Bob Hanlon
• Member for 7 years, 7 months
• Last seen this week
• Clarksville, MD

f[x_] = (1 - 3^(1/x))/(1 + 3^(1/x)); Plot[f[x], {x, -1, 1}, Exclusions -> {x == 0}] From the documentation for Limit, Direction -> -1 takes variables to approach their limits by ...

Clear[a, f, x] f[a_, x_] = (x + 1)^4 - (a + 3) (x^2 + 2 x) + a^2 + 3 a + 1; Reduce[Unequal @@ (x /. Solve[f[a, x] == 0, x, Reals]), a, Reals] (* -(7/3) < a < -2 || -2 < a < -1 *) ...

Clear[f, f1, f2] f1[z_] = Sum[1/k, {k, z - 1}] (* HarmonicNumber[-1 + z] *) Although this definition is for positive integers, the result is valid for all z except non-positive integers. Plot[f1[...

A numeric verification Clear["Global*"] The integrand of the lhs can be written as 1/(x^2 + 1) - 1/((x^2 + 1)(x^2019 + 1)) x^2019/((x^2 + 1) (x^2019 + 1)) == (1+x^2019)/((x^2+1)(x^2019+...

Currently, using Solve, Reduce, or FindInstance, Mathematica can neither solve the problem nor prove that a solution does not exist. Clear["Global*"] Format[d[s_List]] := Subscript[d, StringJoin[...

Clear["Global*"] To bound the problem let 1 <= {a, b, c} <= 100 and to avoid equivalent results that are just reorderings of {a, b, c} let {a, b, c} be ordered such that a <= b &...

Using Mathematica eq = 2 (2^x - 1) x^2 + (2^(x^2) - 2) x == 2^(x + 1) - 2; sol = Solve[eq, x, Reals] (* {{x -> -1}, {x -> 0}, {x -> 1}} *) Verifying the solutions And @@ (eq /. sol) (* ...

Solve[{100 <= n <= 999, ! Element[n/3, Integers], ! Element[n/5, Integers]}, n, Integers] // Length (* 480 *)

eqn = (60 - a) (60 - b) (60 - c) (60 - d) (60 - e) == 1025; Assuming that all variables are restricted to positive integers solns = FindInstance[{eqn, a > 0, b > 0, c > 0, d > 0, e &...

To find all solutions within a range for n {Reduce[{991 n^2 + 1 == m^2, 0 < n < 10^100, m > 0}, {n, m}, Integers] // ToRules} (* {{n -> 12055735790331359447442538767, m -> ...

Using Mathematica: Clear[a, n, m, even]; For n dice, half are initially even and become odd, the other half are flipped and half of those flipped end up even. So 1/4 are even after the first ...

Think of it as a tree. The first choice generates n branches, the second choice generates n branches from each of the original n branches (n^2 total), the third choice generates n branches from each ...

Clear["Global*"] Format[x[n_]] := Subscript[x, n] Format[y[n_]] := Subscript[y, n] eqn = (x - a)^2 + (y - b)^2 == R^2; eqns = eqn /. Transpose[Thread[# -> Array[#, 3]] & /@ {x, y}]...

2^(x + 2) + 2^(x - 1) == 18 2^3*2^(x - 1) + 2^(x - 1) == 18 8*2^(x - 1) + 2^(x - 1) == 18 9*2^(x - 1) == 18 2^(x - 1) == 2 x - 1 == 1 x == 2

You can approximate it using RootApproximant at92 = (ArcTan[9]/π // RootApproximant[#, 2] &) π (* ((89579 + Sqrt[12802312961]) π)/436180 *) The relative error is (at92 - ArcTan[9])/ArcTan[9] //...

f[x_] = 8/(x^2 + x + 2); The tangents are located at x values soln = Solve[f'[x] x + 5 == f[x], x, Reals] // ToRadicals (* {{x -> -2}, {x -> (-45 + 2 Sqrt[510])^(1/3)/15^(2/3) - 1/(15 (-...

Since you appear to be using Mathematica logf[x_, a_, b_] := Log[a*x/(1 - b*x)] With[{a = 1}, Plot3D[logf[x, a, b], {x, 0, 1}, {b, 0, 1}, ClippingStyle -> None, PlotPoints -> 100]] The ...

Using Mathematica poly = z^6 + 7 z^3 - 8; Factor the polynomial and solve the equations formed by setting each factor equal to zero factoredPoly = Factor[poly] (* (-1 + z)*(2 + z)*(4 - 2*z + z^2)*...

Using Mathematica Reduce[Abs[x] - Abs[x - 1] == 1/2, x, Reals] (* x == 3/4 *) With ContourPlot (http://reference.wolfram.com/language/ref/ContourPlot.html) ContourPlot[Abs[x] - Abs[x - 1] == y, {...

Using Mathematica: There are two solutions: a b c = +/- 2 Sqrt[2/15] Clear[a, b, c] eqns = { a b c/(a + b) == -1/2, a b c/(b + c) == 1/2, a b c/(a + c) == 1}; There are two solutions ...

The area is 19/3 f1[x_] = 4 x - x^2; f2[x_] = 4 - x; The two functions intersect for soln = Solve[f1[x] == f2[x], x] {{x -> 1}, {x -> 4}} The intersection points are pts = {x, f1[x]} /. soln ...

Using Mathematica to display the region region = With[{r = Sqrt[x^2 + y^2], theta = ArcTan[x, y]}, ImplicitRegion[-Pi/2 <= theta <= Pi/2 && 1 <= r <= 1 + Cos[theta] &...

eqn = 2^x 4^(x - 1) == 70; For real solution using Mathematica soln = Solve[eqn, x, Reals][[1]] (* {x -> (3*Log[2] + Log[5] + Log[7])/ (3*Log[2])} *) Verifying that soln satisfies eqn ...

Use WolframAlpha command in Mathematica for a step-by-step solution (or go to Wolfram|Alpha at http://www.wolframalpha.com) WolframAlpha["limit ((1+3x)/(1+4x^2+3x^4))^3 as x-> 1", {{"Limit", 2}, ...