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I'm trying to find a root of the following equation $$\theta^\frac{t}{s}-\theta-\sqrt{1-p},$$ where $0<t\leq1\leq s$ and $0<p<1$. I've tried FindRoot and Reduce, but somehow they don't process the output.

Using reduce function, what I tried is

Reduce[theta^(t/s) - theta - Sqrt[1 - p] == 0 && theta > 0 && theta < 1 &&0 < t<=  1<=s && 0 < p < 1, theta]

Any suggestions on how to solve for a symbolic function?

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    $\begingroup$ Related: mathematica.stackexchange.com/questions/139881/… $\endgroup$ – Michael E2 Dec 9 '18 at 17:52
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    $\begingroup$ @MichaelE2 has given you the answer. Probably the best you can do is solve explicitly for the exponent: $\frac{r}{s}\to \frac{\log \left(\theta +\sqrt{1-p}\right)}{\log (\theta )}$. $\endgroup$ – JimB Dec 9 '18 at 18:58
  • $\begingroup$ What is the unknown? $\endgroup$ – Αλέξανδρος Ζεγγ Dec 10 '18 at 3:18
  • $\begingroup$ $\theta$ is the unknown and I want to find roots (probably there will be two of them) of the above equation. All the rest is just parameters. I've tried by simplifuying the equation into $\theta^a-\theta-b$, where $a=\frac{t}{s}$ and $b=\sqrt{1-p}$, but still just don't know how to solve for roots... $\endgroup$ – Greenteamaniac Dec 10 '18 at 5:08
  • $\begingroup$ This is not the type of equations for which I would expect symbolic solutions.Try numerics. $\endgroup$ – Mariusz Iwaniuk Dec 10 '18 at 7:30
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The t/s is somehow a nasty in this question. The inequality chain make t/s limited by 0 and 1. 1 means t==s. At this point the equation reduces to Sqrt[1-p] and is satisfied only for p==1.

Following this path replace u=t/s and solve the equation.

p[theta_, u_] := 1 - (theta^u - theta)^2

Plot3D[p[theta, u], {theta, 0, 1}, {u, 0, 1}, AxesLabel -> Automatic]

enter image description here enter image description here

The quotient t/s makes the equation unnecessary complicated. u makes it trivial.

It is too meaningful to replace the Sqrt[1-p] by q for example and

Plot3D[theta^u - theta, {u, 0, 1}, {theta, 0, 1}]

Plot3D

For very large s the quotient t/s->0 and the equation gets

1-theta-Sqrt[1-p]

Reduce[1 - theta - Sqrt[1 - p] == 0, {p, theta}]

(* theta == 1 - Sqrt[1 - p] *)

Plot3D[theta == 1 - Sqrt[1 - p], {p, 0, 1}, {theta, 0, 1}, 
 AxesLabel -> Automatic, PlotRange -> All]

Plot3D

Reduce[theta^u - theta - Sqrt[1 - p] == 0 && theta > 0 && theta < 1 &&
   0 < p < 1, u]
(* Element[C[1], Integers] && 0 < p < 1 && 0 < theta < 1 && 
  u == (2*I*Pi*C[1] + Log[Sqrt[1 - p] + theta])/Log[theta] *)

Plot3D[Log[Sqrt[1 - p] + theta]/Log[theta], {p, 0, .95}, {theta, 0, .9}, 
 AxesLabel -> Automatic, PlotRange -> All]

Plot3D

This is not so informativ and comfortable to look at if p->1 and theta->1.

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