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When $e<a<b<e^{2}$, $ \ln ^{2} b-\ln ^{2} a>\frac{4}{e^{2}}(b-a)$ should always hold.

Resolve[ForAll[{a, b}, E < a < b < E^2, 
  Log[b]^2 - Log[a]^2 > 4/E^2 (b - a)], Reals]

However, the above code is returned as it is. How can I modify it to get true or prove that the inequality always is true?

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    $\begingroup$ Ask it at MSE. This is rather math than Mathematica. $\endgroup$
    – user64494
    Aug 19, 2020 at 4:21
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    $\begingroup$ The function f[x_]:=Log[x]^2 - 4/Exp[2]*x is strictly increasing on the interval Interval[{E, E^2}]. Hope the rest is clear, however, don't hesitate to ask for further explanation in need. $\endgroup$
    – user64494
    Aug 19, 2020 at 6:19
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    $\begingroup$ @user64494 No need to get snippy here. This is a perfectly reasonable question about how to use Mathematica's capabilities for problems like this. $\endgroup$ Sep 18, 2020 at 9:04
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    $\begingroup$ @SjoerdSmit: Here is the realization of my proposal. The command Minimize[{D[Log[x]^2 - 4/Exp[2]*x, x], x > E && x < E^2}, x] produces the warning "Minimize::wksol: Warning: there is no minimum in the region in which the objective function is defined and the constraints are satisfied; returning a result on the boundary." and {0, {x -> E^2}}. This implies an affirmative answer to the question. Also Resolve[ForAll[x, x > E && x < E^2, D[Log[x]^2 - 4/Exp[2]*x, x] > 0], Reals] produces True and an affirmative answer to the question. $\endgroup$
    – user64494
    Sep 18, 2020 at 9:58
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    $\begingroup$ However, both the above proposals need the Lagrange's Theorem to finish the proof, but the Lagrange's Theorem is not implemented in Mathematica. In view of it I don't give an answer. $\endgroup$
    – user64494
    Sep 18, 2020 at 15:09

2 Answers 2

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Not a proof, but a numerical verification

RegionPlot shows that the conditions holds.

RegionPlot[ 
E < a < b < E^2 && Log[b]^2 - Log[a]^2 > 4/E^2 (b - a)
, {a, 0,10}, {b, 0, 10}, FrameLabel -> {a, b},GridLines -> {{E, E^2}, {E, E^2}},MaxRecursion -> 5]

enter image description here

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  • $\begingroup$ This is not even a numeric verification. $\endgroup$
    – user64494
    Aug 19, 2020 at 6:44
  • $\begingroup$ If your comment would include a suggestion to do it better it could be helpful. $\endgroup$ Aug 19, 2020 at 7:23
  • $\begingroup$ See my second comment to the question to this end. $\endgroup$
    – user64494
    Aug 19, 2020 at 7:32
  • $\begingroup$ Your comment gives an idea for the mathematical proof, not a numerical verification! $\endgroup$ Aug 19, 2020 at 7:37
  • $\begingroup$ I wonder the .upvoters.It is a bad practice to approve a wrong answer $\endgroup$
    – user64494
    Sep 18, 2020 at 15:03
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Not a mathematical proof, but you can use NMinimize to find the point that's most in danger of violating the inequality or see if there are obvious counter-examples:

objective = Log[b]^2 - Log[a]^2 - 4/E^2 (b - a);
cons = E < a < b < E^2
sol = NMinimize[{objective, cons}, {a, b}, WorkingPrecision -> 20]

{6.4205140396121489208*10^-12, {a -> 2.7182895476079523458, b -> 2.7182895476409766661}}

As you can see, it's a pretty close call. Normally I'd use Reduce, but it seems like it gets stuck:

Reduce[E < a < b < E^2 && Log[b]^2 - Log[a]^2 > 4/E^2 (b - a)]
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  • $\begingroup$ What is the absolute error of NMinimize by default? The absolute error of the analogous command of Maple DirectSearch:-Search equals $10^{-6}$. $\endgroup$
    – user64494
    Sep 18, 2020 at 7:25
  • $\begingroup$ @user64494 It depends on your working precision and accuracy/precision goal specifications. I suggest looking it up in the documentation. However, since a == b == E makes the objective 0, I suspect the objective goes to zero for better and better precision. $\endgroup$ Sep 18, 2020 at 8:11
  • $\begingroup$ SioerdSmit (@ does not work.): Sorry, you avoid answering my question. $\endgroup$
    – user64494
    Sep 18, 2020 at 8:21
  • $\begingroup$ @user64494 The default accuracy and precision goals for NMinimize depend on the working precision. "AccuracyGoa -> Automatic normally yields an accuracy goal equal to half the setting for WorkingPrecision." reference.wolfram.com/language/ref/AccuracyGoal.html $\endgroup$ Sep 18, 2020 at 9:01
  • $\begingroup$ Sioerd Smit: Thank you. BTW, for each $a\in [e,e^2]$ and for each $\epsilon >0, \,\epsilon<1$ there exists $b\in [e,e^2], b>a,$ s.t. $\left|-\frac{4 (b-a)}{e^2}-\log ^2(a)+\log ^2(b)\right|< \epsilon$ so you result is meaningless. $\endgroup$
    – user64494
    Sep 18, 2020 at 9:46

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