User Portlet
| Discussions |
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| It is tricky to stop `Sec` turning up. Here is a way: Factor[f'[x] // Simplify] % /. Sec[x] :> 1/Inactive[Cos][x] // Together |
| Another way, using `Scale`: Show[potential1, Graphics3D[Scale[potential1[[1]], {1, 1, 0}, {0, 0, -2*10^8}]], PlotRange -> All] |
| I suppose it is a condition for convergence of the integral. A simple example of divergence: With[{t = 0, u = 0, a = 0}, Integrate[Exp[I x t - a (u - x)], {x, -Infinity, u}]] We may find more examples this way: sol =... |
| How to crunch numbers in Mathematica. How to properly use it to test if a given method solves faster The symbolic solution is found easily: Solve[1 == (((((pnp^2/x) + x^2))/x)/pnp), x] Solve[1 == (((((pnp^2/x) + x^2))/x)/pnp), x, Reals] I have no idea how fast it would be for number crunching. |
| You can remove the conditions by extracting the first element of the ConditionalExpression: Solve[ForAll[{x, y}, Equivalent[ m (x/a) + n (y/b) == 1, ((t1 - t2) (b (-1 + t1 t2) x + a (b + b t1 t2... |
| We must force a symbolic manipulation: Log[4]/Log[2] // PowerExpand Log[4]/Log[2] // FullSimplify Log[4]/Log[2] // Floor // PowerExpand Log[4]/Log[2] // Floor // FullSimplify |
| Using DSolve butFailing to obtain a list of equations expected instead of True in the first argument You may try rewriting the ode as a pde with this trick: instead of `h[t]` you write `h[t,x]` with the additional equation `D[h[t,x],x]==0`. |
| This is a way: Solve[ForAll[{x, y}, Equivalent[l1, l2]], {n, m}] |
| Infinite sums also have this problem, for example Sum[1, {n, 1, Infinity}] should give `Infinity` as output, arguably. |
| Have you tried setting `ImageSize` to larger than the default of `Plot`? |