https://community.wolfram.com RSS Feed for Wolfram Community showing questions tagged with Mathematics sorted by active. https://community.wolfram.com/groups/-/m/t/3756631 Limit[(t^2 (-1 + t^k))/(-1 + t), k -> +Infinity] // FullSimplify[#, 0 < t < 1] & The code is not correctly calculating the right limit value, where the parameter t ranges from 0 < t < 1. The correct result should be t²/(1 - t). How can this be obtained? Wen Dao 2026-07-12T02:50:44Z https://community.wolfram.com/groups/-/m/t/2952778 I would like to be able to explore if certain variable coefficient differential operator commute with integral operators. As a toy example Grünbaum (Grünbaum, F. Alberto. "A remark on Hilbert's matrix." Linear Algebra and its Applications 43 (1982): 119-124.) mentions that on L^2[0.1] the integral operator K with kernel 1/(x+y) commutes with the differential operator P where Pu (t) = ((1-t^2)t^2 u')' -2 t^2 u I wanted to make a pure function for the operators for example K = Function[f, Function[t, Integrate[f[s]/(t + s), {s, 0, 1}]]] and similarly P and then to be able to evaluate the commutator with something like K@*P - P@*K By hand it is fairly easy and uses integration by parts, but obviously I want to do much more complicated cases! The difficulties I have are that Mathematica will not simplify the composition of operators, unless applied to a function, and then only a specific function. And it does not understand subtraction of operators. Am I going about this the wrong way? Is it something Mathematica can do. So far I verified in on a Fourier basis in L^2 with Mathematica, it generates two pages of output but FullSimplify gets it to zero. William Lionheart 2023-07-06T11:48:29Z https://community.wolfram.com/groups/-/m/t/3021398 In Python or Matlab, the eigensolver can solve the system by giving the linear function (for example, in Matlab we can use eigs(Afun,...)), this speeds up the calculation largely as we do not need to construct the whole matrix. I can not find a similar solver in Mathematica. I wonder if there is a way to achieve this within Mathematica. Any ideas would be appreciated. Frank Xu 2023-09-26T06:00:42Z https://community.wolfram.com/groups/-/m/t/3729334 I guessed the following area formula for a (planar) quadrilateral using two opposite sides and four angles:(When $a=AB$ and $c=CD$,) $S = \frac{a^2}{2(\cot A + \cot B)} + \frac{c^2}{2(\cot C + \cot D)}$ ![enter image description here][1] This formula can be applied to the quadrilateral satisfying $A+B \ne \pi$ and $C+D \ne \pi$, including convex, concave, and even self-intersecting cases. I'd like to know the applicable case , inapplicable case and reasons for the validity of those cases. Are there any known papers or more elegant ways to derive this formula?I guess there are many ways to explain those reasons. It greatly resembles $S = \frac{1}{2}(ab \sin B + cd \sin D)$. It seems that $b$ and $d$ can be determined using $a, c, A, B, C, D$. [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=Screenshot2026-06-0811.18.44.png&userId=3725129 Ryo Takayama 2026-06-08T02:26:22Z https://community.wolfram.com/groups/-/m/t/3730754 f[x_] := E^x /; -2 <= x < 0 f[x_] := f[x - 2] /; x > 0 f[x_] := f[x + 2] /; x < -2 Plot[{f[x], -x/8 + 1/2}, {x, -6, 4}] The graph of the function drawn by the above code is as follows: ![enter image description here][1] The precise and correct graph is as shown in the image below: ![enter image description here][2] ![enter image description here][3] The issues with the graph plotted by the code are mainly as follows: 1. At the boundary points of the piecewise intervals, the graph is not a vertical line segment but rather has no image at all. See the parts circled in red in the figure above. 2. The right endpoint of each interval is not included; the graph should use an open circle at that point to indicate this, as shown by the yellow circles in the figure. However, it does not. How should the code be modified so that the generated graph matches the correct, normal function graph shown above? [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=007.jpg&userId=3593842 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-06-10_183157.png&userId=3593842 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=7904007.jpg&userId=3593842 Bill Blair 2026-06-10T10:42:44Z https://community.wolfram.com/groups/-/m/t/3730733 Reduce[ForAll[x, 0 < x < \[Pi]/2, 2 a Sin[a x] >= 0], a, Reals] ![enter image description here][1] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-06-10_074718.png&userId=3593842 Bill Blair 2026-06-09T23:47:41Z https://community.wolfram.com/groups/-/m/t/3727912 a = {a1, a2, a3}; b = {b1, b2, b3, b4}; Join[Tuples[{a, b}], Tuples[{b, a}]] ![enter image description here][1] Given the operation of choosing one element respectively from sets a and b to carry out permutation, what other solution methods are available? [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-06-05_161935.png&userId=3593842 Bill Blair 2026-06-05T08:21:05Z https://community.wolfram.com/groups/-/m/t/3641330 Recently, I found that Google Copilot can solve a nonlinear but solvable equation like coulomb damping[1], just by searching for "r''(t)=-r(t)-sgn(r'(t))/10, r(0)=0, r'(0)=1" [2], whether Copilot's time-dependent answer is correct or not aside. Is it possible to refine the following command DSolve[{r''[t] == -r[t] - Sign[r'[t]]/10, r[0] == 0, r'[0] == 1}, r[t], t] to obtain the same analytic result in Mathematica v.14.0 ? [1] https://en.wikipedia.org/wiki/Coulomb\_damping [2] https://www.google.com/search?q=r%27%27%28t%29%3D+-r%28t%29+-sgn%28r%27%28t%29%29%2F10%2C+r%280%29%3D0%2C+r%27%280%29%3D1 Isao Maruyama 2026-02-18T09:26:00Z https://community.wolfram.com/groups/-/m/t/3304085 Could you help me to understand the following? In[1]:= $VersionNumber Out[1]= 14. In[2]:= DSolve[{2.3*r''[t]==1-r'[t],r[0]==0,r'[0]==0},r[t],t]//FullSimplify Out[2]= {{r[t] -> -2. + 2.3 E^(-0.434783 t)}} In[3]:= DSolve[{m*r''[t]==f-r'[t],r[0]==0,r'[0]==0},r[t],t]/.{m->2.3,f->1}//FullSimplify Out[3]= {{r[t] -> -2.3 + 2.3 E^(-0.434783 t) + t}} 2.3 seems to be a magic number. <del>Is this a bug?</del> **(2024-10-23 ADD)** Thank you for replies. I found version dependence. I report the result in the following by using Gianluca Gorni's DiscretePlot. derivativeAtZero[a_] := DSolveValue[{a*r''[t] == 2 - r'[t], r[0] == 0, r'[0] == 0}, r, t]'[0]; fncAtZero[a_] := DSolveValue[{a*r''[t] == 2 - r'[t], r[0] == 0, r'[0] == 0}, r, t][0]; DiscretePlot[{fncAtZero[a], derivativeAtZero[a]}, {a, Subdivide[1., 3, 200]}, PlotLabel -> $VersionNumber] ![enter image description here][1] Correct answer(right figure) : version=13.1, 13.0, 12.2? Wrong answer(left figure) : version=14.1, 14 **(2024-10-23 ADD2)** Thanks to Michael Rogers' reply, I noticed a way to reveal version-dependent numerical instability. The equation 2.3 r’’(t)=1 - r’(t) can be generalized to 2.3 r’’(t)=exp(k t) - r’(t). The following command DSolveValue[{2.3*r''[t]==Exp[k*t]-r'[t],r[0]==0,r'[0]==0},r[t],t]//FullSimplify//FullForm//Print gives version-dependent outputs version 14: Times[-1,Plus[0.4347826086956521`,Times[-0.43478260869565216`,Power[E,Times[k,t]]],Times[0.9999999999999999`,k],Times[-0.9999999999999999`,Power[E,Times[-0.43478260869565216`,t]],k]],Power[Plus[Times[0.43478260869565216`,k],Times[1.`,Power[k,2]]],-1]] version 13: Times[-1,Plus[0.4347826086956522`,Times[-0.4347826086956522`,Power[E,Times[1.`,k,t]]],Times[1.`,k],Times[-1.`,Power[E,Times[-0.4347826086956522`,t]],k]],Power[Plus[Times[0.4347826086956522`,k],Power[k,2]],-1]] One can see small deviations of the same number 0.4347..., that is 1/2.3. In version 14, the numerical instability 0.4347826086956521 - 0.43478260869565216 = 5.55112*10^-17 appears in the solution shown by Michael Rogers. This numerical instability depends on the number 2.3. For example, we obtain the correct answer and no numerical instability for the number 2.31 even if we use version 14. DSolveValue[{2.31*r''[t]==1-r'[t],r[0]==0,r'[0]==0},r[t],t]//FullSimplify (* -2.31 + 2.31 E^(-0.4329 t) + 1. t *) DSolveValue[{2.31*r''[t]==Exp[k*t]-r'[t],r[0]==0,r'[0]==0},r[t],t]//FullSimplify//FullForm//Print (* Times[-1,Plus[0.4329004329004329`,Times[-0.4329004329004329`,Power[E,Times[k,t]]],Times[1.`,k],Times[-1.`,Power[E,Times[-0.4329004329004329`,t]],k]],Power[Plus[Times[0.4329004329004329`,k],Power[k,2]],-1]] *) **(2024-10-24 ADD)** In conclusion, I have understood that this unexpected result from DSolve was due to the numerical instability of a round-off error existing only in latest versions(>=14) of Mathematica. In the following, I report a new magic number with large round-off error (worst case), `aWorst`, by showing version-dependent solutions of $a\times v'(t)=e^{k t} - v(t) $ with using the following commands. aWorst=5750000000002614783/2500000000000000000; eq[a_,k_]:=a*v'[t]==Exp[10^-k*t]-v[t]; check[a_]:=Block[{}, Print[TraditionalForm[eq[a,k]],"\n a=",FullForm[a]," with Version",$VersionNumber,"\n", TableForm[{#,DSolveValue[eq[a,#],v[t],t]}&/@Range[20],TableHeadings->{None,{"k","v(t)"}}]]]; check[2.3] check[N[aWorst]] check[2.31] The result for version 14 shows unexpected round-off error while that for version 13 is reasonable. ![enter image description here][2] ![enter image description here][3] To avoid it, one can use the following quick hack. dSolveValue[eqs_,fnc_,arg_]:=Expand[DSolveValue[Rationalize[eqs,0],fnc,arg]]/.n_?NumberQ:>N[n]; I guess something has happened during the version update, and it might be not intended by the developer. I will report this to Wolfram support. I would like to thank you for your help and replies. [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=DSolve.png&userId=1875183 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=v14.png&userId=1875183 [3]: https://community.wolfram.com//c/portal/getImageAttachment?filename=v13.png&userId=1875183 Isao Maruyama 2024-10-21T23:35:13Z https://community.wolfram.com/groups/-/m/t/3717077 How to divide numerator and denominator by the numerator? in = -((8 m^2)/(3 + 10 m^2 + 3 m^4)) Numerator[in] in /. a_/b_ -> a/Numerator[in]/(b/Numerator[in]) ![enter image description here][1] Why is there no change in the result when using the code above? I aim to derive this expression in fractional form via identical transformation: -(1/(5/4 + 3/(8 m^2) + (3 m^2)/8)) ![enter image description here][2] [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-17_083943.png&userId=3593842 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-17_214336.png&userId=3593842 Bill Blair 2026-05-17T00:40:45Z https://community.wolfram.com/groups/-/m/t/3717122 Hello everyone, In the Basic Math Assistant palette, there's a Row icon (two squares) that's equivalent to {{ a,b}}. Is there a way to perform the inverse operation? That is, to convert {{a,b}} to that format? Regards, Sinval Sinval Santos 2026-05-16T16:13:32Z https://community.wolfram.com/groups/-/m/t/3716316 ![enter image description here][1] I solved the first problem this way, but I don't know how to approach the second part. a[n_] := a1 + (n - 1) d b[n_] := b1*2^(n - 1) eq1 = a[2] - b[2] == a[3] - b[3]; eq2 = a[3] - b[3] == b[4] - a[4]; sol = Solve[{eq1, eq2}, {d, a1}] // Simplify ![enter image description here][2] How should I tackle the second question? [1]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-15_190410.png&userId=3593842 [2]: https://community.wolfram.com//c/portal/getImageAttachment?filename=2026-05-15_190929.png&userId=3593842 Bill Blair 2026-05-15T11:10:28Z https://community.wolfram.com/groups/-/m/t/3716598 I would like to share a compact deterministic arithmetic program that produces four numerical values lying within the CODATA 2022 one-standard-uncertainty windows for: - the inverse fine-structure constant - the proton/electron mass ratio - the muon/electron mass ratio - the tau/electron mass ratio The code was motivated by a broader physical framework, but the point of this post is not to argue for that framework. The narrower question is this: Can this compact computational object be explained as ordinary numerical fitting, and if so, where is the fitted information stored? The program contains: - no empirical target constants as inputs - no decimal fit coefficients - no continuous optimization loop - no output-specific adjustable parameters In addition, the three mass-ratio outputs share the same denominator ladder. Wolfram Language code: ```wolfram u = 1; r = 3; A = r*(r - u); B = 2; n = A + B - u; rel = A + B + r; ph = n + B; h = (rel + u) + n; nVal = r + rel + ph + h; cVal = 2*nVal; atten = 1 - A/n^2; Phi = nVal*r + rel + r/cVal + (rel - ph)/cVal^2 + (18/(n*Pi))*atten/cVal^3 + (4*Pi)*atten/cVal^4; d1 = r + ph + u; d2 = h*(nVal - (r + u)); d3 = d1*d2; d4 = d2*(nVal - (r + u))*A*(rel - r); Psi1 = cVal*(h + r) - (ph + r) + (rel - ph)/d1 - u/d2 + (rel - ph)/d3 - u/d4; Psi2 = h*rel - r + (rel - u)/d1 - u/d2 + (r + u)/d3 + (h - r - u)/d4; Psi3 = nVal*(cVal - u) - ph + (rel - ph)/d1 - (rel - ph)/d2 + (r + u)/d3 - (r*h)/d4; NumberForm[N[{Phi, Psi1, Psi2, Psi3}, 15], 15] Output: {137.035999165800, 1836.15267343627, 206.768283284455, 3477.15145895483} For comparison, the CODATA 2022 reference values are approximately: alpha^-1 = 137.035999177(21) proton/electron = 1836.152673426(32) muon/electron = 206.7682827(46) tau/electron = 3477.23(23) All four outputs lie within the corresponding one-standard-uncertainty windows. Under a conservative unit-range normalization, the four simultaneous window hits roughly correspond to ~66 bits of target-window information. So my question is simple: **If this is ordinary numerical fitting, what is the information channel?** For completeness, I include a short note containing the description-length analysis: [https://doi.org/10.5281/zenodo.20158423][1] [1]: https://doi.org/10.5281/zenodo.20158423 Tetsuya Momose 2026-05-15T11:09:21Z https://community.wolfram.com/groups/-/m/t/2187926 Where can I find an explanation of the options for FourierParameters? The "Get Help" documentation just gives some "common choices" without explanation. There are apparently two parameters that control how Mathematica calculates the expression returned. Jay Gourley 2021-02-08T03:33:52Z 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/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/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