Community RSS Feed https://community.wolfram.com RSS Feed for Wolfram Community showing any discussions in tag Mathematics sorted by active How to create a list from 100 to 0 decrementing? https://community.wolfram.com/groups/-/m/t/2368904 Hi there !! I should feel embarrassed to ask such a simple question but i can not find any answers on the web....maybe because it is too simple... I use Nestlist[ floor[]] as in the lines below, but if I increase the number it won´t give an answer, is there a simpler way to list from 100 to 0 {100,99,98,97,96...,3,2,1,0}? or from n-&gt;0? n=1223 a=NestList[Floor[n--]&amp;,n,n] Luis Felipe Massena Misiec 2021-09-17T16:17:27Z Problem with plotting Sign[ ]: unexpected output https://community.wolfram.com/groups/-/m/t/2367885 Hello, 1) I want to plot the sign of the square root of some function. One expects to obtain values {1} and {0} in the domain where the expression under the square root is positive or zero respectively. In addition to that, I have several pieces on the plot with values continuously ranging from -1 to 1. What is it and how can one avoid that? The second question is similar. 2) I want to plot the sign of a quite complicated function. But in addition to domains with values {-1,0,1} I have a contribution with value continuously ranging from -1 to 1. What is it and how can one avoid that? The notebook is enclosed. : https://www.wolframcloud.com/obj/e8383d34-d7a5-4383-a82f-b831b60a6a18 Ilya Shabat 2021-09-16T15:14:37Z Convert equation to matrix form? https://community.wolfram.com/groups/-/m/t/2362897 I want to convert set of equation into required matrix for my vibrational analysis. I am attaching herewith the two files. One with set of equation (named as questio1) and other one with required format(Named as required format). Kindly help me out to get the final output. input file/code (A11 Amn \[Alpha]^2 + C11 Dmn \[Alpha]^2 - B11 Cmn \[Alpha]^3 + A12 Bmn \[Alpha] \[Beta] + A66 Bmn \[Alpha] \[Beta] + C12 Emn \[Alpha] \[Beta] + C66 Emn \[Alpha] \[Beta] + A66 Amn \[Beta]^2 + C66 Dmn \[Beta]^2 - B12 Cmn \[Alpha] \[Beta]^2 - 2 B66 Cmn \[Alpha] \[Beta]^2 - Amn I0 \[Omega]^2 - Dmn I3 \[Omega]^2 + Cmn I1 \[Alpha] \[Omega]^2) Cos[x \[Alpha]] Sin[y \[Beta]] == 0 (A12 Amn \[Alpha] \[Beta] + C12 Dmn \[Alpha] \[Beta] - B12 Cmn \[Alpha]^2 \[Beta] - 2 B66 Cmn \[Alpha]^2 \[Beta] + A22 Bmn \[Beta]^2 + C22 Emn \[Beta]^2 - B22 Cmn \[Beta]^3 + A66 \[Alpha] (Bmn \[Alpha] + Amn \[Beta]) + C66 \[Alpha] (Emn \[Alpha] + Dmn \[Beta]) - Bmn I0 \[Omega]^2 - Emn I3 \[Omega]^2 + Cmn I1 \[Beta] \[Omega]^2) Cos[y \[Beta]] Sin[ x \[Alpha]] == 0 Required output ( { {A11 \[Alpha] ^2 + A66 \[Beta]^2, A12 \[Alpha] \[Beta] + A66 \[Alpha] \[Beta], -B11 \[Alpha]^3 - B12 \[Alpha] \[Beta]^2 - 2 B66 \[Alpha] \[Beta]^2, C11 \[Alpha]^2 + C66 \[Beta]^2, C12 \[Alpha] \[Beta] + C66 \[Alpha] \[Beta]}, {A12 \[Alpha] \[Beta] + A66 \[Alpha] \[Beta], A22 \[Beta]^2 + A66 \[Alpha]^2, -B12 \[Alpha]^2 \[Beta] - B22 \[Beta]^3 - 2 B66 \[Alpha]^2 \[Beta], C12 \[Alpha] \[Beta] + C66 \[Alpha] \[Beta], C22 \[Beta]^2 + C66 \[Alpha]^2} } ) - \[Omega]^2 ( { {-I0, 0, \[Alpha] I1, -I4, 0}, {0, -I0, I1 \[Beta], 0, -I3} } ) ( { {Amn}, {Bmn}, {Cmn}, {Dmn}, {Emn} } ) == 0 vin Bha 2021-09-09T10:14:52Z [WSG21] Daily study group on creating custom user interfaces https://community.wolfram.com/groups/-/m/t/2355272 On September 7th we will begin our next Daily Study Group series that will focus on &#034;**Creating Custom User Interfaces**&#034;. Attendees will learn to develop graphical user interfaces using the Wolfram Language through short live lessons hosted by Wolfram-certified instructors, and also work on practice problems and mini projects for a hands-on experience. A certificate of program completion will be available. Register [here]. : https://www.bigmarker.com/series/daily-study-group-creating-custom-user-interfaces/series_details?utm_bmcr_source=community Abrita Chakravarty 2021-08-30T19:33:28Z Create the large size of SparseArray https://community.wolfram.com/groups/-/m/t/2368242 Hi Communities I am trying to create a large size of sparsearray as following base = Permutations[{1, 1, 1, 1, 1, 0, 0, 0, 0, 0}]; dim = Length[base] AbsoluteTiming[ test = SparseArray[{i_, j_} /; (Total@Abs[base[[i]] - base[[j]]] == 0 || Total@Abs[base[[i]] - base[[j]]] == 2 || Total@Abs[base[[i]] - base[[j]]] == 4) -&gt; 1, dim {1, 1}]] The ideal of above code is that only the element at diagonal of the matrix and only two and four element differ between two bases have the value of 1. The above code took only ~0.6 second to complete. However, if I change the base to following: base = Select[ Permutations[{1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0}], 5 &lt;= Total@#[[1 ;; 10]] &lt;= 7 &amp;&amp; 5 &lt;= Total@#[[11 ;; 15]] &lt;= 5 &amp;] It ran out all of my computer&#039;s memory (~6GB) and makes the such computation being impossible. The length of this new base is only 34122. It is not very large. So did I make any thing wrong? Are there any solutions to this ? Thank you for reading this. ddtty kuso 2021-09-17T03:42:28Z Defining range of variables for the whole notebook https://community.wolfram.com/groups/-/m/t/2367980 Hi all, At the start of a notebook, is there a way to define that a parameter belongs to a specific range such as [0,1]? I need Mathematica to know about the sign of the values under radicals and simplify them without being forced to repeat the Assuming function for the whole notebook over and over. Arash Roghani 2021-09-17T06:10:16Z How can I turn a system of equations into a matrix? https://community.wolfram.com/groups/-/m/t/2366895 Hello! I&#039;m working on a program using Wolfram Mathematica that can give us the exponential matrix as a result. I have a 3x3 system as a base (but it may work with a nxn one), which gives me 3 linear equations that involve alphas and lambdas with their respective &#034;t&#039;s&#034;. So far, I don&#039;t know how to turn that system into a matrix to get this form: **e^( λi) = A* α** . I would like to know if there is a way where I can take those α&#039;s away from my equations (attached image) and put them into an independent vector, so I can proceed to to this: **α = A^-1 * e^( λi)**; &#034;A&#034; would be the matrix formed from the Eigenvalues and alphas polynomial equation showed below. I would appreciate if you could help me with this situation. Greetings! P.S. I&#039;m not allowed to use any shortcut like Solve or Matrix Exp. ![image] Thank you! : https://community.wolfram.com//c/portal/getImageAttachment?filename=ForoWolfram.jpg&amp;userId=2366880 Karla Reyes Lomelí 2021-09-15T00:55:46Z Optimal climate policy: behind the scenes solving ODEs model https://community.wolfram.com/groups/-/m/t/2255854 *SUPPLEMENTARY WOLFRAM MATERIALS for ARTICLE:* &gt; Charles F. Manski, Alan H. Sanstad, and Stephen J. DeCanio &gt; &#034;Addressing partial identification in climate modeling and policy analysis&#034;. &gt; Proceedings of the National Academy of Sciences, Vol 118, No. 15 (2021) &gt; https://doi.org/10.1073/pnas.2022886118 &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/79b1a40f-2a7e-4e46-b5be-e171b3702b15 Stephen DeCanio 2021-04-28T17:35:05Z Speed up FindRoot for system of equations https://community.wolfram.com/groups/-/m/t/2367061 Hello, I&#039;m solving a system of m by m equations in two variants. When m=7, the second system takes me approximately 15 seconds. Since I want to solve a more complex system for much larger m, I&#039;m looking for any help on speeding up the solution. Thanks! In:= Clear[PPi, pi, m, n, TT, T, TT0, T0, NN, n0, N0, A, R, e]; m = 7; v[i_, j_] := A[j] Sum[n[k, j], {k, m}]/Sum[n[i, l], {l, m}] PPi = Table[pi[i, j], {i, m}, {j, m}]; pi[i_, j_] := T[i, j] v[i, j]^(e)/Sum[T[k, l] v[k, l]^e, {k, m}, {l, m}]; In:= btab = Table[RandomReal[{1, 1.2}], {j, m}]; A[j_] := btab[[j]] R = Table[RandomReal[], {i, m}, {j, m}]; n0[i_, j_] := R[[i, j]]/Total[Flatten[R]] In:= Clear[T, TT, TTm, n, NN, NNm]; N0 = Table[n0[i, j], {i, m}, {j, m}]; TT0 = Table[1, {i, m}, {j, m}]; T0[i_, j_] := TT0[[i, j]] TT = Table[T[i, j], {i, m}, {j, m}]; NN = Table[n[i, j], {i, m}, {j, m}]; NNm = Drop[Flatten[NN], -1]; NNm0 = Drop[Flatten[N0], -1]; TTm = Drop[Flatten[TT], -1]; TTm0 = Drop[Flatten[TT0], -1]; In:= newmat = Drop[Flatten[PPi - NN], -1]; In:= (* baseline: solve for T given n for given e*) Clear[T, n]; e = 5; n[i_, j_] := N0[[i, j]] T[m, m] = 1; soln1 = FindRoot[{newmat},Transpose[{Flatten[{TTm}], Flatten[{TTm0}]}]] // AbsoluteTiming; %[] Out= 0.500039 In:= TT = TT /. soln1[]; T[i_, j_] := TT[[i, j]] In:= (* new eq : solve for n given T with new e value *) e = 3; Clear[NN, n]; n [m, m] = 1 - Total[NNm]; soln2 = FindRoot[{newmat},Transpose[{Flatten[{NNm}], Flatten[{NNm0}]}]] // AbsoluteTiming; %[] Out= 14.5182 Rainald Borck 2021-09-15T19:46:06Z Real, complex, and infinite results from Integrate[] for like inputs https://community.wolfram.com/groups/-/m/t/2367151 I&#039;ve been working with Mathematica for symbolic calculations, primarily Integrations. The following is a function I am trying to integrate with the shown assumptions: Integrate[(50 (0.000258748 + 0.117362 Vch^2) (23 + (84 Vch)/Sqrt[ 1 + 776.46 Vch^2]) (1 + ( 0.76354 (2 + 2329.38 Vch^2 + 1.20578*10^6 Vch^4))/(1 + 776.46 Vch^2)^(3/2)))/(1 + Vch^2), Vch, Assumptions -&gt; {Vch != 0, Element[{Vch}, Reals]}] However, I am getting various results for the output of the function depending on multiple criteria: 1)If I expand the above function before applying it to Integrate[], I get 1/0^2 infinity errors. 2) If I do not expand the function and leave it as shown above, the result are complex numbers 3) If I expand the above function and manually integrate each term (since the terms are added to each other) I will get a real result, which is what it should be. I have to take multiple integrals similar to the one above. Taking each term after the expansions and Integrating each one is fine, but I think it&#039;s a roundabout solution. Does anyone have any ideas as to what is happening and what steps I could take to get a real solution? Thanks! John Mappes 2021-09-15T15:58:06Z Total Derivative error https://community.wolfram.com/groups/-/m/t/2367219 Hi does anyone know why my total derivative command isn&#039;t working? I&#039;ve never encountered this error message before. ![enter image description here] : https://community.wolfram.com//c/portal/getImageAttachment?filename=mathematicabullshit.PNG&amp;userId=2276986 Nikolas Kwok 2021-09-15T12:18:34Z Mnemonic conversion between kilometers and miles using Fibonacci ratios https://community.wolfram.com/groups/-/m/t/2366694 &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/c18fb267-3634-4a6e-8bbc-d8d49d4616fb Bill Gosper 2021-09-14T16:52:14Z The reciprocal Fibonacci constant https://community.wolfram.com/groups/-/m/t/2365201 &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/f4c3bd95-fcd4-4bb1-9dca-93f43c9e9ee0 Jonathan Kinlay 2021-09-11T09:38:30Z How to draw several contours label in separate colors? https://community.wolfram.com/groups/-/m/t/2362787 Good morning. I have two equations (or more) that I want plot with ContourPlot colored with different colours and labelled according to equation order, in a automatic way. I have tried several solutions without success. My plot should remain as g1 = x + 2 y; g2 = 2 x + y; b = {2, 2}; coloresRest = {Blue, Magenta}; labelRest = Table[Text[Style[&#034;(&#034; &lt;&gt; ToString@i &lt;&gt; &#034;)&#034;, coloresRest[[i]]]], {i, Length@coloresRest}]; ContourPlot[{g1 == b[], g2 == b[]}, {x, -2, 4}, {y, -2, 4}, Frame -&gt; False, Axes -&gt; True, ContourStyle -&gt; Thread[{coloresRest, grosor}], Epilog -&gt; {Text[Style[&#034;(1)&#034;, color3], Offset[{20, 0}, {-2, 2}]], Text[Style[&#034;(2)&#034;, color4], Offset[{-10, -10}, {-1, 4}]]}, ImageSize -&gt; 400] ![enter image description here] I have tried with Table, ContourLabel, Riffle or Epilog, without success. For example: ContourPlot[{g1 == b[], g2 == b[]}, {x, -2, 4}, {y, -2, 4}, Frame -&gt; False, Axes -&gt; True, ContourStyle -&gt; Thread[{coloresRest, grosor}], ContourLabels -&gt; Table[Text[Style[labelRest[[i]], Offset[{10, 0}, {#1, #2} &amp;]]], {i, 1, Length@coloresRest}], ImageSize -&gt; 400] or ContourPlot[{g1 == b[], g2 == b[]}, {x, -2, 4}, {y, -2, 4}, Frame -&gt; False, Axes -&gt; True, ContourStyle -&gt; Thread[{coloresRest, grosor}], ContourLabels -&gt; {Riffle[coloresRest, Inset[labelRest, Offset[{10, 0}, {#1, #2} &amp;]]]}, ImageSize -&gt; 400] I realize that expression {#1, #2}&amp; yield coordinates for labels, chosen by Mathematica. I&#039;d prefer that and not to choose such coordinates. Thank in advance : https://community.wolfram.com//c/portal/getImageAttachment?filename=ContourLabelparapreguntacomunidadWolfram.jpg&amp;userId=1867172 Rafael Rodriguez 2021-09-09T08:28:50Z Solve[ ] gives an empty output? https://community.wolfram.com/groups/-/m/t/2362911 Hello, I need c1,c2....c9 values. Why is Solve returning an empty list? How I can fix it? Thanks for any advice. Solve[{uj == aj*c1 + bj*c2 + cj*c3 + ui, vj == aj*c4 + bj*c5 + cj*c6 + vi, tj == aj*c7 + bj*c8 + cj*c9 + ti, uk == ak*c1 + bk*c2 + ck*c3 + ui, vk == ak*c4 + bk*c5 + ck*c6 + vi, tk == ak*c7 + bk*c8 + ck*c9 + ti, ul == al*c1 + bl*c2 + cl*c3 + ui, vl == al*c4 + bl*c5 + cl*c6 + vi, tl == al*c7 + bl*c8 + cl*c9 + ti, um == am*c1 + bm*c2 + cm*c3 + ui, vm == am*c4 + bm*c5 + cm*c6 + vi, tm == am*c7 + bm*c8 + cm*c9 + ti, un == an*c1 + bn*c2 + cn*c3 + ui, vn == an*c4 + bn*c5 + cn*c6 + vi, tn == an*c7 + bn*c8 + cn*c9 + ti, uo == ao*c1 + bo*c2 + co*c3 + ui, vo == ao*c4 + bo*c5 + co*c6 + vi, to == ao*c7 + bo*c8 + co*c9 + ti, up == ap*c1 + bp*c2 + cp*c3 + ui, vp == ap*c4 + bp*c5 + cp*c6 + vi, tp == ap*c7 + bp*c8 + cp*c9 + ti}, {c1, c2, c3, c4, c5, c6, c7, c8, c9}] Umut Cemre Can 2021-09-09T09:27:27Z Animating Mathematica surfaces: Wolfram and Blender integration 5 https://community.wolfram.com/groups/-/m/t/2366292 ![enter image description here]![enter image description here] &amp;[Wolfram Notebook] : https://community.wolfram.com//c/portal/getImageAttachment?filename=8129imageGif1.gif&amp;userId=20103 : https://community.wolfram.com//c/portal/getImageAttachment?filename=2208imageGif2.gif&amp;userId=20103 : https://www.wolframcloud.com/obj/8ea43d6f-9377-46ad-8d4d-5c361a2c18bb Guenther Gsaller 2021-09-13T15:01:56Z Estimating parameters of a stable distribution https://community.wolfram.com/groups/-/m/t/2364156 Hi! I tried to estimated a stable distribution from the dataset hereunder. don = {-0.03394489884326527, -0.03394489884326527, \ -0.03394489884326527, 0.017575499123739236, -0.02922087436589889, 0.010555301952384558, -0.0004591035505102574, \ -0.010672854855081883, -0.010672854855081883, \ -0.010672854855081883, -0.08128446482267304, 0.0032508155059952823, -0.02538460889511576, \ -0.024967155569905922, -0.03200434753397141, -0.03200434753397141, \ -0.03200434753397141, 0.06752654532202125, -0.05355714737031551, 0.023927179959644322, -0.01692677371352175, 0.004213861657695463, 0.004213861657695463, 0.004213861657695463, 0.06477832885486146, 0.047395621108916, -0.013796405278464848, 0.017297775562395895, -0.0156695124441822, -0.0156695124441822, \ -0.0156695124441822, 0.015427766859393002, -0.07003501198980705, 0.022493946849797, -0.010375566995092983, 0.018904524972064116, 0.018904524972064116, 0.018904524972064116, 0.010076845370000675, 0.007618949872832066, -0.016456882934399578, 0.007208123141758686, 0.011401391548995866, 0.011401391548995866, 0.011401391548995866, -0.0253288097675204, \ -0.058520237459744974, -0.02375294625331719, 0.017375496159210194, 0.034068174572563635, 0.034068174572563635, 0.034068174572563635, 0.027290444945758614, 0.04156935324646256, -0.003750649699042637, 0.037454954921030306, -0.09162567061098696, \ -0.09162567061098696, -0.09162567061098696, 0.10097428521409842, 0.06268162221233843, -0.010062863532103786, \ -0.011450360453901243, 0.023602452463382263, 0.023602452463382263, 0.023602452463382263, 0.051080593533900966, 0.013565832731434217, 0.0023202018721298267, -0.02578344291194117, \ -0.003982473432848802, -0.003982473432848802, \ -0.003982473432848802, 0.021815347172571033, 0.05031452199325336, -0.006703957641941638, \ -0.012443436269867542, 0.0018818646288593615, 0.0018818646288593615, 0.0018818646288593615, -0.029845016829764696, 0.04160475193396266, -0.005227759187617618, 0.005939136002577116, 0.008100174424780083, 0.008100174424780083, 0.008100174424780083, -0.011169013468163798, \ -0.008258286074726324, 0.012235814625092312, -0.0219780557330319, \ -0.018525610718695348, -0.018525610718695348, \ -0.018525610718695348, 0.010313197133002959, 0.01652889231101857, -0.006807824169006668, -0.00916036980725222, 0.005692639452927531, 0.005692639452927531, 0.005692639452927531, -0.05865805690492846, 0.030007756501598502, 0.047866401371804805, 0.009191211435466534, 0.009191211435466534, 0.009191211435466534, 0.009191211435466534, 0.012704204855465558, -0.004374802694069071, 0.006159405798702674, 0.010398007713170057, 0.010398007713170057, 0.010398007713170057, 0.010398007713170057, -0.005407391955942583, \ -0.01425961090609437, 0.05396061417482968, 0.0075523767867888125, -0.004829312771382072, \ -0.004829312771382072, -0.004829312771382072, \ -0.011867336746311232, 0.036002650842475184, 0.03631654440886033, -0.0032531650183932883, 0.02381710046507902, 0.02381710046507902, 0.02381710046507902, 0.013162035573157968, -0.007896429640234936, 0.001261888587978409, -0.004754391454896683, \ -0.0015872621812589682, -0.0015872621812589682, \ -0.0015872621812589682, -0.023724470246995304, 0.038136912638033475, -0.022044653284554773, 0.009806991072314142, -0.012167821949641078, \ -0.012167821949641078, -0.012167821949641078, 0.029822985875977166, 0.022471874318293367, 0.012297578413894355, 0.013317501793784084, 0.023974565929419284, 0.023974565929419284, 0.023974565929419284, 0.06432436525081472, 0., 0.07081869836272542, 0.0012540446542990394, 0.0032500028610229492, 0.0032500028610229492, 0.0032500028610229492, -0.0012515931374596837, 0.0012500286102294922, 0.01234572552595136, -0.007212206652050459, 0.009361938080480808, 0.009361938080480808, 0.009361938080480808, 0.00024628757015877857, -0.006445221468641928, 0.0032122588743909337, 0.0004939133482378811, -0.0032210139170059915, \ -0.0032210139170059915, -0.0032210139170059915, 0.003702874076738646, 0.0068643976356639125, 0.014734268663654605, -0.01272016831139334, 0.0014655696538679252, 0.0014655696538679252, 0.0014655696538679252, 0.0067929047282415875, -0.0046307708131043636, 0.0017031555365026954, -0.0029282781303375826, 0.0009751795927657461, 0.0009751795927657461, 0.0009751795927657461, -0.010344900079711672, \ -0.002469074462651643, 0.0046693481580596325, -0.000245803235037827, \ -0.004940706922891259, -0.004940706922891259, \ -0.004940706922891259, 0.00024695667411894057, -0.0012363099455563766, \ -0.0009901452866180843, 0.0004947687115533803, -0.0014866308960014083, \ -0.0014866308960014083, -0.0014866308960014083, \ -0.0034808402625094177, 0.007403802929277001, 0.007349273541704096, -0.00024502022965067265, \ -0.00024502022965067265, -0.00024502022965067265, \ -0.00024502022965067265, -0.00024502022965067265, \ -0.008899820581918147, 0.0004941574191833465, -0.004218292963291941, 0.0004959957860462927, 0.0004959957860462927, 0.0004959957860462927, 0.0014858944534322119, 0., 0.007618496501533813, -0.004443260558482116, 0.0022167055790065743, 0.0022167055790065743, 0.0022167055790065743, 0.0019665430969144602, -0.00394861644264421, 0.0004933041663618312, -0.00024671293544038965, 0., 0., 0., 0.003442325858585361, 0.002941918101143018, 0.002201522729608241, -0.003190187006766056, \ -0.00270661397991699, -0.00270661397991699, -0.00270661397991699, 0.000983212940151308, -0.002217251660530477, 0.004415041761553593, -0.0027054000107892325, \ -0.00024598462648870844, -0.00024598462648870844, \ -0.00024598462648870844, 0.00877829222293795, 0.009420298451695876, 0.0024096936510914803, -0.0031423764032808975, \ -0.01696171260034035, -0.01696171260034035, -0.01696171260034035, 0.0024522385270853556, 0.000735060688952018, 0.00024496020944389947, 0.00024501697600935783, 0.0009786420816095091, 0.0009786420816095091, 0.0009786420816095091, -0.001715678826793574, \ -0.0009812830896425734, -0.0029527764617041344, \ -0.0009852677282126817, -0.0004928178901589532, \ -0.0004928178901589532, -0.0004928178901589532, \ -0.0004930608793812498, 0., -0.00970874593139338, -0.00970874593139338, \ -0.0032467559533485603, -0.0032467559533485603, \ -0.0032467559533485603, -0.001000046730041504, 0., 0., 0., 0., 0., 0., 0.0004997139215246588, -0.0004999637603759766, 0., 0., 0.0007493836435673616, 0.0007493836435673616, 0.0007493836435673616, 0., -0.0004998388097305292, 0., 0., 0.0004995890957115743, 0.0004995890957115743, 0.0004995890957115743, -0.0007499456405639648, 0.0002499194048652646, -0.0002499818801879883, 0.0029910477747560134, -0.0027493518125115825, \ -0.0027493518125115825, -0.0027493518125115825, 0., -0.0002499818801879883, 0., 0., 0.0002499194048652646, 0.0002499194048652646, 0.0002499194048652646, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.} It seems impossible toe estimate the corresponding parameter : endless computation. In:= EstimatedDistribution[don, StableDistribution[\[Alpha], \[Beta], \[Mu], \[Sigma]]] Out= $Aborted Probably because an excess of zeroes? Deleting zeroes, it works more or less... In:= EstimatedDistribution[Select[don, # =!= 0. &amp;], StableDistribution[\[Alpha], \[Beta], \[Mu], \[Sigma]]] During evaluation of In:= FindMaximum::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations. Out= StableDistribution[1, 1.01354, 0.128295, 0.0363144, \ 0.00597755] What should I do to process the whole dataset? Thanks in advance, Claude Claude Mante 2021-09-10T13:54:46Z Try to beat these MRB constant records! https://community.wolfram.com/groups/-/m/t/366628 The MRB constant: ALL ABOARD! ----------------------------- POSTED BY: Marvin Ray Burns. ![CMRB] On 9/5/2021 I added the following strange MRB constant integral ![strange] See proof [here]. On Pi Day, 2021, 2:40 pm EST, I added a new MRB constant integral. ![CMRB] ![=] ![integral to sum] Map: ---- - First, we have formal identities and theory for **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. - Then, at the end of this initial posting, we have world records of the maximum number of digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; computations by date. - Then we have some hints for anyone serious about breaking my record. - Followed by speed records, - a program Richard Crandall wrote to check his code for the computing record number of digits - and a conversation about whether Mathematica uses the same algorithm for computing **C**&lt;sub&gt;*MRB*&lt;/sub&gt; by a couple of different methods. - Then, for a few replies, we compute **C**&lt;sub&gt;*MRB*&lt;/sub&gt; from Crandall&#039;s eta derivative formulas and see records there. - There are three replies about &#034;NEW RECORD ATTEMPTS OF 4,000,000 DIGITS!&#034; and the computation is now complete!!!!!. - We see where I am on a 5,000,000 digit calculation. **(Just recently completed!!!!!!!!!!!!)** - I describe the MRB supercomputer!!!!!! (faster than some computers with dual platinum Xeon processors) It was used for the 5,000,000 digit calculation. - Then it comes time for the 6 million digit computation of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. (put on hold, but taken up again at the end) - We compute **C**&lt;sub&gt;*MRB*&lt;/sub&gt; sum via an integral, which certifies the accuracy of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; calculations!!!!! (since the sum and integral are vastly different in every way they are computed) - Then we take up the task of calculating 6,000,000 (6 million) digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; (for the &lt;s&gt;fourth, fifth time, sixth, seventh time!) (will finish Thu 25 Feb 2021 06:48:06).&lt;/s&gt; - We look for closed forms and find nontrivial, arbitrarily close approximations of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. - The latest updates on the MRB constant supercomputer 2 (with a GIF of it) and how I&#039;m using it to break new records -- up to 7,500,000 digits -- are in a few replies. - **On my 8th try I finally report that I successfully computed 6,000,000 digits of **C**&lt;sub&gt;*MRB*** ! - We work at proving the accuracy of my computations. - For a few replies, we find a few more formulas for **C**&lt;sub&gt;*MRB*,&lt;/sub&gt;. - We take up the challenge of computing 6,500,000 digits for the second time. ---------- **C**&lt;sub&gt;*MRB*&lt;/sub&gt; is defined below. See http://mathworld.wolfram.com/MRBConstant.html. I developed this informal catalog of formulas for the MRB constant with over 20 years of research and ideas from users like you. ![enter image description here] based on **C**&lt;sub&gt;*MRB*&lt;/sub&gt; ![eta equals] ![enter image description here] is proven below by an internet scholar going by the moniker &#034;Dark Malthorp.&#034; ![Dark Marthorp&#039;s proof] ---------- ---------- ---------- ![eta sums] denoting the kth derivative of the Dirichlet eta function of k and 0 respectively, was first discovered in 2012 by Richard Crandall of Apple Computer. The left half is proven below by Gottfried Helms and it is proven more rigorously![(]considering the conditionally convergent sum,![enter image description here]![)] below that. Then the right half is a Taylor expansion of ?(s) around s = 0. &gt; ![n^(1/n)-1] At [https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal], it has been noted that &#034;even though one has cause to be a little bit wary around formal rearrangements of conditionally convergent sums (see the [Riemann series theorem]), it&#039;s not very difficult to validate the formal manipulation of Helms. The idea is to cordon off a big chunk of the infinite double summation (all the terms from the second column on) that we know is absolutely convergent, which we are then free to rearrange with impunity. (Most relevantly for our purposes here, see pages 80-85 of this [document], culminating with the Fubini theorem which is essentially the manipulation Helms is using.)&#034; &gt; ![argrument 1] ![argrument 2] ---------- ---------- ---------- We see many more integrals for **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. We can expand ![1/x] into the following. ![xx = 25.656654035] xx = 25.65665403510586285599072933607445153794770546058072048626118194\ 90097321718621288009944007124739159792146480733342667100.; g[x_] = x^(xx/ x); I NIntegrate[(g[(-t I + 1)] - g[(t I + 1)])/(Exp[Pi t] - Exp[-Pi t]), {t, 0, Infinity}, WorkingPrecision -&gt; 100] (* 0.18785964246206712024851793405427323005590309490013878617200468408947\ 72315646602137032966544331074969.*) **Expanding upon the previously mentioned** ![enMRB sinh] we get the following set of formulas that all equal **C**&lt;sub&gt;*MRB*&lt;/sub&gt;: Let x= 25.656654035105862855990729 ... along with the following constants (approximate values given) {u = -3.20528124009334715662802858}, {u = -1.975955817063408761652299}, {u = -1.028853359952178482391753}, {u = 0.0233205964164237996087020}, {u = 1.0288510656792879404912390}, {u = 1.9759300365560440110320579}, {u = 3.3776887945654916860102506}, {u = 4.2186640662797203304551583} or$ u = \infty .$Another set follows. let x = 1 and along with the following {approximations} {u = 2.451894470180356539050514}, {u = 1.333754341654332447320456} or$ u = \infty $then ![enter image description here] See [this notebook from the wolfram cloud] for justification. ---------- ---------- Also, in terms of the Euler-Riemann zeta function, **C**&lt;sub&gt;*MRB*&lt;/sub&gt; =![enter image description here] Furthermore, as ![enter image description here], according to [user90369] at StackExchange, **C**&lt;sub&gt;*MRB*&lt;/sub&gt; can be written as the sum of zeta derivatives similar to the eta derivatives discovered by Crandall. ![zeta hint ]Informations about ?&lt;sup&gt;(j)&lt;/sup&gt;(k) please see e.g. [here], formulas (11)+(16)+(19).![credit] In the light of the parts above, where **C**&lt;sub&gt;*MRB*&lt;/sub&gt; = ![k^(1/k)-1] = ![eta&#039;(k)] = ![sum from 0] ![enter image description here] as well as ![double equals RHS] an internet scholar going by the moniker &#034;Dark Malthorp&#034; wrote: &gt; ![eta *z^k] ---------- Here is proof of a faster converging integral for its integrated analog (The MKB constant) by Ariel Gershon. g(x)=x^(1/x), M1=![hypothesis] Which is the same as ![enter image description here] because changing the upper limit to 2N + 1 increases MI by 2i/?. MKB constant calculations have been moved to their own discussion at [http://community.wolfram.com/groups/-/m/t/1323951?p_p_auth=W3TxvEwH] . ![Iimofg-&gt;1] ![Cauchy&#039;s Integral Theorem] ![Lim surface h gamma r=0] ![Lim surface h beta r=0] ![limit to 2n-1] ![limit to 2n-] Plugging in equations  and  into equation  gives us: ![left]![right] Now take the limit as N?? and apply equations  and  : ![QED] He went on to note that ![enter image description here] I wondered about the relationship between CMRB and its integrated analog and asked the following. ![enter image description here] So far I came up with &amp;[Wolfram Notebook] Another relationship between the sum and integral that remains more unproven than I would like is ![CMRB(1-i)] f[x_] = E^(I \[Pi] x) (1 - (1 + x)^(1/(1 + x))); CMRB = NSum[f[n], {n, 0, Infinity}, WorkingPrecision -&gt; 30, Method -&gt; &#034;AlternatingSigns&#034;]; M2 = NIntegrate[f[t], {t, 0, Infinity I}, WorkingPrecision -&gt; 50]; part = NIntegrate[(Im[2 f[(-t)]] + (f[(-t)] - f[(t)]))/(-1 + E^(-2 I \[Pi] t)), {t, 0, Infinity I}, WorkingPrecision -&gt; 50]; CMRB (1 - I) - (M2 - part) gives &gt; 6.103779*10^-23 - 6.103779*10^-23 I. ---------- ---------- ---------- As with any scientific paper, this post contains only reproducible results with methods. These records represent the advancement of consumer-level computers and clever programming over the past 20 years. I see others breaking these records, even after I die! Here are some record computations. If you know of any others let me know. - On or about Dec 31, 1998, I computed 1 digit of the (additive inverse of ) **C**&lt;sub&gt;*MRB*&lt;/sub&gt; with my TI-92s, by adding 1-sqrt(2)+3^(1/3)-4^(1/4)+... as far as I could. That first digit, by the way, is just 0. Then by using the sum feature, in approximate mode, to compute$\sum _{n=1}^{1000 } (-1)^n \left(n^{1/n}\right),$I computed the first correct decimal of$\text{CMRB}=\sum _{n=1}^{\infty } (-1)^n \left(n^{1/n}-1\right)$i.e. (.1). It gave (.1_91323989714) which is close to what Mathematica gives for summing to only an upper limit of 1000. - On Jan 11, 1999, I computed 4 decimals(.1878) of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; with the Inverse Symbolic Calculator, with the command evalf( 0.1879019633921476926565342538468+sum((-1)^n* (n^(1/n)-1),n=140001..150000)); where 0.1879019633921476926565342538468 was the running total of t=sum((-1)^n* (n^(1/n)-1),n=1..10000), then t= t+the sum from (10001.. 20000), then t=t+the sum from (20001..30000) ... up to t=t+the sum from (130001..140000). - In Jan of 1999, I computed 5 correct decimals (rounded to .18786)of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95. - Shortly afterward I tried to compute 9 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; using Mathcad 7 professional on the Pentium II mentioned below, by summing (-1)^x x^(1/x) for x=1 to 10,000,000, 20,000,000, and a many more, then linearly approximating the sum to a what a few billion terms would have given. - On Jan 23, 1999, I computed 500 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; with an online tool called Sigma. Remarkably the sum in 4. was correct to 6 of the 9 decimal places! See [http://marvinrayburns.com/Original_MRB_Post.html] if you can read the printed and scanned copy there. - In September of 1999, I computed the first 5,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; on a 350 MHz Pentium II with 64 Mb of RAM using the simple PARI commands \p 5000;sumalt(n=1,((-1)^n*(n^(1/n)-1))), after allocating enough memory. - On June 10-11, 2003 over a period, of 10 hours, on a 450 MHz P3 with an available 512 MB RAM, I computed 6,995 accurate digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. - Using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, at 2:04 PM 3/25/2004, I finished computing 8000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. - On March 01, 2006, with a 3 GHz PD with 2 GB RAM available, I computed the first 11,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. - On Nov 24, 2006, I computed 40, 000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 33 hours and 26 min via my program written in Mathematica 5.2. The computation was run on a 32-bit Windows 3 GHz PD desktop computer using 3.25 GB of Ram. The program was something like this: Block[{a, b = -1, c = -1 - d, d = (3 + Sqrt)^n, n = 131 Ceiling[40000/100], s = 0}, a = 1; d = (d + 1/d)/2; For[m = 1, m &lt; n, a[m] = (1 + m)^(1/(1 + m)); m++]; For[k = 0, k &lt; n, c = b - c; b = b (k + n) (k - n)/((k + 1/2) (k + 1)); s = s + c*a[k]; k++]; N[1/2 - s/d, 40000]] - Finishing on July 29, 2007, at 11:57 PM EST, I computed 60,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. Computed in 50.51 hours on a 2.6 GHz AMD Athlon with 64 bit Windows XP. Max memory used was 4.0 GB of RAM. - Finishing on Aug 3, 2007, at 12:40 AM EST, I computed 65,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. Computed in only 50.50 hours on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 5.0 GB of RAM. - Finishing on Aug 12, 2007, at 8:00 PM EST, I computed 100,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. They were computed in 170 hours on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 11.3 GB of RAM. The typical daily record of memory used was 8.5 GB of RAM. - Finishing on Sep 23, 2007, at 11:00 AM EST, I computed 150,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. They were computed in 330 hours on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 22 GB of RAM. The typical daily record of memory used was 17 GB of RAM. - Finishing on March 16, 2008, at 3:00 PM EST, I computed 200,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; using Mathematica 5.2. They were computed in 845 hours on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 47 GB of RAM. The typical daily record of memory used was 28 GB of RAM. - Washed away by Hurricane Ike -- on September 13, 2008 sometime between 2:00 PM - 8:00 PM EST an almost complete computation of 300,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; was destroyed. Computed for a long 4015. Hours (23.899 weeks or 1.4454*10^7 seconds) on a 2.66 GHz Core 2 Duo using 64 bit Windows XP. Max memory used was 91 GB of RAM. The Mathematica 6.0 code used follows: Block[{$MaxExtraPrecision = 300000 + 8, a, b = -1, c = -1 - d, d = (3 + Sqrt)^n, n = 131 Ceiling[300000/100], s = 0}, a = 1; d = (d + 1/d)/2; For[m = 1, m &lt; n, a[m] = (1 + m)^(1/(1 + m)); m++]; For[k = 0, k &lt; n, c = b - c; b = b (k + n) (k - n)/((k + 1/2) (k + 1)); s = s + c*a[k]; k++]; N[1/2 - s/d, 300000]] - On September 18, 2008, computation of 225,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; was started with a 2.66 GHz Core 2 Duo using 64 bit Windows XP. It was completed in 1072 hours. Memory usage is recorded in the attachment pt 225000.xls, near the bottom of this post. - 250,000 digits were attempted but failed to be completed to a serious internal error that restarted the machine. The error occurred sometime on December 24, 2008, between 9:00 AM and 9:00 PM. The computation began on November 16, 2008, at 10:03 PM EST. Like the 300,000 digit computation, this one was almost complete when it failed. The Max memory used was 60.5 GB. - On Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, I finished computing 250,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. with a multiple-step Mathematica command running on a dedicated 64 bit XP using 4 GB DDR2 RAM onboard and 36 GB virtual. The computation took only 333.102 hours. The digits are at http://marvinrayburns.com/250KMRB.txt. The computation is completely documented in the attached 250000.PD at bottom of this post. - On Sun 28 Mar 2010 21:44:50 (UTC-0500) EST, I started a computation of 300000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; using an i7 with 8.0 GB of DDR3 RAM onboard, but it failed due to hardware problems. - I computed 299,998 Digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. The computation began Fri 13 Aug 2010 10:16:20 pm EDT and ended 2.23199*10^6 seconds later | Wednesday, September 8, 2010. I used Mathematica 6.0 for Microsoft Windows (64-bit) (June 19, 2007) That is an average of 7.44 seconds per digit. I used my Dell Studio XPS 8100 i7 860 @ 2.80 GHz with 8GB physical DDR3 RAM. Windows 7 reserved an additional 48.929 GB virtual Ram. - I computed exactly 300,000 digits to the right of the decimal point of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; from Sat 8 Oct 2011 23:50:40 to Sat 5 Nov 2011 19:53:42 (2.405*10^6 seconds later). This run was 0.5766 seconds per digit slower than the 299,998 digit computation even though it used 16 GB physical DDR3 RAM on the same machine. The working precision and accuracy goal combination were maximized for exactly 300,000 digits, and the result was automatically saved as a file instead of just being displayed on the front end. Windows reserved a total of 63 GB of working memory of which 52 GB were recorded being used. The 300,000 digits came from the Mathematica 7.0 command Quit; DateString[] digits = 300000; str = OpenWrite[]; SetOptions[str, PageWidth -&gt; 1000]; time = SessionTime[]; Write[str, NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]}, WorkingPrecision -&gt; digits + 3, AccuracyGoal -&gt; digits, Method -&gt; &#034;AlternatingSigns&#034;]]; timeused = SessionTime[] - time; here = Close[str] DateString[] - 314159 digits of the constant took 3 tries due to hardware failure. Finishing on September 18, 2012, I computed 314159 digits, taking 59 GB of RAM. The digits came from the Mathematica 8.0.4 code DateString[] NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]}, WorkingPrecision -&gt; 314169, Method -&gt; &#034;AlternatingSigns&#034;] // Timing DateString[] - Sam Noble of Apple computed 1,000,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 18 days 9 hours 11 minutes 34.253417 seconds. - Finishing on Dec 11, 2012, Richard Crandall, an Apple scientist, computed 1,048,576 digits in a lightning-fast 76.4 hours computation time (from the timing command). That&#039;s on a 2.93 GHz 8-core Nehalem. - In Aug of 2018, I computed 1,004,993 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 53.5 hours with 10 DDR4 RAM (of up to 3000 MHz) supported processor cores overclocked up to 4.7 GHz! Search this post for &#034;53.5&#034; for documentation. - Sept 21, 2018: I computed 1,004,993 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 50.37 hours of absolute time (35.4 hours computation time) with 18 (DDR3 and DDR4) processor cores! Search this post for &#034;50.37 hours&#034; for documentation.** - On May 11, 2019, I computed over 1,004,993 digits, using 28 kernels on 18 DDR4 RAM (of up to 3200 MHz) supported cores overclocked up to 5.1 GHz in 45,5 hours of absolute time and only 32.5 hours of computation time! Search &#039;Documented in the attached &#034;:3 fastest computers together 3.nb.&#034; &#039; for the post that has the attached documenting notebook. - On 10/19/20, using 3/4 of the MRB constant supercomputer 2, I finished an over 1,004,993 digits computation of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 44 hours of absolute time -- see [https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb] for documentation. - I computed a little over 1,200,000 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; in 11 days, 21 hours, 17 minutes, and 41 seconds (finishing on March 31, 2013). I used a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz. - On May 17, 2013, I finished a 2,000,000 or more digit computation of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;, using only around 10GB of RAM. It took 37 days 5 hours 6 minutes 47.1870579 seconds. I used my six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz. - A previous world record computation of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; was finished on Sun 21 Sep 2014 at 18:35:06. It took 1 month 27 days 2 hours 45 minutes 15 seconds. The processor time from the 3,000,000+ digit computation was 22 days. I computed the 3,014,991 digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt; with Mathematica 10.0. I Used my new version of Richard Crandall&#039;s code in the attached 3M.nb, optimized for my platform and large computations. I also used a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz with 64 GB of RAM of which only 16 GB was used. Can you beat it (in more number of digits, less memory used, or less time taken)? This confirms that my previous &#034;2,000,000 or more digit computation&#034; was accurate to 2,009,993 digits. they were used to check the first several digits of this computation. See attached 3M.nb for the full code and digits. - Finished on Wed 16 Jan 2019 19:55:20, I computed over 4 million digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. It took 4 years of continuous tries. This successful run took 65.13 days computation time, with a processor time of 25.17 days, on a 3.7 GHz overclocked up to 4.7 GHz on all cores Intel 6 core computer with 3000 MHz RAM. According to this computation, the previous record, 3,000,000+ digit computation, was accurate to 3,014,871 decimals, as this computation used my algorithm for computing n^(1/n) as found in chapter 3 in the paper at https://www.sciencedirect.com/science/article/pii/0898122189900242 and the 3 million+ computation used Crandall&#039;s algorithm. Both algorithms outperform Newton&#039;s method per calculation and iteration. See attached [notebook]. M R Burns&#039; algorithm: x = SetPrecision[x, pr]; y = x^n; z = (n - y)/y; t = 2 n - 1; t2 = t^2; x = x*(1 + SetPrecision[4.5, pr] (n - 1)/t2 + (n + 1) z/(2 n t) - SetPrecision[13.5, pr] n (n - 1) 1/(3 n t2 + t^3 z)); (*N[Exp[Log[n]/n],pr]*) Example: ClearSystemCache[]; n = 123456789; (*n is the n in n^(1/n)*) x = N[n^(1/n),100]; (*x starts out as a relatively small precision approximation to n^(1/n)*) pc = Precision[x]; pr = 10000000; (*pr is the desired precision of your n^(1/n)*) Print[t0 = Timing[While[pc &lt; pr, pc = Min[4 pc, pr]; x = SetPrecision[x, pc]; y = x^n; z = (n - y)/y; t = 2 n - 1; t2 = t^2; x = x*(1 + SetPrecision[4.5, pc] (n - 1)/t2 + (n + 1) z/(2 n t) - SetPrecision[13.5, pc] n (n - 1)/(3 n t2 + t^3 z))]; (*You get a much faster version of N[n^(1/n),pr]*) N[n - x^n, 10]](*The error*)]; ClearSystemCache[]; n = 123456789; Print[t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]] Gives {25.5469,0.*10^-9999984} {101.359,0.*10^-9999984} R Crandall&#039;s algorithm: While[pc &lt; pr, pc = Min[3 pc, pr]; x = SetPrecision[x, pc]; y = x^n - n; x = x (1 - 2 y/((n + 1) y + 2 n n));]; (*N[Exp[Log[n]/ n],pr]*) Example: ClearSystemCache[]; n = 123456789; (*n is the n in n^(1/n)*) x = N[n^(1/n)]; (*x starts out as a machine precision approximation to n^(1/n)*) pc = Precision[x]; pr = 10000000; (*pr is the desired precision of your n^(1/n)*) Print[t0 = Timing[While[pc &lt; pr, pc = Min[3 pc, pr]; x = SetPrecision[x, pc]; y = x^n - n; x = x (1 - 2 y/((n + 1) y + 2 n n));]; (*N[Exp[Log[n]/n],pr]*) N[n - x^n, 10]](* The error*)]; Print[ t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]] Gives {32.1406,0.*10^-9999984} {104.516,0.*10^-9999984} More information available upon request. ---------- - Finished on Fri 19 Jul 2019 18:49:02, I computed over 5 million digits of **C**&lt;sub&gt;*MRB*&lt;/sub&gt;. Methods described in the reply below that starts with &#034;Attempts at a 5,000,000 digit calculation .&#034; For this 5 million calculation of MRB using the 3 node MRB supercomputer: processor time was 40 days. and the actual time was 64 days. That is faster than the 4 million digit computation using just one node. - I finally computed 6,000,000 digits of the MRB constant after 8 tries in 19 months. (Search &#034;8/24/2019 It&#039;s time for more digits!&#034; below.) finishing on Tue 30 Mar 2021 22:02:49 in 160 days. The MRB constant supercomputer 2 said the following: Finished on Tue 30 Mar 2021 22:02:49. Processor and actual time were 5.28815859375*10^6 and 1.38935720536301*10^7 s. respectively Enter MRB1 to print 6029991 digits. The error from a 5,000,000 or more digit calculation that used a different method is 0.*10^-5024993 That means that the 5,000,000 digit computation Was actually accurate to 5024993 decimals!!! ---------- ---------- ---------- ---------- Here is my mini-cluster of the fastest 3 computers (the MRB constant supercomputer 0) mentioned below: The one to the left is my custom-built extreme edition 6 core and later with an 8 core Xeon processor. The one in the center is my fast little 4 core Asus with 2400 MHz RAM. Then the one on the right is my fastest -- a Digital Storm 6 core overclocked to 4.7 GHz on all cores and with 3000 MHz RAM. ![first 3 way cluster] : https://community.wolfram.com//c/portal/getImageAttachment?filename=1ac.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=10878Capture.JPG&amp;userId=366611 : https://www.wolframcloud.com/obj/bmmmburns/Published/Sept_5_2021.nb : https://community.wolfram.com//c/portal/getImageAttachment?filename=8785Capture1.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=10043Capture2.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=75504.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=1593Capture1.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=5191Capture3.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=47625a.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=6657Capture12.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=2081Capture14.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=left.gif&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=3959Capture7.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=right.gif&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=10297Capture.JPG&amp;userId=366611 : https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal : https://en.wikipedia.org/wiki/Riemann_series_theorem : https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch4.pdf : https://community.wolfram.com//c/portal/getImageAttachment?filename=7211Capture.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=10837Capture.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=Capture21%281%29.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=9644Capture.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=Capture23.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=115335.JPG&amp;userId=366611 : https://www.wolframcloud.com/obj/bmmmburns/Published/double%20zeroes%20of%20CMRB.nb : https://community.wolfram.com//c/portal/getImageAttachment?filename=11i.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=6033d.JPG&amp;userId=366611 : https://math.stackexchange.com/users/332823/user90369 : https://community.wolfram.com//c/portal/getImageAttachment?filename=1558a.JPG&amp;userId=366611 : https://digitalcommons.wku.edu/cgi/viewcontent.cgi?referer=http://www.google.de/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=27&amp;ved=2ahUKEwjMx5SuxbnjAhVLLpoKHcBPBWo4FBAWMAZ6BAgAEAI&amp;url=http://digitalcommons.wku.edu/cgi/viewcontent.cgi?article=2093&amp;context=theses&amp;usg=AOvVaw0gQx0dl_Nw4esC2IQc0LEo&amp;httpsredir=1&amp;article=2093&amp;context=theses : https://community.wolfram.com//c/portal/getImageAttachment?filename=2965b.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=11ka.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=84481.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=1869Capture.JPG&amp;userId=366611 : https://community.wolfram.com//c/portal/getImageAttachment?filename=6878p1o2.jpg&amp;userId=366611 : 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https://community.wolfram.com/groups?p_auth=zWk1Qjoj&amp;p_p_auth=r1gPncLu&amp;p_p_id=19&amp;p_p_lifecycle=1&amp;p_p_state=exclusive&amp;p_p_mode=view&amp;p_p_col_id=column-1&amp;p_p_col_count=6&amp;_19_struts_action=/message_boards/get_message_attachment&amp;_19_messageId=1593151&amp;_19_attachment=4%20million%2011%202018.nb : http://community.wolfram.com//c/portal/getImageAttachment?filename=ezgif.com-video-to-gif.gif&amp;userId=366611 Marvin Ray Burns 2014-10-09T18:08:49Z Intersection points of two planes? https://community.wolfram.com/groups/-/m/t/2365163 I am trying to get intersecting points of the two planes. How can I obtain it? ![enter image description here] : https://community.wolfram.com//c/portal/getImageAttachment?filename=11.PNG&amp;userId=2269999 Kyaw Hla Chak 2021-09-12T03:22:05Z Error solving polynomial equation using Solve[ ]? https://community.wolfram.com/groups/-/m/t/2363066 I have got two complex equations after some derivation with log and polynomial. But Solve and NSolve function cannot able to solve them stating &#034;Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.&#034; Can anyone make any suggestions how can I solve that? Attached the mathematica file. &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/e6a786eb-2e7d-49f3-ae3d-e23ea82daf24 Kyaw Hla Chak 2021-09-09T19:07:49Z After using manipulate every manipulate computes the same graph https://community.wolfram.com/groups/-/m/t/2364701 Hi there, anyone... I used an example of manipulate with parametric plot, the initial graph I could alter making changes to the value of x, but now any value of x shows the same graph, including the initial example that gave me only a circle, now the initial example shows the graph that I modified...every manipulate with different values now gives me the same graph...what is happening? I used clearall and it remains the same...I am not being able to clear the memory to renew the graph even to its first graph example: Manipulate[ ParametricPlot[{r Cos[x], r Sin[x]}, {x, 1, m}, {r, 0, k}, Mesh -&gt; Full, PlotRange -&gt; {{-2, 2}, {-2, 2}}], {m, 0, 20 Pi}, {k, 1, 3}] Luis Felipe Massena Misiec 2021-09-10T16:16:58Z Finding the first digits of 2^2^2^2^2^2 https://community.wolfram.com/groups/-/m/t/2359839 &amp;[Wolfram Notebook] : https://www.wolframcloud.com/obj/34c78076-8e13-471b-b527-cd6b87932374 James Choi 2021-09-04T11:55:11Z Use Manipulate in a ParametricPlot ? https://community.wolfram.com/groups/-/m/t/896287 Hi, I&#039;m trying to manipulate a Parametric Plot, but I clearly see its not working because I&#039;m not able to fix it, that&#039;s why I would love some help. Regards Manipulate[ParametricPlot[{r Cos[x], r Sin[y]}, {y,1,k}, {x, 0, n}, Mesh -&gt; Full], {n, 0, 2 Pi}, {k, 1, 2}] Muhammad Afzal 2016-07-30T07:33:09Z