https://community.wolfram.com RSS Feed for Wolfram Community showing questions from all groups sorted by active. https://community.wolfram.com/groups/-/m/t/3715203 &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/082dedb0-4fa3-48a1-aed3-87e91737913d Alejandro Latorre Chirot 2026-05-12T18:39:25Z https://community.wolfram.com/groups/-/m/t/3713745 three constraints are: l1 Tan@a + l2 == Sin@a/Cos[a]^2 + 1/Sin@a, l1 + l2/Tan@a == 1/(Cos@a (Sin@a)^2), 0 < a < \[Pi]/2 Why can't the following code solve the range of l1+l2? How to compute it correctly? Reduce[{l1 Tan@a + l2 == Sin@a/Cos[a]^2 + 1/Sin@a, l1 + l2/Tan@a == 1/(Cos@a (Sin@a)^2), 0 < a < \[Pi]/2, t == l1 + l2}, t, {l1, l2, a}, Reals] // FullSimplify ![enter image description here][1] update 01： FunctionRange[{l1 + l2, Reduce[{l1 Tan@a + l2 == Sin@a/Cos[a]^2 + 1/Sin@a, l1 + l2/Tan@a == 1/(Cos@a (Sin@a)^2), 0 < a < \[Pi]/2, {l1, l2} > 0}, l2]}, {l1, a, l2}, y] Why is the exact value unattainable? How to obtain the exact value? ![enter image description here][2] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-10_075121.png&userId=3593842 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-12_145527.png&userId=3593842 Bill Blair 2026-05-09T23:52:05Z https://community.wolfram.com/groups/-/m/t/3714960 Solve[ForAll[{x, y}, Equivalent[y == Sqrt[3]/2 - x, y + 2 (x - x0) Sin[\[Pi]/6 + 2 x0] == Cos[\[Pi]/6 + 2 x0]]], x0, Reals] ![enter image description here][1] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-12_214254.png&userId=3593842 Bill Blair 2026-05-12T13:43:27Z https://community.wolfram.com/groups/-/m/t/3714447 Hello everyone, Is it possible to create a shortcut (preferably with only one key) to insert a function that I created? For example: units [ex]. Regards, Sinval Sinval Santos 2026-05-11T16:30:31Z https://community.wolfram.com/groups/-/m/t/3713611 Hi everyone, I’m a 3rd-year Mathematics student and I’m reaching a point where I finally start to understand what math really is. I’m fascinated by the Theory of Everything. I have a specific perspective: I believe that while physics can change based on new data, mathematics is absolute. For me, physics is like a subset of a huge axiomatic mathematical structure. I really want to study this connection, but I don’t have much background in physics yet. I live in a small city where it’s hard to find experts in this specific field, and sometimes it's difficult to stay disciplined when you study everything alone. I’m looking for a study partner or a mentor who is interested in: Mathematical Physics and Axiomatic systems. String Theory or Quantum Mechanics from a math perspective. Deep scientific discussions (not just methodology, but pure science). I want to learn fast and I'm ready to dive into complex topics. If you are interested in exploring these "intersection points" of the universe together, please let me know! Naira naira.mndlyan2005@gmail.com 2026-05-08T17:17:40Z https://community.wolfram.com/groups/-/m/t/2437685 I'm very happy with the continued improvement of the Dicom import and export functionality and speed. However, there is one issue with the current implementation that makes it very unuseful if you do quantitative image analysis. It might be that I miss an option if not I think this should be fixed. In the Mathematica documentation and examples, Dicom data is typically shown and imported as images which I understand for display purposes. But I consider the information in Dicom files as data, very well curated and standardized. For many (MRI) applications, the actual quantitative values of voxels stored in Dicom actually have meaning, values, and even units. For example in the image below each voxel value is actually a quantitative measure of T2 relaxation time in the heart, where the values are stored in milliseconds as voxel values as is also mentioned in the metadata. ![enter image description here][1] ![enter image description here][2] To get from the stored values to quantitative values (WV, DV or FP) the fields from the header and the equations are well defined. Header values: - SV = stored value of DICOM PIXEL DATA without scaling - WS = RealWorldValue slope (0040,9225) "RWVSlope" - WI = RealWorldValue intercept (0040,9224) "RWVIntercept" - RS = rescale slope (0028,1053) "RescaleSlope" - RI = rescale intercept (0028,1052) "RescaleIntercept" - SS = scale slope (2005,100E) "ScaleSlope" Outputs: - WV = real world value - FP = precise value - DV = displayed value Formulas: - WV = SV * WS + WI - DV = SV * RS + RI - FP = DV / (RS * SS) So my first try was that I want to obtain the "RawData" to access the SV pixel data, which does not output anything. ![enter image description here][3] Eventually, if I import this Dicom file into Mathematica I have to use a lot of tricks to get to the correct stored values. The Dicom images I use are stored as 12-bit Integers but are converted by Mathematica to a Numerical array with type Int16. Also, I have to specifically specify that I don't want any "DataTransformation" which by default rescales the data and actually changes some voxel values!!!! In[1]:= << QMRITools` {meta, data, bd} = Import[file, {"dicom", {"MetaInformation", "Data", "BitDepth"}}, "DataTransformation" -> None]; dataT = Import[file, {"dicom", {"Data"}}]; {dd = ToExpression[ StringJoin @@ StringCases[NumericArrayType[data], DigitCharacter]], bd} {ss, rs, ri} = meta /@ {"2005_100e", "RescaleSlope", "RescaleIntercept"} {data, dataT} = Normal@{data, dataT}; svT = 2.^bd (dataT/(2.^dd)); pfT = (rs svT + ri)/(rs ss); sv = 2.^bd (data/(2.^dd)); pf = (rs sv + ri)/(rs ss); PlotData[pfT, pf] Out[3]= {16, 12} Out[4]= {1.99854, 0.500366, -2.} For my current research, I need the PF values and to get them correctly I have to: 1. Use "DataTransformation"->None, which is not really well documented what it actually does. But based on what I see actual values of the data are changed by clipping the histogram, which for default handling of medical data is never OK!! ![enter image description here][4] 2. As far as I am aware the only way to obtain the actual stored values of my Dicom data (the actual binary values stored in the file itself) I have to find the actual BiteDepth and the imported data type and rescale my data accordingly. Below are the obtained PF valued data I need with and without DataTransformation. Although the image on the right might look less appealing with default range and scaling it is actually correct when scaled apropriately. ![enter image description here][5] ![enter image description here][6] Am I missing a correct option for getting the actual data stored? If not this should definitely be changed. Dicom is the international standard to transmit, store, retrieve, print, process, and display medical imaging information. With the current implementation, it is impossible to Import and Export such files without actually changing the stored data. Thanks, Martijn [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=dcmtag.png&userId=1332602 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=T2map.png&userId=1332602 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=raw.png&userId=1332602 [4]: https://community.wolfram.com//c/portal/getImageAttachment?filename=hist.png&userId=1332602 [5]: https://community.wolfram.com//c/portal/getImageAttachment?filename=dataTrans1.png&userId=1332602 [6]: https://community.wolfram.com//c/portal/getImageAttachment?filename=dataTrans2.png&userId=1332602 Martijn Froeling 2022-01-05T09:42:43Z https://community.wolfram.com/groups/-/m/t/3711906 Maybe the theorem has a set of measure 0 for which it fails? I have 3 nested functions, each quadratic in its inputs. A standard multi-layer linear plus logistic NN with ADAM optimizer can't seem to minimize the loss (or it gets stuck in a local minimum). I've tried varying the number of layers, and their width, but to no avail. What's going on? &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/47d5251d-896e-4401-97eb-d9794b91e871 Iuval Clejan 2026-05-04T19:31:25Z https://community.wolfram.com/groups/-/m/t/3700131 Does anyone know when Mathematica Version 15.0 will be released? Lan Wei 2026-04-22T01:55:15Z https://community.wolfram.com/groups/-/m/t/3644908 **TL;DR** I'm looking for someone to refactor a Mathematica notebook I've got from a fellow researcher so I can use it as a reliable reference implementation. I expect it's a few hours of work for the right person. I can offer financial compensation and/or my own technical expertise. Hello Wolfram community! I hope this is the right place for this kind of request. If not, my apologies! I'm a PhD student in the final stage of my project, an attempt at closed-loop control of water jets from firefighting robots using UAV imagery as feedback. The controller design is based on the Smith predictor architecture, which requires a predictive model to compensate for the long dead time of the system. Accurately predicting the trajectory of water jets is far from trivial. One of the most promising models I could find is described in https://link.springer.com/article/10.1007/s10694-021-01175-1. The model is formulated as a system of ordinary differential equations. I tried implementing it in Python so I can integrate it with my other components. It's almost complete, but despite several months of debugging I haven't been able to resolve the remaining issues. So I contacted the corresponding author. They confirmed some errors I found in the printed versions of the equations, and kindly provided their original Mathematica implementation. This helped, but my own implementation is still incomplete. The issues could stem from additional errors in the printed equations I/we haven't found yet, mistakes in my implementation, or differences in solver behavior (Mathematica's vs. SciPy's solve_ivp() function). Unfortunately, the notebook is hard for me to follow and differs quite a bit from the published paper (structure, variable naming, angle conventions, etc.). I've never worked with Mathematica and don't have the time nor patience to properly learn it before my deadline. The author is currently unable to provide further support, but since I'm getting more and more desperate to finish this subproject, I'm now seeking third-party help. I'm looking for someone to refactor the notebook into a clean, well-structured reference implementation. Specifically, I'd like them to - remove unused and redundant code (many expressions are duplicated) - improve structure - improve documentation - add small quality-of-life improvements if appropriate - flag any noticeable discrepancies The refactored version must reproduce the original results, in particular the figures shown in the paper. Ideally, it should make it easy to experiment with the equations and parameters. One specific goal is to verify whether the rearranged equation forms I use in Python (to match SciPy's solver interface) produce the same results as the original formulation. If you're interested, I'll obtain the author's permission and share the notebook privately so you can assess the scope before we discuss compensation. Bonus points if you have experience with physics-based simulations and are open to occasional follow-up questions :) Many thanks and regards! Merlin Stampa 2026-02-24T20:34:23Z https://community.wolfram.com/groups/-/m/t/3675919 **Intro** A fundamental question keeps coming up for me: If biological evolution operates in astronomically large spaces, why is search computationally tractable at all? Even a modest protein corresponds to a combinatorial space that is effectively impossible to exhaustively explore. Yet evolution does not behave like an unconstrained random search. So what makes the space navigable? **Essay** In 1859, two different perspectives on complexity emerged. Bernhard Riemann revealed deep structural order underlying the distribution of prime numbers. Charles Darwin introduced a dynamical process of variation and selection. Modern biology has successfully developed Darwin’s framework. However, something is often left implicit: the assumption that the search space is already structured in a way that makes local exploration effective. From a purely combinatorial perspective, this is problematic. Under simple assumptions (independent variation, no bias), expected search time grows exponentially with the amount of required information. In that regime, evolution would be computationally intractable. But real systems do not operate in that regime. Instead, they appear to evolve within a highly structured, constrained subspace, where: functional states are not isolated viable configurations form connected regions local mutations can traverse meaningful paths This suggests that evolution can be framed as a constrained search problem, rather than a purely stochastic process. Evolution is not merely a process acting within a space — it is a process shaped by the structure of the space it can access. This shifts the central question: What determines that accessible space? **A Minimal Computational Model** To make this concrete, consider a simple toy model. We define: a sequence space a mutation operator a constraint that restricts transitions **Basic setup** L = 20; randomSeq[] := RandomInteger[{0, 1}, L]; mutate[s_] := ReplacePart[s, RandomInteger[{1, L}] -> 1 - #] & @ s; **Fitness function** fitness[s_] := Boole[Total[s] > 12]; **Constraint energy** energy[s_] := Total[ Map[If[# === {1, 1}, 0, 1] &, Partition[s, 2, 1]] ]; **Dynamics: constrained vs unconstrained** stepConstrained[s_] := Module[{s2 = mutate[s]}, If[constraint[s, s2], s2, s] ]; stepRandom[s_] := mutate[s]; **Search experiment** findFunctional[step_, max_] := Module[ {s = randomSeq[], t = 0}, While[t < max && !TrueQ[fitness[s] == 1], s = step[s]; t++; ]; t ]; trialsConstrained = Table[ findFunctional[stepConstrained, 1000], {50} ]; trialsRandom = Table[ findFunctional[stepRandom, 1000], {50} ]; **Visualization** Histogram[ {trialsRandom, trialsConstrained}, ChartLegends -> {"Random", "Constrained"}, PlotTheme -> "Scientific", Frame -> True ] **Interpretation** In many runs, the constrained dynamics reaches functional states faster — not because the system is explicitly guided toward a target, but because the structure of the space itself has changed. Even in this minimal model, a key effect emerges: Pure random mutation behaves like unstructured search Even a simple constraint dramatically reshapes accessibility The constraint does not “guide” the system toward solutions. Instead, it reshapes the space such that functional paths become possible in the first place. **Open Questions** This raises several structural questions: - How can we formally define a constraint operator in general systems? - Can constraint-induced subspaces be measured or classified? - How does connectivity emerge in high-dimensional spaces under constraints? - Do constrained systems exhibit characteristic spectral signatures (e.g., non-random eigenvalue statistics)? **Closing Thought** The difference between intractable search and effective evolution may not lie in time or randomness — but in the geometry of the accessible space itself. Maurice Crutzen 2026-04-07T09:31:01Z https://community.wolfram.com/groups/-/m/t/3710008 (8 + 4 x1 + 2 x2 + x1 x2)/(-8 - 2 x1 + 4 x2 + x1 x2) How to transform the fraction above into each of these three equivalent forms? 1: (8 + 4 (x1 + x2) - 2 x2 + x1 x2)/(-8 - 2 (x1 + x2) + 6 x2 + x1 x2) 2: (8 + 2 (x1 + x2) + 2 x1 + x1 x2)/(-8 - 2 (x1 + x2) + 6 x2 + x1 x2) 3: (8 + 2 (x1 + x2) + 2 x1 + x1 x2)/(-8 + 4 (x1 + x2) - 6 x1 + x1 x2) The rule for algebraic manipulation is to rewrite x₁ or x₂ in the numerator and denominator as coefficient multiples of (x₁ + x₂), using addition and subtraction identities based on their fixed coefficients, while keeping all other terms unchanged. Bill Blair 2026-05-01T00:40:45Z https://community.wolfram.com/groups/-/m/t/3710069 Script: ```wls Do[ queryresult = RunProcess[ { "bq", "query", "--format=json", "--nouse_legacy_sql", "--max_rows=86400", "SELECT DISTINCT ST_X(`geography`) AS longitude, ST_Y(`geography`) AS latitude FROM `personalinformatics.locations` WHERE EXTRACT(DATE from `timestamp` AT TIME ZONE 'America/New_York') = '"<>DateString[mapdate,"ISODate"]<>"'" }, "StandardOutput" ]; data = ImportString[queryresult,"RawJSON"]; table = Table[ GeoPosition[ { ToExpression@data[[i,"latitude"]], ToExpression@data[[i,"longitude"]] } ], {i, Length[data]} ]; map = GeoListPlot[ table, ImageSize->{1080,1080}, GeoServer->"https://tiles.openstreetmap.us/vectiles/`1`/`2`/`3`.pbf", GeoScaleBar->{"Imperial","Metric"}, PlotLabel->Style["Daily GPS Map | " <> DateString[mapdate,"LocaleDateFull"],"Subtitle"] ]; Export["/home/steven/public_html/dailymaps/"<>DateString[mapdate,"ISODate"]<>"_map.png",map]; Print["Completed map export for "<>DateString[mapdate,"ISODate"]]; ,{mapdate, DateRange[DateObject[{2026, 4, 27}], DateObject[{2026, 4, 30}]]}] ``` The following error is returned: ``` GeoServer::styleres: Cannot download vector style resources. ``` I have tried the following URLs for GeoServer and am getting this error (note, there should be backticks around the `1`, `2`, and `3`—this forum composer removes them). - https://tile.openstreetmap.org/`1`/`2`/`3`.pbf (the current OpenStreetMap vector tile URL) - https://tile.openstreetmap.org/`1`/`2`/`3`.png (the current OpenStreetMap raster tile URL) Steven Buehler 2026-05-01T12:48:58Z https://community.wolfram.com/groups/-/m/t/3709739 &[Wolfram Notebook][1] DOI: 10.1103/PhysRevLett.120.183603 ![enter image description here][2] I have read a paper about Dissipation-Induced Multicritical Phenomena, the relevant DOI is above.Its [21] reference is supplemental material,the sixth part gives a set of ten equations,and we want to gain its steady state, so the lhs is set to zero, thus,we obtain a ten-dimensional linear equtions. How could i get this analytical result by Mathematic, you can download my notebook, help me solve this. Shall i think about block matrix? [1]: https://www.wolframcloud.com/obj/0c30b0b9-968f-4ebd-be9d-442681df2e7c [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=picture_page-0001%281%29.jpg&userId=3460873 Tree Clean 2026-04-30T11:57:16Z https://community.wolfram.com/groups/-/m/t/3707706 If anyone has time, can you help simplify the code in Section 4.4.2 (pg. 62-82) of [the attatchment][1] using the math equations in Theorem 12 (pg. 16-21)? Ignore the beginning and go straight to the Code 1 (pg. 63). I need to simplify the Q-functions (Code 26, pg. 73-75) and the functions BoldM1 and BoldM2 (Code 28, pg. 76) I tried `FullySimplify` on the Q-functions, but it returns an error message. (See [this post][2] for a minimal example.) I need to simplify the code into an elegant mathematical equation in terms of Theorem 12 (pg. 16-21). If this question is innapropriate for the site, please let me know. [1]: https://www.researchgate.net/publication/403194504_Averaging_an_Explicit_Non-Lebesgue_Integrable_and_Unbounded_Function_That_Is_Defined_Without_Axiom_of_Choice [2]: https://mathematica.stackexchange.com/questions/319306/how-do-we-simplify-the-inequality-q3-at-the-bottom-of-this-post Bharath Krishnan 2026-04-29T01:59:46Z https://community.wolfram.com/groups/-/m/t/3706662 **Question:** At the end of the post, how do we simplify all cases of `Q3[r,1,2]` and combine them into one expression? I provided working examples at the end of this post. ---------- **Preliminaries:** Suppose, we have the following code (i.e., `f[r]` and `g[r]` can be any function, where `f[r+c]<V[r]<g[r+d]` or `g[r+d]<V[r]<f[r+c]` and `c`, `d` are small constants; e.g., between $-10$ and $10$.) Clear["Global`*"] V[r_]:=V[r]=r!+1 f[r]:=f[r]= g[r_]:=g[r]= LengthS[r_] := LengthS[r] = {f[r],g[r]} LengthS1[r_, x_] := LengthS1[r, x] = LengthS[r][[x]] LengthS2[j_, y_] := LengthS2[j, y] = LengthS[j][[y]] We approximate the constants, in the code below, using this equation: > If $F:\mathbb{N}\to\mathbb{R}$ and $G:\mathbb{N}\to\mathbb{R}$ are > arbitrary functions, we want to calculate this equation with > Mathematica: > > $$\small{\begin{equation} c=\inf\left\{|1-\mathbf{c_1}|:\forall(\epsilon>0)\exists(\mathbf{c_1}>0)\forall(r\in\mathbb{N})\exists(v\in\mathbb{N})\left(\left|\frac{F(r)}{G(v)}-\mathbf{c_1}\right|<\varepsilon\right)\right\} \tag{1}\label{eq:1} \end{equation}}$$ cV1[r_, 1, 2] := cV1[r, 1, 2] = N[Min[Table[ RealAbs[1 - V[r]/LengthS1[v, x]], {v, 1, 30000}]]] cV1[r_, 2, 1 ] := cV1[r, 2, 1] = N[Min[Table[ RealAbs[1 - LengthS1[r, x]]/V[v], {v, 1, 30000}]]] cV2[r_, 1, 2] := cV2[r, 1, 2] = N[Min[Table[ RealAbs[1 - V[r]/LengthS2[v, y]], {v, 1, 30000}]]] cV2[r_, 2, 1] := cV2[r, 2, 1] = N[Min[Table[ RealAbs[1 - LengthS2[r, y]/V[v]], {v, 1, 30000}]]] c[r_, 1, 2] := c[r, 1, 2] = N[Min[Table[ RealAbs[1 - LengthS1[r, x]/LengthS2[v, y]], {v, 1, 30000}]]] c[r_, 2, 1] := c[r, 2, 1] = N[Min[Table[ RealAbs[1 - LengthS2[r, y]/LengthS1[v, x]], {v, 1, 30000}]]] In case the outputs are incorrect, use Equation \eqref{eq:1} to solve the exact value. In addition, w.r.t. the inequality `f[r+c]<V[r]<g[r+d]`, we adjust `r+c` *and* `r+d` into `r+c1` *and* `r+d1`, using `P1` in the code below (i.e., `c1==c+1||c1==c||c1==c-1` and `d1==d+1||d1==d||d1==d-1`). If this makes no sense, then attempt to analyze the code. P1 = 3 (*P1 can be any constant positive integer*) Min11[r_, x_] := Min11[r, x] = Max[r1 /. FindInstance[LengthS1[r1, x] <= V[r] < LengthS1[r1 + 1, x], {r1}, PositiveIntegers]] Min12[r_, x_] := Min12[r, x] = ArgMin[{RealAbs[LengthS1[r2, x] - V[r]], r - P1 <= r2 <= r + P1}, r2, PositiveIntegers] Min21[r_, y_] := Min21[r, y] = Max[r3 /. FindInstance[LengthS2[r3, y] <= V[r] < LengthS2[r3 + 1, y], {r3}, PositiveIntegers]] Min22[r_, y_] := Min22[r, y] = ArgMin[{RealAbs[LengthS1[r4, y] - V[r]], r - P1 <= r4 <= r + P1}, r4, PositiveIntegers] rMin1[r_, x_] := rMin1[r, x] = Min12[r, x] + Sign[Floor[RealAbs[2 r - Min11[r, x] - Min12[r, x]]/2]] rMin2[r_, y_] := rMin2[r, y] = Min22[r, y] + Sign[Floor[RealAbs[2 r - Min21[r, y] - Min22[r, y]]/2]] Putting it together, this is what I want: For any `LengthS1[r,x]=f[r+c1]` and `LengthS2[r,y]==g[r+d1]` such that for any function `f` and `g`, where `f[r+c1]<V[r]<g[r+d1]` or `g[r+d1]<V[r]<f[r+c1]`, consider the following case (we use this case to simplify the other cases): 1. Case 3: `V[r] >= LengthS1[rMin1[r, x], x] && LengthS1[rMin1[r, x], x] > LengthS2[rMin2[r, y], y]` and extra criteria including `cV1`, `cV2` and `c` (see Equation \ref{eq:1}) To define `Q3`, we use Case 3 and define the following: Sign3[r_, x_, y_] := Sign3[r, x, y] = (Sign[ Sign[cV2[r, x, y] c[r, x, y]] + Sign[cV2[r, x, y] cV2[r, y, x]] + Sign[1 - cV2[r, y, x] c[r, y, x]]]) Sign3S[r_, x_, y_] := Sign3S[r, x, y] = Sign[c[r, x, y] - cV2[r, y, x]] Sign[ cV1[r, y, x] - cV2[r, y, x]] Sign[cV1[r, x, y] - c[r, y, x]] Sign3SS[r_, x_, y_] := Sign3SS[r, x, y] = Sign[1 + Sign[c[r, y, x] - c[r, x, y]]] where `Q3[r,x,y]` is one out of the 16 criteria of the original code (i.e., I'm applying this example to simplify the other cases): Q3[r_, x_, y_] := Q3[r, x, y] = Sign3[r, x, y] Sign3S[r, x, y] cV2[r, x, y] >= Sign3SS[r, x, y] c[r, x, y] && Sign3SS[r, x, y] c[r, x, y] >= Sign3[r, x, y] Sign3SS[r, x, y] c[r, y, x] && V[r] >= LengthS1[rMin1[r, x], x] && LengthS1[rMin1[r, x], x] > LengthS2[rMin2[r, y], y] **Question (Reasked):** How do we simplify `Q3[r,1,2]` in terms of the equations in the previous code? Here are some examples: 1. `LengthS[r_]:=LengthS[r]={5r!/7, 3r!/11}` (it should return `{Q3[r,1,2]}=={True}`) 3. `LengthS[r_]:=LengthS[r]={4r!/5, 2r!/3}` (it should return `{Q3[r,1,2]}=={True}`) 4. `LengthS[r_]:=LengthS[r]={r!, (3/(Pi^2))r^2}` (it should return `{Q3[r,1,2]}=={True}`) 5. `LengthS[r_]:=LengthS[r]={2^r, (3/(Pi^2))r^2}` (it should return `{Q3[r,1,2]}=={True}`) 6. `LengthS[r_]:=LengthS[r]={2^r, (2^r)/3}` (it should return `{Q3[r,1,2]}=={True}`) We wish to simplify examples 1.-6. and every other case satisfying Case 3, then combine them into one expression written in terms of the equations in the previous codes. **Attempt:** For all Cases satisfying Case 3 (e.g., examples 1-6), we want to simplify `Q3[r,1,2]` into one expression. FullSimplify[Q3[r, 1, 2], Element[r, PositiveIntegers]] However, with all the examples (e.g., example 6), I get the following error message: FindInstance::exvar: The system contains a nonconstant expression r! independent of variables {r1}. ReplaceAll::reps: {FindInstance[2^r1<=1+r!<2^(1+r1),{r1},Subscript[\[DoubleStruckCapitalZ], >\[ThinSpace]0]]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. FindInstance::exvar: The system contains a nonconstant expression r! independent of variables {r3}. ReplaceAll::reps: {FindInstance[2^r3/3<=1+r!<2^(1+r3)/3,{r3},Subscript[\[DoubleStruckCapitalZ], >\[ThinSpace]0]]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. **Sub-Question:** How do we fix the attempt to answer the question? Bharath Krishnan 2026-04-28T01:57:08Z https://community.wolfram.com/groups/-/m/t/3702416 I would appreciate help in getting Mathematica to solve an indefinite integral that is closely related to two that it can solve, or at least to understand why it cannot. Consider an indefinite integral that includes the half-integer Macdonald function, also known as the “modified spherical Bessel function of the second kind” and the “reduced Bessel function,” multiplied by $x^{\frac{1}{2}-h} e^{-a x^2}$. If $n=2$ and $h=0$ or $2$, Mathematica 7 and 13 can solve the integral: $$\int e^{-a x^2} x^{\frac{1}{2}-0} K_{\frac{1}{2}+2}(x b) b^{\frac{1}{2}+2} dx = \sqrt{\frac{\pi }{2}} \left(-\frac{3 e^{-x (b+a x)}}{x}-\frac{\left(6 a-b^2\right) e^{\frac{b^2}{4 a}} \sqrt{\pi } \text{erf}\left(\frac{b+2 a x}{2 \sqrt{a}}\right)}{2 \sqrt{a}}\right) \quad (1)$$ and $$\int e^{-a x^2} x^{\frac{1}{2}-2} K_{\frac{1}{2}+2}(x b) b^{\frac{1}{2}+2} dx =$$ $$\sqrt{\frac{\pi }{2}} \left(e^{-b x-a x^2} \left(-\frac{1}{x^3}-\frac{b}{x^2}+\frac{2 a}{x}\right)+2 a^{3/2} e^{\frac{b^2}{4 a}} \sqrt{\pi } \text{erf}\left(\frac{b+2 a x}{2 \sqrt{a}}\right)\right) \quad, \quad (2)$$ respectively. But (now shifting to InputForm) Integrate[ E^(- a (x^2)) x^(1/2 - h) BesselK[1/2 + n, x b] b^(1/2 + n), x] /. n -> 2 /. h -> 1 yields the response $$b^{5/2} \sqrt{\frac{\pi }{2}} \int \frac{e^{-b x-a x^2} \left(1+\frac{3}{b^2 x^2}+\frac{3}{b x}\right)}{\sqrt{x} \sqrt{b x}} \, dx $$ that indicates a failure to solve the integral. One gets the same response if one directly integrates the above series expression for the half-integer Macdonald function b^2*Sqrt[Pi/2]*Integrate[(E^((-b)*x - a*x^2)*(1 + 3/(b^2*x^2) + 3/(b*x)))/x, x] (or attempts to integrate the series term-by-term) or uses the Meijer-G function form of the half-integer Macdonald function: Integrate[(x^2 b^2)^(1/4)/(Sqrt[x] Sqrt[b]) E^(-x b)/E^-Sqrt[x^2 b^2] E^(- a (x^2)) x^(1/2 - h) *1/2 MeijerG[{{}, {}}, {{1/2 (1/2 + n), 1/2 (-(1/2) - n)}, {}}, 1/4 x^2 b^2] * b^(1/2 + n), x] /. n -> 2 /. h -> 1 /. {1/Sqrt[x^2 b^2] -> 1/(x b)} where one needs the two factors of one on the first line, `(x^2 b^2)^(1/4)/(Sqrt[x] Sqrt[b])` and `E^(-x b)/E^-Sqrt[x^2 b^2]`, and the substitution `/. {1/Sqrt[x^2 b^2] -> 1/(x b)}` on the last line in order for Mathematica to yield the right-hand sides of equations (1) and (2) when $h$ is set to 0 and 2, resp. So my question is, since Mathematica can numerically integrate these expressions when $h=0$, 1, or 2, and since it can solve for the indefinite integral when $h$ takes on the bracketing values $h=0$ and $h=2$, why will it not solve for the indefinite integral at the intermediate value $h=1$? Is there some way to get it to do so? (It also fails to do so for even values $h=4$ and larger but can for positive powers of x in the series given by $h=-2$.) Perhaps I should instead be thankful that Mathematica *can* solve for the indefinite integral for the bracketing values $h=0$ and $h=2$ since equations (1) and (2) are not tabled in Gradshteyn and Ryzhik (Section 5.5) nor Prudnikov, Brychkov, and Marichev (Section 1.12.2) and ask how it manages to do so given that it cannot integrate the series term-by-term. Thanks much, Jack Jack Straton 2026-04-24T18:11:13Z https://community.wolfram.com/groups/-/m/t/3704725 ![enter image description here][1] p1 = (1 - (1 - p)^3) q^3 // Expand p2 = (1 - (1 - q)^3) p^3 // Expand p1 - p2 // Simplify // Factor Simplify[Sign[p1 - p2], 0 < p < q] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-04-25_211049.png&userId=3593842 Bill Blair 2026-04-25T13:14:13Z https://community.wolfram.com/groups/-/m/t/3673820 I’m having some startup difficulty with the first example in Guide to Modern Physics (James W. Rohlf). The book has all the right topics but I cannot enter the examples successfully. I am running Mathematica 14.3 on a Mac Book Pro with M5 Pro, OS Tahoe and 24 GB of memory. The file myFile.nb shows what I typed. Notice that the N[UnitConvert[e, C],4] does not give the expected answer. I am not sure what to type for e. If I use the Basic Math Paclet, e is that transcendental number, not the ElementaryCharge definition. I can get the correct answer with a copy/paste from the original Quantity definition. But notice that I copied 1 C, not just C. Quantity["ElementaryCharge"] returns e Quantity["Coulambs"] returns 1 C N[UnitConvert[e,C] returns UnitConvert[e,1 C],4] N[UnitConvert[e,1 C],4] returns 1.602 x 10-19 Then I went on Rohlf’s site and bought the files that contain the example Mathmatica code. They were not what I expected. It was just the code without any further explanation. I want to be able to type the code in, not use copy/paste.But if I copy/paste the code from this file, it works fine. And yes, the e is in italic, indicating it is the definition of ElementaryCharge. I don’t know how to input that number. Because I bought the file from Rohlf, I don't think it is appropriate to include it. I may be breaking some copyright rule by posting a file that I bought in a forum. When I copy/paste I get N[UnitConvert[e,C],4] returns 1.602 x 10-19 I can also get the example to work by explicitly using Quantity to define ElementaryCharge and Coulambs, but that’s not what th example does. N[UnitConvert[Quantity["ElementaryCharge"],Quantity["Coulombs"]],4 returns 1.602 x 10-19 So I don't know how to run this example the way that the book does. This is just the first example and I fear I may experience the same problem with the others. What am I doing wrong? Theodore Kubaska 2026-04-03T02:09:35Z https://community.wolfram.com/groups/-/m/t/3701399 The circles of Apollonius are formed by two pencils: The first is the ratio r of the distances from one point P to two fixed points C and D, denoted as follows: { P{x, y} | dPC / dPD = r } The second is the geometric locus of the P{x, y} that form an inscribed angle theta with the fixed points C and D, it is denoted as follows: { P{x, y} | Angle CPD = theta }. &[Wolfram Notebook][1] [1]: https://www.wolframcloud.com/obj/306781fc-2688-49e4-9044-01ac3ab01339 Alejandro Latorre Chirot 2026-04-23T22:03:58Z https://community.wolfram.com/groups/-/m/t/3701601 Hello everyone, I’m working on a mathematical modeling problem: The Art of Reflection. The problem is about designing a chaotic pattern on a flat paper such that, when viewed in a vertical ideal cylindrical mirror, it reflects a clear portrait/text/pattern. Core Problem (Simplified) • Paper: flat A4 plane • Mirror: ideal vertical cylindrical surface • Given a target reflected image (in the mirror), find the original paper pattern • Also determine: cylinder radius, position, orientation My Questions : 1. How to build the coordinate transformation model between the paper plane and the cylindrical mirror reflection? 2. How to implement image unwarping/warping from the mirror image to the paper pattern in Mathematica? 3. Given a binary/logo image (like a face or text), how to compute and export the distorted paper pattern automatically? Thank you very much for your help! ZnCu LI 2026-04-23T13:44:24Z