2 = Ac mod D > Therefore, Ac is a quadratic residue mod D. thus: ah, sort of a solopsistic monad. I am reminded of hte Many Words Interpretation of the Copenhagenskool: to be or not to be that is the shape of the box or the bubble -- inside or out? > Q. What is a tautological space? > A. A tautological space. thus: speaking of Young's anhialation of Newton's photons, here is the earlier elaboration on light by Fermat: http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf > as we know from Newton's iron-poor corpuscles. --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U. Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com === Subject: Re: More with Quadratic Diophantine Theorem days. My association with the Department is that of an alumnus. [...] >What I don't see is how a proof of non-existence fits into the >constructive/non-constructive classification. For instance, consider >the classic proof that sqrt(2) is irrational: Assume it is rational, say a/b, with at least one of a, b odd. 2 = a^2/b^2 > 2 b^2 = a^2 > Therefore, a must be even, say a = 2r > 2 b^2 = 4 r^2 > b^2 = 2 r^2 > Therefore, b must be even. > This is a contradiction, therefore sqrt(2) is irrational. Would a constructionist allow that proof? For most meanings of constructionist (they may vary a bit), the implication is acceptable. It is the implication not(for all x (P(x))) ==> exists x (not(P(x))) that is considered unacceptable in the absence of a construction of x. (Constructionists also reject not(not(P)) --> P, so you also have to be a bit careful there). -- === Subject: Re: What binary operation lambda quantifier corresponds to? I think Tegiri Nenashi is on to something here. > Observation: all quantuifiers correspond to some binary algebraic > operation. I would say instead all binding constructs. Lambda and limit are not quantifiers, but they are binding constructs. > Consider: > 1. Sigma summation and integral are iterative forms of binary plus. > 2. Pi-capital product is iterative form of multiplication. > 3. Lattice supremum is iterative form binary meet. > 5. Lattice infinum is iterative form binary join. > 6. Existential and universal quatifiers are lattice supremum and > infinum. > Now, there must be a binary operation that lambda is iterative form as > well? I agree. Let {e1 -> e2} be the partial function that does nothing except mapping e1 to e2. If f1 and f2 are partial functions with disjoint domains, let f1 /+/ f2 be the union of the partial functions; i.e. the partial function whose graph is the union of the graph of f1 and the graph of f2. Then lambda x . e is indeed {e1 -> e[x/e1]} /+/ {e2 -> e[x/e2]} /+/ {e3 -> e[x/e3]} /+/ ... for some enumeration e1, e2, e3, ... of all the closed terms. I also think Tegiri was on the mark with the comment about limit and the second projection function. Let (a /$/ b) be b, and let /$/ be left-associative. Then limit (x -> infty) e is indeed (in some sense that can probably be made more precise) e[x/0] /$/ e[x/1] /$/ e[x/2] /$/ ... assuming we are talking about x as an integer. If x is a real number, then there is probably some interesting generalization. --Jamie. (efil4dreN) andrews .uwo } Merge these two lines to obtain my e-mail address. @csd .ca } (Unsolicited bulk e-mail costs everyone.) === Subject: Re: What binary operation lambda quantifier corresponds to? [F'up again to clf] >> Now, there must be a binary operation that lambda is iterative form as >> well? > I agree. Let {e1 -> e2} be the partial function that does > nothing except mapping e1 to e2. If f1 and f2 are partial > functions with disjoint domains, let f1 /+/ f2 be the union of > the partial functions; i.e. the partial function whose graph is > the union of the graph of f1 and the graph of f2. Then > > lambda x . e > > is indeed > > {e1 -> e[x/e1]} /+/ {e2 -> e[x/e2]} /+/ {e3 -> e[x/e3]} /+/ ... > > for some enumeration e1, e2, e3, ... of all the closed terms. You have only said that a function is the union of all single-point mappings. That's pretty trivial, and such a scheme will work for every construct that's binding a variable. And it's quite different from what happens with (finite) sums, products, and lattice operations. These are all non-trivial folds. If I'm inventive enough, I can of course find some binary operation everywhere. For example, every natural number is connected to the binary operation (+), because I can e.g. write 3 = 1 + 1 + 1, or 5 = 1 + 1 + 1 + 1 + 1. It's also connected to the binary operation (*), because it can be written as a product of prime powers. Nevertheless, we normally think of numbers as a whole. As we do of functions. > I also think Tegiri was on the mark with the comment about > limit and the second projection function. Let (a /$/ b) be b, > and let /$/ be left-associative. Then > > limit (x -> infty) e > > is indeed (in some sense that can probably be made more precise) > > e[x/0] /$/ e[x/1] /$/ e[x/2] /$/ ... > > assuming we are talking about x as an integer. The trick is of course that you have to made it more precise, because then you see that it doesn't work. The limit is *not* somehow the last element of a sequence. Not even for integers. Sorry. The whole point of the limit is that you can make some statement about what infinite sequence is approximating. So you need at least some notion of proximity. Which you won't get with this operation. And dealing with the infinite can be subtle, so you *must* make the step going from finite to infinite explicit. Just writing ... is not enough. - Dirk === Subject: Formula for quarter of a circle/or half of a parabola midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter in the first quadrant. Alternatively, I also attempting to derive the formula for a half a parabola positioned in the first quadrant with a y-intercept of 5 and an x-intercept of 5. I don't have a particular focal point in mind. Keith === Subject: Re: Formula for quarter of a circle/or half of a parabola > midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter > in the first quadrant. y = sqrt( 25 - sqrt(x)^4 ) > Alternatively, I also attempting to derive the formula for a half a > parabola positioned in the first quadrant with a y-intercept of 5 and > an x-intercept of 5. I don't have a particular focal point in mind. y = sqrt( 5 - sqrt(x)^4 / 5 )^2 Dave === Subject: Re: Formula for quarter of a circle/or half of a parabola > midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter > in the first quadrant. y = sqrt( 25 - sqrt(x)^4 ) ******** What's with sqrt(x)^4 instead of x^2 , and y = 25 - x^2 ??? > Alternatively, I also attempting to derive the formula for a half a > parabola positioned in the first quadrant with a y-intercept of 5 and > an x-intercept of 5. I don't have a particular focal point in mind. y = sqrt( 5 - sqrt(x)^4 / 5 )^2 ********* Ahh, of course, there is only one solution in this case as well, because all parabolas look the same except for scale, as do circles. Only one free variable. But why now the redundant squaring and square root, instead of simply y= 5 - x^2 /5 === Subject: Re: Formula for quarter of a circle/or half of a parabola > midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter > in the first quadrant. y = sqrt( 25 - sqrt(x)^4 ) ******** What's with sqrt(x)^4 instead of x^2 , and y = 25 - x^2 Ê ??? Because 25 - x^2 is defined for -5 <= x <= 5, so it gives the top half of the circle, but sqrt(x)^4 is defined only for 0 <= x <= 5, giving only the top half of the circle. With y = 25 - x^2, you have to add the domain restriction 0 <= x <= 5, but that is implicit in my definition. > Alternatively, I also attempting to derive the formula for a half a > parabola positioned in the first quadrant with a y-intercept of 5 and > an x-intercept of 5. I don't have a particular focal point in mind. y = sqrt( 5 - sqrt(x)^4 / 5 )^2 ********* Ahh, of course, there is only one solution in this case as well, > because all parabolas look the same except for scale, as do circles. Only > one free variable. But why now the redundant squaring and square root, > instead of simply y= 5 - x^2 /5 The inner square root makes sure that x >= 0, and the outer one makes sure that x <= 5. Again, it avoids domain restrictions, because the function is defined and real only in the first quadrant. Dave === Subject: Re: Formula for quarter of a circle/or half of a parabola > midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter > in the first quadrant. Alternatively, I also attempting to derive the formula for a half a > parabola positioned in the first quadrant with a y-intercept of 5 and > an x-intercept of 5. I don't have a particular focal point in mind. For the quarter circle, how about: y = Sqrt[25 - x^2] where {x, 0, 5} === Subject: Re: Formula for quarter of a circle/or half of a parabola > midpoint 0,0 and radius 5. Specifically, I'm looking for the quarter > in the first quadrant. y = sqrt(25-x^2) for 0<=x<=5 Alternatively, I also attempting to derive the formula for a half a > parabola positioned in the first quadrant with a y-intercept of 5 and > an x-intercept of 5. I don't have a particular focal point in mind. There are a lot of possibilities. Would y = 5 - x^2/5 for for 0<=x<=5 do? > Keith === Subject: Re: Set Theory Notation Question >I am translating a soma puzzle solution into laymans terms and am >not familiar with all of the notation used here: >http://www.cs.dartmouth.edu/~mckeeman/references/soma/sld013.htm Which notation are you unfamiliar with? As far as I can tell, it is > all reasonably standard. The first line says, given u, and given s a subset of u, a Gamma, and > a P 0. The second line says let P be the union of all Gamma(p), with p > running over all elements of P 0 (that is, P consists exactly of all > elements x that lie in some Gamma(p), with p in P 0). The third line says and if we let R be the set that conists of all V > which are: > Ê Ê (i) subsets of P; > Ê Ê(ii) the union of V is an element of Gamma(s) > Ê Ê Ê Ê [that is, take all elements y of V, and then consider the > Ê Ê Ê Ê collection of all Êx that are in at least one of those y; this > Ê Ê Ê Ê should be an element of Gamma(s), where s is the specified > Ê Ê Ê Ê subset of u given in the first line]; and > Ê (iii) the union of V is disjoint from V itself (that is, there is no > Ê Ê Ê Ê element y of V which is also an element of an element of V). And the final line says that S (the set of solutions) consists > exactly of those Gamma(V) for which V is an element of the R defined > in the third line. If the problem is the meaning of the Gamma(p) and so on, then the > question is not about set theory notation, but about the meanings > given to the symbols in the context of space fillings. and I guess I had missed the subset notation notes. === Subject: Re: Set Theory Notation Question days. My association with the Department is that of an alumnus. [...] >and I guess I had missed the subset notation notes. Do yourself two favors: First, get yourself an actual book on set theory that will introduce you to the necessary particulars. My recommendation would be Paul Halmos's Naive Set Theory. Second: realize taht wikipedia may (emphasis on but it may not) be a good place to start, but it should never be the place to end your investigation. It is not always reliable, and it is always incomplete. I would not trust an encylopedia which underwent real, serious vetting as the final word, even if just published. Why give that kind of final word to wikipedia? -- === Subject: Re: Set Theory Notation Question >and I guess I had missed the subset notation notes. Do yourself two favors: First, get yourself an actual book on set theory that will introduce > you to the necessary particulars. My recommendation would be Paul > Halmos's Naive Set Theory. Second: realize taht wikipedia may (emphasis on but it may not) be a > good place to start, but it should never be the place to end your > investigation. It is not always reliable, and it is always > incomplete. I would not trust an encylopedia which underwent real, > serious vetting as the final word, even if just published. Why give > that kind of final word to wikipedia? === Subject: Re: traceless matrix unitarily similar to matrix with 0 diagonals? > Yes, there is a 2x2 matrix with trace 0 but the two columns are not > orthogonal. For example [ 1 1 ; 0 -1]. > A is any complex matrix. The way to solution is to find a unit vector > v with following property : v* A v = 0 > Rest of the problem is induction. > Where * is hermitian transpose. > But how do we guarantee existence of such v ? (Maybe I am late with this, but who is supposed to wade through unrelated > stuff?) Also, it may be a big tool for a small problem, but why not. The set of all (v* A v) over all unit vectors v is the numerical range, > which is known to contain all eigenvalues of A, and to be convex in the > complex plane. If zero is the trace of A then the sum of the eigenvalues of A is 0, and > so is their average, which is in the convex hull of these eigenvalues. > In short, 0 is in the numerical range of A. Hence the existence of a unit vector v such that (v* A v) = 0. > If A is a real matrix, and the unitary is required to be real (i.e. orthogonal), one has to look at the real parts of the eigenvalues when A is brought (by orthogonal similarity) to a real Schur form first, that is, block diagonal with at most 2-by-2 blocks corresponding to non-real complex conjugate pairs of eigenvalues. It's a little messier. Anyone with a shortcut? === Subject: Re: an interesting sum <10027060.1220971993506.JavaMail.jakarta@nitrogen.mathforum.org>, > Well, calculation of several hundred roots and elementary calculus give a > head-start. Really? Exactly how did you calculate that 4 is a root? I didn't check my work very carefully, but when I calculate S(4) I get approximately 0.0181922290846 . (Anyone else have a numeric value for S(4)?) > Using induction and derivatives we have peaks +/- at every prime. > That's sort of built into the sum I think. -- === Subject: Re: an interesting sum You are correct. I looked harder at large values because I thought the function was more likely go awry there. Some of those values are off on the order of 10^-20 but off they are. It's an interesting function. The part about a zero between successive primes (the part that one can prove using derivatives)is apparently true. And it mimics the conjectured property in a pretty audacious way. I did look at hundreds of zeros, just ones so close I took the difference to be computational error. === Subject: Re: an interesting sum > You are correct. I looked harder at large values because I thought the function was more likely go awry there. Some of those values are off on the order of 10^-20 but off they are. It's an interesting function. The part about a zero between successive primes (the part that Êone can prove using derivatives)is apparently true. And it mimics the conjectured property in a pretty audacious way. Ê I did look at hundreds of zeros, just ones so close I took the difference to be computational error. > Just a note that when you reply to posts and remove the original message - the readers have no clue to which message you are replying. You are welcome! ~Amzoti === Subject: Re: an interesting sum Well, before I throw this away, I did take the trouble to take derivatives and so on, and I have that S'(x)=0 when (x-prime[k])=0 for all k. Taken together with the sign changes, can we safely say the function has max/min at each prime and a zero between successive primes? > Just a note that when you reply to posts and remove > the original > message - the readers have no clue to which message > you are replying. > > You are welcome! > > ~Amzoti === Subject: Re: an interesting sum posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > <10027060.1220971993506.JavaMail.jaka...@nitrogen.mathforum.org>, > Well, calculation of several hundred roots and elementary calculus give a > head-start. Really? Exactly how did you calculate that 4 is a root? > I didn't check my work very carefully, but when I calculate > S(4) I get approximately 0.0181922290846 . (Anyone else have a numeric value for S(4)?) > Using induction and derivatives we have peaks +/- at every prime. > That's sort of built into the sum I think. -- > Here are some exact values - (you have correct magnitude - but missing negative sign). e = exponential k = 1 {s(1), s(2), s(3), s(4)} = {-1/e, -1, -1/e, -1/e^4} k = 2 {s(1), s(2), s(3), s(4)} = {1/e^4, 1/e, 1, 1/e} k = 3 {s(1), s(2), s(3), s(4)} = {-1/e^16, -1/e^9, -1/e^4, -1/e} k=4 {s(1), s(2), s(3), s(4)} = {1/e^36, 1/e^25, 1/e^16, 1/e^9} . . . k=100 {s(1), s(2), s(3), s(4)} = {1/e^291600, 1/e^290521, 1/e^289444, 1/ e^288369} HTH ~A === Subject: Re: an interesting sum >> Well, calculation of several hundred roots and elementary calculus give a >> head-start. >> Really? Exactly how did you calculate that 4 is a root? >> I didn't check my work very carefully, but when I calculate >> S(4) I get approximately 0.0181922290846 . >> (Anyone else have a numeric value for S(4)?) >> Using induction and derivatives we have peaks +/- at every prime. >> That's sort of built into the sum I think. >> -- >> >Here are some exact values - (you have correct magnitude - but missing >negative sign). >e = exponential >k = 1 >{s(1), s(2), s(3), s(4)} = {-1/e, -1, -1/e, -1/e^4} Huh? S(x) is a function of one variable - there's no > such thing as S(1) for k = 1 and S(1) for k = 2. Maybe your s(x) is not supposed to be the same as S(x); > if so I don't know what you mean by s(x). The definition > of S(x) was this: S(x) = Sum [(-1)^k * Exp[-(x-prime[k])^2], We're summing over k there. >k = 2 >{s(1), s(2), s(3), s(4)} = {1/e^4, 1/e, 1, 1/e} >k = 3 >{s(1), s(2), s(3), s(4)} = {-1/e^16, -1/e^9, -1/e^4, -1/e} >k=4 >{s(1), s(2), s(3), s(4)} = {1/e^36, 1/e^25, 1/e^16, 1/e^9} >. >. >. >k=100 >{s(1), s(2), s(3), s(4)} = {1/e^291600, 1/e^290521, 1/e^289444, 1/ >e^288369} >HTH ~A Understanding Godel isn't about following his formal proof. > That would make a mockery of everything Godel was up to. > (John Jones, My talk about Godel to the post-grads. > in sci.logic.) Can you please provide an algorithm for finding Prime[k] for k ranging from 1 to infinity over that sum? How did you find your value of S(4)? (Given that we got the same result for it.) === Subject: Re: an interesting sum > Can you please provide an algorithm for finding Prime[k] for k ranging > from 1 to infinity over that sum? If we had an algorithm for finding prime[k] for all k, we'd be a very happy bunch of people, here, believe me :-) I don't know how David did it, but I am guessing that he used an approximation of the original function with k running from 1 to some large value, say 1000 or something. > How did you find your value of S(4)? (Given that we got the same > result for it.) -- I.N. Galidakis === Subject: Re: an interesting sum If you look at Amzoti's calculations, as k increases the value of S(4) decreases quickly. At k=100 he's at 1/e^288369. k doesn't stop there, does it?? === Subject: Re: an interesting sum <10669042.1220999228176.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > If you look at Amzoti's calculations, as k increases the value of S(4) decreases quickly. At k=100 he's at 1/e^288369. k doesn't stop there, does it?? Certainly not - want more - here you go. k = 100000 {s(1), s(2), s(3), s(4)} = {1/e^1689240885264, 1/e^1689238285849, 1/ e^1689235686436, 1/e^1689233087025} . . . k = 1000000 {s(1), s(2), s(3), s(4)} = {1/e^239811921883044, 1/e^239811890911321, 1/e^239811859939600, 1/e^239811828967881} Enough? HTH ~A === Subject: Re: an interesting sum === Subject: Re: an interesting sum <10669042.1220999228176.JavaMail.jakarta@nitrogen.mathforum.org> posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX Gecko/2008070208 Firefox/3.0.1,gzip(gfe),gzip(gfe) > If you look at Amzoti's calculations, as k increases the value of S(4) decreases quickly. At k=100 he's at 1/e^288369. k doesn't stop there, does it?? Certainly not - want more - here you go. k = 100000 {s(1), s(2), s(3), s(4)} = {1/e^1689240885264, 1/e^1689238285849, 1/ > e^1689235686436, 1/e^1689233087025} . > . > . k = 1000000 {s(1), s(2), s(3), s(4)} = {1/e^239811921883044, 1/e^239811890911321, > 1/e^239811859939600, > Ê Ê 1/e^239811828967881} Enough? HTH ~A Okay - last one. k = 10^8 {s(1), s(2), s(3), s(4)} = {1/e^4153748653978366564, 1/ e^4153748649902217081, 1/e^4153748645826067600, 1/ e^4153748641749918121} === Subject: Re: minimum cycle length a^k mod n As examples, consider the minumum cycle length of 3^k mod 10^i, for 1 <= i <= 8; according to my computations, these are, respectively, 4, 20, 100, 500, 5000, 50000, 500000, 5000000. I don't see anything posted so far that explains all of these values. In the last five cases (i = 4, 5, 6, 7, 8) the minimum cycle length is 10^i/20 -- how would one determine that fact other than by brute force? Is the formula 10^i/20 valid for all higher moduli 10^i? Not coincidentally, this relates to some properties of power towers mentioned in the section Decimal digits of Graham's number at http://en.wikipedia.org/wiki/Graham's_number (and the discussion page). === Subject: Re: Longevity of Math Books (was Re: Quadratic Diophantine Theorem) <6996c4lha799djvtrktcfnld533gdkbnl0@4ax.com> posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) Apostol begins the course with integration, as per Liebniz; there's a special name for that, per Liebniz: primitive equation or prime equation? this was the subject ot the keynote speaker at the Ninth Annual Nonlinear Science Conference at UCLA, although I no longer blame Newton exclusively for the controversy with Liebniz. anyway, apparently, Hooke did not do the math in a formalistic way; after all, he was the Royal Society's experimenter. the other part of the joke is that the '99 2-pound coin is inscribed, On the shoulders of giants. on the edge; at least, Newton was effective in his office-of-reward. > stole the solution from Hooke, to boot -- on the shoulders of a giant, > who was physically a dwarf; hence, the joke. Wrong again, but excusably, since that meme has spread widely. Read the > shoulders bit in context, and you can see it was not an insult. thus: ah, Lord Berty's thing was called the Unity of Sciences; well, here's a reference: http://www.larouchepub.com/other/2002/2949moonification.html The two operations of Wells and Russell from which Moon sprung are these: The Moral Re-Armament Movement, founded at a 1921 meeting between a wacky Lutheran preacher from Philadelphia and two British delegates to the Washington Disarmament Conference, Lord Arthur Balfour and H.G. Wells. Moral Re-Armament became the mass organizational vehicle for implementation of Wells' 1928 call in The Open Conspiracy, for a worldwide movement for draft resistance. The environment of Moon's Korean ministry was under control of Moral Re-Armament when he was picked up as an intelligence asset during the Korean War. The Unity of the Sciences movement. Founded in 1935 under the supervision of Lord Bertrand Russell and John Dewey, it brought together Trotskyite academics Albert Wohlstetter (mentor of current Defense Policy Board Chairman Richard Perle), Sidney Hook, and Ernest Nagel, with members of the radical-positivist Vienna Circle. Merging with Robert M. Hutchins at the University of Chicago in the 1950s, this operation took over the teaching of science in the United States. Thomas Kuhn's widely read fraud, The Structure of Scientific Revolutions, was published as the second volume of their Encyclopedia of Unified Sciences. In 1972, the Moonies were given the Unity of Sciences franchise, sponsoring the first of their still-ongoing International Conferences of the Unity of Sciences. Their early sessions featured such notables as Manhattan Project physicist Eugene Wigner, the lifelong ally of that truly mad scientist Leo Szilard (the model for Dr. Strangelove, in Stanley Kubrick's film of that name), and environmental fascists Alexander King and Aurelio Peccei, founders of the no-growth Club of Rome. Before looking back to the history of these projects, let us first briefly dispense with the person of Rev. Sun Myung Moon. Moon as a personality is of very little importance, in himself. The real Reverend Moon is a pathetic, if nonetheless nasty, victim of Japanese internment and North Korean torture sessions. He is what the.... thus: an aether simply provides a prealistical model of the phenomenon of waves; one need not deal with photons at all, since they are just formally dual to waves, not at all pardoxical, as shown by Dirac. far too much of theory rests upon Einstein, saying that's impossible, which is the reality behind Michelson-Morley and those who carried the examination further -- but, herr doktor-professor Albert said, Nein!, when he was once in his dysused office at Caltech. let us remember him for those things which he was really good at, not the Big Bang Constant and its infernal klingons (fridge magnets) ?! > I don't need any aether to stick a magnet to a fridge, light up an thus: so, to ask an ignorant question of personal interest, how much of this could be used toward a new or old proof of quadratic reciprocity? > That Ac and -Dc must be quadratic residues, mod D and A, respectively, > is a trivial result: > Let Ax^2 - Dy^2 = c. [1] > Take this mod A: > -Dy^2 = c mod A > (Dy)^2 = -Dc mod A > Therefore, -Dc is a quadratic residue mod A. > Similarly, taking [1] mod D: > Ax^2 = c mod D > (Ax)^2 = Ac mod D > Therefore, Ac is a quadratic residue mod D. thus: ah, sort of a solopsistic monad. I am reminded of hte Many Words Interpretation of the Copenhagenskool: to be or not to be that is the shape of the box or the bubble -- inside or out? > Q. What is a tautological space? > A. A tautological space. thus: speaking of Young's anhialation of Newton's photons, here is the earlier elaboration on light by Fermat: http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf > as we know from Newton's iron-poor corpuscles. --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U. Press! http://larouchepub.com/pr/2008/080813moloch_brown.html http://wlym.com === so, to ask an ignorant question of personal interest, how much of this could be used toward a new or old proof of quadratic reciprocity? > That Ac and -Dc must be quadratic residues, mod D and A, respectively, > is a trivial result: > Let Ax^2 - Dy^2 = c. [1] > Take this mod A: > -Dy^2 = c mod A > (Dy)^2 = -Dc mod A > Therefore, -Dc is a quadratic residue mod A. > Similarly, taking [1] mod D: > Ax^2 = c mod D > (Ax)^2 = Ac mod D > Therefore, Ac is a quadratic residue mod D. thus: ah, sort of a solopsistic monad. I am reminded of hte Many Words Interpretation of the Copenhagenskool: to be or not to be that is the shape of the box or the bubble -- inside or out? > Q. What is a tautological space? > A. A tautological space. thus: speaking of Young's anhialation of Newton's photons, here is the earlier elaboration on light by Fermat: http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf > as we know from Newton's iron-poor corpuscles. --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U. Press! http://larouchepub.com/pr/2008/080813moloch_brown.html http://wlym.com === Subject: Why does MySpace condone child rape and pedophiles? They keep withholding vital evidence from the District Attorneys and continue to let sexual offenders pour into and operate on Myspace. http://www.youtube.com/watch?v=8PV9FrNLjto === Subject: probability question I have a probablity question: Suppose we throw a dice 100 times. What is the probability that at least 1 number appears exactly once. I would be grateful for your help, You can do it by inclusion-exclusion. Let E_k be the event that k appears only once in 100 rolls. You want P(E_1 U ... U E_6), which equals P(E_1) + ... + P(E_6) - sum P(E_j n E_k) + sum P(E_i n E_j n E_k) - sum P(E_i n E_j n E_k n E_l) + P(E_i n E_j n E_k n E_l n E_m) This equals [ 6*100*5^99 - 15*100*99*4^98 + 20*100*99*98*3^97 - 15*100*99*98*97*2^96 + 6*100*99*98*97*96 ]/6^100 I'm getting 0.00000144896 in Excel. === Subject: Re: probability question Have you learned about the binomial distribution yet? === Subject: Re: probability question > > I have a probablity question: > > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. Very, very close to zero. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: probability question posting-account=eFTX7goAAACJBRoHHOnDC9IuTZeZT1_H 1.1.4322),gzip(gfe),gzip(gfe) I have a probablity question: Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. I would be grateful for your help, > eugene I'll guess For a particular die face (doesn't matter when it happens) it would have to not happen 99 times, thus (5/6)^99 for one die face (i.e.number) But there are six die faces (same odds for each and they're independent)... 6 ( 5/6)^99 =~ 8.4 * 10^-8 Am I right? What do I win? === I'll guess For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus (5/6)^99 for one die face (i.e.number) But there are six die faces (same odds for each and they're > independent)... 6 ( 5/6)^99 =~ 8.4 * 10^-8 Am I right? ÊWhat do I win? Nope, I'm not right - I'm high by a factor of 6.... (See why?) so the answer is .....1.4*E-8 === Subject: Re: probability question > > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =~ 8.4 * 10^-8 > Am I right? ÊWhat do I win? > > Nope, I'm not right - I'm high by a factor of 6.... (See why?) No, in fact you're low. Think about just the probability p_1 of getting exactly one 1 in 100 rolls. The 1 can go into 100 slots, each time with probability 1/6. Once the 1 is placed, the other 99 slots are randomly chosen from {2, ..., 6}. So p_1 = 100*(1/6)*(5/6)^99. > so the answer is .....1.4*E-8 === Subject: Re: probability question > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =~ 8.4 * 10^-8 > Am I right? ÊWhat do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) No, in fact you're low. Think about just the probability p 1 of > getting exactly one 1 in 100 rolls. The 1 can go into 100 slots, each > time with probability 1/6. Once the 1 is placed, the other 99 slots > are randomly chosen from {2, ..., 6}. So p 1 = 100*(1/6)*(5/6)^99. > so the answer is .....1.4*E-8- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - Yes,..the factor of 100 is needed. 1.4E-6 There are a hundred ways ...so 100 (1/6)(5/6)^99 The OP asked about *at least* one number appears only once, so I think we've covered all the basis with this answer. How the 99 are distributed among the 5 remaining numbers, we could care less. === Subject: Re: probability question > On Sep 9, 9:07Êpm, The World Wide Wade I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =~ 8.4 * 10^-8 > Am I right? ÊWhat do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) > No, in fact you're low. Think about just the probability p_1 of > getting exactly one 1 in 100 rolls. The 1 can go into 100 slots, each > time with probability 1/6. Once the 1 is placed, the other 99 slots > are randomly chosen from {2, ..., 6}. So p_1 = 100*(1/6)*(5/6)^99. > > so the answer is .....1.4*E-8- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text - > > Yes,..the factor of 100 is needed. 1.4E-6 There are a hundred > ways ...so 100 (1/6)(5/6)^99 > > The OP asked about *at least* one number appears only once, so I think > we've covered all the basis with this answer. How the 99 are > distributed among the 5 remaining numbers, we could care less. No certainly not. What about a 2 on the first roll and all 1's thereafter? See my other post in this thread. === Subject: Re: probability question > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =~ 8.4 * 10^-8 > Am I right? ÊWhat do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) > No, in fact you're low. Think about just the probability p 1 of > getting exactly one 1 in 100 rolls. The 1 can go into 100 slots, each > time with probability 1/6. Once the 1 is placed, the other 99 slots > are randomly chosen from {2, ..., 6}. So p 1 = 100*(1/6)*(5/6)^99. > so the answer is .....1.4*E-8- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text - > Yes,..the factor of 100 is needed. Ê 1.4E-6 ÊThere are a hundred > ways ...so 100 (1/6)(5/6)^99 > The OP asked about *at least* one number appears only once, so I think > we've covered all the basis with this answer. ÊHow the 99 are > distributed among the 5 remaining numbers, we could care less. No certainly not. What about a 2 on the first roll and all 1's > thereafter? See my other post in this thread.- Hide quoted text - - Show quoted text - I don't understand. That case is covered by the 100*(1/6)*(5/6)^99 calculation. Once we have calculated the probability of having a particular number appearing only once, the multiplicity of the rest of the 99 doesn't matter. If the OP had asked for the probability of a number appearing only once and then gave conditions on the rest of the 99 then we would further work to do. But the OP didn't say that each number had to appear at least once. === Subject: Re: probability question > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =~ 8.4 * 10^-8 > Am I right? ÊWhat do I win? Nope, I'm not right - I'm high by a factor of 6.... (See why?) so the answer is .....1.4*E-8- Hide quoted text - - Show quoted text - Raphanus, you are still not right. Let us consider rolling a 1 precisely once in 100 tosses. The 1 could be rolled either on the first roll, the second roll, the third roll, etc. The probability of any of these events occuring is, of course, (1/6) * (5/6)^99, but it can happen 100 different (disjoint events) ways, so the probability of rolling precisely one 1 in 100 tosses is 100 * (1/6) *(5/6)^99. Unfortunately the events, rolling precisely one 1, rolling precisely one 2, ..., rolling precisely one 6, are not disjoint events. Therefore, you really have to work harder. I leave it to you to contain working on it because I am much too lazy and have other things to do right at the moment. Achava === Subject: Re: probability question > > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =3D~ 8.4 * 10^-8 > Am I right? =A0What do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) > so the answer is .....1.4*E-8- Hide quoted text - > - Show quoted text - > > Raphanus, you are still not right. Let us consider rolling a 1 > precisely once in 100 tosses. The 1 could be rolled either on the > first roll, the second roll, the third roll, etc. The probability of > any of these events occuring is, of course, (1/6) * (5/6)^99, but it > can happen 100 different (disjoint events) ways, so the probability of > rolling precisely one 1 in 100 tosses is 100 * (1/6) *(5/6)^99. > Unfortunately the events, rolling precisely one 1, rolling precisely > one 2, ..., rolling precisely one 6, are not disjoint events. > Therefore, you really have to work harder. I leave it to you to > contain working on it because I am much too lazy and have other things > to do right at the moment. For each of the six faces, the probability of that face coming up exactly once is a_1 = 100 * (1/6) * (5/6)^99 For each of the (6 choose 2) = 15 pairs, the probability of those coming up exactly once each is a_2 = 100 * 99 * (1/6)^2 * (4/6)^98 For each of the (6 choose 3) = 20 triples, the probability of those coming up exactly once each is a_3 = 100 * 99 * 98 * (1/6)^3 * (3/6)^97 For each of the (6 choose 4) = 15 quadruples, the probability of those coming up exactly once each is a_4 = 100 * 99 * 98 * 97 * (1/6)^4 * (2/6)^96 For each of the 6 quintuples, the probability of those coming up exactly once each is a_5 = 100 * 99 * 98 * 97 * 96 * (1/6)^5 * (1/6)^95 It is impossible for all 6 to come up exactly once each. By the inclusion-exclusion principle, the probability of at least one face coming up exactly once is 6 a_1 - 15 a_2 + 20 a_3 - 15 a_4 + 6 a_5 = 39443044639617518482737726181871382116510270399912518309933101857486225/ 2722160931250295442069542779825240918904765460303978631429466526539571422822 4 or approximately 1.4489607938609723460 * 10^(-6). -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: probability question posting-account=eFTX7goAAACJBRoHHOnDC9IuTZeZT1_H 1.1.4322),gzip(gfe),gzip(gfe) On Sep 10, 12:13Êam, Robert Israel > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =3D~ 8.4 * 10^-8 > Am I right? =A0What do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) > so the answer is .....1.4*E-8- Hide quoted text - > - Show quoted text - > Raphanus, you are still not right. ÊLet us consider rolling a 1 > precisely once in 100 tosses. ÊThe 1 could be rolled either on the > first roll, the second roll, the third roll, etc. ÊThe probability of > any of these events occuring is, of course, (1/6) * (5/6)^99, but it > can happen 100 different (disjoint events) ways, so the probability of > rolling precisely one 1 in 100 tosses is 100 * (1/6) *(5/6)^99. > Unfortunately the events, rolling precisely one 1, rolling precisely > one 2, ..., rolling precisely one 6, are not disjoint events. > Therefore, you really have to work harder. ÊI leave it to you to > contain working on it because I am much too lazy and have other things > to do right at the moment. For each of the six faces, the probability of that face coming up exactly > once is a 1 = 100 * (1/6) * (5/6)^99 Ê > For each of the (6 choose 2) = 15 pairs, the probability of those coming up > exactly once each is a 2 = 100 * 99 * (1/6)^2 * (4/6)^98 > For each of the (6 choose 3) = 20 triples, the probability of those coming up > exactly once each is a 3 = 100 * 99 * 98 * (1/6)^3 * (3/6)^97 > For each of the (6 choose 4) = 15 quadruples, the probability of those coming > up exactly once each is a 4 = 100 * 99 * 98 * 97 * (1/6)^4 * (2/6)^96 > For each of the 6 quintuples, the probability of those coming up exactly once > each is a 5 = 100 * 99 * 98 * 97 * 96 * (1/6)^5 * (1/6)^95 > It is impossible for all 6 to come up exactly once each. > By the inclusion-exclusion principle, the probability of at least one face > coming up exactly once is > 6 a 1 - 15 a 2 + 20 a 3 - 15 a 4 + 6 a 5 > = Ê394430446396175184827377261818713821165102703999125183099331018 57486225/ > 272216093125029544206954277982524091890476546030397863142946652653957142282[ CapitalEth]24 > or approximately 1.4489607938609723460 * 10^(-6). Ê > -- > Robert Israel Ê Ê Ê Ê Ê Ê Êisr...@math.MyUniversitysInitials.ca > Department of Mathematics Ê Ê Ê Êhttp://www.math.ubc.ca/~israel > University of British Columbia Ê Ê Ê Ê Ê ÊVancouver, BC, Canada- Hide quoted text - - Show quoted text - Yeah,...what Robert said. === Subject: Re: probability question posting-account=plkOvwoAAABPOKn1e8t3zJKYVRn7nmuq Gecko/20080702 Firefox/2.0.0.16,gzip(gfe),gzip(gfe) On Sep 10, 6:13 am, Robert Israel > I have a probablity question: > Suppose we throw a dice 100 times. What is the probability that at > least 1 number appears exactly once. > I would be grateful for your help, > eugene > I'll guess > For a particular die face (doesn't matter when it happens) it would > have to not happen 99 times, thus > (5/6)^99 for one die face (i.e.number) > But there are six die faces (same odds for each and they're > independent)... > 6 ( 5/6)^99 =3D~ 8.4 * 10^-8 > Am I right? =A0What do I win? > Nope, I'm not right - I'm high by a factor of 6.... (See why?) > so the answer is .....1.4*E-8- Hide quoted text - > - Show quoted text - > Raphanus, you are still not right. Let us consider rolling a 1 > precisely once in 100 tosses. The 1 could be rolled either on the > first roll, the second roll, the third roll, etc. The probability of > any of these events occuring is, of course, (1/6) * (5/6)^99, but it > can happen 100 different (disjoint events) ways, so the probability of > rolling precisely one 1 in 100 tosses is 100 * (1/6) *(5/6)^99. > Unfortunately the events, rolling precisely one 1, rolling precisely > one 2, ..., rolling precisely one 6, are not disjoint events. > Therefore, you really have to work harder. I leave it to you to > contain working on it because I am much too lazy and have other things > to do right at the moment. For each of the six faces, the probability of that face coming up exactly > once is a_1 = 100 * (1/6) * (5/6)^99 > For each of the (6 choose 2) = 15 pairs, the probability of those coming up > exactly once each is a_2 = 100 * 99 * (1/6)^2 * (4/6)^98 > For each of the (6 choose 3) = 20 triples, the probability of those coming up > exactly once each is a_3 = 100 * 99 * 98 * (1/6)^3 * (3/6)^97 > For each of the (6 choose 4) = 15 quadruples, the probability of those coming > up exactly once each is a_4 = 100 * 99 * 98 * 97 * (1/6)^4 * (2/6)^96 > For each of the 6 quintuples, the probability of those coming up exactly once > each is a_5 = 100 * 99 * 98 * 97 * 96 * (1/6)^5 * (1/6)^95 > It is impossible for all 6 to come up exactly once each. > By the inclusion-exclusion principle, the probability of at least one face > coming up exactly once is > 6 a_1 - 15 a_2 + 20 a_3 - 15 a_4 + 6 a_5 > = 39443044639617518482737726181871382116510270399912518309933101857486225/ > 27221609312502954420695427798252409189047654603039786314294665265395714228224 > or approximately 1.4489607938609723460 * 10^(-6). > -- > Department of Mathematics http://www.math.ubc.ca/~israel === Subject: Re: Quadratic Diophantine Theorem 7. Summary Starting with Euler in [Eul1773], many authors have come up with essentially the same idea: solving the Pell equation X2 - dY 2 = 1 (29) becomes easier by looking at certain auxiliary equations. The first step is writing down Legendre's equations rx2 - sy2 = 1, 2 for d = rs. (30) There is one nontrivial equation among these with a solution, and the smallest solution will have about half as many digits as the smallest solution of (29). The second step is to look at equations whose solutions give rise to a solution of one of the equations (30); but this second step has never been completed in full generality. What we have are specific equations applicable only in special situations; these were first discovered by Euler, rediscovered by Hart, Sylvester, G¬unther, G«erardin, Hardy & Williams, and Arteha, and slightly generalized by Bapoungu«e. HIGHER DESCENT ON PELL CONICS 11 References http://www.fen.bilkent.edu.tr/~franz/publ/pell-b.pdf Tuesday, July 27, 1638 Sir, The receipt of your letter in which you do me the honor of promising your friendship, gave me no less joy than if I had received it from a mistress whose good graces I passionately desired. And your other writings which preceded it reminded me of the Bradamante of our poets, who would take no one as a servant who had not previously proved himself against her in combat. It is not every day that I compare myself to that Roger who was the only one in the world capable of standing against her in combat; but such as I am, I assure you that I honor your merit considerably. And seeing the most recent method that you used to find tangents to curved lines [in your last letter], I have nothing to say in response, other than that it is very good, and that, if you had explained it that way in the beginning, I would have had no disagreement. The remainder of the letter deals specifically with geometry, and has not been included here. http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf thus: AP, you nut.... actually, it is a commonplace hearsay that the founding parental units were atheists, usually called Deists -- I'm the Higher Power of the 13th Step -- ME ?!? ... but it isn't true. if you read LaRouche, you'd know that, but there is some question about Shakespeare as a Third Language. thus: Apostol begins the course with integration, as per Liebniz; there's a special name for that, per Liebniz: primitive equation or prime equation? thus: this was the subject ot the keynote speaker at the Ninth Annual Nonlinear Science Conference at UCLA, although I no longer blame Newton exclusively for the controversy with Liebniz. anyway, apparently, Hooke did not do the math in a formalistic way; after all, he was the Royal Society's experimenter. the other part of the joke is that the '99 2-pound coin is inscribed, On the shoulders of giants. on the edge; at least, Newton was effective in his office-of-reward. > stole the solution from Hooke, to boot -- on the shoulders of a giant, > who was physically a dwarf; hence, the joke. thus: ah, Lord Berty's thing was called the Unity of Sciences; reference: http://www.larouchepub.com/other/2002/2949moonification.html The two operations of Wells and Russell from which Moon sprung are these: The Moral Re-Armament Movement, founded at a 1921 meeting between a wacky Lutheran preacher from Philadelphia and two British delegates to the Washington Disarmament Conference, Lord Arthur Balfour and H.G. Wells. Moral Re-Armament became the mass organizational vehicle for implementation of Wells' 1928 call in The Open Conspiracy, for a worldwide movement for draft resistance. The environment of Moon's Korean ministry was under control of Moral Re-Armament when he was picked up as an intelligence asset during the Korean War. The Unity of the Sciences movement. Founded in 1935 under the supervision of Lord Bertrand Russell and John Dewey, it brought together Trotskyite academics Albert Wohlstetter (mentor of current Defense Policy Board Chairman Richard Perle), Sidney Hook, and Ernest Nagel, with members of the radical-positivist Vienna Circle. Merging with Robert M. Hutchins at the University of Chicago in the 1950s, this operation took over the teaching of science in the United States. Thomas Kuhn's widely read fraud, The Structure of Scientific Revolutions, was published as the second volume of their Encyclopedia of Unified Sciences. In 1972, the Moonies were given the Unity of Sciences franchise, sponsoring the first of their still-ongoing International Conferences of the Unity of Sciences. Their early sessions featured such notables as Manhattan Project physicist Eugene Wigner, the lifelong ally of that truly mad scientist Leo Szilard (the model for Dr. Strangelove, in Stanley Kubrick's film of that name), and environmental fascists Alexander King and Aurelio Peccei, founders of the no-growth Club of Rome. Before looking back to the history of these projects, let us first briefly dispense with the person of Rev. Sun Myung Moon. Moon as a personality is of very little importance, in himself. The real Reverend Moon is a pathetic, if nonetheless nasty, victim of Japanese internment and North Korean torture sessions. He is what the.... thus: an aether simply provides a prealistical model of the phenomenon of waves; one need not deal with photons at all, since they are just formally dual to waves, not at all pardoxical, as shown by Dirac. far too much of theory rests upon Einstein, saying that's impossible, which is the reality behind Michelson-Morley and those who carried the examination further -- but, herr doktor-professor Albert said, Nein!, when he was once in his dysused office at Caltech. let us remember him for those things which he was really good at, not the Big Bang Constant and its infernal klingons (fridge magnets) ?! > I don't need any aether to stick a magnet to a fridge, light up an thus: so, to ask an ignorant question of personal interest, how much of this could be used toward a new or old proof of quadratic reciprocity? > That Ac and -Dc must be quadratic residues, mod D and A, respectively, > is a trivial result: > Let Ax^2 - Dy^2 = c. [1] > Take this mod A: > -Dy^2 = c mod A > (Dy)^2 = -Dc mod A > Therefore, -Dc is a quadratic residue mod A. > Similarly, taking [1] mod D: > Ax^2 = c mod D > (Ax)^2 = Ac mod D > Therefore, Ac is a quadratic residue mod D. thus: ah, sort of a solopsistic monad. I am reminded of hte Many Words Interpretation of the Copenhagenskool: to be or not to be that is the shape of the box or the bubble -- inside or out? > Q. What is a tautological space? > A. A tautological space. thus: speaking of Young's anhialation of Newton's photons, here is the earlier elaboration on light by Fermat: http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf > as we know from Newton's iron-poor corpuscles. --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U. Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com === Subject: Re: Short Mars travel times at high speed. posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) cannot tack into the solar wind? well, you could always burn the sail for reaction-mass! 7. Summary Starting with Euler in [Eul1773], many authors have come up with essentially the same idea: solving the Pell equation X2 - dY 2 = 1 (29) becomes easier by looking at certain auxiliary equations. The first step is writing down Legendre's equations rx2 - sy2 = 1, 2 for d = rs. (30) There is one nontrivial equation among these with a solution, and the smallest solution will have about half as many digits as the smallest solution of (29). The second step is to look at equations whose solutions give rise to a solution of one of the equations (30); but this second step has never been completed in full generality. What we have are specific equations applicable only in special situations; these were first discovered by Euler, rediscovered by Hart, Sylvester, G¬unther, G«erardin, Hardy & Williams, and Arteha, and slightly generalized by Bapoungu«e. HIGHER DESCENT ON PELL CONICS 11 References http://www.fen.bilkent.edu.tr/~franz/publ/pell-b.pdf Tuesday, July 27, 1638 Sir, The receipt of your letter in which you do me the honor of promising your friendship, gave me no less joy than if I had received it from a mistress whose good graces I passionately desired. And your other writings which preceded it reminded me of the Bradamante of our poets, who would take no one as a servant who had not previously proved himself against her in combat. It is not every day that I compare myself to that Roger who was the only one in the world capable of standing against her in combat; but such as I am, I assure you that I honor your merit considerably. And seeing the most recent method that you used to find tangents to curved lines [in your last letter], I have nothing to say in response, other than that it is very good, and that, if you had explained it that way in the beginning, I would have had no disagreement. The remainder of the letter deals specifically with geometry, and has not been included here. http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf thus: AP, you nut.... actually, it is a commonplace hearsay that the founding parental units were atheists, usually called Deists -- I'm the Higher Power of the 13th Step -- ME ?!? ... but it isn't true. if you read LaRouche, you'd know that, but there is some question about Shakespeare as a Third Language. thus: Apostol begins the course with integration, as per Liebniz; there's a special name for that, per Liebniz: primitive equation or prime equation? thus: this was the subject ot the keynote speaker at the Ninth Annual Nonlinear Science Conference at UCLA, although I no longer blame Newton exclusively for the controversy with Liebniz. anyway, apparently, Hooke did not do the math in a formalistic way; after all, he was the Royal Society's experimenter. the other part of the joke is that the '99 2-pound coin is inscribed, On the shoulders of giants. on the edge; at least, Newton was effective in his office-of-reward. > stole the solution from Hooke, to boot -- on the shoulders of a giant, > who was physically a dwarf; hence, the joke. thus: ah, Lord Berty's thing was called the Unity of Sciences; reference: http://www.larouchepub.com/other/2002/2949moonification.html The two operations of Wells and Russell from which Moon sprung are these: The Moral Re-Armament Movement, founded at a 1921 meeting between a wacky Lutheran preacher from Philadelphia and two British delegates to the Washington Disarmament Conference, Lord Arthur Balfour and H.G. Wells. Moral Re-Armament became the mass organizational vehicle for implementation of Wells' 1928 call in The Open Conspiracy, for a worldwide movement for draft resistance. The environment of Moon's Korean ministry was under control of Moral Re-Armament when he was picked up as an intelligence asset during the Korean War. The Unity of the Sciences movement. Founded in 1935 under the supervision of Lord Bertrand Russell and John Dewey, it brought together Trotskyite academics Albert Wohlstetter (mentor of current Defense Policy Board Chairman Richard Perle), Sidney Hook, and Ernest Nagel, with members of the radical-positivist Vienna Circle. Merging with Robert M. Hutchins at the University of Chicago in the 1950s, this operation took over the teaching of science in the United States. Thomas Kuhn's widely read fraud, The Structure of Scientific Revolutions, was published as the second volume of their Encyclopedia of Unified Sciences. In 1972, the Moonies were given the Unity of Sciences franchise, sponsoring the first of their still-ongoing International Conferences of the Unity of Sciences. Their early sessions featured such notables as Manhattan Project physicist Eugene Wigner, the lifelong ally of that truly mad scientist Leo Szilard (the model for Dr. Strangelove, in Stanley Kubrick's film of that name), and environmental fascists Alexander King and Aurelio Peccei, founders of the no-growth Club of Rome. Before looking back to the history of these projects, let us first briefly dispense with the person of Rev. Sun Myung Moon. Moon as a personality is of very little importance, in himself. The real Reverend Moon is a pathetic, if nonetheless nasty, victim of Japanese internment and North Korean torture sessions. He is what the.... thus: an aether simply provides a prealistical model of the phenomenon of waves; one need not deal with photons at all, since they are just formally dual to waves, not at all pardoxical, as shown by Dirac. far too much of theory rests upon Einstein, saying that's impossible, which is the reality behind Michelson-Morley and those who carried the examination further -- but, herr doktor-professor Albert said, Nein!, when he was once in his dysused office at Caltech. let us remember him for those things which he was really good at, not the Big Bang Constant and its infernal klingons (fridge magnets) ?! > I don't need any aether to stick a magnet to a fridge, light up an thus: so, to ask an ignorant question of personal interest, how much of this could be used toward a new or old proof of quadratic reciprocity? > That Ac and -Dc must be quadratic residues, mod D and A, respectively, > is a trivial result: > Let Ax^2 - Dy^2 = c. [1] > Take this mod A: > -Dy^2 = c mod A > (Dy)^2 = -Dc mod A > Therefore, -Dc is a quadratic residue mod A. > Similarly, taking [1] mod D: > Ax^2 = c mod D > (Ax)^2 = Ac mod D > Therefore, Ac is a quadratic residue mod D. thus: ah, sort of a solopsistic monad. I am reminded of hte Many Words Interpretation of the Copenhagenskool: to be or not to be that is the shape of the box or the bubble -- inside or out? > Q. What is a tautological space? > A. A tautological space. thus: speaking of Young's anhialation of Newton's photons, here is the earlier elaboration on light by Fermat: http://www.wlym.com/~seattle/dynamis/issues/august08-fermat.pdf > as we know from Newton's iron-poor corpuscles. --ROTC, your summer vacation in the Sahara Desert ( S u d a n ) ; presage the Draft for your middleschool class of '12 -- brought to you by Allstate (tm) and Oxford U. Press! http://larouchepub.com/pr/2008/080813moloch brown.html http://wlym.com === Subject: Re: Short Mars travel times at high speed. http://www.allentwood.com/essays/lordofflies.html My path to the Unification Church began in 1966 when I matriculated at the University of the South in Sewanee, Tennessee. I entered college with little concrete notion of what I wanted to do in life. I had rejected the establishment rewards that my secondary education promised. To me the academic challenge at Sewanee was meaningless. Absorbed in a search for values by which I could live, the Peace steps of the burned out ROTC building at the University of the South in the spring of 1968, I was the first student to speak publicly against the War at Sewanee. That summer in New Jersey I worked for the Presidential campaign of Senator Eugene J. McCarthy. In August I went to The National Democratic convention in Chicago as a messenger for the New Jersey delegation. The ferocity of the police riots during the convention destroyed what lingering faith I had in the political process, and I returned to college in the fall bewildered by our nati onal leaders' refusal to respond to the people's wish to end the war in Vietnam. Unable to concentrate on my studies, I began to explore Eastern religions, specifically the religious synthesism of RamaKrishna. In the spring of 1969 I dropped out of the University of the South and hitchhiked to California, a stop on my way to India. In May I arrived in Berkeley, with a suitcase and a navy duffel bag and a copy of the Gospel According to Rama Krishna. I didn't know it, but it was the end of the line. === Subject: Re: Short Mars travel times at high speed. posting-account=BEQ9kgkAAAA_-zTZLFIXg8hu7ZTTi4Q8 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > http://www.allentwood.com/essays/lordofflies.html My path to the Unification Church began in 1966 when I matriculated at > the University of the South in Sewanee, Tennessee. I entered college > with little concrete notion of what I wanted to do in life. I had > rejected the establishment rewards that my secondary education > promised. To me the academic challenge at Sewanee was meaningless. > Absorbed in a search for values by which I could live, the Peace > steps of the burned out ROTC building at the University of the South > in the spring of 1968, I was the first student to speak publicly > against the War at Sewanee. That summer in New Jersey I worked for the > Presidential campaign of Senator Eugene J. McCarthy. In August I went > to The National Democratic convention in Chicago as a messenger for > the New Jersey delegation. The ferocity of the police riots during the > convention destroyed what lingering faith I had in the political > process, and I returned to college in the fall bewildered by our nati > onal leaders' refusal to respond to the people's wish to end the war > in Vietnam. Unable to concentrate on my studies, I began to explore Eastern > religions, specifically the religious synthesism of RamaKrishna. In > the spring of 1969 I dropped out of the University of the South and > hitchhiked to California, a stop on my way to India. In May I arrived > in Berkeley, with a suitcase and a navy duffel bag and a copy of the > Gospel According to Rama Krishna. I didn't know it, but it was the end > of the line. Okay, when the effects of the LSD wore off, what happened next? === Subject: Re: Short Mars travel times at high speed. posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1 AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.2 Safari/525.20.1,gzip(gfe),gzip(gfe) wow, maybe you could advise me. so, which is better, LSD or mild ergot poisoning -- will the latter flashforward to cure flashbacks from the former? most of which I did not read & that I just noticed, is, that little statement about synthesizm; no synesthezia required? also, Moog just came-out with a guitar (saw the ad; at 6500US, it's time to roll one's own .-) >http://www.allentwood.com/essays/lordofflies.html > religions, specifically the religious synthesism of RamaKrishna. In > Okay, when the effects of the LSD wore off, what happened next? thus quoth: Fannie Mae was created by President Franklin D. Roosevelt in 1938, as a government agency to buy mortgages from lenders, as a way of funding the purchase of homes in the Great Depression. In more recent years, Fannie Mae and its sibling Freddie Mac, were taken over by what FDR attacked as the economic royalists, and turned into vehicles for derivatives speculation. Under the great Greenspan bubble, Fannie and Freddie were turned into money machines to feed the run-up in real estate values to provide assetsÑin the form of mortgage debtÑas fuel to the derivatives markets. This scheme was bound to fail, as it spectacularly has, leaving Fannie and Freddie, and the U.S. banking system, utterly bankrupt. However, Fannie and Freddie are at the heart of Treasury Secretary Henry Paulson's and Federal Reserve chairman Ben Bernanke's insane scheme to bail out the banks by dumping all their bad mortgage paper into the two government-sponsored enterprises, effectively transferring the banks' losses to the government, and ultimately to the taxpayer. The government is not really bailing out Fannie and Freddie, but merely funding their conversion into the largest toxic waste dumps in history. Far from being saved, Fannie Mae and Freddie Mac are being destroyed. http://larouchepub.com/other/editorials/2008/3536tantamount treason.html === Subject: Re: Analysis with integral before.. >> |1 log(x) >> | ------ dx >> | 0 1-x >> |oo t e^{-t} >> = - | -------- dt >> | 0 1-e^{-t} >> |oo -t -2t -3t -4t >> = - | t ( e + e + e + e + ... ) dt >> | 0 >> oo >> --- 1 |oo -t >> = - > --- | t e dt --------(*****) >> --- k^2 | 0 >> k=1 >> For that the substitution u = kt was used. >> Of course, this only explains why is it that >> oo >> --- |oo -kt >> - > | t e dt (*) >> --- | 0 >> k=1 >> is equal to >> oo >> --- 1 |oo -t >> - > --- | t e dt >> --- k^2 | 0 >> k=1 >> But it doesn't explain why (*) is also equal to: >> |oo -t -2t -3t -4t >> - | t ( e + e + e + e + ... ) dt >> | 0 > Hm... it's not easy. > S(x) = Sum{n=1 to oo} x.e^{-nx} is not uniformly convergence on [0, oo). > Because, > If x = 0 , then S(x) = 0 > If x =/= 0 , then S(x) = x / (e^x - 1) --> 1 as x --> 0. > so, not continuous. > so, not uniformly convergence. > But > int{0 to 1} log[x] / (1-x) dx = -(pi)^2 / 6 is true by mathematica. > > Ok, I already showed that int{0 to 1} log(x) / (1-x) dx exists. > > so, int{0 to 1} log(x) / (1-x) dx > = lim{b->1-} int{0 to b} log(x) / (1-x) dx > Let x = e^{-t}. > = lim{b->1-} (-)int{-log(b) to oo} {t.e^(-t)}/{1-e^(-t)} dt > = lim{b->1-} (-)int{-log(b) to oo} t[e^(-t) + e^(-2t) + e^(-3t) + ...] dt > = lim{b->1-} (-)int{-log(b) to oo} sum{n=1 to oo} t.e^(-nt) dt > Of course, -log(b) > 0. > > Sum{n=1 to oo} x.e^(-nx) is uniformly convergence on [-log(b), oo) > by Weierstrass M-test. > so, int <==> sum (exchangeable) False. Uniform convergence on an infinite interval does not imply int <==> sum can be interchanged. > so, we can calculate it as Rob Johnson. > > ---------------------------------------------------------------------- > OR) > int{0 to 1} log(x) / (1-x) dx > = lim{a->0+, b->1-} int{a to b} log(x) / (1-x) dx > = lim{a->0+, b->1-} int{a to b} sum{k=0 to oo} (x^k).log(x) dx > by log(x) / (1-x) = sum{k=0 to oo} (x^k).log(x), > > Sum{k=0 to oo} (x^k).log(x) is uniformly convergence on [a, b] > by Weierstrass M-test. > > so, int <==> sum (exchangeable) > > Calculate the int{a to b} (x^k).log(x) dx. > Integrating by parts, > = [{x^(k+1)/(k+1)}*log(x)]_{a to b} - int{a to b} (x^k)/(k+1) dx > = -1/(k+1)^2 as a -> 0 , b -> 1. > > Consequently, > answer is -sum{k=1 to oo} 1/k^2 No, you still have to show sum and limit can be interchanged. A nice way to see int{0 to 1} log(x) /(1-x) dx = sum{k=0 to oo} int{0 to 1} x^k log(x) dx is to use the dominated convergence theorem. === Subject: Re: Analysis with integral before.. >> |1 log(x) >> | ------ dx >> | 0 1-x >> |oo t e^{-t} >> = - | -------- dt >> | 0 1-e^{-t} >> |oo -t -2t -3t -4t >> = - | t ( e + e + e + e + ... ) dt >> | 0 >> oo >> --- 1 |oo -t >> = - > --- | t e dt --------(*****) >> --- k^2 | 0 >> k=1 >> For that the substitution u = kt was used. >> Of course, this only explains why is it that >> oo >> --- |oo -kt >> - > | t e dt (*) >> --- | 0 >> k=1 >> is equal to >> oo >> --- 1 |oo -t >> - > --- | t e dt >> --- k^2 | 0 >> k=1 >> But it doesn't explain why (*) is also equal to: >> |oo -t -2t -3t -4t >> - | t ( e + e + e + e + ... ) dt >> | 0 > Hm... it's not easy. > S(x) = Sum{n=1 to oo} x.e^{-nx} is not uniformly convergence on [0, oo). > Because, > If x = 0 , then S(x) = 0 > If x =/= 0 , then S(x) = x / (e^x - 1) --> 1 as x --> 0. > so, not continuous. > so, not uniformly convergence. > But > int{0 to 1} log[x] / (1-x) dx = -(pi)^2 / 6 is true by mathematica. > > Ok, I already showed that int{0 to 1} log(x) / (1-x) dx exists. > > so, int{0 to 1} log(x) / (1-x) dx > = lim{b->1-} int{0 to b} log(x) / (1-x) dx > Let x = e^{-t}. > = lim{b->1-} (-)int{-log(b) to oo} {t.e^(-t)}/{1-e^(-t)} dt > = lim{b->1-} (-)int{-log(b) to oo} t[e^(-t) + e^(-2t) + e^(-3t) + ...] dt > = lim{b->1-} (-)int{-log(b) to oo} sum{n=1 to oo} t.e^(-nt) dt > Of course, -log(b) > 0. > > Sum{n=1 to oo} x.e^(-nx) is uniformly convergence on [-log(b), oo) > by Weierstrass M-test. > so, int <==> sum (exchangeable) False. Uniform convergence on an infinite interval does not imply int <==> sum can be interchanged. > so, we can calculate it as Rob Johnson. > > ---------------------------------------------------------------------- > OR) > int{0 to 1} log(x) / (1-x) dx > = lim{a->0+, b->1-} int{a to b} log(x) / (1-x) dx > = lim{a->0+, b->1-} int{a to b} sum{k=0 to oo} (x^k).log(x) dx > by log(x) / (1-x) = sum{k=0 to oo} (x^k).log(x), > > Sum{k=0 to oo} (x^k).log(x) is uniformly convergence on [a, b] > by Weierstrass M-test. > > so, int <==> sum (exchangeable) > > Calculate the int{a to b} (x^k).log(x) dx. > Integrating by parts, > = [{x^(k+1)/(k+1)}*log(x)]_{a to b} - int{a to b} (x^k)/(k+1) dx > = -1/(k+1)^2 as a -> 0 , b -> 1. > > Consequently, > answer is -sum{k=1 to oo} 1/k^2 No, you still have to show sum and limit can be interchanged. A nice way to see int{0 to 1} log(x) /(1-x) dx = sum{k=0 to oo} int{0 to 1} x^k log(x) dx is to use the dominated convergence theorem. === Subject: Re: Analysis with integral before.. > |1 log(x) > | ------ dx > | 0 1-x >> |oo t e^{-t} > = - | -------- dt > | 0 1-e^{-t} >> |oo -t -2t -3t -4t > = - | t ( e + e + e + e + ... ) dt > | 0 > oo > --- 1 |oo -t > = - > --- | t e dt --------(*****) > --- k^2 | 0 > k=1 >> For that the substitution u = kt was used. >> Of course, this only explains why is it that >> oo > --- |oo -kt > - > | t e dt (*) > --- | 0 > k=1 >> is equal to >> oo > --- 1 |oo -t > - > --- | t e dt > --- k^2 | 0 > k=1 >> But it doesn't explain why (*) is also equal to: >> |oo -t -2t -3t -4t > - | t ( e + e + e + e + ... ) dt > | 0 >> Hm... it's not easy. >> S(x) = Sum{n=1 to oo} x.e^{-nx} is not uniformly convergence on [0, >> oo). >> Because, >> If x = 0 , then S(x) = 0 >> If