0
$\begingroup$

As per the answer to this question: SolveAlways rejects inequality, I tried the following commands:

Clear[f, x]
D[Erf[x  f[x]], x]
Reduce[D[Erf[x  f[x]], x] <= 0, Reals]

Which gives the following output:

(2 E^(-x^2 f[x]^2) (f[x] + x f'[x]))/Sqrt[[Pi]]

Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>

I would like to use the relation D[Erf[x f[x]], x] <= 0 to find a relation between f[x] and f'[x] over the domain x >= 0.

I know that x is positive and real, and that f[x] is a function from Reals to Reals, and that it is positive, continuous, and differentiable, but the exact form of the function is unknown.

In this case, it is clear that the derivative will be less than or equal to zero whenever f[x] <= -x f'[x] but I can't get Mathematica to work with me.

Any ideas on how I can use Mathematica to find the relation f[x] <= -x f'[x]?

$\endgroup$
1
$\begingroup$

Since the term Sqrt[Pi] in the denominator is a positive constant, the equation is negative exactly when the numerator is negative. Hence

Numerator[D[Erf[x f[x]], x]] <= 0

which gives

enter image description here

This is the same as the inequality you are looking for.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.