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I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

Resolve[
 ForAll[x, (a/x^2)*(Log[x - b] + 1) 
             - (a/x)*(1/(x - b)) -
              (1/x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious either). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer to this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
   x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001, 
0.01}]

does not seem to be correct syntax.

Thanks a lot

I am trying to determine the regions parameters (a, b) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

Resolve[
 ForAll[x, (a/x^2)*(Log[x - b] + 1) 
             - (a/x)*(1/(x - b)) -
              (1/x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious either). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer to this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
   x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001, 
0.01}]

does not seem to be correct syntax.

Thanks a lot

I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

       Resolve[
        ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
        x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious neither). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
   x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001, 
0.01}]

does not seem to be correct syntax.

Thanks a lot

I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

Resolve[
 ForAll[x, (a/x^2)*(Log[x - b] + 1) 
             - (a/x)*(1/(x - b)) -
              (1/x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious either). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer to this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
   x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001, 
0.01}]

does not seem to be correct syntax.

Thanks a lot

I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

       Resolve[
        ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
        x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious neither). Is there a maybe more clever way, fully exploiting Mathematica's capabilities? Thanks

I am trying to determine the regions parameters (Te, ymin) must belong to for a function to be negative for any argument x greater than zero. I have tried using Resolve

       Resolve[
        ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
        x^2)*(-b^3 + b) < 0] && x > 0, Reals]
 

but no expression could be provided. This is not surprising, as I suppose such expression would involve the solution of trascendental equations.

On the other hand, is there a way to get a numerical output from Resolve? For example, a set of (a, b) parameters values on the boundary of the region for which the function is positive?

I thought maybe Nsolve could take an inequality as an expression, but could not find any examples and could not make it work.

A "brute force" alternative I thought about is to create a grid of (a,b) values, and for each couple of values verify if the function is negative along all the positive x-axis (the last step does not seem obvious neither). Is there a maybe more clever way, fully exploiting Mathematica's capabilities?

I got intrigued by the answer this post, Reduce a complex inequality but I could not make it work for my case, for example

RegionPlot[
ForAll[x, (a/x^2)*(Log[x - b] + 1) - (a/x)*(1/(x - b)) - (1/
   x^2)*(-b^3 + b) < 0], {a, 0.01, 0.5}, {b, 0.001, 
0.01}]

does not seem to be correct syntax.

Thanks a lot