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I'm trying to study the sign of an expression, having information on some of my parameters. I don't know how to code these conditions. For example, I want to know when this expression:

k (2(a'[x])^2 - a[x]a''[x]) / (a[x]^2) 

is positive given that k > 0, x > 0 and that my function a is positive and increasing for all x > 0 (k is a constant, a is a function)

So basically, since a''[x] is the only term of the equation that I don't know anything about, I want Mathematica to give me:

(* a''[x] < a'[x]^2/a[x] *)
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  • $\begingroup$ Elsa, welcome to Mathematica.SE. Please use backticks to format your code so that your question is more readable; see mathematica.stackexchange.com/editing-help#code $\endgroup$ – Mr.Wizard May 2 '14 at 5:50
  • $\begingroup$ @Elsa I did not clearly understand, what is your aim. Think about adding an example of a function in question along with the answer in the form you wish it to be. You can do it using the option "edit" below your question. $\endgroup$ – Alexei Boulbitch May 2 '14 at 7:07
  • $\begingroup$ Hello Elsa, my apologies that your edit went unnoticed and you had to wait so long... I've formatted the question a bit and reopened it :) I've also changed " to '' since " is used only for strings and it looks like you meant the second derivative. Please also clarify if k here is a function or just a quantity. Your usage of [], when interpreted as Mathematica syntax indicates that k is a function, but if interpreted purely as a grouping, then perhaps not. $\endgroup$ – rm -rf May 19 '14 at 23:06
  • $\begingroup$ Hello, thank you so much for your edit! (indeed, I meant the 2nd derivative and k is just a quantity). Cheers $\endgroup$ – Elsa May 20 '14 at 0:20
  • $\begingroup$ Ok this is good enough actually, thank you. Is there any way I wouldn't have to repeat k>0 in every function but settle it beforehand? (I tried different things on the same idea as testParam = {k -> 3}but it doesn't work with inequalities) $\endgroup$ – Elsa May 20 '14 at 18:00
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Questions need to be answered, so I guess I will shoot:

Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && k > 0]

$a'(x)\in \mathbb{R}\land \left(\left(a(x)<0\land a''(x)>\frac{2 a'(x)^2}{a(x)}\land k>0\right)\lor \left(a(x)>0\land a''(x)<\frac{2 a'(x)^2}{a(x)}\land k>0\right)\right)$

which is the result that you expected.


Regarding your comment:

Is there any way I wouldn't have to repeat k>0 in every function but settle it beforehand?

You can do:

assum = k > 0;
Reduce[k (2 (a'[x])^2 - a[x] a''[x])/(a[x]^2) > 0 && assum]
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  • $\begingroup$ That's awesome! thank you $\endgroup$ – Elsa May 20 '14 at 22:51

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